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Eigenvalue Problem and Unbounded Connected Branch of Positive Solutions to a Class of Singular Elastic Beam Equations

Eigenvalue Problem and Unbounded Connected Branch of Positive Solutions to a Class of Singular... Eigenvalue Problem and Unbounded Connected Branch of Positive Solutions to a Class of Singular Elastic Beam Equations //// Hindawi Publishing Corporation Home Journals About Us About this Journal Submit a Manuscript Table of Contents Journal Menu Abstracting and Indexing Aims and Scope Article Processing Charges Articles in Press Author Guidelines Bibliographic Information Contact Information Editorial Board Editorial Workflow Free eTOC Alerts Reviewers Acknowledgment Subscription Information Open Special Issues Published Special Issues Special Issue Guidelines Abstract Full-Text PDF Full-Text HTML Linked References How to Cite this Article Boundary Value Problems Volume 2011 (2011), Article ID 594128, 21 pages doi:10.1155/2011/594128 Research Article <h2>Eigenvalue Problem and Unbounded Connected Branch of Positive Solutions to a Class of Singular Elastic Beam Equations</h2> Huiqin Lu School of Mathematical Sciences, Shandong Normal University, Jinan, 250014 Shandong, China Received 16 October 2010; Revised 22 December 2010; Accepted 27 January 2011 Academic Editor: Kanishka Perera Copyright © 2011 Huiqin Lu. 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 This paper investigates the eigenvalue problem for a class of singular elastic beam equations where one end is simply supported and the other end is clamped by sliding clamps. Firstly, we establish a necessary and sufficient condition for the existence of positive solutions, then we prove that the closure of positive solution set possesses an unbounded connected branch which bifurcates from ( 0 , 𝜃 ) . Our nonlinearity 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) may be singular at 𝑢 , 𝑣 , 𝑡 = 0 and/or 𝑡 = 1 . 1. Introduction Singular differential equations arise in the fields of gas dynamics, Newtonian fluid mechanics, the theory of boundary layer, and so on. Therefore, singular boundary value problems have been investigated extensively in recent years (see [ 1 – 4 ] and references therein). This paper investigates the following fourth-order nonlinear singular eigenvalue problem: 𝑢 ( 4 )  ( 𝑡 ) = 𝜆 𝑓 𝑡 , 𝑢 ( 𝑡 ) , 𝑢  ( 𝑡 ) , 𝑢    ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) , 𝑢 ( 0 ) = 𝑢  ( 1 ) = 𝑢   ( 0 ) = 𝑢    ( 1 ) = 0 , ( 1 . 1 ) where 𝜆 ∈ ( 0 , + ∞ ) is a parameter and 𝑓 satisfies the following hypothesis: ( 𝐻 ) 𝑓 ∈ 𝐶 ( ( 0 , 1 ) × ( 0 , + ∞ ) × ( 0 , + ∞ ) × ( − ∞ , 0 ] , [ 0 , + ∞ ) ) , and there exist constants 𝛼 𝑖 , 𝛽 𝑖 , 𝑁 𝑖 , 𝑖 = 1 , 2 , 3 ( − ∞ < 𝛼 1 ≤ 0 ≤ 𝛽 1 < + ∞ , − ∞ < 𝛼 2 ≤ 0 ≤ 𝛽 2 < + ∞ , 0 ≤ 𝛼 3 ≤ 𝛽 3 < 1 , ∑ 3 𝑖 = 1 𝛽 𝑖 < 1 ; 0 < 𝑁 𝑖 ≤ 1 , 𝑖 = 1 , 2 , 3 ) such that for any 𝑡 ∈ ( 0 , 1 ) , 𝑢 , 𝑣 ∈ ( 0 , + ∞ ) , 𝑤 ∈ ( − ∞ , 0 ] , 𝑓 satisfies 𝑐 𝛽 1 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑐 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑐 𝛼 1 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 0 < 𝑐 ≤ 𝑁 1 , 𝑐 𝛽 2 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑢 , 𝑐 𝑣 , 𝑤 ) ≤ 𝑐 𝛼 2 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 0 < 𝑐 ≤ 𝑁 2 , 𝑐 𝛽 3 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑐 𝑤 ) ≤ 𝑐 𝛼 3 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 0 < 𝑐 ≤ 𝑁 3 . ( 1 . 2 ) Typical functions that satisfy the above sublinear hypothesis ( 𝐻 ) are those taking the form 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) = 𝑚 1  𝑚 𝑖 = 1 2  𝑚 𝑗 = 1 3  𝑘 = 1 𝑝 𝑖 , 𝑗 , 𝑘 ( 𝑡 ) 𝑢 𝑟 𝑖 𝑣 𝑠 𝑗 𝑤 𝜎 𝑘 , ( 1 . 3 ) where 𝑝 𝑖 , 𝑗 , 𝑘 ( 𝑡 ) ∈ 𝐶 [ ( 0 , 1 ) , ( 0 , + ∞ ) ] , 𝑟 𝑖 , 𝑠 𝑗 ∈ 𝑅 , 0 ≤ 𝜎 𝑘 < 1 , m a x { 𝑟 𝑖 , 0 } + m a x { 𝑠 𝑗 } + 𝜎 𝑘 < 1 , 𝑖 = 1 , 2 , … , 𝑚 1 , 𝑗 = 1 , 2 , … , 𝑚 2 , 𝑘 = 1 , 2 , … , 𝑚 3 . The hypothesis ( 𝐻 ) is similar to that in [ 5 , 6 ]. Because of the extensive applications in mechanics and engineering, nonlinear fourth-order two-point boundary value problems have received wide attentions (see [ 7 – 12 ] and references therein). In mechanics, the boundary value problem ( 1.1 ) (BVP ( 1.1 ) for short) describes the deformation of an elastic beam simply supported at left and clamped at right by sliding clamps. The term 𝑢   in 𝑓 represents bending effect which is useful for the stability analysis of the beam. BVP ( 1.1 ) has two special features. The first one is that the nonlinearity 𝑓 may depend on the first-order derivative of the unknown function 𝑢 , and the second one is that the nonlinearity 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) may be singular at 𝑢 , 𝑣 , 𝑡 = 0 and/or 𝑡 = 1 . In this paper, we study the existence of positive solutions and the structure of positive solution set for the BVP ( 1.1 ). Firstly, we construct a special cone and present a necessary and sufficient condition for the existence of positive solutions, then we prove that the closure of positive solution set possesses an unbounded connected branch which bifurcates from ( 0 , 𝜃 ) . Our analysis mainly relies on the fixed point theorem in a cone and the fixed point index theory. By singularity of 𝑓 , we mean that the function 𝑓 in ( 1.1 ) is allowed to be unbounded at the points 𝑢 = 0 , 𝑣 = 0 , 𝑡 = 0 , and/or 𝑡 = 1 . A function 𝑢 ( 𝑡 ) ∈ 𝐶 2 [ 0 , 1 ] ∩ 𝐶 4 ( 0 , 1 ) is called a (positive) solution of the BVP ( 1.1 ) if it satisfies the BVP ( 1.1 ) ( 𝑢 ( 𝑡 ) > 0 , − 𝑢   ( 𝑡 ) > 0 for 𝑡 ∈ ( 0 , 1 ] and 𝑢  ( 𝑡 ) > 0 for 𝑡 ∈ [ 0 , 1 ) ). For some 𝜆 ∈ ( 0 , + ∞ ) , if the B V P ( 1.1 ) has a positive solution 𝑢 , then 𝜆 is called an eigenvalue and 𝑢 is called corresponding eigenfunction of the BVP ( 1.1 ). The existence of positive solutions of BVPs has been studied by several authors in the literature; for example, see [ 7 – 20 ] and the references therein. Yao [ 15 , 18 ] studied the following BVP: 𝑢 ( 4 )  ( 𝑡 ) = 𝑓 𝑡 , 𝑢 ( 𝑡 ) , 𝑢   [ ] ( 𝑡 ) , 𝑡 ∈ 0 , 1 ⧵ 𝐸 , 𝑢 ( 0 ) = 𝑢  ( 0 ) = 𝑢   ( 1 ) = 𝑢    ( 1 ) = 0 , ( 1 . 4 ) where 𝐸 ⊂ [ 0 , 1 ] is a closed subset and m e s 𝐸 = 0 , 𝑓 ∈ 𝐶 ( ( [ 0 , 1 ] ⧵ 𝐸 ) × [ 0 , + ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) . In [ 15 ], he obtained a sufficient condition for the existence of positive solutions of B V P ( 1.4 ) by using the monotonically iterative technique. In [ 13 , 18 ], he applied Guo-Krasnosel'skii's fixed point theorem to obtain the existence and multiplicity of positive solutions of BVP ( 1.4 ) and the following BVP: 𝑢 ( 4 ) [ ] , ( 𝑡 ) = 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) , 𝑡 ∈ 0 , 1 𝑢 ( 0 ) = 𝑢  ( 0 ) = 𝑢 ( 1 ) = 𝑢   ( 1 ) = 0 . ( 1 . 5 ) These differ from our problem because 𝑓 ( 𝑡 , 𝑢 , 𝑣 ) in ( 1.4 ) cannot be singular at 𝑢 = 0 , 𝑣 = 0 and the nonlinearity 𝑓 in ( 1.5 ) does not depend on the derivatives of the unknown functions. In this paper, we first establish a necessary and sufficient condition for the existence of positive solutions of BVP ( 1.1 ) for any 𝜆 > 0 by using the following Lemma 1.1 . Efforts to obtain necessary and sufficient conditions for the existence of positive solutions of BVPs by the lower and upper solution method can be found, for example, in [ 5 , 6 , 21 – 23 ]. In [ 5 , 6 , 22 , 23 ] they considered the case that 𝑓 depends on even order derivatives of 𝑢 . Although the nonlinearity 𝑓 in [ 21 ] depends on the first-order derivative, where the nonlinearity 𝑓 is increasing with respect to the unknown function 𝑢 . Papers [ 24 , 25 ] derived the existence of positive solutions of BVPs by the lower and upper solution method, but the nonlinearity 𝑓 ( 𝑡 , 𝑢 ) does not depend on the derivatives of the unknown functions, and 𝑓 ( 𝑡 , 𝑢 ) is decreasing with respect to 𝑢 . Recently, the global structure of positive solutions of nonlinear boundary value problems has also been investigated (see [ 26 – 28 ] and references therein). Ma and An [ 26 ] and Ma and Xu [ 27 ] discussed the global structure of positive solutions for the nonlinear eigenvalue problems and obtained the existence of an unbounded connected branch of positive solution set by using global bifurcation theorems (see [ 29 , 30 ]). The terms 𝑓 ( 𝑢 ) in [ 26 ] and 𝑓 ( 𝑡 , 𝑢 , 𝑢   ) in [ 27 ] are not singular at 𝑡 = 0 , 1 , 𝑢 = 0 , 𝑢   = 0 . Yao [ 14 ] obtained one or two positive solutions to a singular elastic beam equation rigidly fixed at both ends by using Guo-Krasnosel'skii's fixed point theorem, but the global structure of positive solutions was not considered. Since the nonlinearity 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) in BVP ( 1.1 ) may be singular at 𝑢 , 𝑣 , 𝑡 = 0 and/or 𝑡 = 1 , the global bifurcation theorems in [ 29 , 30 ] do not apply to our problem here. In Section 4 , we also investigate the global structure of positive solutions for BVP ( 1.1 ) by applying the following Lemma 1.2 . The paper is organized as follows: in the rest of this section, two known results are stated. In Section 2 , some lemmas are stated and proved. In Section 3 , we establish a necessary and sufficient condition for the existence of positive solutions. In Section 4 , we prove that the closure of positive solution set possesses an unbounded connected branch which comes from ( 0 , 𝜃 ) . Finally we state the following results which will be used in Sections 3 and 4 , respectively. Lemma 1.1 (see [ 31 ]). Let 𝑋 be a real Banach space, let 𝐾 be a cone in 𝑋 , and let Ω 1 , Ω 2 be bounded open sets of 𝐸 , 𝜃 ∈ Ω 1 ⊂ Ω 1 ⊂ Ω 2 . Suppose that 𝑇 ∶ 𝐾 ∩ ( Ω 2 ⧵ Ω 1 ) → 𝐾 is completely continuous such that one of the following two conditions is satisfied: (1) ‖ 𝑇 ( 𝑥 ) ‖ ≤ ‖ 𝑥 ‖ , 𝑥 ∈ 𝐾 ∩ 𝜕 Ω 1 ; ‖ 𝑇 ( 𝑥 ) ‖ ≥ ‖ 𝑥 ‖ , 𝑥 ∈ 𝐾 ∩ 𝜕 Ω 2 . (2) ‖ 𝑇 ( 𝑥 ) ‖ ≥ ‖ 𝑥 ‖ , 𝑥 ∈ 𝐾 ∩ 𝜕 Ω 1 ; ‖ 𝑇 ( 𝑥 ) ‖ ≤ ‖ 𝑥 ‖ , 𝑥 ∈ 𝐾 ∩ 𝜕 Ω 2 . Then, 𝑇 has a fixed point in 𝐾 ∩ ( Ω 2 ⧵ Ω 1 ) . Lemma 1.2 (see [ 32 ]). Let 𝑀 be a metric space and ( 𝑎 , 𝑏 ) ⊂ 𝑅 1 . Let { 𝑎 𝑛 } ∞ 𝑛 = 1 and { 𝑏 𝑛 } ∞ 𝑛 = 1 satisfy 𝑎 < ⋯ < 𝑎 𝑛 < ⋯ < 𝑎 1 < 𝑏 1 < ⋯ < 𝑏 𝑛 < ⋯ < 𝑏 , l i m 𝑛 → + ∞ 𝑎 𝑛 = 𝑎 , l i m 𝑛 → + ∞ 𝑏 𝑛 = 𝑏 . ( 1 . 6 ) Suppose also that ∑ = { 𝐶 𝑛 ∶ 𝑛 = 1 , 2 , … } is a family of connected subsets of 𝑅 1 × 𝑀 , satisfying the following conditions: ( 1 ) 𝐶 𝑛 ∩ ( { 𝑎 𝑛 } × 𝑀 ) ≠ ∅ and 𝐶 𝑛 ∩ ( { 𝑏 𝑛 } × 𝑀 ) ≠ ∅ for each 𝑛 . (2) For any two given numbers 𝛼 and 𝛽 with 𝑎 < 𝛼 < 𝛽 < 𝑏 , ( ⋃ ∞ 𝑛 = 1 𝐶 𝑛 ) ∩ ( [ 𝛼 , 𝛽 ] × 𝑀 ) is a relatively compact set of 𝑅 1 × 𝑀 . Then there exists a connected branch 𝐶 of l i m s u p 𝑛 → + ∞ 𝐶 𝑛 such that 𝐶 ∩ ( { 𝜆 } × 𝑀 ) ≠ ∅ , ∀ 𝜆 ∈ ( 𝑎 , 𝑏 ) , ( 1 . 7 ) where l i m s u p 𝑛 → + ∞ 𝐶 𝑛 = { 𝑥 ∈ 𝑀 ∶ there exists a sequence 𝑥 𝑛 𝑖 ∈ 𝐶 𝑛 𝑖 such that 𝑥 𝑛 𝑖 → 𝑥 , ( 𝑖 → ∞ ) } . 2. Some Preliminaries and Lemmas Let 𝐸 = { 𝑢 ∈ 𝐶 2 [ 0 , 1 ] ∶ 𝑢 ( 0 ) = 0 , 𝑢  ( 1 ) = 0 , 𝑢   ( 0 ) = 0 } , ‖ 𝑢 ‖ 2 = m a x { ‖ 𝑢 ‖ , ‖ 𝑢  ‖ , ‖ 𝑢   ‖ } , then ( 𝐸 , ‖ ⋅ ‖ 2 ) is a Banach space, where ‖ 𝑢 ‖ = m a x 𝑡 ∈ [ 0 , 1 ] | 𝑢 ( 𝑡 ) | . Define   𝑡 𝑃 = 𝑢 ∈ 𝐸 ∶ 𝑢 ( 𝑡 ) ≥ 𝑡 − 2 2  ‖ 𝑢 ‖ , 𝑢  1 ( 𝑡 ) ≥ 2 ‖ ‖ 𝑢 ( 1 − 𝑡 )  ‖ ‖ , − 𝑢   ‖ ‖ 𝑢 ( 𝑡 ) ≥ 𝑡   ‖ ‖ [ ]  , 𝑡 ∈ 0 , 1 . ( 2 . 1 ) It is easy to conclude that 𝑃 is a cone of 𝐸 . Denote 𝑃 𝑟 =  𝑢 ∈ 𝑃 ∶ ‖ 𝑢 ‖ 2  < 𝑟 ; 𝜕 𝑃 𝑟 =  𝑢 ∈ 𝑃 ∶ ‖ 𝑢 ‖ 2  = 𝑟 . ( 2 . 2 ) Let 𝐺 0   ( 𝑡 , 𝑠 ) = 𝑠 , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , 𝑡 , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 , 𝐺 ( 𝑡 , 𝑠 ) = 1 0 𝐺 0 ( 𝑡 , 𝑟 ) 𝐺 0 ( 𝑟 , 𝑠 ) 𝑑 𝑟 . ( 2 . 3 ) Then 𝐺 ( 𝑡 , 𝑠 ) is the Green function of homogeneous boundary value problem 𝑢 ( 4 ) ( 𝑡 ) = 0 , 𝑡 ∈ ( 0 , 1 ) , 𝑢 ( 0 ) = 𝑢  ( 1 ) = 𝑢   ( 0 ) = 𝑢    ( ⎧ ⎪ ⎨ ⎪ ⎩ 𝑠 1 ) = 0 , 𝐺 ( 𝑡 , 𝑠 ) = 3 3 + 𝑠  𝑡 2 − 𝑠 2  2 𝑡 + 𝑠 𝑡 ( 1 − 𝑡 ) , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , 3 3 + 𝑡  𝑠 2 − 𝑡 2  2 𝐺 + 𝑡 𝑠 ( 1 − 𝑠 ) , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 , 1 ( 𝑡 , 𝑠 ) = ∶ 𝐺  𝑡 ⎧ ⎪ ⎨ ⎪ ⎩ 𝑠 ( 𝑡 , 𝑠 ) = 𝑠 ( 1 − 𝑡 ) , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , 2 2 − 𝑡 2 2 𝐺 + 𝑠 ( 1 − 𝑠 ) , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 , 2 ( 𝑡 , 𝑠 ) = ∶ − 𝐺 𝑡    ( 𝑡 , 𝑠 ) = 𝑠 , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , 𝑡 , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 . ( 2 . 4 ) Lemma 2.1. 𝐺 ( 𝑡 , 𝑠 ) , 𝐺 1 ( 𝑡 , 𝑠 ) , and 𝐺 2 ( 𝑡 , 𝑠 ) have the following properties: (1) 𝐺 ( 𝑡 , 𝑠 ) > 0 , 𝐺 𝑖 ( 𝑡 , 𝑠 ) > 0 , 𝑖 = 1 , 2 , for all 𝑡 , 𝑠 ∈ ( 0 , 1 ) . (2) 𝐺 ( 𝑡 , 𝑠 ) ≤ 𝑠 ( 𝑡 − 𝑡 2 / 2 ) , 𝐺 1 ( 𝑡 , 𝑠 ) ≤ 𝑠 ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≤ 𝑡 (or 𝑠 ), for all 𝑡 , 𝑠 ∈ [ 0 , 1 ] . (3) m a x 𝑡 ∈ [ 0 , 1 ] 𝐺 ( 𝑡 , 𝑠 ) ≤ ( 1 / 2 ) 𝑠 , m a x 𝑡 ∈ [ 0 , 1 ] 𝐺 𝑖 ( 𝑡 , 𝑠 ) ≤ 𝑠 , 𝑖 = 1 , 2 , for all 𝑠 ∈ [ 0 , 1 ] . (4) 𝐺 ( 𝑡 , 𝑠 ) ≥ ( 𝑠 / 2 ) ( 𝑡 − 𝑡 2 / 2 ) , 𝐺 1 ( 𝑡 , 𝑠 ) ≥ ( 𝑠 / 2 ) ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≥ 𝑠 𝑡 , for all 𝑡 , 𝑠 ∈ [ 0 , 1 ] . Proof. From ( 2.4 ), it is easy to obtain the property ( 2.18 ). We now prove that property (2) is true. For 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , by ( 2.4 ), we have 𝑠 𝐺 ( 𝑡 , 𝑠 ) = 3 3 + 𝑠 𝑡 2 2 − 𝑠 3 2 + 𝑠 𝑡 − 𝑠 𝑡 2 ≤ 𝑠 𝑡 − 𝑠 𝑡 2 2  𝑡 = 𝑠 𝑡 − 2 2  , 𝐺 1 ( 𝑡 , 𝑠 ) = 𝑠 ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≤ 𝑡 ( o r 𝑠 ) . ( 2 . 5 ) For 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 , by ( 2.4 ), we have 𝑡 𝐺 ( 𝑡 , 𝑠 ) = 3 3 − 𝑡 3 2 + 𝑡 𝑠 − 𝑡 𝑠 2 2 ≤ 𝑠 𝑡 − 𝑠 𝑡 2 2  𝑡 = 𝑠 𝑡 − 2 2  , 𝐺 1 𝑡 ( 𝑡 , 𝑠 ) = 𝑠 − 2 2 − 𝑠 2 2 ≤ 𝑠 − 𝑡 𝑠 = 𝑠 ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≤ 𝑡 ( o r 𝑠 ) . ( 2 . 6 ) Consequently, property (2) holds. From property (2), it is easy to obtain property (3). We next show that property (4) is true. From ( 2.4 ), we know that property (4) holds for 𝑠 = 0 . For 0 < 𝑠 ≤ 1 , if 𝑠 ≤ 𝑡 ≤ 1 , then 𝐺 ( 𝑡 , 𝑠 ) 𝑠 𝑡 = 𝑡 − 2 2 − 𝑠 2 6 = 1 2  𝑡 𝑡 − 2 2 +  𝑡 𝑡 − 2 2 − 𝑠 2 3 ≥ 1   2  𝑡 𝑡 − 2 2 +  𝑡 𝑡 − 2 2 − 𝑡 2 3 > 1   2  𝑡 𝑡 − 2 2  , 𝐺 1 ( 𝑡 , 𝑠 ) 𝑠 1 = ( 1 − 𝑡 ) ≥ 2 ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≥ 𝑠 𝑡 ; ( 2 . 7 ) if 0 ≤ 𝑡 ≤ 𝑠 , then 𝐺 ( 𝑡 , 𝑠 ) 𝑠 𝑡 ≥ 𝑡 − 2 6 − 𝑡 𝑠 2 = 1 2  𝑡 𝑡 − 2 3  ≥ 1 + ( 𝑡 − 𝑡 𝑠 ) 2  𝑡 𝑡 − 2 3  ≥ 1 2  𝑡 𝑡 − 2 2  , 𝐺 1 ( 𝑡 , 𝑠 ) 𝑠 𝑡 ≥ 1 − 2 − 𝑠 2 ≥ 1 2 ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≥ 𝑠 𝑡 . ( 2 . 8 ) Therefore, property (4) holds. Lemma 2.2. Assume that 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , then ‖ 𝑢 ‖ 2 = ‖ 𝑢   ‖ and 1 4 ‖ ‖ 𝑢  ‖ ‖ ≤ ‖ ‖ 𝑢 ‖ 𝑢 ‖ ≤  ‖ ‖ , 1 2 ‖ ‖ 𝑢   ‖ ‖ ≤ ‖ ‖ 𝑢  ‖ ‖ ≤ ‖ ‖ 𝑢   ‖ ‖ . 1 ( 2 . 9 ) 8  𝑡 𝑡 − 2 2  ‖ 𝑢 ‖ 2  𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑡 − 2 2  ‖ 𝑢 ‖ 2 , 1 4 ( 1 − 𝑡 ) ‖ 𝑢 ‖ 2 ≤ 𝑢  ( 𝑡 ) ≤ ( 1 − 𝑡 ) ‖ 𝑢 ‖ 2 , 𝑡 ‖ 𝑢 ‖ 2 ≤ − 𝑢   ( 𝑡 ) ≤ ‖ 𝑢 ‖ 2 [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 2 . 1 0 ) Proof. Assume that 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , then 𝑢  ( 𝑡 ) ≥ 0 , − 𝑢   ( 𝑡 ) ≥ 0 , 𝑡 ∈ [ 0 , 1 ] , so ‖ 𝑢 ‖ = m a x [ ] 𝑡 ∈ 0 , 1  𝑡 0 𝑢   ( 𝑠 ) 𝑑 𝑠 = 1 0 𝑢  ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤  ‖ ‖ , ‖ 𝑢 ‖ = m a x [ ] 𝑡 ∈ 0 , 1  𝑡 0 𝑢  (  𝑠 ) 𝑑 𝑠 = 1 0 𝑢  ( 1 𝑠 ) 𝑑 𝑠 ≥ 2 ‖ ‖ 𝑢  ‖ ‖  1 0 ( 1 1 − 𝑠 ) 𝑑 𝑠 = 4 ‖ ‖ 𝑢  ‖ ‖ , ‖ ‖ 𝑢  ‖ ‖ = m a x [ ] 𝑡 ∈ 0 , 1  1 𝑡 − 𝑢    ( 𝑠 ) 𝑑 𝑠 = 1 0 − 𝑢   ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤   ‖ ‖ , ‖ ‖ 𝑢  ‖ ‖ = m a x [ ] 𝑡 ∈ 0 , 1  1 𝑡 − 𝑢   (  𝑠 ) 𝑑 𝑠 = 1 0 − 𝑢   (  𝑠 ) 𝑑 𝑠 ≥ 1 0 𝑠 ‖ ‖ 𝑢   ‖ ‖ 1 𝑑 𝑠 = 2 ‖ ‖ 𝑢   ‖ ‖ . ( 2 . 1 1 ) Therefore, ( 2.9 ) holds. From ( 2.9 ), we get ‖ 𝑢 ‖ 2  ‖ ‖ 𝑢 = m a x ‖ 𝑢 ‖ ,  ‖ ‖ , ‖ ‖ 𝑢   ‖ ‖  = ‖ ‖ 𝑢   ‖ ‖ . ( 2 . 1 2 ) By ( 2.9 ) and the definition of 𝑃 , we can obtain that  𝑢 ( 𝑡 ) = 1 0 𝐺 0  ( 𝑡 , 𝑠 ) − 𝑢      ( 𝑠 ) 𝑑 𝑠 ≤ 𝑡 0  𝑠 𝑑 𝑠 + 1 𝑡  ‖ ‖ 𝑢 𝑡 𝑑 𝑠   ‖ ‖ =  𝑡 𝑡 − 2 2  ‖ ‖ 𝑢   ‖ ‖ =  𝑡 𝑡 − 2 2  ‖ 𝑢 ‖ 2 , [ ] ,  𝑡 ∀ 𝑡 ∈ 0 , 1 𝑢 ( 𝑡 ) ≥ 𝑡 − 2 2  1 ‖ 𝑢 ‖ ≥ 8  𝑡 𝑡 − 2 2  ‖ 𝑢 ‖ 2 [ ] , 𝑢 , ∀ 𝑡 ∈ 0 , 1   ( 𝑡 ) = 1 𝑡 − 𝑢   ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ ( 1 − 𝑡 )   ‖ ‖ = ( 1 − 𝑡 ) ‖ 𝑢 ‖ 2 , 𝑢  1 ( 𝑡 ) ≥ 2 ‖ ‖ 𝑢 ( 1 − 𝑡 )  ‖ ‖ ≥ 1 4 ( 1 − 𝑡 ) ‖ 𝑢 ‖ 2 [ ] , , ∀ 𝑡 ∈ 0 , 1 𝑡 ‖ 𝑢 ‖ 2 ‖ ‖ 𝑢 = 𝑡   ‖ ‖ ≤ − 𝑢   ‖ ‖ 𝑢 ( 𝑡 ) ≤   ‖ ‖ = ‖ 𝑢 ‖ 2 [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 2 . 1 3 ) Thus, ( 2.10 ) holds. For any fixed 𝜆 ∈ ( 0 , + ∞ ) , define an operator 𝑇 𝜆 by  𝑇 𝜆 𝑢   ( 𝑡 ) = ∶ 𝜆 1 0  𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ( 𝑠 ) 𝑑 𝑠 , ∀ 𝑢 ∈ 𝑃 ⧵ { 𝜃 } . ( 2 . 1 4 ) Then, it is easy to know that  𝑇 𝜆 𝑢    ( 𝑡 ) = 𝜆 1 0 𝐺 1  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     𝑇 ( 𝑠 ) 𝑑 𝑠 , ∀ 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , ( 2 . 1 5 ) 𝜆 𝑢     ( 𝑡 ) = − 𝜆 1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ( 𝑠 ) 𝑑 𝑠 , ∀ 𝑢 ∈ 𝑃 ⧵ { 𝜃 } . ( 2 . 1 6 ) Lemma 2.3. Suppose that ( 𝐻 ) and  0 < 1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ ( 2 . 1 7 ) hold. Then 𝑇 𝜆 ( 𝑃 ⧵ { 𝜃 } ) ⊂ 𝑃 . Proof. From ( 𝐻 ), for any 𝑡 ∈ ( 0 , 1 ) , 𝑢 , 𝑣 ∈ ( 0 , + ∞ ) , 𝑤 ∈ ( − ∞ , 0 ] , we easily obtain the following inequalities: 𝑐 𝛼 1 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑐 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑐 𝛽 1 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 𝑐 ≥ 𝑁 1 − 1 , 𝑐 𝛼 2 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑢 , 𝑐 𝑣 , 𝑤 ) ≤ 𝑐 𝛽 2 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 𝑐 ≥ 𝑁 2 − 1 , 𝑐 𝛼 3 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑐 𝑤 ) ≤ 𝑐 𝛽 3 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 𝑐 ≥ 𝑁 3 − 1 . ( 2 . 1 8 ) For every 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , 𝑡 ∈ [ 0 , 1 ] , choose positive numbers 𝑐 1 ≤ m i n { 𝑁 1 , ( 1 / 8 ) 𝑁 1 ‖ 𝑢 ‖ 2 } , 𝑐 2 ≤ m i n { 𝑁 2 , ( 1 / 4 ) 𝑁 2 ‖ 𝑢 ‖ 2 } , 𝑐 3 ≥ m a x { 𝑁 3 − 1 , 𝑁 3 − 1 ‖ 𝑢 ‖ 2 } . It follows from ( 𝐻 ), ( 2.10 ), Lemma 2.1 , and ( 2.17 ) that  𝑇 𝜆 𝑢   ( 𝑡 ) = 𝜆 1 0  𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ≤ 1 ( 𝑠 ) 𝑑 𝑠 2 𝜆  1 0  𝑠 𝑓 𝑠 , 𝑐 1 𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   𝑠 / 2 𝑠 − 2 2  , 𝑐 2 𝑢  ( 𝑠 ) 𝑐 2 ( 1 − 𝑠 ) ( 1 − 𝑠 ) , ( − 1 ) 𝑐 3 𝑢   ( 𝑠 ) − 𝑐 3  ≤ 1 𝑑 𝑠 2 𝜆  1 0 𝑠 𝑐 𝛼 1 1  𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   / 2 𝛽 1 𝑐 𝛼 2 2  𝑢  ( 𝑠 ) 𝑐 2  ( 1 − 𝑠 ) 𝛽 2 𝑐 𝛽 3 3  𝑢   ( 𝑠 ) − 𝑐 3  𝛼 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  ≤ 1 , 1 − 𝑠 , − 1 𝑑 𝑠 2 𝜆  1 0 𝑠 𝑐 𝛼 1 1  ‖ 𝑢 ‖ 2 𝑐 1  𝛽 1 𝑐 𝛼 2 2  ‖ 𝑢 ‖ 2 𝑐 2  𝛽 2 𝑐 𝛽 3 3  ‖ 𝑢 ‖ 2 𝑐 3  𝛼 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  ≤ 1 , 1 − 𝑠 , − 1 𝑑 𝑠 2 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 + 𝛼 3 2 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ . ( 2 . 1 9 ) Similar to ( 2.19 ), from ( 𝐻 ), ( 2.10 ), Lemma 2.1 , and ( 2.17 ), for every 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , 𝑡 ∈ [ 0 , 1 ] , we have  𝑇 𝜆 𝑢    ( 𝑡 ) = 𝜆 1 0 𝐺 1  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑐 1 𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   𝑠 / 2 𝑠 − 2 2  , 𝑐 2 𝑢  ( 𝑠 ) 𝑐 2 ( 1 − 𝑠 ) ( 1 − 𝑠 ) , ( − 1 ) 𝑐 3 𝑢   ( 𝑠 ) − 𝑐 3  𝑑 𝑠 ≤ 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 + 𝛼 3 2 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  −  𝑇 , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ . 𝜆 𝑢     ( 𝑡 ) = 𝜆 1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑐 1 𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   𝑠 / 2 𝑠 − 2 2  , 𝑐 2 𝑢  ( 𝑠 ) 𝑐 2 ( 1 − 𝑠 ) ( 1 − 𝑠 ) , ( − 1 ) 𝑐 3 𝑢   ( 𝑠 ) − 𝑐 3  𝑑 𝑠 ≤ 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 + 𝛼 3 2 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ . ( 2 . 2 0 ) Thus, 𝑇 𝜆 is well defined on 𝑃 ⧵ { 𝜃 } . From ( 2.4 ) and ( 2.14 )–( 2.16 ), it is easy to know that  𝑇 𝜆 𝑢   𝑇 ( 0 ) = 0 , 𝜆 𝑢    𝑇 ( 1 ) = 0 , 𝜆 𝑢     𝑇 ( 0 ) = 0 , 𝜆 𝑢   ( 𝑡 ) = 𝜆 1 0  𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ≥  𝑡 ( 𝑠 ) 𝑑 𝑠 𝑡 − 2 2  𝜆  1 0 1 2  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ≥  𝑡 ( 𝑠 ) 𝑑 𝑠 𝑡 − 2 2  𝜆  1 0 m a x [ ] 𝜏 ∈ 0 , 1  𝐺 ( 𝜏 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    =  𝑡 ( 𝑠 ) 𝑑 𝑠 𝑡 − 2 2  ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ [ ]  𝑇 , ∀ 𝑡 ∈ 0 , 1 , 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , 𝜆 𝑢    ( 𝑡 ) = 𝜆 1 0 𝐺 1  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ≥ 1 ( 𝑠 ) 𝑑 𝑠 2  ( 1 − 𝑡 ) 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( s ) , 𝑢    ≥ 1 ( 𝑠 ) 𝑑 𝑠 2  ( 1 − 𝑡 ) 𝜆 1 0 m a x [ ] 𝜏 ∈ 0 , 1 𝐺 1  ( 𝜏 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    = 1 ( 𝑠 ) 𝑑 𝑠 2 ‖ ‖  𝑇 ( 1 − 𝑡 ) 𝜆 𝑢   ‖ ‖ [ ] −  𝑇 , ∀ 𝑡 ∈ 0 , 1 , 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , 𝜆 𝑢     ( 𝑡 ) = 𝜆 1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≥ 𝑡 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≥ 𝑡 𝜆 1 0 m a x [ ] 𝜏 ∈ 0 , 1 𝐺 2 (  𝜏 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢   (  ‖ ‖  𝑇 𝑠 ) 𝑑 𝑠 = 𝑡 𝜆 𝑢    ‖ ‖ [ ] , ∀ 𝑡 ∈ 0 , 1 , 𝑢 ∈ 𝑃 ⧵ { 𝜃 } . ( 2 . 2 1 ) Therefore, 𝑇 ( 𝑃 ⧵ { 𝜃 } ) ⊂ 𝑃 follows from ( 2.21 ). Obviously, 𝑢 ∗ is a positive solution of BVP ( 1.1 ) if and only if 𝑢 ∗ is a positive fixed point of the integral operator 𝑇 𝜆 in 𝑃 . Lemma 2.4. Suppose that ( 𝐻 ) and ( 2.17 ) hold. Then for any 𝑅 > 𝑟 > 0 , 𝑇 𝜆 ∶ 𝑃 𝑅 ⧵ 𝑃 𝑟 → 𝑃 is completely continuous. Proof. First of all, notice that 𝑇 𝜆 maps 𝑃 𝑅 ⧵ 𝑃 𝑟 into 𝑃 by Lemma 2.3 . Next, we show that 𝑇 𝜆 is bounded. In fact, for any 𝑢 ∈ 𝑃 𝑅 ⧵ 𝑃 𝑟 , by ( 2.10 ) we can get 𝑟 8  𝑡 𝑡 − 2 2   𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑡 − 2 2  𝑟 𝑅 , 4 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ ( 1 − 𝑡 ) 𝑅 , 𝑟 𝑡 ≤ − 𝑢   [ ] ( 𝑡 ) ≤ 𝑅 , ∀ 𝑡 ∈ 0 , 1 . ( 2 . 2 2 ) Choose positive numbers 𝑐 1 ≤ m i n { 𝑁 1 , ( 𝑟 / 8 ) 𝑁 1 } , 𝑐 2 ≤ m i n { 𝑁 2 , ( 𝑟 / 4 ) 𝑁 2 } , 𝑐 3 ≥ m a x { 𝑁 3 − 1 , 𝑁 3 − 1 𝑅 } . This, together with ( 𝐻 ), ( 2.22 ), ( 2.16 ), and Lemma 2.1 yields that | | |  𝑇 𝜆 𝑢    | | |  ( 𝑡 ) = 𝜆 1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑐 1 𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   𝑠 / 2 𝑠 − 2 2  , 𝑐 2 𝑢  ( 𝑠 ) 𝑐 2 ( 1 − 𝑠 ) ( 1 − 𝑠 ) , ( − 1 ) 𝑐 3 𝑢   ( 𝑠 ) − 𝑐 3   𝑑 𝑠 ≤ 𝜆 1 0 𝑠 𝑐 𝛼 1 1  𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   / 2 𝛽 1 𝑐 𝛼 2 2  𝑢  ( 𝑠 ) 𝑐 2  ( 1 − 𝑠 ) 𝛽 2 𝑐 𝛽 3 3  𝑢   ( 𝑠 ) − 𝑐 3  𝛼 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 s ≤ 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 𝑅 𝛽 1 + 𝛽 2 + 𝛼 3 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  [ ] , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ , ∀ 𝑡 ∈ 0 , 1 , 𝑢 ∈ 𝑃 𝑅 ⧵ 𝑃 𝑟 . ( 2 . 2 3 ) Thus, 𝑇 𝜆 is bounded on 𝑃 𝑅 ⧵ 𝑃 𝑟 . Now we show that 𝑇 𝜆 is a compact operator on 𝑃 𝑅 ⧵ 𝑃 𝑟 . By ( 2.23 ) and Ascoli-Arzela theorem, it suffices to show that 𝑇 𝜆 𝑉 is equicontinuous for arbitrary bounded subset 𝑉 ⊂ 𝑃 𝑅 ⧵ 𝑃 𝑟 . Since for each 𝑢 ∈ 𝑉 , ( 2.22 ) holds, we may choose still positive numbers 𝑐 1 ≤ m i n { 𝑁 1 , ( 𝑟 / 8 ) 𝑁 1 } , 𝑐 2 ≤ m i n { 𝑁 2 , ( 𝑟 / 4 ) 𝑁 2 } , 𝑐 3 ≥ m a x { 𝑁 3 − 1 , 𝑁 3 − 1 𝑅 } . Then | | |  𝑇 𝜆 𝑢     | | |  ( 𝑡 ) = 𝜆 1 𝑡 𝑓  𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ( 𝑠 ) 𝑑 𝑠 ≤ 𝐶 0  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 = ∶ 𝐻 ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) , ( 2 . 2 4 ) where 𝐶 0 = 𝜆 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 𝑅 𝛽 1 + 𝛽 2 + 𝛼 3 . Notice that  1 0 𝐻 ( 𝑡 ) 𝑑 𝑡 = 𝐶 0  1 0  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 𝑑 𝑡 = 𝐶 0  1 0  𝑠 0 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑡 𝑑 𝑠 = 𝐶 0  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ . ( 2 . 2 5 ) Thus for any given 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ] with 𝑡 1 ≤ 𝑡 2 and for any 𝑢 ∈ 𝑉 , we get | | |  𝑇 𝜆 𝑢     𝑡 2  −  𝑇 𝜆 𝑢     𝑡 1  | | | ≤  𝑡 2 𝑡 1 | | |  𝑇 𝜆 𝑢     ( | | |  𝑡 ) 𝑑 𝑡 ≤ 𝑡 2 𝑡 1 𝐻 ( 𝑡 ) 𝑑 𝑡 . ( 2 . 2 6 ) From ( 2.25 ), ( 2.26 ), and the absolute continuity of integral function, it follows that 𝑇 𝜆 𝑉 is equicontinuous. Therefore, 𝑇 𝜆 𝑉 is relatively compact, that is, 𝑇 𝜆 is a compact operator on 𝑃 𝑅 ⧵ 𝑃 𝑟 . Finally, we show that 𝑇 𝜆 is continuous on 𝑃 𝑅 ⧵ 𝑃 𝑟 . Suppose 𝑢 𝑛 , 𝑢 ∈ 𝑃 𝑅 ⧵ 𝑃 𝑟 , 𝑛 = 1 , 2 , … and ‖ 𝑢 𝑛 − 𝑢 ‖ 2 → 0 , ( 𝑛 → + ∞ ) . Then 𝑢 𝑛   ( 𝑡 ) → 𝑢   ( 𝑡 ) , 𝑢  𝑛 ( 𝑡 ) → 𝑢 ′ ( 𝑡 ) and 𝑢 𝑛 ( 𝑡 ) → 𝑢 ( 𝑡 ) as 𝑛 → + ∞ uniformly, with respect to 𝑡 ∈ [ 0 , 1 ] . From ( 𝐻 ) , choose still positive numbers 𝑐 1 ≤ m i n { 𝑁 1 , ( 𝑟 / 8 ) 𝑁 1 } , 𝑐 2 ≤ m i n { 𝑁 2 , ( 𝑟 / 4 ) 𝑁 2 } , 𝑐 3 ≥ m a x { 𝑁 3 − 1 , 𝑁 3 − 1 𝑅 } . Then  0 ≤ 𝑓 𝑡 , 𝑢 𝑛 ( 𝑡 ) , 𝑢  𝑛 ( 𝑡 ) , 𝑢 𝑛    ( 𝑡 ) ≤ 𝐶 0 𝑓  𝑡 𝑡 , 𝑡 − 2 2  , 1 − 𝑡 , − 1 , 𝑡 ∈ ( 0 , 1 ) , 0 ≤ 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝑛 ( 𝑠 ) , 𝑢  𝑛 ( 𝑠 ) , 𝑢 𝑛    ( 𝑠 ) ≤ 𝐶 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  [ ] , 1 − 𝑠 , − 1 , 𝑡 ∈ 0 , 1 , 𝑠 ∈ ( 0 , 1 ) . ( 2 . 2 7 ) By ( 2.17 ), we know that 𝑠 𝑓 ( 𝑠 , 𝑠 − 𝑠 2 / 2 , 1 − 𝑠 , − 1 ) is integrable on [ 0 , 1 ] . Thus, from the Lebesgue dominated convergence theorem, it follows that l i m 𝑛 → + ∞ ‖ ‖  𝑇 𝜆 𝑢 𝑛  −  𝑇 𝜆 𝑢  ‖ ‖ 2 = l i m 𝑛 → + ∞ ‖ ‖  T 𝜆 𝑢 𝑛    −  𝑇 𝜆 𝑢    ‖ ‖ ≤ l i m 𝑛 → + ∞ 𝜆  1 0 𝑠 | | 𝑓  𝑠 , 𝑢 𝑛 ( 𝑠 ) , 𝑢  𝑛 ( 𝑠 ) , 𝑢 𝑛     ( 𝑠 ) − 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    | |  ( 𝑠 ) 𝑑 𝑠 = 𝜆 1 0 𝑠 | | | l i m 𝑛 → + ∞  𝑓  𝑠 , 𝑢 𝑛 ( 𝑠 ) , 𝑢  𝑛 ( 𝑠 ) , 𝑢 𝑛     ( 𝑠 ) − 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢 𝑛   | | | ( 𝑠 )   𝑑 𝑠 = 0 . ( 2 . 2 8 ) Thus, 𝑇 𝜆 is continuous on 𝑃 𝑅 ⧵ 𝑃 𝑟 . Therefore, 𝑇 𝜆 ∶ 𝑃 𝑅 ⧵ 𝑃 𝑟 → 𝑃 is completely continuous. 3. A Necessary and Sufficient Condition for Existence of Positive Solutions In this section, by using the fixed point theorem of cone, we establish the following necessary and sufficient condition for the existence of positive solutions for BVP ( 1.1 ). Theorem 3.1. Suppose ( 𝐻 ) holds, then BVP ( 1.1 ) has at least one positive solution for any 𝜆 > 0 if and only if the integral inequality ( 2.17 ) holds. Proof. Suppose first that 𝑢 ( 𝑡 ) be a positive solution of BVP ( 1.1 ) for any fixed 𝜆 > 0 . Then there exist constants 𝐼 𝑖 ( 𝑖 = 1 , 2 , 3 , 4 ) with 0 < 𝐼 𝑖 < 1 < 𝐼 𝑖 + 1 , 𝑖 = 1 , 3 such that 𝐼 1  𝑡 𝑡 − 2 2  ≤ 𝑢 ( 𝑡 ) ≤ 𝐼 2  𝑡 𝑡 − 2 2  , 𝐼 3 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ 𝐼 4 [ ] ( 1 − 𝑡 ) , 𝑡 ∈ 0 , 1 . ( 3 . 1 ) In fact, it follows from 𝑢 ( 4 ) ( 𝑡 ) ≥ 0 , 𝑡 ∈ ( 0 , 1 ) and 𝑢 ( 0 ) = 𝑢  ( 1 ) = 𝑢   ( 0 ) = 𝑢    ( 1 ) = 0 , that 𝑢    ( 𝑡 ) ≤ 0 for 𝑡 ∈ ( 0 , 1 ] and 𝑢   ( 𝑡 ) ≤ 0 , 𝑢  ( 𝑡 ) ≥ 0 for 𝑡 ∈ [ 0 , 1 ] . By the concavity of 𝑢 ( 𝑡 ) and 𝑢  ( 𝑡 ) , we have  𝑡 𝑢 ( 𝑡 ) ≥ 𝑡 𝑢 ( 1 ) + ( 1 − 𝑡 ) 𝑢 ( 0 ) = 𝑡 ‖ 𝑢 ‖ ≥ 𝑡 − 2 2  𝑢 ‖ 𝑢 ‖ ,  ( 𝑡 ) ≥ 𝑡 𝑢  ( 1 ) + ( 1 − 𝑡 ) 𝑢  ( ‖ ‖ 𝑢 0 ) = ( 1 − 𝑡 )  ‖ ‖ [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 3 . 2 ) On the other hand,  𝑢 ( 𝑡 ) = 1 0 𝐺 0  ( 𝑡 , 𝑠 ) − 𝑢     ( 𝑠 ) 𝑑 𝑠 = 𝑡 0 𝑠  − 𝑢     ( 𝑠 ) 𝑑 𝑠 + 1 𝑡 𝑡  − 𝑢    ≤ 𝑡 ( 𝑠 ) 𝑑 𝑠 2 2 ‖ ‖ 𝑢   ‖ ‖ ‖ ‖ 𝑢 + 𝑡 ( 1 − 𝑡 )   ‖ ‖ =  𝑡 𝑡 − 2 2  ‖ ‖ 𝑢   ‖ ‖ , 𝑢   ( 𝑡 ) = 1 𝑡 − 𝑢   ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ ( 1 − 𝑡 )   ‖ ‖ [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 3 . 3 ) Let 𝐼 1 = m i n { ‖ 𝑢 ‖ , 1 / 2 } , let 𝐼 2 = 𝐼 4 = m a x { ‖ 𝑢   ‖ , 2 } , and let 𝐼 3 = m i n { ‖ 𝑢 ′ ‖ , 1 / 2 } , then ( 3.1 ) holds. Choose positive numbers 𝑐 1 ≤ 𝑁 1 𝐼 2 − 1 , 𝑐 2 ≤ 𝑁 2 𝐼 4 − 1 , 𝑐 3 ≥ m a x { 𝑁 3 − 1 , 𝑁 3 − 1 ‖ 𝑢 ‖ 2 } . This, together with ( 𝐻 ), ( 1.2 ), and ( 2.18 ) yields that 𝑓  𝑡 𝑡 , 𝑡 − 2 2   , 1 − 𝑡 , − 1 = 𝑓 𝑡 , 𝑐 1 𝑡 − 𝑡 2 / 2 𝑐 1 𝑢 ( 𝑡 ) 𝑢 ( 𝑡 ) , 𝑐 2 1 − 𝑡 𝑐 2 𝑢  𝑢 ( 𝑡 )  1 ( 𝑡 ) , 𝑐 3 𝑐 3 − 𝑢   𝑢 ( 𝑡 )    ( 𝑡 ) ≤ 𝑐 𝛼 1 1  𝑡 − 𝑡 2 / 2 𝑐 1 𝑢  ( 𝑡 ) 𝛽 1 𝑐 𝛼 2 2  1 − 𝑡 𝑐 2 𝑢   ( 𝑡 ) 𝛽 2  1 𝑐 3  𝛼 3  𝑐 3 − 𝑢    ( 𝑡 ) 𝛽 3 𝑓  𝑡 , 𝑢 ( 𝑡 ) , 𝑢  ( 𝑡 ) , 𝑢    ( 𝑡 ) ≤ 𝑐 𝛼 1 1  1 𝑐 1 𝐼 1  𝛽 1 𝑐 𝛼 2 2  1 𝑐 2 𝐼 3  𝛽 2  1 𝑐 3  𝛼 3  − 𝑐 3 𝑢    ( 𝑡 ) 𝛽 3 𝑓  𝑡 , 𝑢 ( 𝑡 ) , 𝑢  ( 𝑡 ) , 𝑢   (  𝑡 ) = 𝐶 ∗  − 𝑢   (  𝑡 ) − 𝛽 3 𝑓  𝑡 , 𝑢 ( 𝑡 ) , 𝑢  ( 𝑡 ) , 𝑢   (  𝑡 ) , 𝑡 ∈ ( 0 , 1 ) , ( 3 . 4 ) where 𝐶 ∗ = 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 𝐼 − 𝛽 1 1 𝐼 − 𝛽 2 3 . Hence, integrating ( 3.4 ) from 𝑡 to 1, we obtain 𝜆  1 𝑡  − 𝑢    ( 𝑠 ) 𝛽 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝐶 ∗  − 𝑢     ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) . ( 3 . 5 ) Since − 𝑢   ( 𝑡 ) increases on [ 0 , 1 ] , we get  − 𝑢    ( 𝑡 ) 𝛽 3 𝜆  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝐶 ∗  − 𝑢     ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) , ( 3 . 6 ) that is, 𝜆  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝐶 ∗ − 𝑢    ( 𝑡 ) ( − 𝑢   ( 𝑡 ) ) 𝛽 3 , 𝑡 ∈ ( 0 , 1 ) . ( 3 . 7 ) Notice that 𝛽 3 < 1 , integrating ( 3.7 ) from 0 to 1, we have 𝜆  1 0  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 𝑑 𝑡 ≤ 𝐶 ∗  1 − 𝛽 3  − 1  − 𝑢    ( 1 ) 1 − 𝛽 3 . ( 3 . 8 ) That is, 𝜆  1 0  𝑠 0 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑡 𝑑 𝑠 ≤ 𝐶 ∗  1 − 𝛽 3  − 1  − 𝑢    ( 1 ) 1 − 𝛽 3 . ( 3 . 9 ) Thus,  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ . ( 3 . 1 0 ) By an argument similar to the one used in deriving ( 3.5 ), we can obtain 𝜆  1 𝑡  − 𝑢    ( 𝑠 ) 𝛼 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝐶 ∗  − 𝑢     ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) , ( 3 . 1 1 ) where 𝐶 ∗ = 𝑐 𝛽 1 − 𝛼 1 1 𝑐 𝛽 2 − 𝛼 2 2 𝑐 𝛼 3 − 𝛽 3 3 𝐼 − 𝛼 1 2 𝐼 − 𝛼 2 4 . So, 𝜆  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝐶 ∗ ‖ 𝑢 ‖ − 𝛼 3 2  − 𝑢     ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) . ( 3 . 1 2 ) Integrating ( 3.12 ) from 0 to 1, we have 𝜆  1 0  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 𝑑 𝑡 ≥ 𝐶 ∗ ‖ 𝑢 ‖ − 𝛼 3 2  − 𝑢    . ( 1 ) ( 3 . 1 3 ) That is, 𝜆  1 0  𝑠 0 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑡 𝑑 𝑠 ≥ 𝐶 ∗ ‖ 𝑢 ‖ − 𝛼 3 2  − 𝑢    . ( 1 ) ( 3 . 1 4 ) So,  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 > 0 . ( 3 . 1 5 ) This and ( 3.10 ) imply that ( 2.17 ) holds. Now assume that ( 2.17 ) holds, we will show that BVP ( 1.1 ) has at least one positive solution for any 𝜆 > 0 . By ( 2.17 ), there exists a sufficient small 𝛿 > 0 such that  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 > 0 . ( 3 . 1 6 ) For any fixed 𝜆 > 0 , first of all, we prove ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ 2 ≥ ‖ 𝑢 ‖ 2 , ∀ 𝑢 ∈ 𝜕 𝑃 𝑟 , ( 3 . 1 7 ) where 0 < 𝑟 ≤ m i n { 𝑁 1 , 𝑁 2 , 𝑁 3 , ( 𝜆 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 ) ∫ 𝛿 1 − 𝛿 𝑠 𝑓 ( 𝑠 , 𝑠 − 𝑠 2 / 2 , 1 − 𝑠 , − 1 ) 𝑑 𝑠 ) 1 / ( 1 − ( 𝛽 1 + 𝛽 2 + 𝛽 3 ) ) } . Let 𝑢 ∈ 𝜕 𝑃 𝑟 , then 𝑟 8  𝑡 𝑡 − 2 2   𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑟 𝑡 − 2 2  ≤ 𝑁 1  𝑡 𝑡 − 2 2  , 𝑟 4 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ 𝑟 ( 1 − 𝑡 ) ≤ 𝑁 2 ( 1 − 𝑡 ) , 𝛿 𝑟 ≤ 𝑟 𝑡 ≤ − 𝑢   ( 𝑡 ) ≤ 𝑟 ≤ 𝑁 3 [ ] . , ∀ 𝑡 ∈ 𝛿 , 1 − 𝛿 ( 3 . 1 8 ) From Lemma 2.1 , ( 3.18 ), and ( 𝐻 ), we get ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ 2 = ‖ ‖  𝑇 𝜆 𝑢    ‖ ‖ ≥ 𝜆 m a x [ ] 𝑡 ∈ 𝛿 , 1 − 𝛿  1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≥ 𝛿 𝜆 𝛿 1 − 𝛿  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  𝑠 / 2 𝑠 − 2 2  , 𝑢  ( 𝑠 ) (  1 − 𝑠 1 − 𝑠 ) , ( − 1 ) − 𝑢   (    𝑠 ) 𝑑 𝑠 ≥ 𝛿 𝜆 𝛿 1 − 𝛿 𝑠  𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  / 2 𝛽 1  𝑢  ( 𝑠 )  1 − 𝑠 𝛽 2  − 𝑢    ( 𝑠 ) 𝛽 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2   𝑟 , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝛿 8  𝛽 1  𝑟 4  𝛽 2 ( 𝛿 𝑟 ) 𝛽 3 𝜆  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 ) 𝑟 𝛽 1 + 𝛽 2 + 𝛽 3 𝜆  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝑟 = ‖ 𝑢 ‖ 2 , 𝑢 ∈ 𝜕 𝑃 𝑟 . ( 3 . 1 9 ) Thus, ( 3.17 ) holds. Next, we claim that ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ 2 ≤ ‖ 𝑢 ‖ 2 , ∀ 𝑢 ∈ 𝜕 𝑃 𝑅 , ( 3 . 2 0 ) where 𝑅 ≥ m a x { 8 𝑁 1 − 1 , 4 𝑁 2 − 1 , ( 𝜆 𝑁 𝛼 3 − 𝛽 3 3 ∫ 1 0 𝑠 𝑓 ( 𝑠 , 𝑠 − 𝑠 2 / 2 , 1 − 𝑠 , − 1 ) 𝑑 𝑠 ) 1 / ( 1 − ( 𝛽 1 + 𝛽 2 + 𝛽 3 ) ) } . Let 𝑐 = 𝑁 3 / 𝑅 , then for 𝑢 ∈ 𝜕 𝑃 𝑅 , we get 𝑁 1 − 1  𝑡 𝑡 − 2 2  ≤ 𝑅 8  𝑡 𝑡 − 2 2   𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑅 𝑡 − 2 2  , 𝑁 2 − 1 𝑅 ( 1 − 𝑡 ) ≤ 4 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ 𝑅 ( 1 − 𝑡 ) , − 𝑐 𝑢   ( 𝑡 ) ≤ 𝑐 ‖ 𝑢 ‖ 2 = 𝑐 𝑅 = 𝑁 3 [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 3 . 2 1 ) Therefore, by Lemma 2.1 and ( 𝐻 ), it follows that | | |  𝑇 𝜆 𝑢    | | |  ( 𝑡 ) = 𝜆 1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  𝑠 / 2 𝑠 − 2 2  , 𝑢  ( 𝑠 )  1 1 − 𝑠 ( 1 − 𝑠 ) , ( − 1 ) 𝑐   − 𝑐 𝑢      ( s ) 𝑑 𝑠 ≤ 𝜆 1 0  𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  / 2 𝛽 1  𝑢  ( 𝑠 )  1 − 𝑠 𝛽 2  1 𝑐  𝛽 3  − 𝑐 𝑢    ( 𝑠 ) 𝛼 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝑅 𝛽 1 + 𝛽 2  𝑁 3 𝑅  𝛼 3 − 𝛽 3 𝑅 𝛼 3 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 = 𝑅 𝛽 1 + 𝛽 2 + 𝛽 3  𝑁 3  𝛼 3 − 𝛽 3 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝑅 = ‖ 𝑢 ‖ 2 , 𝑢 ∈ 𝜕 𝑃 𝑅 . ( 3 . 2 2 ) This implies that ( 3.20 ) holds. By Lemmas 1.1 and 2.4 , ( 3.17 ), and ( 3.20 ), we obtain that 𝑇 𝜆 has a fixed point in 𝑃 𝑅 ⧵ 𝑃 𝑟 . Therefore, BVP ( 1.1 ) has a positive solution in 𝑃 𝑅 ⧵ 𝑃 𝑟 for any 𝜆 > 0 . 4. Unbounded Connected Branch of Positive Solutions In this section, we study the global continua results under the hypotheses ( 𝐻 ) and ( 2.17 ). Let 𝐿 = { ( 𝜆 , 𝑢 ) ∈ ( 0 , + ∞ ) × ( 𝑃 ⧵ { 𝜃 } ) ∶ ( 𝜆 , 𝑢 ) s a t i s fi e s B V P ( 1 . 1 ) } , ( 4 . 1 ) then, by Theorem 3.1 , 𝐿 ∩ ( { 𝜆 } × 𝑃 ) ≠ ∅ for any 𝜆 > 0 . Theorem 4.1. Suppose ( 𝐻 ) and ( 2.17 ) hold, then the closure 𝐿 of positive solution set possesses an unbounded connected branch 𝐶 which comes from ( 0 , 𝜃 ) such that (i) for any 𝜆 > 0 , 𝐶 ∩ ( { 𝜆 } × 𝑃 ) ≠ ∅ , and (ii) l i m ( 𝜆 , 𝑢 𝜆 ) ∈ 𝐶 , 𝜆 → 0 + ‖ 𝑢 𝜆 ‖ 2 = 0 , l i m ( 𝜆 , 𝑢 𝜆 ) ∈ 𝐶 , 𝜆 → + ∞ ‖ 𝑢 𝜆 ‖ 2 = + ∞ . Proof. We now prove our conclusion by the following several steps. First, we prove that for arbitrarily given 0 < 𝜆 1 < 𝜆 2 < + ∞ , 𝐿 ∩ ( [ 𝜆 1 , 𝜆 2 ] × 𝑃 ) is bounded. In fact, let  𝑅 = 2 m a x 8 𝑁 1 − 1 , 4 𝑁 2 − 1 ,  𝜆 2 𝑁 𝛼 3 − 𝛽 3 3  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2   , 1 − 𝑠 , − 1 𝑑 𝑠 1 / ( 1 − ( 𝛽 1 + 𝛽 2 + 𝛽 3 ) )  , ( 4 . 2 ) then for 𝑢 ∈ 𝑃 ⧵ { 𝜃 } and ‖ 𝑢 ‖ 2 ≥ 𝑅 , we get 𝑁 1 − 1  𝑡 𝑡 − 2 2  ≤ 𝑅 8  𝑡 𝑡 − 2 2   𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑡 − 2 2  ‖ 𝑢 ‖ 2 , 𝑁 2 − 1 ( 𝑅 1 − 𝑡 ) ≤ 4 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ ( 1 − 𝑡 ) ‖ 𝑢 ‖ 2 [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 4 . 3 ) Therefore, by Lemma 2.1 and ( 𝐻 ), it follows that ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ 2 ≤ ‖ ‖ 𝑇 𝜆 2 𝑢 ‖ ‖ 2 ≤ 𝜆 2  1 0  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 2  1 0  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  𝑠 / 2 𝑠 − 2 2  , 𝑢  ( 𝑠 ) 1 − s ( 1 − 𝑠 ) , ( − 1 ) ‖ 𝑢 ‖ 2 𝑁 3 𝑁 3 ‖ 𝑢 ‖ 2  − 𝑢     ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 2 ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 2  𝑁 3 ‖ 𝑢 ‖ 2  𝛼 3 − 𝛽 3 ‖ 𝑢 ‖ 𝛼 3 2  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 = 𝜆 2 ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 + 𝛽 3 2  𝑁 3  𝛼 3 − 𝛽 3  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < ‖ 𝑢 ‖ 2  𝜆 , ∀ 𝜆 ∈ 1 , 𝜆 2  . ( 4 . 4 ) Let 1 𝑟 = 2  𝑁 m i n 1 , 𝑁 2 , 𝑁 3 ,  𝜆 1 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 )  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2   , 1 − 𝑠 , − 1 𝑑 𝑠 1 / ( 1 − ( 𝛽 1 + 𝛽 2 + 𝛽 3 ) )  , ( 4 . 5 ) where 𝛿 is given by ( 3.16 ). Then for 𝑢 ∈ 𝑃 ⧵ { 𝜃 } and ‖ 𝑢 ‖ 2 ≤ 𝑟 , we get ‖ 𝑢 ‖ 2 8  𝑡 𝑡 − 2 2   𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑟 𝑡 − 2 2  ≤ 𝑁 1  𝑡 𝑡 − 2 2  ; ‖ 𝑢 ‖ 2 4 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ 𝑟 ( 1 − 𝑡 ) ≤ 𝑁 2 ( 1 − 𝑡 ) , 𝛿 ‖ 𝑢 ‖ 2 ≤ 𝑡 ‖ 𝑢 ‖ 2 ≤ − 𝑢   ( 𝑡 ) ≤ 𝑟 ≤ 𝑁 3 [ ] . , ∀ 𝑡 ∈ 𝛿 , 1 − 𝛿 ( 4 . 6 ) Therefore, by Lemma 2.1 and ( 𝐻 ), it follows that ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ ≥ ‖ ‖ 𝑇 𝜆 1 𝑢 ‖ ‖ ≥ 𝜆 1 m a x [ ] 𝑡 ∈ 𝛿 , 1 − 𝛿  1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ( 𝑠 ) 𝑑 𝑠 ≥ 𝛿 𝜆 1  𝛿 1 − 𝛿  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  𝑠 / 2 𝑠 − 2 2  , 𝑢  ( 𝑠 ) (  1 − 𝑠 1 − 𝑠 ) , ( − 1 ) − 𝑢   (   𝑠 ) 𝑑 𝑠 ≥ 𝛿 𝜆 1  𝛿 1 − 𝛿 𝑠  𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  / 2 𝛽 1  𝑢  ( 𝑠 )  1 − 𝑠 𝛽 2  − 𝑢    ( 𝑠 ) 𝛽 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2   , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝛿 ‖ 𝑢 ‖ 2 ) 8  𝛽 1  ‖ 𝑢 ‖ 2 4  𝛽 2  𝛿 ‖ 𝑢 ‖ 2  𝛽 3 𝜆 1  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 ) ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 + 𝛽 3 2 𝜆 1  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 > ‖ 𝑢 ‖ 2 , 𝑢 ∈ 𝜕 𝑃 𝑟 . ( 4 . 7 ) Therefore, 𝑢 = 𝑇 𝜆 𝑢 has no positive solution in ( [ 𝜆 1 , 𝜆 2 ] × ( 𝑃 ⧵ 𝑃 𝑅 ) ) ∪ ( [ 𝜆 1 , 𝜆 2 ] × 𝑃 𝑟 ) . As a consequence, 𝐿 ∩ ( [ 𝜆 1 , 𝜆 2 ] × 𝑃 ) is bounded. By the complete continuity of 𝑇 𝜆 , 𝐿 ∩ ( [ 𝜆 1 , 𝜆 2 ] × 𝑃 ) is compact. Second, we choose sequences { 𝑎 𝑛 } ∞ 𝑛 = 1 and { 𝑏 𝑛 } ∞ 𝑛 = 1 satisfy 0 < ⋯ < 𝑎 𝑛 < ⋯ < 𝑎 1 < 𝑏 1 < ⋯ < 𝑏 𝑛 < ⋯ , l i m 𝑛 → + ∞ 𝑎 𝑛 = 0 , l i m 𝑛 → + ∞ 𝑏 𝑛 = + ∞ . ( 4 . 8 ) We are to prove that for any positive integer 𝑛 , there exists a connected branch 𝐶 𝑛 of 𝐿 satisfying 𝐶 𝑛 ∩ 𝑎   𝑛   × 𝑃 ≠ ∅ , 𝐶 𝑛 ∩ 𝑏   𝑛   × 𝑃 ≠ ∅ . ( 4 . 9 ) Let 𝑛 be fixed, suppose that for any ( 𝑏 𝑛 , 𝑢 ) ∈ 𝐿 ∩ ( { 𝑏 𝑛 } × 𝑃 ) , the connected branch 𝐶 𝑢 of 𝐿 ∩ ( [ 𝑎 𝑛 , 𝑏 𝑛 ] × 𝑃 ) , passing through ( 𝑏 𝑛 , 𝑢 ) , leads to 𝐶 𝑢 ∩ ( { 𝑎 𝑛 } × 𝑃 ) = ∅ . Since 𝐶 𝑢 is compact, there exists a bounded open subset Ω 1 of [ 𝑎 𝑛 , 𝑏 𝑛 ] × 𝑃 such that 𝐶 𝑢 ⊂ Ω 1 , Ω 1 ∩ ( { 𝑎 𝑛 } × 𝑃 ) = ∅ , and Ω 1 ∩ ( [ 𝑎 𝑛 , 𝑏 𝑛 ] × { 𝜃 } ) = ∅ , where Ω 1 and later 𝜕 Ω 1 denote the closure and boundary of Ω 1 with respect to [ 𝑎 𝑛 , 𝑏 𝑛 ] × 𝑃 . If 𝐿 ∩ 𝜕 Ω 1 ≠ ∅ , then 𝐶 u and 𝐿 ∩ 𝜕 Ω 1 are two disjoint closed subsets of 𝐿 ∩ Ω 1 . Since 𝐿 ∩ Ω 1 is a compact metric space, there are two disjoint compact subsets 𝑀 1 and 𝑀 2 of 𝐿 ∩ Ω 1 such that 𝐿 ∩ Ω 1 = 𝑀 1 ∪ 𝑀 2 , 𝐶 𝑢 ⊂ 𝑀 1 , and 𝐿 ∩ 𝜕 Ω 1 ⊂ 𝑀 2 . Evidently, 𝛾 = ∶ d i s t ( 𝑀 1 , 𝑀 2 ) > 0 . Denoting by 𝑉 the 𝛾 / 3 -neighborhood of 𝑀 1 and letting Ω 𝑢 = Ω 1 ∩ 𝑉 , then it follows that 𝐶 𝑢 ⊂ Ω 𝑢 , Ω 𝑢 ∩ 𝑎    𝑛   ∪ 𝑎 × 𝑃   𝑛 , 𝑏 𝑛  × { 𝜃 }   = ∅ , 𝐿 ∩ 𝜕 Ω 𝑢 = ∅ . ( 4 . 1 0 ) If 𝐿 ∩ 𝜕 Ω 1 = ∅ , then taking Ω 𝑢 = Ω 1 . It is obvious that in { 𝑏 𝑛 } × 𝑃 , the family of { Ω 𝑢 ∩ ( { 𝑏 𝑛 } × 𝑃 ) ∶ ( 𝑏 𝑛 , 𝑢 ) ∈ 𝐿 } makes up an open covering of 𝐿 ∩ ( { 𝑏 𝑛 } × 𝑃 ) . Since 𝐿 ∩ ( { 𝑏 𝑛 } × 𝑃 ) is a compact set, there exists a finite subfamily { Ω 𝑢 𝑖 ∩ ( { 𝑏 𝑛 } × 𝑃 ) ∶ ( 𝑏 𝑛 , 𝑢 𝑖 ) ∈ 𝐿 } 𝑘 𝑖 = 1 which also covers 𝐿 ∩ ( { 𝑏 𝑛 } × 𝑃 ) . Let ⋃ Ω = 𝑘 𝑖 = 1 Ω 𝑢 𝑖 , then 𝑏 𝐿 ∩   𝑛   × 𝑃 ⊂ Ω , 𝑎 Ω ∩    𝑛   ∪ 𝑎 × 𝑃   𝑛 , 𝑏 𝑛  × { 𝜃 }   = ∅ , 𝐿 ∩ 𝜕 Ω = ∅ . ( 4 . 1 1 ) Hence, by the homotopy invariance of the fixed point index, we obtain 𝑖  𝑇 𝑏 𝑛 𝑏 , Ω ∩   𝑛     𝑇 × 𝑃 , 𝑃 = 𝑖 𝑎 𝑛 𝑎 , Ω ∩   𝑛    × 𝑃 , 𝑃 = 0 . ( 4 . 1 2 ) By the first step of this proof, the construction of Ω , ( 4.4 ), and ( 4.7 ), it follows easily that there exist 0 < 𝑟 𝑛 < 𝑅 𝑛 such that  𝑏 Ω ∩   𝑛    ∩ 𝑏 × 𝑃   𝑛  × 𝑃 𝑟 𝑛   = ∅ , 𝑏 Ω ∩   𝑛    ⊂ 𝑏 × 𝑃   𝑛  × 𝑃 𝑅 𝑛  𝑖  𝑇 , ( 4 . 1 3 ) 𝑏 𝑛 , 𝑃 𝑟 𝑛  𝑖  𝑇 , 𝑃 = 0 , ( 4 . 1 4 ) 𝑏 𝑛 , 𝑃 𝑅 𝑛  , 𝑃 = 1 . ( 4 . 1 5 ) However, by the excision property and additivity of the fixed point index, we have from ( 4.12 ) and ( 4.14 ) that 𝑖 ( 𝑇 𝑏 𝑛 , 𝑃 𝑅 𝑛 , 𝑃 ) = 0 , which contradicts ( 4.15 ). Hence, there exists some ( 𝑏 𝑛 , 𝑢 ) ∈ 𝐿 ∩ ( { 𝑏 𝑛 } × 𝑃 ) such that the connected branch 𝐶 𝑢 of 𝐿 ∩ ( [ 𝑎 𝑛 , 𝑏 𝑛 ] × 𝑃 ) containing ( 𝑏 𝑛 , 𝑢 ) satisfies that 𝐶 𝑢 ∩ ( { 𝑎 𝑛 } × 𝑃 ) ≠ ∅ . Let 𝐶 𝑛 be the connected branch of 𝐿 including 𝐶 𝑢 , then this 𝐶 𝑛 satisfies ( 4.9 ). By Lemma 1.2 , there exists a connected branch 𝐶 ∗ of l i m s u p 𝑛 → + ∞ 𝐶 𝑛 such that 𝐶 ∗ ∩ ( { 𝜆 } × 𝑃 ) ≠ ∅ for any 𝜆 > 0 . Noticing l i m s u p 𝑛 → + ∞ 𝐶 𝑛 ⊂ 𝐿 , we have 𝐶 ∗ ⊂ 𝐿 . Let 𝐶 be the connected branch of 𝐿 including 𝐶 ∗ , then 𝐶 ∩ ( { 𝜆 } × 𝑃 ) ≠ ∅ for any 𝜆 > 0 . Similar to ( 4.4 ) and ( 4.7 ), for any 𝜆 > 0 , ( 𝜆 , 𝑢 𝜆 ) ∈ 𝐶 , we have, by ( 𝐻 ), ( 4.2 ), ( 4.3 ), ( 4.5 ), ( 4.6 ), and Lemma 2.1 , ‖ ‖ 𝑢 𝜆 ‖ ‖ 2 = ‖ ‖ 𝑇 𝜆 𝑢 𝜆 ‖ ‖ 2  ≤ 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑢 𝜆 ( 𝑠 ) , 𝑢  𝜆 ( s ) , 𝑢 𝜆    ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 𝜆 ‖ ‖ 𝛽 1 + 𝛽 2 2  𝑁 3 ‖ ‖ 𝑢 𝜆 ‖ ‖ 2  𝛼 3 − 𝛽 3 ‖ ‖ 𝑢 𝜆 ‖ ‖ 𝛼 3 2  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  ‖ ‖ 𝑢 , 1 − 𝑠 , − 1 𝑑 𝑠 = 𝜆 𝜆 ‖ ‖ 𝛽 1 + 𝛽 2 + 𝛽 3 2  𝑁 3  𝛼 3 − 𝛽 3  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝜆 𝑅 𝛽 1 + 𝛽 2 + 𝛽 3  𝑁 3  𝛼 3 − 𝛽 3  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  ‖ ‖ 𝑢 , 1 − 𝑠 , − 1 𝑑 𝑠 , ( 4 . 1 6 ) 𝜆 ‖ ‖ 2 = ‖ ‖ 𝑇 𝜆 𝑢 𝜆 ‖ ‖ 2 ≥ 𝜆 m a x [ ] 𝑡 ∈ 𝛿 , 1 − 𝛿  1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝜆 ( 𝑠 ) , 𝑢  𝜆 ( 𝑠 ) , 𝑢 𝜆     ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≥ 𝜆 𝛿 𝜆 ‖ ‖ 2 8  𝛽 1  ‖ ‖ 𝑢 𝜆 ‖ ‖ 2 4  𝛽 2  𝛿 ‖ ‖ 𝑢 𝜆 ‖ ‖ 2  𝛽 3  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝜆 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 ) ‖ ‖ 𝑢 𝜆 ‖ ‖ 𝛽 1 + 𝛽 2 + 𝛽 3 2  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝜆 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 ) 𝑟 𝛽 1 + 𝛽 2 + 𝛽 3  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 , ( 4 . 1 7 ) where 𝛿 is given by ( 3.16 ). Let 𝜆 → 0 + in ( 4.16 ) and 𝜆 → + ∞ in ( 4.17 ), we have l i m  𝜆 , 𝑢 𝜆  ∈ 𝐶 , 𝜆 → 0 + ‖ ‖ 𝑢 𝜆 ‖ ‖ 2 = 0 , l i m  𝜆 , 𝑢 𝜆  ∈ 𝐶 , 𝜆 → + ∞ ‖ ‖ 𝑢 𝜆 ‖ ‖ 2 = + ∞ . ( 4 . 1 8 ) Therefore, Theorem 4.1 holds and the proof is complete. Acknowledgments This work is carried out while the author is visiting the University of New England. The author thanks Professor Yihong Du for his valuable advices and the Department of Mathematics for providing research facilities. The author also thanks the anonymous referees for their carefully reading of the first draft of the manuscript and making many valuable suggestions. Research is supported by the NSFC (10871120) and HESTPSP (J09LA08). <h4>References</h4> R. P. Agarwal and D. O'Regan, “ Nonlinear superlinear singular and nonsingular second order boundary value problems ,” Journal of Differential Equations , vol. 143, no. 1, pp. 60–95, 1998. L. Liu, P. Kang, Y. Wu, and B. 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Rabinowitz, “ Some aspects of nonlinear eigenvalue problems ,” The Rocky Mountain Journal of Mathematics , vol. 3, no. 2, pp. 161–202, 1973. D. J. Guo and V. Lakshmikantham, Nonlinear Problems in Abstract Cones , vol. 5 of Notes and Reports in Mathematics in Science and Engineering , Academic Press, Boston, Mass, USA, 1988. J. X. Sun, “A theorem in point set topology,” Journal of Systems Science & Mathematical Sciences , vol. 7, no. 2, pp. 148–150, 1987. // http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Boundary Value Problems Hindawi Publishing Corporation

Eigenvalue Problem and Unbounded Connected Branch of Positive Solutions to a Class of Singular Elastic Beam Equations

Boundary Value Problems , Volume 2011 (2011) – Mar 14, 2011

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Eigenvalue Problem and Unbounded Connected Branch of Positive Solutions to a Class of Singular Elastic Beam Equations //// Hindawi Publishing Corporation Home Journals About Us About this Journal Submit a Manuscript Table of Contents Journal Menu Abstracting and Indexing Aims and Scope Article Processing Charges Articles in Press Author Guidelines Bibliographic Information Contact Information Editorial Board Editorial Workflow Free eTOC Alerts Reviewers Acknowledgment Subscription Information Open Special Issues Published Special Issues Special Issue Guidelines Abstract Full-Text PDF Full-Text HTML Linked References How to Cite this Article Boundary Value Problems Volume 2011 (2011), Article ID 594128, 21 pages doi:10.1155/2011/594128 Research Article <h2>Eigenvalue Problem and Unbounded Connected Branch of Positive Solutions to a Class of Singular Elastic Beam Equations</h2> Huiqin Lu School of Mathematical Sciences, Shandong Normal University, Jinan, 250014 Shandong, China Received 16 October 2010; Revised 22 December 2010; Accepted 27 January 2011 Academic Editor: Kanishka Perera Copyright © 2011 Huiqin Lu. 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 This paper investigates the eigenvalue problem for a class of singular elastic beam equations where one end is simply supported and the other end is clamped by sliding clamps. Firstly, we establish a necessary and sufficient condition for the existence of positive solutions, then we prove that the closure of positive solution set possesses an unbounded connected branch which bifurcates from ( 0 , 𝜃 ) . Our nonlinearity 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) may be singular at 𝑢 , 𝑣 , 𝑡 = 0 and/or 𝑡 = 1 . 1. Introduction Singular differential equations arise in the fields of gas dynamics, Newtonian fluid mechanics, the theory of boundary layer, and so on. Therefore, singular boundary value problems have been investigated extensively in recent years (see [ 1 – 4 ] and references therein). This paper investigates the following fourth-order nonlinear singular eigenvalue problem: 𝑢 ( 4 )  ( 𝑡 ) = 𝜆 𝑓 𝑡 , 𝑢 ( 𝑡 ) , 𝑢  ( 𝑡 ) , 𝑢    ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) , 𝑢 ( 0 ) = 𝑢  ( 1 ) = 𝑢   ( 0 ) = 𝑢    ( 1 ) = 0 , ( 1 . 1 ) where 𝜆 ∈ ( 0 , + ∞ ) is a parameter and 𝑓 satisfies the following hypothesis: ( 𝐻 ) 𝑓 ∈ 𝐶 ( ( 0 , 1 ) × ( 0 , + ∞ ) × ( 0 , + ∞ ) × ( − ∞ , 0 ] , [ 0 , + ∞ ) ) , and there exist constants 𝛼 𝑖 , 𝛽 𝑖 , 𝑁 𝑖 , 𝑖 = 1 , 2 , 3 ( − ∞ < 𝛼 1 ≤ 0 ≤ 𝛽 1 < + ∞ , − ∞ < 𝛼 2 ≤ 0 ≤ 𝛽 2 < + ∞ , 0 ≤ 𝛼 3 ≤ 𝛽 3 < 1 , ∑ 3 𝑖 = 1 𝛽 𝑖 < 1 ; 0 < 𝑁 𝑖 ≤ 1 , 𝑖 = 1 , 2 , 3 ) such that for any 𝑡 ∈ ( 0 , 1 ) , 𝑢 , 𝑣 ∈ ( 0 , + ∞ ) , 𝑤 ∈ ( − ∞ , 0 ] , 𝑓 satisfies 𝑐 𝛽 1 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑐 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑐 𝛼 1 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 0 < 𝑐 ≤ 𝑁 1 , 𝑐 𝛽 2 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑢 , 𝑐 𝑣 , 𝑤 ) ≤ 𝑐 𝛼 2 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 0 < 𝑐 ≤ 𝑁 2 , 𝑐 𝛽 3 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑐 𝑤 ) ≤ 𝑐 𝛼 3 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 0 < 𝑐 ≤ 𝑁 3 . ( 1 . 2 ) Typical functions that satisfy the above sublinear hypothesis ( 𝐻 ) are those taking the form 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) = 𝑚 1  𝑚 𝑖 = 1 2  𝑚 𝑗 = 1 3  𝑘 = 1 𝑝 𝑖 , 𝑗 , 𝑘 ( 𝑡 ) 𝑢 𝑟 𝑖 𝑣 𝑠 𝑗 𝑤 𝜎 𝑘 , ( 1 . 3 ) where 𝑝 𝑖 , 𝑗 , 𝑘 ( 𝑡 ) ∈ 𝐶 [ ( 0 , 1 ) , ( 0 , + ∞ ) ] , 𝑟 𝑖 , 𝑠 𝑗 ∈ 𝑅 , 0 ≤ 𝜎 𝑘 < 1 , m a x { 𝑟 𝑖 , 0 } + m a x { 𝑠 𝑗 } + 𝜎 𝑘 < 1 , 𝑖 = 1 , 2 , … , 𝑚 1 , 𝑗 = 1 , 2 , … , 𝑚 2 , 𝑘 = 1 , 2 , … , 𝑚 3 . The hypothesis ( 𝐻 ) is similar to that in [ 5 , 6 ]. Because of the extensive applications in mechanics and engineering, nonlinear fourth-order two-point boundary value problems have received wide attentions (see [ 7 – 12 ] and references therein). In mechanics, the boundary value problem ( 1.1 ) (BVP ( 1.1 ) for short) describes the deformation of an elastic beam simply supported at left and clamped at right by sliding clamps. The term 𝑢   in 𝑓 represents bending effect which is useful for the stability analysis of the beam. BVP ( 1.1 ) has two special features. The first one is that the nonlinearity 𝑓 may depend on the first-order derivative of the unknown function 𝑢 , and the second one is that the nonlinearity 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) may be singular at 𝑢 , 𝑣 , 𝑡 = 0 and/or 𝑡 = 1 . In this paper, we study the existence of positive solutions and the structure of positive solution set for the BVP ( 1.1 ). Firstly, we construct a special cone and present a necessary and sufficient condition for the existence of positive solutions, then we prove that the closure of positive solution set possesses an unbounded connected branch which bifurcates from ( 0 , 𝜃 ) . Our analysis mainly relies on the fixed point theorem in a cone and the fixed point index theory. By singularity of 𝑓 , we mean that the function 𝑓 in ( 1.1 ) is allowed to be unbounded at the points 𝑢 = 0 , 𝑣 = 0 , 𝑡 = 0 , and/or 𝑡 = 1 . A function 𝑢 ( 𝑡 ) ∈ 𝐶 2 [ 0 , 1 ] ∩ 𝐶 4 ( 0 , 1 ) is called a (positive) solution of the BVP ( 1.1 ) if it satisfies the BVP ( 1.1 ) ( 𝑢 ( 𝑡 ) > 0 , − 𝑢   ( 𝑡 ) > 0 for 𝑡 ∈ ( 0 , 1 ] and 𝑢  ( 𝑡 ) > 0 for 𝑡 ∈ [ 0 , 1 ) ). For some 𝜆 ∈ ( 0 , + ∞ ) , if the B V P ( 1.1 ) has a positive solution 𝑢 , then 𝜆 is called an eigenvalue and 𝑢 is called corresponding eigenfunction of the BVP ( 1.1 ). The existence of positive solutions of BVPs has been studied by several authors in the literature; for example, see [ 7 – 20 ] and the references therein. Yao [ 15 , 18 ] studied the following BVP: 𝑢 ( 4 )  ( 𝑡 ) = 𝑓 𝑡 , 𝑢 ( 𝑡 ) , 𝑢   [ ] ( 𝑡 ) , 𝑡 ∈ 0 , 1 ⧵ 𝐸 , 𝑢 ( 0 ) = 𝑢  ( 0 ) = 𝑢   ( 1 ) = 𝑢    ( 1 ) = 0 , ( 1 . 4 ) where 𝐸 ⊂ [ 0 , 1 ] is a closed subset and m e s 𝐸 = 0 , 𝑓 ∈ 𝐶 ( ( [ 0 , 1 ] ⧵ 𝐸 ) × [ 0 , + ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) . In [ 15 ], he obtained a sufficient condition for the existence of positive solutions of B V P ( 1.4 ) by using the monotonically iterative technique. In [ 13 , 18 ], he applied Guo-Krasnosel'skii's fixed point theorem to obtain the existence and multiplicity of positive solutions of BVP ( 1.4 ) and the following BVP: 𝑢 ( 4 ) [ ] , ( 𝑡 ) = 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) , 𝑡 ∈ 0 , 1 𝑢 ( 0 ) = 𝑢  ( 0 ) = 𝑢 ( 1 ) = 𝑢   ( 1 ) = 0 . ( 1 . 5 ) These differ from our problem because 𝑓 ( 𝑡 , 𝑢 , 𝑣 ) in ( 1.4 ) cannot be singular at 𝑢 = 0 , 𝑣 = 0 and the nonlinearity 𝑓 in ( 1.5 ) does not depend on the derivatives of the unknown functions. In this paper, we first establish a necessary and sufficient condition for the existence of positive solutions of BVP ( 1.1 ) for any 𝜆 > 0 by using the following Lemma 1.1 . Efforts to obtain necessary and sufficient conditions for the existence of positive solutions of BVPs by the lower and upper solution method can be found, for example, in [ 5 , 6 , 21 – 23 ]. In [ 5 , 6 , 22 , 23 ] they considered the case that 𝑓 depends on even order derivatives of 𝑢 . Although the nonlinearity 𝑓 in [ 21 ] depends on the first-order derivative, where the nonlinearity 𝑓 is increasing with respect to the unknown function 𝑢 . Papers [ 24 , 25 ] derived the existence of positive solutions of BVPs by the lower and upper solution method, but the nonlinearity 𝑓 ( 𝑡 , 𝑢 ) does not depend on the derivatives of the unknown functions, and 𝑓 ( 𝑡 , 𝑢 ) is decreasing with respect to 𝑢 . Recently, the global structure of positive solutions of nonlinear boundary value problems has also been investigated (see [ 26 – 28 ] and references therein). Ma and An [ 26 ] and Ma and Xu [ 27 ] discussed the global structure of positive solutions for the nonlinear eigenvalue problems and obtained the existence of an unbounded connected branch of positive solution set by using global bifurcation theorems (see [ 29 , 30 ]). The terms 𝑓 ( 𝑢 ) in [ 26 ] and 𝑓 ( 𝑡 , 𝑢 , 𝑢   ) in [ 27 ] are not singular at 𝑡 = 0 , 1 , 𝑢 = 0 , 𝑢   = 0 . Yao [ 14 ] obtained one or two positive solutions to a singular elastic beam equation rigidly fixed at both ends by using Guo-Krasnosel'skii's fixed point theorem, but the global structure of positive solutions was not considered. Since the nonlinearity 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) in BVP ( 1.1 ) may be singular at 𝑢 , 𝑣 , 𝑡 = 0 and/or 𝑡 = 1 , the global bifurcation theorems in [ 29 , 30 ] do not apply to our problem here. In Section 4 , we also investigate the global structure of positive solutions for BVP ( 1.1 ) by applying the following Lemma 1.2 . The paper is organized as follows: in the rest of this section, two known results are stated. In Section 2 , some lemmas are stated and proved. In Section 3 , we establish a necessary and sufficient condition for the existence of positive solutions. In Section 4 , we prove that the closure of positive solution set possesses an unbounded connected branch which comes from ( 0 , 𝜃 ) . Finally we state the following results which will be used in Sections 3 and 4 , respectively. Lemma 1.1 (see [ 31 ]). Let 𝑋 be a real Banach space, let 𝐾 be a cone in 𝑋 , and let Ω 1 , Ω 2 be bounded open sets of 𝐸 , 𝜃 ∈ Ω 1 ⊂ Ω 1 ⊂ Ω 2 . Suppose that 𝑇 ∶ 𝐾 ∩ ( Ω 2 ⧵ Ω 1 ) → 𝐾 is completely continuous such that one of the following two conditions is satisfied: (1) ‖ 𝑇 ( 𝑥 ) ‖ ≤ ‖ 𝑥 ‖ , 𝑥 ∈ 𝐾 ∩ 𝜕 Ω 1 ; ‖ 𝑇 ( 𝑥 ) ‖ ≥ ‖ 𝑥 ‖ , 𝑥 ∈ 𝐾 ∩ 𝜕 Ω 2 . (2) ‖ 𝑇 ( 𝑥 ) ‖ ≥ ‖ 𝑥 ‖ , 𝑥 ∈ 𝐾 ∩ 𝜕 Ω 1 ; ‖ 𝑇 ( 𝑥 ) ‖ ≤ ‖ 𝑥 ‖ , 𝑥 ∈ 𝐾 ∩ 𝜕 Ω 2 . Then, 𝑇 has a fixed point in 𝐾 ∩ ( Ω 2 ⧵ Ω 1 ) . Lemma 1.2 (see [ 32 ]). Let 𝑀 be a metric space and ( 𝑎 , 𝑏 ) ⊂ 𝑅 1 . Let { 𝑎 𝑛 } ∞ 𝑛 = 1 and { 𝑏 𝑛 } ∞ 𝑛 = 1 satisfy 𝑎 < ⋯ < 𝑎 𝑛 < ⋯ < 𝑎 1 < 𝑏 1 < ⋯ < 𝑏 𝑛 < ⋯ < 𝑏 , l i m 𝑛 → + ∞ 𝑎 𝑛 = 𝑎 , l i m 𝑛 → + ∞ 𝑏 𝑛 = 𝑏 . ( 1 . 6 ) Suppose also that ∑ = { 𝐶 𝑛 ∶ 𝑛 = 1 , 2 , … } is a family of connected subsets of 𝑅 1 × 𝑀 , satisfying the following conditions: ( 1 ) 𝐶 𝑛 ∩ ( { 𝑎 𝑛 } × 𝑀 ) ≠ ∅ and 𝐶 𝑛 ∩ ( { 𝑏 𝑛 } × 𝑀 ) ≠ ∅ for each 𝑛 . (2) For any two given numbers 𝛼 and 𝛽 with 𝑎 < 𝛼 < 𝛽 < 𝑏 , ( ⋃ ∞ 𝑛 = 1 𝐶 𝑛 ) ∩ ( [ 𝛼 , 𝛽 ] × 𝑀 ) is a relatively compact set of 𝑅 1 × 𝑀 . Then there exists a connected branch 𝐶 of l i m s u p 𝑛 → + ∞ 𝐶 𝑛 such that 𝐶 ∩ ( { 𝜆 } × 𝑀 ) ≠ ∅ , ∀ 𝜆 ∈ ( 𝑎 , 𝑏 ) , ( 1 . 7 ) where l i m s u p 𝑛 → + ∞ 𝐶 𝑛 = { 𝑥 ∈ 𝑀 ∶ there exists a sequence 𝑥 𝑛 𝑖 ∈ 𝐶 𝑛 𝑖 such that 𝑥 𝑛 𝑖 → 𝑥 , ( 𝑖 → ∞ ) } . 2. Some Preliminaries and Lemmas Let 𝐸 = { 𝑢 ∈ 𝐶 2 [ 0 , 1 ] ∶ 𝑢 ( 0 ) = 0 , 𝑢  ( 1 ) = 0 , 𝑢   ( 0 ) = 0 } , ‖ 𝑢 ‖ 2 = m a x { ‖ 𝑢 ‖ , ‖ 𝑢  ‖ , ‖ 𝑢   ‖ } , then ( 𝐸 , ‖ ⋅ ‖ 2 ) is a Banach space, where ‖ 𝑢 ‖ = m a x 𝑡 ∈ [ 0 , 1 ] | 𝑢 ( 𝑡 ) | . Define   𝑡 𝑃 = 𝑢 ∈ 𝐸 ∶ 𝑢 ( 𝑡 ) ≥ 𝑡 − 2 2  ‖ 𝑢 ‖ , 𝑢  1 ( 𝑡 ) ≥ 2 ‖ ‖ 𝑢 ( 1 − 𝑡 )  ‖ ‖ , − 𝑢   ‖ ‖ 𝑢 ( 𝑡 ) ≥ 𝑡   ‖ ‖ [ ]  , 𝑡 ∈ 0 , 1 . ( 2 . 1 ) It is easy to conclude that 𝑃 is a cone of 𝐸 . Denote 𝑃 𝑟 =  𝑢 ∈ 𝑃 ∶ ‖ 𝑢 ‖ 2  < 𝑟 ; 𝜕 𝑃 𝑟 =  𝑢 ∈ 𝑃 ∶ ‖ 𝑢 ‖ 2  = 𝑟 . ( 2 . 2 ) Let 𝐺 0   ( 𝑡 , 𝑠 ) = 𝑠 , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , 𝑡 , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 , 𝐺 ( 𝑡 , 𝑠 ) = 1 0 𝐺 0 ( 𝑡 , 𝑟 ) 𝐺 0 ( 𝑟 , 𝑠 ) 𝑑 𝑟 . ( 2 . 3 ) Then 𝐺 ( 𝑡 , 𝑠 ) is the Green function of homogeneous boundary value problem 𝑢 ( 4 ) ( 𝑡 ) = 0 , 𝑡 ∈ ( 0 , 1 ) , 𝑢 ( 0 ) = 𝑢  ( 1 ) = 𝑢   ( 0 ) = 𝑢    ( ⎧ ⎪ ⎨ ⎪ ⎩ 𝑠 1 ) = 0 , 𝐺 ( 𝑡 , 𝑠 ) = 3 3 + 𝑠  𝑡 2 − 𝑠 2  2 𝑡 + 𝑠 𝑡 ( 1 − 𝑡 ) , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , 3 3 + 𝑡  𝑠 2 − 𝑡 2  2 𝐺 + 𝑡 𝑠 ( 1 − 𝑠 ) , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 , 1 ( 𝑡 , 𝑠 ) = ∶ 𝐺  𝑡 ⎧ ⎪ ⎨ ⎪ ⎩ 𝑠 ( 𝑡 , 𝑠 ) = 𝑠 ( 1 − 𝑡 ) , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , 2 2 − 𝑡 2 2 𝐺 + 𝑠 ( 1 − 𝑠 ) , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 , 2 ( 𝑡 , 𝑠 ) = ∶ − 𝐺 𝑡    ( 𝑡 , 𝑠 ) = 𝑠 , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , 𝑡 , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 . ( 2 . 4 ) Lemma 2.1. 𝐺 ( 𝑡 , 𝑠 ) , 𝐺 1 ( 𝑡 , 𝑠 ) , and 𝐺 2 ( 𝑡 , 𝑠 ) have the following properties: (1) 𝐺 ( 𝑡 , 𝑠 ) > 0 , 𝐺 𝑖 ( 𝑡 , 𝑠 ) > 0 , 𝑖 = 1 , 2 , for all 𝑡 , 𝑠 ∈ ( 0 , 1 ) . (2) 𝐺 ( 𝑡 , 𝑠 ) ≤ 𝑠 ( 𝑡 − 𝑡 2 / 2 ) , 𝐺 1 ( 𝑡 , 𝑠 ) ≤ 𝑠 ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≤ 𝑡 (or 𝑠 ), for all 𝑡 , 𝑠 ∈ [ 0 , 1 ] . (3) m a x 𝑡 ∈ [ 0 , 1 ] 𝐺 ( 𝑡 , 𝑠 ) ≤ ( 1 / 2 ) 𝑠 , m a x 𝑡 ∈ [ 0 , 1 ] 𝐺 𝑖 ( 𝑡 , 𝑠 ) ≤ 𝑠 , 𝑖 = 1 , 2 , for all 𝑠 ∈ [ 0 , 1 ] . (4) 𝐺 ( 𝑡 , 𝑠 ) ≥ ( 𝑠 / 2 ) ( 𝑡 − 𝑡 2 / 2 ) , 𝐺 1 ( 𝑡 , 𝑠 ) ≥ ( 𝑠 / 2 ) ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≥ 𝑠 𝑡 , for all 𝑡 , 𝑠 ∈ [ 0 , 1 ] . Proof. From ( 2.4 ), it is easy to obtain the property ( 2.18 ). We now prove that property (2) is true. For 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , by ( 2.4 ), we have 𝑠 𝐺 ( 𝑡 , 𝑠 ) = 3 3 + 𝑠 𝑡 2 2 − 𝑠 3 2 + 𝑠 𝑡 − 𝑠 𝑡 2 ≤ 𝑠 𝑡 − 𝑠 𝑡 2 2  𝑡 = 𝑠 𝑡 − 2 2  , 𝐺 1 ( 𝑡 , 𝑠 ) = 𝑠 ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≤ 𝑡 ( o r 𝑠 ) . ( 2 . 5 ) For 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 , by ( 2.4 ), we have 𝑡 𝐺 ( 𝑡 , 𝑠 ) = 3 3 − 𝑡 3 2 + 𝑡 𝑠 − 𝑡 𝑠 2 2 ≤ 𝑠 𝑡 − 𝑠 𝑡 2 2  𝑡 = 𝑠 𝑡 − 2 2  , 𝐺 1 𝑡 ( 𝑡 , 𝑠 ) = 𝑠 − 2 2 − 𝑠 2 2 ≤ 𝑠 − 𝑡 𝑠 = 𝑠 ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≤ 𝑡 ( o r 𝑠 ) . ( 2 . 6 ) Consequently, property (2) holds. From property (2), it is easy to obtain property (3). We next show that property (4) is true. From ( 2.4 ), we know that property (4) holds for 𝑠 = 0 . For 0 < 𝑠 ≤ 1 , if 𝑠 ≤ 𝑡 ≤ 1 , then 𝐺 ( 𝑡 , 𝑠 ) 𝑠 𝑡 = 𝑡 − 2 2 − 𝑠 2 6 = 1 2  𝑡 𝑡 − 2 2 +  𝑡 𝑡 − 2 2 − 𝑠 2 3 ≥ 1   2  𝑡 𝑡 − 2 2 +  𝑡 𝑡 − 2 2 − 𝑡 2 3 > 1   2  𝑡 𝑡 − 2 2  , 𝐺 1 ( 𝑡 , 𝑠 ) 𝑠 1 = ( 1 − 𝑡 ) ≥ 2 ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≥ 𝑠 𝑡 ; ( 2 . 7 ) if 0 ≤ 𝑡 ≤ 𝑠 , then 𝐺 ( 𝑡 , 𝑠 ) 𝑠 𝑡 ≥ 𝑡 − 2 6 − 𝑡 𝑠 2 = 1 2  𝑡 𝑡 − 2 3  ≥ 1 + ( 𝑡 − 𝑡 𝑠 ) 2  𝑡 𝑡 − 2 3  ≥ 1 2  𝑡 𝑡 − 2 2  , 𝐺 1 ( 𝑡 , 𝑠 ) 𝑠 𝑡 ≥ 1 − 2 − 𝑠 2 ≥ 1 2 ( 1 − 𝑡 ) , 𝐺 2 ( 𝑡 , 𝑠 ) ≥ 𝑠 𝑡 . ( 2 . 8 ) Therefore, property (4) holds. Lemma 2.2. Assume that 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , then ‖ 𝑢 ‖ 2 = ‖ 𝑢   ‖ and 1 4 ‖ ‖ 𝑢  ‖ ‖ ≤ ‖ ‖ 𝑢 ‖ 𝑢 ‖ ≤  ‖ ‖ , 1 2 ‖ ‖ 𝑢   ‖ ‖ ≤ ‖ ‖ 𝑢  ‖ ‖ ≤ ‖ ‖ 𝑢   ‖ ‖ . 1 ( 2 . 9 ) 8  𝑡 𝑡 − 2 2  ‖ 𝑢 ‖ 2  𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑡 − 2 2  ‖ 𝑢 ‖ 2 , 1 4 ( 1 − 𝑡 ) ‖ 𝑢 ‖ 2 ≤ 𝑢  ( 𝑡 ) ≤ ( 1 − 𝑡 ) ‖ 𝑢 ‖ 2 , 𝑡 ‖ 𝑢 ‖ 2 ≤ − 𝑢   ( 𝑡 ) ≤ ‖ 𝑢 ‖ 2 [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 2 . 1 0 ) Proof. Assume that 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , then 𝑢  ( 𝑡 ) ≥ 0 , − 𝑢   ( 𝑡 ) ≥ 0 , 𝑡 ∈ [ 0 , 1 ] , so ‖ 𝑢 ‖ = m a x [ ] 𝑡 ∈ 0 , 1  𝑡 0 𝑢   ( 𝑠 ) 𝑑 𝑠 = 1 0 𝑢  ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤  ‖ ‖ , ‖ 𝑢 ‖ = m a x [ ] 𝑡 ∈ 0 , 1  𝑡 0 𝑢  (  𝑠 ) 𝑑 𝑠 = 1 0 𝑢  ( 1 𝑠 ) 𝑑 𝑠 ≥ 2 ‖ ‖ 𝑢  ‖ ‖  1 0 ( 1 1 − 𝑠 ) 𝑑 𝑠 = 4 ‖ ‖ 𝑢  ‖ ‖ , ‖ ‖ 𝑢  ‖ ‖ = m a x [ ] 𝑡 ∈ 0 , 1  1 𝑡 − 𝑢    ( 𝑠 ) 𝑑 𝑠 = 1 0 − 𝑢   ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤   ‖ ‖ , ‖ ‖ 𝑢  ‖ ‖ = m a x [ ] 𝑡 ∈ 0 , 1  1 𝑡 − 𝑢   (  𝑠 ) 𝑑 𝑠 = 1 0 − 𝑢   (  𝑠 ) 𝑑 𝑠 ≥ 1 0 𝑠 ‖ ‖ 𝑢   ‖ ‖ 1 𝑑 𝑠 = 2 ‖ ‖ 𝑢   ‖ ‖ . ( 2 . 1 1 ) Therefore, ( 2.9 ) holds. From ( 2.9 ), we get ‖ 𝑢 ‖ 2  ‖ ‖ 𝑢 = m a x ‖ 𝑢 ‖ ,  ‖ ‖ , ‖ ‖ 𝑢   ‖ ‖  = ‖ ‖ 𝑢   ‖ ‖ . ( 2 . 1 2 ) By ( 2.9 ) and the definition of 𝑃 , we can obtain that  𝑢 ( 𝑡 ) = 1 0 𝐺 0  ( 𝑡 , 𝑠 ) − 𝑢      ( 𝑠 ) 𝑑 𝑠 ≤ 𝑡 0  𝑠 𝑑 𝑠 + 1 𝑡  ‖ ‖ 𝑢 𝑡 𝑑 𝑠   ‖ ‖ =  𝑡 𝑡 − 2 2  ‖ ‖ 𝑢   ‖ ‖ =  𝑡 𝑡 − 2 2  ‖ 𝑢 ‖ 2 , [ ] ,  𝑡 ∀ 𝑡 ∈ 0 , 1 𝑢 ( 𝑡 ) ≥ 𝑡 − 2 2  1 ‖ 𝑢 ‖ ≥ 8  𝑡 𝑡 − 2 2  ‖ 𝑢 ‖ 2 [ ] , 𝑢 , ∀ 𝑡 ∈ 0 , 1   ( 𝑡 ) = 1 𝑡 − 𝑢   ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ ( 1 − 𝑡 )   ‖ ‖ = ( 1 − 𝑡 ) ‖ 𝑢 ‖ 2 , 𝑢  1 ( 𝑡 ) ≥ 2 ‖ ‖ 𝑢 ( 1 − 𝑡 )  ‖ ‖ ≥ 1 4 ( 1 − 𝑡 ) ‖ 𝑢 ‖ 2 [ ] , , ∀ 𝑡 ∈ 0 , 1 𝑡 ‖ 𝑢 ‖ 2 ‖ ‖ 𝑢 = 𝑡   ‖ ‖ ≤ − 𝑢   ‖ ‖ 𝑢 ( 𝑡 ) ≤   ‖ ‖ = ‖ 𝑢 ‖ 2 [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 2 . 1 3 ) Thus, ( 2.10 ) holds. For any fixed 𝜆 ∈ ( 0 , + ∞ ) , define an operator 𝑇 𝜆 by  𝑇 𝜆 𝑢   ( 𝑡 ) = ∶ 𝜆 1 0  𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ( 𝑠 ) 𝑑 𝑠 , ∀ 𝑢 ∈ 𝑃 ⧵ { 𝜃 } . ( 2 . 1 4 ) Then, it is easy to know that  𝑇 𝜆 𝑢    ( 𝑡 ) = 𝜆 1 0 𝐺 1  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     𝑇 ( 𝑠 ) 𝑑 𝑠 , ∀ 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , ( 2 . 1 5 ) 𝜆 𝑢     ( 𝑡 ) = − 𝜆 1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ( 𝑠 ) 𝑑 𝑠 , ∀ 𝑢 ∈ 𝑃 ⧵ { 𝜃 } . ( 2 . 1 6 ) Lemma 2.3. Suppose that ( 𝐻 ) and  0 < 1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ ( 2 . 1 7 ) hold. Then 𝑇 𝜆 ( 𝑃 ⧵ { 𝜃 } ) ⊂ 𝑃 . Proof. From ( 𝐻 ), for any 𝑡 ∈ ( 0 , 1 ) , 𝑢 , 𝑣 ∈ ( 0 , + ∞ ) , 𝑤 ∈ ( − ∞ , 0 ] , we easily obtain the following inequalities: 𝑐 𝛼 1 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑐 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑐 𝛽 1 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 𝑐 ≥ 𝑁 1 − 1 , 𝑐 𝛼 2 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑢 , 𝑐 𝑣 , 𝑤 ) ≤ 𝑐 𝛽 2 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 𝑐 ≥ 𝑁 2 − 1 , 𝑐 𝛼 3 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) ≤ 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑐 𝑤 ) ≤ 𝑐 𝛽 3 𝑓 ( 𝑡 , 𝑢 , 𝑣 , 𝑤 ) , ∀ 𝑐 ≥ 𝑁 3 − 1 . ( 2 . 1 8 ) For every 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , 𝑡 ∈ [ 0 , 1 ] , choose positive numbers 𝑐 1 ≤ m i n { 𝑁 1 , ( 1 / 8 ) 𝑁 1 ‖ 𝑢 ‖ 2 } , 𝑐 2 ≤ m i n { 𝑁 2 , ( 1 / 4 ) 𝑁 2 ‖ 𝑢 ‖ 2 } , 𝑐 3 ≥ m a x { 𝑁 3 − 1 , 𝑁 3 − 1 ‖ 𝑢 ‖ 2 } . It follows from ( 𝐻 ), ( 2.10 ), Lemma 2.1 , and ( 2.17 ) that  𝑇 𝜆 𝑢   ( 𝑡 ) = 𝜆 1 0  𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ≤ 1 ( 𝑠 ) 𝑑 𝑠 2 𝜆  1 0  𝑠 𝑓 𝑠 , 𝑐 1 𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   𝑠 / 2 𝑠 − 2 2  , 𝑐 2 𝑢  ( 𝑠 ) 𝑐 2 ( 1 − 𝑠 ) ( 1 − 𝑠 ) , ( − 1 ) 𝑐 3 𝑢   ( 𝑠 ) − 𝑐 3  ≤ 1 𝑑 𝑠 2 𝜆  1 0 𝑠 𝑐 𝛼 1 1  𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   / 2 𝛽 1 𝑐 𝛼 2 2  𝑢  ( 𝑠 ) 𝑐 2  ( 1 − 𝑠 ) 𝛽 2 𝑐 𝛽 3 3  𝑢   ( 𝑠 ) − 𝑐 3  𝛼 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  ≤ 1 , 1 − 𝑠 , − 1 𝑑 𝑠 2 𝜆  1 0 𝑠 𝑐 𝛼 1 1  ‖ 𝑢 ‖ 2 𝑐 1  𝛽 1 𝑐 𝛼 2 2  ‖ 𝑢 ‖ 2 𝑐 2  𝛽 2 𝑐 𝛽 3 3  ‖ 𝑢 ‖ 2 𝑐 3  𝛼 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  ≤ 1 , 1 − 𝑠 , − 1 𝑑 𝑠 2 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 + 𝛼 3 2 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ . ( 2 . 1 9 ) Similar to ( 2.19 ), from ( 𝐻 ), ( 2.10 ), Lemma 2.1 , and ( 2.17 ), for every 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , 𝑡 ∈ [ 0 , 1 ] , we have  𝑇 𝜆 𝑢    ( 𝑡 ) = 𝜆 1 0 𝐺 1  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑐 1 𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   𝑠 / 2 𝑠 − 2 2  , 𝑐 2 𝑢  ( 𝑠 ) 𝑐 2 ( 1 − 𝑠 ) ( 1 − 𝑠 ) , ( − 1 ) 𝑐 3 𝑢   ( 𝑠 ) − 𝑐 3  𝑑 𝑠 ≤ 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 + 𝛼 3 2 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  −  𝑇 , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ . 𝜆 𝑢     ( 𝑡 ) = 𝜆 1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑐 1 𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   𝑠 / 2 𝑠 − 2 2  , 𝑐 2 𝑢  ( 𝑠 ) 𝑐 2 ( 1 − 𝑠 ) ( 1 − 𝑠 ) , ( − 1 ) 𝑐 3 𝑢   ( 𝑠 ) − 𝑐 3  𝑑 𝑠 ≤ 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 + 𝛼 3 2 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ . ( 2 . 2 0 ) Thus, 𝑇 𝜆 is well defined on 𝑃 ⧵ { 𝜃 } . From ( 2.4 ) and ( 2.14 )–( 2.16 ), it is easy to know that  𝑇 𝜆 𝑢   𝑇 ( 0 ) = 0 , 𝜆 𝑢    𝑇 ( 1 ) = 0 , 𝜆 𝑢     𝑇 ( 0 ) = 0 , 𝜆 𝑢   ( 𝑡 ) = 𝜆 1 0  𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ≥  𝑡 ( 𝑠 ) 𝑑 𝑠 𝑡 − 2 2  𝜆  1 0 1 2  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ≥  𝑡 ( 𝑠 ) 𝑑 𝑠 𝑡 − 2 2  𝜆  1 0 m a x [ ] 𝜏 ∈ 0 , 1  𝐺 ( 𝜏 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    =  𝑡 ( 𝑠 ) 𝑑 𝑠 𝑡 − 2 2  ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ [ ]  𝑇 , ∀ 𝑡 ∈ 0 , 1 , 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , 𝜆 𝑢    ( 𝑡 ) = 𝜆 1 0 𝐺 1  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ≥ 1 ( 𝑠 ) 𝑑 𝑠 2  ( 1 − 𝑡 ) 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( s ) , 𝑢    ≥ 1 ( 𝑠 ) 𝑑 𝑠 2  ( 1 − 𝑡 ) 𝜆 1 0 m a x [ ] 𝜏 ∈ 0 , 1 𝐺 1  ( 𝜏 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    = 1 ( 𝑠 ) 𝑑 𝑠 2 ‖ ‖  𝑇 ( 1 − 𝑡 ) 𝜆 𝑢   ‖ ‖ [ ] −  𝑇 , ∀ 𝑡 ∈ 0 , 1 , 𝑢 ∈ 𝑃 ⧵ { 𝜃 } , 𝜆 𝑢     ( 𝑡 ) = 𝜆 1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≥ 𝑡 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≥ 𝑡 𝜆 1 0 m a x [ ] 𝜏 ∈ 0 , 1 𝐺 2 (  𝜏 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢   (  ‖ ‖  𝑇 𝑠 ) 𝑑 𝑠 = 𝑡 𝜆 𝑢    ‖ ‖ [ ] , ∀ 𝑡 ∈ 0 , 1 , 𝑢 ∈ 𝑃 ⧵ { 𝜃 } . ( 2 . 2 1 ) Therefore, 𝑇 ( 𝑃 ⧵ { 𝜃 } ) ⊂ 𝑃 follows from ( 2.21 ). Obviously, 𝑢 ∗ is a positive solution of BVP ( 1.1 ) if and only if 𝑢 ∗ is a positive fixed point of the integral operator 𝑇 𝜆 in 𝑃 . Lemma 2.4. Suppose that ( 𝐻 ) and ( 2.17 ) hold. Then for any 𝑅 > 𝑟 > 0 , 𝑇 𝜆 ∶ 𝑃 𝑅 ⧵ 𝑃 𝑟 → 𝑃 is completely continuous. Proof. First of all, notice that 𝑇 𝜆 maps 𝑃 𝑅 ⧵ 𝑃 𝑟 into 𝑃 by Lemma 2.3 . Next, we show that 𝑇 𝜆 is bounded. In fact, for any 𝑢 ∈ 𝑃 𝑅 ⧵ 𝑃 𝑟 , by ( 2.10 ) we can get 𝑟 8  𝑡 𝑡 − 2 2   𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑡 − 2 2  𝑟 𝑅 , 4 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ ( 1 − 𝑡 ) 𝑅 , 𝑟 𝑡 ≤ − 𝑢   [ ] ( 𝑡 ) ≤ 𝑅 , ∀ 𝑡 ∈ 0 , 1 . ( 2 . 2 2 ) Choose positive numbers 𝑐 1 ≤ m i n { 𝑁 1 , ( 𝑟 / 8 ) 𝑁 1 } , 𝑐 2 ≤ m i n { 𝑁 2 , ( 𝑟 / 4 ) 𝑁 2 } , 𝑐 3 ≥ m a x { 𝑁 3 − 1 , 𝑁 3 − 1 𝑅 } . This, together with ( 𝐻 ), ( 2.22 ), ( 2.16 ), and Lemma 2.1 yields that | | |  𝑇 𝜆 𝑢    | | |  ( 𝑡 ) = 𝜆 1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑐 1 𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   𝑠 / 2 𝑠 − 2 2  , 𝑐 2 𝑢  ( 𝑠 ) 𝑐 2 ( 1 − 𝑠 ) ( 1 − 𝑠 ) , ( − 1 ) 𝑐 3 𝑢   ( 𝑠 ) − 𝑐 3   𝑑 𝑠 ≤ 𝜆 1 0 𝑠 𝑐 𝛼 1 1  𝑢 ( 𝑠 ) 𝑐 1  𝑠 − 𝑠 2   / 2 𝛽 1 𝑐 𝛼 2 2  𝑢  ( 𝑠 ) 𝑐 2  ( 1 − 𝑠 ) 𝛽 2 𝑐 𝛽 3 3  𝑢   ( 𝑠 ) − 𝑐 3  𝛼 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 s ≤ 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 𝑅 𝛽 1 + 𝛽 2 + 𝛼 3 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  [ ] , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ , ∀ 𝑡 ∈ 0 , 1 , 𝑢 ∈ 𝑃 𝑅 ⧵ 𝑃 𝑟 . ( 2 . 2 3 ) Thus, 𝑇 𝜆 is bounded on 𝑃 𝑅 ⧵ 𝑃 𝑟 . Now we show that 𝑇 𝜆 is a compact operator on 𝑃 𝑅 ⧵ 𝑃 𝑟 . By ( 2.23 ) and Ascoli-Arzela theorem, it suffices to show that 𝑇 𝜆 𝑉 is equicontinuous for arbitrary bounded subset 𝑉 ⊂ 𝑃 𝑅 ⧵ 𝑃 𝑟 . Since for each 𝑢 ∈ 𝑉 , ( 2.22 ) holds, we may choose still positive numbers 𝑐 1 ≤ m i n { 𝑁 1 , ( 𝑟 / 8 ) 𝑁 1 } , 𝑐 2 ≤ m i n { 𝑁 2 , ( 𝑟 / 4 ) 𝑁 2 } , 𝑐 3 ≥ m a x { 𝑁 3 − 1 , 𝑁 3 − 1 𝑅 } . Then | | |  𝑇 𝜆 𝑢     | | |  ( 𝑡 ) = 𝜆 1 𝑡 𝑓  𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ( 𝑠 ) 𝑑 𝑠 ≤ 𝐶 0  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 = ∶ 𝐻 ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) , ( 2 . 2 4 ) where 𝐶 0 = 𝜆 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 𝑅 𝛽 1 + 𝛽 2 + 𝛼 3 . Notice that  1 0 𝐻 ( 𝑡 ) 𝑑 𝑡 = 𝐶 0  1 0  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 𝑑 𝑡 = 𝐶 0  1 0  𝑠 0 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑡 𝑑 𝑠 = 𝐶 0  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ . ( 2 . 2 5 ) Thus for any given 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ] with 𝑡 1 ≤ 𝑡 2 and for any 𝑢 ∈ 𝑉 , we get | | |  𝑇 𝜆 𝑢     𝑡 2  −  𝑇 𝜆 𝑢     𝑡 1  | | | ≤  𝑡 2 𝑡 1 | | |  𝑇 𝜆 𝑢     ( | | |  𝑡 ) 𝑑 𝑡 ≤ 𝑡 2 𝑡 1 𝐻 ( 𝑡 ) 𝑑 𝑡 . ( 2 . 2 6 ) From ( 2.25 ), ( 2.26 ), and the absolute continuity of integral function, it follows that 𝑇 𝜆 𝑉 is equicontinuous. Therefore, 𝑇 𝜆 𝑉 is relatively compact, that is, 𝑇 𝜆 is a compact operator on 𝑃 𝑅 ⧵ 𝑃 𝑟 . Finally, we show that 𝑇 𝜆 is continuous on 𝑃 𝑅 ⧵ 𝑃 𝑟 . Suppose 𝑢 𝑛 , 𝑢 ∈ 𝑃 𝑅 ⧵ 𝑃 𝑟 , 𝑛 = 1 , 2 , … and ‖ 𝑢 𝑛 − 𝑢 ‖ 2 → 0 , ( 𝑛 → + ∞ ) . Then 𝑢 𝑛   ( 𝑡 ) → 𝑢   ( 𝑡 ) , 𝑢  𝑛 ( 𝑡 ) → 𝑢 ′ ( 𝑡 ) and 𝑢 𝑛 ( 𝑡 ) → 𝑢 ( 𝑡 ) as 𝑛 → + ∞ uniformly, with respect to 𝑡 ∈ [ 0 , 1 ] . From ( 𝐻 ) , choose still positive numbers 𝑐 1 ≤ m i n { 𝑁 1 , ( 𝑟 / 8 ) 𝑁 1 } , 𝑐 2 ≤ m i n { 𝑁 2 , ( 𝑟 / 4 ) 𝑁 2 } , 𝑐 3 ≥ m a x { 𝑁 3 − 1 , 𝑁 3 − 1 𝑅 } . Then  0 ≤ 𝑓 𝑡 , 𝑢 𝑛 ( 𝑡 ) , 𝑢  𝑛 ( 𝑡 ) , 𝑢 𝑛    ( 𝑡 ) ≤ 𝐶 0 𝑓  𝑡 𝑡 , 𝑡 − 2 2  , 1 − 𝑡 , − 1 , 𝑡 ∈ ( 0 , 1 ) , 0 ≤ 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝑛 ( 𝑠 ) , 𝑢  𝑛 ( 𝑠 ) , 𝑢 𝑛    ( 𝑠 ) ≤ 𝐶 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  [ ] , 1 − 𝑠 , − 1 , 𝑡 ∈ 0 , 1 , 𝑠 ∈ ( 0 , 1 ) . ( 2 . 2 7 ) By ( 2.17 ), we know that 𝑠 𝑓 ( 𝑠 , 𝑠 − 𝑠 2 / 2 , 1 − 𝑠 , − 1 ) is integrable on [ 0 , 1 ] . Thus, from the Lebesgue dominated convergence theorem, it follows that l i m 𝑛 → + ∞ ‖ ‖  𝑇 𝜆 𝑢 𝑛  −  𝑇 𝜆 𝑢  ‖ ‖ 2 = l i m 𝑛 → + ∞ ‖ ‖  T 𝜆 𝑢 𝑛    −  𝑇 𝜆 𝑢    ‖ ‖ ≤ l i m 𝑛 → + ∞ 𝜆  1 0 𝑠 | | 𝑓  𝑠 , 𝑢 𝑛 ( 𝑠 ) , 𝑢  𝑛 ( 𝑠 ) , 𝑢 𝑛     ( 𝑠 ) − 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    | |  ( 𝑠 ) 𝑑 𝑠 = 𝜆 1 0 𝑠 | | | l i m 𝑛 → + ∞  𝑓  𝑠 , 𝑢 𝑛 ( 𝑠 ) , 𝑢  𝑛 ( 𝑠 ) , 𝑢 𝑛     ( 𝑠 ) − 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢 𝑛   | | | ( 𝑠 )   𝑑 𝑠 = 0 . ( 2 . 2 8 ) Thus, 𝑇 𝜆 is continuous on 𝑃 𝑅 ⧵ 𝑃 𝑟 . Therefore, 𝑇 𝜆 ∶ 𝑃 𝑅 ⧵ 𝑃 𝑟 → 𝑃 is completely continuous. 3. A Necessary and Sufficient Condition for Existence of Positive Solutions In this section, by using the fixed point theorem of cone, we establish the following necessary and sufficient condition for the existence of positive solutions for BVP ( 1.1 ). Theorem 3.1. Suppose ( 𝐻 ) holds, then BVP ( 1.1 ) has at least one positive solution for any 𝜆 > 0 if and only if the integral inequality ( 2.17 ) holds. Proof. Suppose first that 𝑢 ( 𝑡 ) be a positive solution of BVP ( 1.1 ) for any fixed 𝜆 > 0 . Then there exist constants 𝐼 𝑖 ( 𝑖 = 1 , 2 , 3 , 4 ) with 0 < 𝐼 𝑖 < 1 < 𝐼 𝑖 + 1 , 𝑖 = 1 , 3 such that 𝐼 1  𝑡 𝑡 − 2 2  ≤ 𝑢 ( 𝑡 ) ≤ 𝐼 2  𝑡 𝑡 − 2 2  , 𝐼 3 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ 𝐼 4 [ ] ( 1 − 𝑡 ) , 𝑡 ∈ 0 , 1 . ( 3 . 1 ) In fact, it follows from 𝑢 ( 4 ) ( 𝑡 ) ≥ 0 , 𝑡 ∈ ( 0 , 1 ) and 𝑢 ( 0 ) = 𝑢  ( 1 ) = 𝑢   ( 0 ) = 𝑢    ( 1 ) = 0 , that 𝑢    ( 𝑡 ) ≤ 0 for 𝑡 ∈ ( 0 , 1 ] and 𝑢   ( 𝑡 ) ≤ 0 , 𝑢  ( 𝑡 ) ≥ 0 for 𝑡 ∈ [ 0 , 1 ] . By the concavity of 𝑢 ( 𝑡 ) and 𝑢  ( 𝑡 ) , we have  𝑡 𝑢 ( 𝑡 ) ≥ 𝑡 𝑢 ( 1 ) + ( 1 − 𝑡 ) 𝑢 ( 0 ) = 𝑡 ‖ 𝑢 ‖ ≥ 𝑡 − 2 2  𝑢 ‖ 𝑢 ‖ ,  ( 𝑡 ) ≥ 𝑡 𝑢  ( 1 ) + ( 1 − 𝑡 ) 𝑢  ( ‖ ‖ 𝑢 0 ) = ( 1 − 𝑡 )  ‖ ‖ [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 3 . 2 ) On the other hand,  𝑢 ( 𝑡 ) = 1 0 𝐺 0  ( 𝑡 , 𝑠 ) − 𝑢     ( 𝑠 ) 𝑑 𝑠 = 𝑡 0 𝑠  − 𝑢     ( 𝑠 ) 𝑑 𝑠 + 1 𝑡 𝑡  − 𝑢    ≤ 𝑡 ( 𝑠 ) 𝑑 𝑠 2 2 ‖ ‖ 𝑢   ‖ ‖ ‖ ‖ 𝑢 + 𝑡 ( 1 − 𝑡 )   ‖ ‖ =  𝑡 𝑡 − 2 2  ‖ ‖ 𝑢   ‖ ‖ , 𝑢   ( 𝑡 ) = 1 𝑡 − 𝑢   ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ ( 1 − 𝑡 )   ‖ ‖ [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 3 . 3 ) Let 𝐼 1 = m i n { ‖ 𝑢 ‖ , 1 / 2 } , let 𝐼 2 = 𝐼 4 = m a x { ‖ 𝑢   ‖ , 2 } , and let 𝐼 3 = m i n { ‖ 𝑢 ′ ‖ , 1 / 2 } , then ( 3.1 ) holds. Choose positive numbers 𝑐 1 ≤ 𝑁 1 𝐼 2 − 1 , 𝑐 2 ≤ 𝑁 2 𝐼 4 − 1 , 𝑐 3 ≥ m a x { 𝑁 3 − 1 , 𝑁 3 − 1 ‖ 𝑢 ‖ 2 } . This, together with ( 𝐻 ), ( 1.2 ), and ( 2.18 ) yields that 𝑓  𝑡 𝑡 , 𝑡 − 2 2   , 1 − 𝑡 , − 1 = 𝑓 𝑡 , 𝑐 1 𝑡 − 𝑡 2 / 2 𝑐 1 𝑢 ( 𝑡 ) 𝑢 ( 𝑡 ) , 𝑐 2 1 − 𝑡 𝑐 2 𝑢  𝑢 ( 𝑡 )  1 ( 𝑡 ) , 𝑐 3 𝑐 3 − 𝑢   𝑢 ( 𝑡 )    ( 𝑡 ) ≤ 𝑐 𝛼 1 1  𝑡 − 𝑡 2 / 2 𝑐 1 𝑢  ( 𝑡 ) 𝛽 1 𝑐 𝛼 2 2  1 − 𝑡 𝑐 2 𝑢   ( 𝑡 ) 𝛽 2  1 𝑐 3  𝛼 3  𝑐 3 − 𝑢    ( 𝑡 ) 𝛽 3 𝑓  𝑡 , 𝑢 ( 𝑡 ) , 𝑢  ( 𝑡 ) , 𝑢    ( 𝑡 ) ≤ 𝑐 𝛼 1 1  1 𝑐 1 𝐼 1  𝛽 1 𝑐 𝛼 2 2  1 𝑐 2 𝐼 3  𝛽 2  1 𝑐 3  𝛼 3  − 𝑐 3 𝑢    ( 𝑡 ) 𝛽 3 𝑓  𝑡 , 𝑢 ( 𝑡 ) , 𝑢  ( 𝑡 ) , 𝑢   (  𝑡 ) = 𝐶 ∗  − 𝑢   (  𝑡 ) − 𝛽 3 𝑓  𝑡 , 𝑢 ( 𝑡 ) , 𝑢  ( 𝑡 ) , 𝑢   (  𝑡 ) , 𝑡 ∈ ( 0 , 1 ) , ( 3 . 4 ) where 𝐶 ∗ = 𝑐 𝛼 1 − 𝛽 1 1 𝑐 𝛼 2 − 𝛽 2 2 𝑐 𝛽 3 − 𝛼 3 3 𝐼 − 𝛽 1 1 𝐼 − 𝛽 2 3 . Hence, integrating ( 3.4 ) from 𝑡 to 1, we obtain 𝜆  1 𝑡  − 𝑢    ( 𝑠 ) 𝛽 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝐶 ∗  − 𝑢     ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) . ( 3 . 5 ) Since − 𝑢   ( 𝑡 ) increases on [ 0 , 1 ] , we get  − 𝑢    ( 𝑡 ) 𝛽 3 𝜆  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝐶 ∗  − 𝑢     ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) , ( 3 . 6 ) that is, 𝜆  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝐶 ∗ − 𝑢    ( 𝑡 ) ( − 𝑢   ( 𝑡 ) ) 𝛽 3 , 𝑡 ∈ ( 0 , 1 ) . ( 3 . 7 ) Notice that 𝛽 3 < 1 , integrating ( 3.7 ) from 0 to 1, we have 𝜆  1 0  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 𝑑 𝑡 ≤ 𝐶 ∗  1 − 𝛽 3  − 1  − 𝑢    ( 1 ) 1 − 𝛽 3 . ( 3 . 8 ) That is, 𝜆  1 0  𝑠 0 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑡 𝑑 𝑠 ≤ 𝐶 ∗  1 − 𝛽 3  − 1  − 𝑢    ( 1 ) 1 − 𝛽 3 . ( 3 . 9 ) Thus,  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < + ∞ . ( 3 . 1 0 ) By an argument similar to the one used in deriving ( 3.5 ), we can obtain 𝜆  1 𝑡  − 𝑢    ( 𝑠 ) 𝛼 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝐶 ∗  − 𝑢     ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) , ( 3 . 1 1 ) where 𝐶 ∗ = 𝑐 𝛽 1 − 𝛼 1 1 𝑐 𝛽 2 − 𝛼 2 2 𝑐 𝛼 3 − 𝛽 3 3 𝐼 − 𝛼 1 2 𝐼 − 𝛼 2 4 . So, 𝜆  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝐶 ∗ ‖ 𝑢 ‖ − 𝛼 3 2  − 𝑢     ( 𝑡 ) , 𝑡 ∈ ( 0 , 1 ) . ( 3 . 1 2 ) Integrating ( 3.12 ) from 0 to 1, we have 𝜆  1 0  1 𝑡 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 𝑑 𝑡 ≥ 𝐶 ∗ ‖ 𝑢 ‖ − 𝛼 3 2  − 𝑢    . ( 1 ) ( 3 . 1 3 ) That is, 𝜆  1 0  𝑠 0 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑡 𝑑 𝑠 ≥ 𝐶 ∗ ‖ 𝑢 ‖ − 𝛼 3 2  − 𝑢    . ( 1 ) ( 3 . 1 4 ) So,  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 > 0 . ( 3 . 1 5 ) This and ( 3.10 ) imply that ( 2.17 ) holds. Now assume that ( 2.17 ) holds, we will show that BVP ( 1.1 ) has at least one positive solution for any 𝜆 > 0 . By ( 2.17 ), there exists a sufficient small 𝛿 > 0 such that  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 > 0 . ( 3 . 1 6 ) For any fixed 𝜆 > 0 , first of all, we prove ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ 2 ≥ ‖ 𝑢 ‖ 2 , ∀ 𝑢 ∈ 𝜕 𝑃 𝑟 , ( 3 . 1 7 ) where 0 < 𝑟 ≤ m i n { 𝑁 1 , 𝑁 2 , 𝑁 3 , ( 𝜆 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 ) ∫ 𝛿 1 − 𝛿 𝑠 𝑓 ( 𝑠 , 𝑠 − 𝑠 2 / 2 , 1 − 𝑠 , − 1 ) 𝑑 𝑠 ) 1 / ( 1 − ( 𝛽 1 + 𝛽 2 + 𝛽 3 ) ) } . Let 𝑢 ∈ 𝜕 𝑃 𝑟 , then 𝑟 8  𝑡 𝑡 − 2 2   𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑟 𝑡 − 2 2  ≤ 𝑁 1  𝑡 𝑡 − 2 2  , 𝑟 4 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ 𝑟 ( 1 − 𝑡 ) ≤ 𝑁 2 ( 1 − 𝑡 ) , 𝛿 𝑟 ≤ 𝑟 𝑡 ≤ − 𝑢   ( 𝑡 ) ≤ 𝑟 ≤ 𝑁 3 [ ] . , ∀ 𝑡 ∈ 𝛿 , 1 − 𝛿 ( 3 . 1 8 ) From Lemma 2.1 , ( 3.18 ), and ( 𝐻 ), we get ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ 2 = ‖ ‖  𝑇 𝜆 𝑢    ‖ ‖ ≥ 𝜆 m a x [ ] 𝑡 ∈ 𝛿 , 1 − 𝛿  1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≥ 𝛿 𝜆 𝛿 1 − 𝛿  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  𝑠 / 2 𝑠 − 2 2  , 𝑢  ( 𝑠 ) (  1 − 𝑠 1 − 𝑠 ) , ( − 1 ) − 𝑢   (    𝑠 ) 𝑑 𝑠 ≥ 𝛿 𝜆 𝛿 1 − 𝛿 𝑠  𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  / 2 𝛽 1  𝑢  ( 𝑠 )  1 − 𝑠 𝛽 2  − 𝑢    ( 𝑠 ) 𝛽 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2   𝑟 , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝛿 8  𝛽 1  𝑟 4  𝛽 2 ( 𝛿 𝑟 ) 𝛽 3 𝜆  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 ) 𝑟 𝛽 1 + 𝛽 2 + 𝛽 3 𝜆  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝑟 = ‖ 𝑢 ‖ 2 , 𝑢 ∈ 𝜕 𝑃 𝑟 . ( 3 . 1 9 ) Thus, ( 3.17 ) holds. Next, we claim that ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ 2 ≤ ‖ 𝑢 ‖ 2 , ∀ 𝑢 ∈ 𝜕 𝑃 𝑅 , ( 3 . 2 0 ) where 𝑅 ≥ m a x { 8 𝑁 1 − 1 , 4 𝑁 2 − 1 , ( 𝜆 𝑁 𝛼 3 − 𝛽 3 3 ∫ 1 0 𝑠 𝑓 ( 𝑠 , 𝑠 − 𝑠 2 / 2 , 1 − 𝑠 , − 1 ) 𝑑 𝑠 ) 1 / ( 1 − ( 𝛽 1 + 𝛽 2 + 𝛽 3 ) ) } . Let 𝑐 = 𝑁 3 / 𝑅 , then for 𝑢 ∈ 𝜕 𝑃 𝑅 , we get 𝑁 1 − 1  𝑡 𝑡 − 2 2  ≤ 𝑅 8  𝑡 𝑡 − 2 2   𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑅 𝑡 − 2 2  , 𝑁 2 − 1 𝑅 ( 1 − 𝑡 ) ≤ 4 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ 𝑅 ( 1 − 𝑡 ) , − 𝑐 𝑢   ( 𝑡 ) ≤ 𝑐 ‖ 𝑢 ‖ 2 = 𝑐 𝑅 = 𝑁 3 [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 3 . 2 1 ) Therefore, by Lemma 2.1 and ( 𝐻 ), it follows that | | |  𝑇 𝜆 𝑢    | | |  ( 𝑡 ) = 𝜆 1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢     ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  𝑠 / 2 𝑠 − 2 2  , 𝑢  ( 𝑠 )  1 1 − 𝑠 ( 1 − 𝑠 ) , ( − 1 ) 𝑐   − 𝑐 𝑢      ( s ) 𝑑 𝑠 ≤ 𝜆 1 0  𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  / 2 𝛽 1  𝑢  ( 𝑠 )  1 − 𝑠 𝛽 2  1 𝑐  𝛽 3  − 𝑐 𝑢    ( 𝑠 ) 𝛼 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝑅 𝛽 1 + 𝛽 2  𝑁 3 𝑅  𝛼 3 − 𝛽 3 𝑅 𝛼 3 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 = 𝑅 𝛽 1 + 𝛽 2 + 𝛽 3  𝑁 3  𝛼 3 − 𝛽 3 𝜆  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝑅 = ‖ 𝑢 ‖ 2 , 𝑢 ∈ 𝜕 𝑃 𝑅 . ( 3 . 2 2 ) This implies that ( 3.20 ) holds. By Lemmas 1.1 and 2.4 , ( 3.17 ), and ( 3.20 ), we obtain that 𝑇 𝜆 has a fixed point in 𝑃 𝑅 ⧵ 𝑃 𝑟 . Therefore, BVP ( 1.1 ) has a positive solution in 𝑃 𝑅 ⧵ 𝑃 𝑟 for any 𝜆 > 0 . 4. Unbounded Connected Branch of Positive Solutions In this section, we study the global continua results under the hypotheses ( 𝐻 ) and ( 2.17 ). Let 𝐿 = { ( 𝜆 , 𝑢 ) ∈ ( 0 , + ∞ ) × ( 𝑃 ⧵ { 𝜃 } ) ∶ ( 𝜆 , 𝑢 ) s a t i s fi e s B V P ( 1 . 1 ) } , ( 4 . 1 ) then, by Theorem 3.1 , 𝐿 ∩ ( { 𝜆 } × 𝑃 ) ≠ ∅ for any 𝜆 > 0 . Theorem 4.1. Suppose ( 𝐻 ) and ( 2.17 ) hold, then the closure 𝐿 of positive solution set possesses an unbounded connected branch 𝐶 which comes from ( 0 , 𝜃 ) such that (i) for any 𝜆 > 0 , 𝐶 ∩ ( { 𝜆 } × 𝑃 ) ≠ ∅ , and (ii) l i m ( 𝜆 , 𝑢 𝜆 ) ∈ 𝐶 , 𝜆 → 0 + ‖ 𝑢 𝜆 ‖ 2 = 0 , l i m ( 𝜆 , 𝑢 𝜆 ) ∈ 𝐶 , 𝜆 → + ∞ ‖ 𝑢 𝜆 ‖ 2 = + ∞ . Proof. We now prove our conclusion by the following several steps. First, we prove that for arbitrarily given 0 < 𝜆 1 < 𝜆 2 < + ∞ , 𝐿 ∩ ( [ 𝜆 1 , 𝜆 2 ] × 𝑃 ) is bounded. In fact, let  𝑅 = 2 m a x 8 𝑁 1 − 1 , 4 𝑁 2 − 1 ,  𝜆 2 𝑁 𝛼 3 − 𝛽 3 3  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2   , 1 − 𝑠 , − 1 𝑑 𝑠 1 / ( 1 − ( 𝛽 1 + 𝛽 2 + 𝛽 3 ) )  , ( 4 . 2 ) then for 𝑢 ∈ 𝑃 ⧵ { 𝜃 } and ‖ 𝑢 ‖ 2 ≥ 𝑅 , we get 𝑁 1 − 1  𝑡 𝑡 − 2 2  ≤ 𝑅 8  𝑡 𝑡 − 2 2   𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑡 − 2 2  ‖ 𝑢 ‖ 2 , 𝑁 2 − 1 ( 𝑅 1 − 𝑡 ) ≤ 4 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ ( 1 − 𝑡 ) ‖ 𝑢 ‖ 2 [ ] . , ∀ 𝑡 ∈ 0 , 1 ( 4 . 3 ) Therefore, by Lemma 2.1 and ( 𝐻 ), it follows that ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ 2 ≤ ‖ ‖ 𝑇 𝜆 2 𝑢 ‖ ‖ 2 ≤ 𝜆 2  1 0  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 2  1 0  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  𝑠 / 2 𝑠 − 2 2  , 𝑢  ( 𝑠 ) 1 − s ( 1 − 𝑠 ) , ( − 1 ) ‖ 𝑢 ‖ 2 𝑁 3 𝑁 3 ‖ 𝑢 ‖ 2  − 𝑢     ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 2 ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 2  𝑁 3 ‖ 𝑢 ‖ 2  𝛼 3 − 𝛽 3 ‖ 𝑢 ‖ 𝛼 3 2  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 = 𝜆 2 ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 + 𝛽 3 2  𝑁 3  𝛼 3 − 𝛽 3  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 < ‖ 𝑢 ‖ 2  𝜆 , ∀ 𝜆 ∈ 1 , 𝜆 2  . ( 4 . 4 ) Let 1 𝑟 = 2  𝑁 m i n 1 , 𝑁 2 , 𝑁 3 ,  𝜆 1 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 )  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2   , 1 − 𝑠 , − 1 𝑑 𝑠 1 / ( 1 − ( 𝛽 1 + 𝛽 2 + 𝛽 3 ) )  , ( 4 . 5 ) where 𝛿 is given by ( 3.16 ). Then for 𝑢 ∈ 𝑃 ⧵ { 𝜃 } and ‖ 𝑢 ‖ 2 ≤ 𝑟 , we get ‖ 𝑢 ‖ 2 8  𝑡 𝑡 − 2 2   𝑡 ≤ 𝑢 ( 𝑡 ) ≤ 𝑟 𝑡 − 2 2  ≤ 𝑁 1  𝑡 𝑡 − 2 2  ; ‖ 𝑢 ‖ 2 4 ( 1 − 𝑡 ) ≤ 𝑢  ( 𝑡 ) ≤ 𝑟 ( 1 − 𝑡 ) ≤ 𝑁 2 ( 1 − 𝑡 ) , 𝛿 ‖ 𝑢 ‖ 2 ≤ 𝑡 ‖ 𝑢 ‖ 2 ≤ − 𝑢   ( 𝑡 ) ≤ 𝑟 ≤ 𝑁 3 [ ] . , ∀ 𝑡 ∈ 𝛿 , 1 − 𝛿 ( 4 . 6 ) Therefore, by Lemma 2.1 and ( 𝐻 ), it follows that ‖ ‖ 𝑇 𝜆 𝑢 ‖ ‖ ≥ ‖ ‖ 𝑇 𝜆 1 𝑢 ‖ ‖ ≥ 𝜆 1 m a x [ ] 𝑡 ∈ 𝛿 , 1 − 𝛿  1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 ( 𝑠 ) , 𝑢  ( 𝑠 ) , 𝑢    ( 𝑠 ) 𝑑 𝑠 ≥ 𝛿 𝜆 1  𝛿 1 − 𝛿  𝑠 𝑓 𝑠 , 𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  𝑠 / 2 𝑠 − 2 2  , 𝑢  ( 𝑠 ) (  1 − 𝑠 1 − 𝑠 ) , ( − 1 ) − 𝑢   (   𝑠 ) 𝑑 𝑠 ≥ 𝛿 𝜆 1  𝛿 1 − 𝛿 𝑠  𝑢 ( 𝑠 ) 𝑠 − 𝑠 2  / 2 𝛽 1  𝑢  ( 𝑠 )  1 − 𝑠 𝛽 2  − 𝑢    ( 𝑠 ) 𝛽 3 𝑓  𝑠 𝑠 , 𝑠 − 2 2   , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝛿 ‖ 𝑢 ‖ 2 ) 8  𝛽 1  ‖ 𝑢 ‖ 2 4  𝛽 2  𝛿 ‖ 𝑢 ‖ 2  𝛽 3 𝜆 1  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 ) ‖ 𝑢 ‖ 𝛽 1 + 𝛽 2 + 𝛽 3 2 𝜆 1  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 > ‖ 𝑢 ‖ 2 , 𝑢 ∈ 𝜕 𝑃 𝑟 . ( 4 . 7 ) Therefore, 𝑢 = 𝑇 𝜆 𝑢 has no positive solution in ( [ 𝜆 1 , 𝜆 2 ] × ( 𝑃 ⧵ 𝑃 𝑅 ) ) ∪ ( [ 𝜆 1 , 𝜆 2 ] × 𝑃 𝑟 ) . As a consequence, 𝐿 ∩ ( [ 𝜆 1 , 𝜆 2 ] × 𝑃 ) is bounded. By the complete continuity of 𝑇 𝜆 , 𝐿 ∩ ( [ 𝜆 1 , 𝜆 2 ] × 𝑃 ) is compact. Second, we choose sequences { 𝑎 𝑛 } ∞ 𝑛 = 1 and { 𝑏 𝑛 } ∞ 𝑛 = 1 satisfy 0 < ⋯ < 𝑎 𝑛 < ⋯ < 𝑎 1 < 𝑏 1 < ⋯ < 𝑏 𝑛 < ⋯ , l i m 𝑛 → + ∞ 𝑎 𝑛 = 0 , l i m 𝑛 → + ∞ 𝑏 𝑛 = + ∞ . ( 4 . 8 ) We are to prove that for any positive integer 𝑛 , there exists a connected branch 𝐶 𝑛 of 𝐿 satisfying 𝐶 𝑛 ∩ 𝑎   𝑛   × 𝑃 ≠ ∅ , 𝐶 𝑛 ∩ 𝑏   𝑛   × 𝑃 ≠ ∅ . ( 4 . 9 ) Let 𝑛 be fixed, suppose that for any ( 𝑏 𝑛 , 𝑢 ) ∈ 𝐿 ∩ ( { 𝑏 𝑛 } × 𝑃 ) , the connected branch 𝐶 𝑢 of 𝐿 ∩ ( [ 𝑎 𝑛 , 𝑏 𝑛 ] × 𝑃 ) , passing through ( 𝑏 𝑛 , 𝑢 ) , leads to 𝐶 𝑢 ∩ ( { 𝑎 𝑛 } × 𝑃 ) = ∅ . Since 𝐶 𝑢 is compact, there exists a bounded open subset Ω 1 of [ 𝑎 𝑛 , 𝑏 𝑛 ] × 𝑃 such that 𝐶 𝑢 ⊂ Ω 1 , Ω 1 ∩ ( { 𝑎 𝑛 } × 𝑃 ) = ∅ , and Ω 1 ∩ ( [ 𝑎 𝑛 , 𝑏 𝑛 ] × { 𝜃 } ) = ∅ , where Ω 1 and later 𝜕 Ω 1 denote the closure and boundary of Ω 1 with respect to [ 𝑎 𝑛 , 𝑏 𝑛 ] × 𝑃 . If 𝐿 ∩ 𝜕 Ω 1 ≠ ∅ , then 𝐶 u and 𝐿 ∩ 𝜕 Ω 1 are two disjoint closed subsets of 𝐿 ∩ Ω 1 . Since 𝐿 ∩ Ω 1 is a compact metric space, there are two disjoint compact subsets 𝑀 1 and 𝑀 2 of 𝐿 ∩ Ω 1 such that 𝐿 ∩ Ω 1 = 𝑀 1 ∪ 𝑀 2 , 𝐶 𝑢 ⊂ 𝑀 1 , and 𝐿 ∩ 𝜕 Ω 1 ⊂ 𝑀 2 . Evidently, 𝛾 = ∶ d i s t ( 𝑀 1 , 𝑀 2 ) > 0 . Denoting by 𝑉 the 𝛾 / 3 -neighborhood of 𝑀 1 and letting Ω 𝑢 = Ω 1 ∩ 𝑉 , then it follows that 𝐶 𝑢 ⊂ Ω 𝑢 , Ω 𝑢 ∩ 𝑎    𝑛   ∪ 𝑎 × 𝑃   𝑛 , 𝑏 𝑛  × { 𝜃 }   = ∅ , 𝐿 ∩ 𝜕 Ω 𝑢 = ∅ . ( 4 . 1 0 ) If 𝐿 ∩ 𝜕 Ω 1 = ∅ , then taking Ω 𝑢 = Ω 1 . It is obvious that in { 𝑏 𝑛 } × 𝑃 , the family of { Ω 𝑢 ∩ ( { 𝑏 𝑛 } × 𝑃 ) ∶ ( 𝑏 𝑛 , 𝑢 ) ∈ 𝐿 } makes up an open covering of 𝐿 ∩ ( { 𝑏 𝑛 } × 𝑃 ) . Since 𝐿 ∩ ( { 𝑏 𝑛 } × 𝑃 ) is a compact set, there exists a finite subfamily { Ω 𝑢 𝑖 ∩ ( { 𝑏 𝑛 } × 𝑃 ) ∶ ( 𝑏 𝑛 , 𝑢 𝑖 ) ∈ 𝐿 } 𝑘 𝑖 = 1 which also covers 𝐿 ∩ ( { 𝑏 𝑛 } × 𝑃 ) . Let ⋃ Ω = 𝑘 𝑖 = 1 Ω 𝑢 𝑖 , then 𝑏 𝐿 ∩   𝑛   × 𝑃 ⊂ Ω , 𝑎 Ω ∩    𝑛   ∪ 𝑎 × 𝑃   𝑛 , 𝑏 𝑛  × { 𝜃 }   = ∅ , 𝐿 ∩ 𝜕 Ω = ∅ . ( 4 . 1 1 ) Hence, by the homotopy invariance of the fixed point index, we obtain 𝑖  𝑇 𝑏 𝑛 𝑏 , Ω ∩   𝑛     𝑇 × 𝑃 , 𝑃 = 𝑖 𝑎 𝑛 𝑎 , Ω ∩   𝑛    × 𝑃 , 𝑃 = 0 . ( 4 . 1 2 ) By the first step of this proof, the construction of Ω , ( 4.4 ), and ( 4.7 ), it follows easily that there exist 0 < 𝑟 𝑛 < 𝑅 𝑛 such that  𝑏 Ω ∩   𝑛    ∩ 𝑏 × 𝑃   𝑛  × 𝑃 𝑟 𝑛   = ∅ , 𝑏 Ω ∩   𝑛    ⊂ 𝑏 × 𝑃   𝑛  × 𝑃 𝑅 𝑛  𝑖  𝑇 , ( 4 . 1 3 ) 𝑏 𝑛 , 𝑃 𝑟 𝑛  𝑖  𝑇 , 𝑃 = 0 , ( 4 . 1 4 ) 𝑏 𝑛 , 𝑃 𝑅 𝑛  , 𝑃 = 1 . ( 4 . 1 5 ) However, by the excision property and additivity of the fixed point index, we have from ( 4.12 ) and ( 4.14 ) that 𝑖 ( 𝑇 𝑏 𝑛 , 𝑃 𝑅 𝑛 , 𝑃 ) = 0 , which contradicts ( 4.15 ). Hence, there exists some ( 𝑏 𝑛 , 𝑢 ) ∈ 𝐿 ∩ ( { 𝑏 𝑛 } × 𝑃 ) such that the connected branch 𝐶 𝑢 of 𝐿 ∩ ( [ 𝑎 𝑛 , 𝑏 𝑛 ] × 𝑃 ) containing ( 𝑏 𝑛 , 𝑢 ) satisfies that 𝐶 𝑢 ∩ ( { 𝑎 𝑛 } × 𝑃 ) ≠ ∅ . Let 𝐶 𝑛 be the connected branch of 𝐿 including 𝐶 𝑢 , then this 𝐶 𝑛 satisfies ( 4.9 ). By Lemma 1.2 , there exists a connected branch 𝐶 ∗ of l i m s u p 𝑛 → + ∞ 𝐶 𝑛 such that 𝐶 ∗ ∩ ( { 𝜆 } × 𝑃 ) ≠ ∅ for any 𝜆 > 0 . Noticing l i m s u p 𝑛 → + ∞ 𝐶 𝑛 ⊂ 𝐿 , we have 𝐶 ∗ ⊂ 𝐿 . Let 𝐶 be the connected branch of 𝐿 including 𝐶 ∗ , then 𝐶 ∩ ( { 𝜆 } × 𝑃 ) ≠ ∅ for any 𝜆 > 0 . Similar to ( 4.4 ) and ( 4.7 ), for any 𝜆 > 0 , ( 𝜆 , 𝑢 𝜆 ) ∈ 𝐶 , we have, by ( 𝐻 ), ( 4.2 ), ( 4.3 ), ( 4.5 ), ( 4.6 ), and Lemma 2.1 , ‖ ‖ 𝑢 𝜆 ‖ ‖ 2 = ‖ ‖ 𝑇 𝜆 𝑢 𝜆 ‖ ‖ 2  ≤ 𝜆 1 0  𝑠 𝑓 𝑠 , 𝑢 𝜆 ( 𝑠 ) , 𝑢  𝜆 ( s ) , 𝑢 𝜆    ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≤ 𝜆 𝜆 ‖ ‖ 𝛽 1 + 𝛽 2 2  𝑁 3 ‖ ‖ 𝑢 𝜆 ‖ ‖ 2  𝛼 3 − 𝛽 3 ‖ ‖ 𝑢 𝜆 ‖ ‖ 𝛼 3 2  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  ‖ ‖ 𝑢 , 1 − 𝑠 , − 1 𝑑 𝑠 = 𝜆 𝜆 ‖ ‖ 𝛽 1 + 𝛽 2 + 𝛽 3 2  𝑁 3  𝛼 3 − 𝛽 3  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≤ 𝜆 𝑅 𝛽 1 + 𝛽 2 + 𝛽 3  𝑁 3  𝛼 3 − 𝛽 3  1 0  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  ‖ ‖ 𝑢 , 1 − 𝑠 , − 1 𝑑 𝑠 , ( 4 . 1 6 ) 𝜆 ‖ ‖ 2 = ‖ ‖ 𝑇 𝜆 𝑢 𝜆 ‖ ‖ 2 ≥ 𝜆 m a x [ ] 𝑡 ∈ 𝛿 , 1 − 𝛿  1 0 𝐺 2  ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝜆 ( 𝑠 ) , 𝑢  𝜆 ( 𝑠 ) , 𝑢 𝜆     ‖ ‖ 𝑢 ( 𝑠 ) 𝑑 𝑠 ≥ 𝜆 𝛿 𝜆 ‖ ‖ 2 8  𝛽 1  ‖ ‖ 𝑢 𝜆 ‖ ‖ 2 4  𝛽 2  𝛿 ‖ ‖ 𝑢 𝜆 ‖ ‖ 2  𝛽 3  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝜆 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 ) ‖ ‖ 𝑢 𝜆 ‖ ‖ 𝛽 1 + 𝛽 2 + 𝛽 3 2  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 ≥ 𝜆 𝛿 1 + 𝛽 3 2 − 3 ( 𝛽 1 + 𝛽 2 ) 𝑟 𝛽 1 + 𝛽 2 + 𝛽 3  𝛿 1 − 𝛿  𝑠 𝑠 𝑓 𝑠 , 𝑠 − 2 2  , 1 − 𝑠 , − 1 𝑑 𝑠 , ( 4 . 1 7 ) where 𝛿 is given by ( 3.16 ). Let 𝜆 → 0 + in ( 4.16 ) and 𝜆 → + ∞ in ( 4.17 ), we have l i m  𝜆 , 𝑢 𝜆  ∈ 𝐶 , 𝜆 → 0 + ‖ ‖ 𝑢 𝜆 ‖ ‖ 2 = 0 , l i m  𝜆 , 𝑢 𝜆  ∈ 𝐶 , 𝜆 → + ∞ ‖ ‖ 𝑢 𝜆 ‖ ‖ 2 = + ∞ . ( 4 . 1 8 ) Therefore, Theorem 4.1 holds and the proof is complete. Acknowledgments This work is carried out while the author is visiting the University of New England. The author thanks Professor Yihong Du for his valuable advices and the Department of Mathematics for providing research facilities. The author also thanks the anonymous referees for their carefully reading of the first draft of the manuscript and making many valuable suggestions. Research is supported by the NSFC (10871120) and HESTPSP (J09LA08). <h4>References</h4> R. P. Agarwal and D. O'Regan, “ Nonlinear superlinear singular and nonsingular second order boundary value problems ,” Journal of Differential Equations , vol. 143, no. 1, pp. 60–95, 1998. L. Liu, P. Kang, Y. Wu, and B. 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