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Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order m -Point Boundary Value Problem

Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order m -Point... Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order m-Point Boundary Value Problem <meta name="citation_title" content="Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order m -Point Boundary Value Problem" /> 0 (i=1,2,…,m-2), 0<η1<η2<⋯<ηm-2<1 are constants, and f(t,u) can have singularities for t=0 and/or t=1 and for u=0. The main tool is the perturbation technique and Schauder fixed point theorem." /> //// 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 2010 (2010), Article ID 254928, 16 pages doi:10.1155/2010/254928 Research Article Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order 𝑚 -Point Boundary Value Problem Xuezhe Lv and Minghe Pei Department of Mathematics, Beihua University, JiLin City 132013, China Received 25 November 2009; Accepted 10 March 2010 Academic Editor: Ivan T. Kiguradze Copyright © 2010 Xuezhe Lv and Minghe Pei. 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 The existence and uniqueness of positive solution is obtained for the singular second-order 𝑚 -point boundary value problem 𝑢   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 for 𝑡 ∈ ( 0 , 1 ) , 𝑢 ( 0 ) = 0 , ∑ 𝑢 ( 1 ) = 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝑢 ( 𝜂 𝑖 ) , where 𝑚 ≥ 3 , 𝛼 𝑖 > 0 ( 𝑖 = 1 , 2 , … , 𝑚 − 2 ) , 0 < 𝜂 1 < 𝜂 2 < ⋯ < 𝜂 𝑚 − 2 < 1 are constants, and 𝑓 ( 𝑡 , 𝑢 ) can have singularities for 𝑡 = 0 and/or 𝑡 = 1 and for 𝑢 = 0 . The main tool is the perturbation technique and Schauder fixed point theorem. 1. Introduction In this paper, we investigate the existence and uniqueness of positive solution for the singular second-order differential equation 𝑢   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 , 𝑡 ∈ ( 0 , 1 ) ( 1 . 1 ) with the 𝑚 -point boundary conditions 𝑢 ( 0 ) = 0 , 𝑢 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  , ( 1 . 2 ) where 𝑚 ≥ 3 , 𝛼 𝑖 > 0 ( 𝑖 = 1 , 2 , … , 𝑚 − 2 ) , 0 < 𝜂 1 < 𝜂 2 < ⋯ < 𝜂 𝑚 − 2 < 1 are constants, and 𝑓 ( 𝑡 , 𝑢 ) can have singularities for 𝑡 = 0 and/or 𝑡 = 1 and for 𝑢 = 0 . Multipoint boundary value problems for second-order ordinary differential equations arise in many areas of applied mathematics and physics; see [ 1 – 3 ] and references therein. The study of three-point boundary value problems for nonlinear second-order ordinary differential equations was initiated by Lomtatidze [ 4 , 5 ]. Since then, the nonlinear second-order multipoint boundary value problems have been studied by many authors; see [ 1 – 3 , 6 – 29 ] and references therein. Most of all the works in the above mentioned references are nonsingular multipoint boundary value problems; see [ 1 – 3 , 10 – 17 , 20 – 23 , 25 , 26 , 28 , 29 ], but the works on the singularities have been quite rarely seen; see [ 4 – 8 , 18 , 19 , 24 , 27 ]. Recently, Du and Zhao [ 7 ], by constructing lower and upper solutions and together with the maximal principle, proved the existence and uniqueness of positive solutions for the following singular second-order 𝑚 -point boundary value problem: 𝑢   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 , 𝑡 ∈ ( 0 , 1 ) , 𝑢 ( 0 ) = 0 , 𝑢 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  , ( 1 . 3 ) where 𝑚 ≥ 3 , 0 < 𝛼 𝑖 < 1 ( 𝑖 = 1 , 2 , … , 𝑚 − 2 ) , 0 < 𝜂 1 < 𝜂 2 < ⋯ < 𝜂 𝑚 − 2 < 1 are constants, ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 < 1 , 𝑓 ( 𝑡 , 𝑢 ) is singular at 𝑡 = 0 , 𝑡 = 1 and 𝑢 = 0 , under conditions that ( H 1 ) 𝑓 ( 𝑡 , 𝑢 ) ∈ 𝐶 ( ( 0 , 1 ) × ( 0 , + ∞ ) , [ 0 , + ∞ ) ) , and 𝑓 ( 𝑡 , 𝑢 ) is decreasing in 𝑢 ; ( H 2 ) 𝑓 ( 𝑡 , 𝜆 ) ≢ 0 , ∫ 1 0 𝑡 ( 1 − 𝑡 ) 𝑓 ( 𝑡 , 𝜆 𝑡 ( 1 − 𝑡 ) ) 𝑑 𝑡 < + ∞ , for all 𝜆 > 0 . The purpose of this paper is to establish existence and uniqueness result of positive solution to SBVP( 1.1 ), ( 1.2 ) under conditions that are weaker than conditions in [ 7 ] and hence improve the result in [ 7 ] by using perturbation technique and Schauder fixed point theorem [ 30 ]. Throughout this paper, we make the following assumptions: ( C 0 ) 𝛼 𝑖 > 0 , 𝑖 = 1 , 2 , … , 𝑚 − 2 and ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ≤ 1 ; ( C 1 ) 𝑓 ∶ ( 0 , 1 ) × ( 0 , + ∞ ) → [ 0 , + ∞ ) is continuous and nonincreasing in 𝑢 for each fixed 𝑡 ∈ ( 0 , 1 ) ; ( C 2 ) ∫ 0 < 1 0 𝑠 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 0 ) 𝑑 𝑠 < + ∞ for each constant 𝑢 0 ∈ ( 0 , + ∞ ) . 2. Preliminary We consider the perturbation problems that are given by 𝑢   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 , 𝑡 ∈ ( 0 , 1 ) , 𝑢 ( 0 ) = ℎ , 𝑢 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  +  1 − 𝑚 − 2  𝑖 = 1 𝛼 𝑖  ℎ , ( ( 2 . 1 ) ℎ ) where ℎ is any nonnegative constant. Definition 2.1. For each fixed constant ℎ ≥ 0 , a function 𝑢 ( 𝑡 ) is said to be a positive solution of BVP( 2 . 1 ℎ ) if 𝑢 ∈ 𝐶 [ 0 , 1 ] ∩ 𝐶 2 ( 0 , 1 ) with 𝑢 ( 𝑡 ) > 0 on ( 0 , 1 ] such that 𝑢   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 holds for all 𝑡 ∈ ( 0 , 1 ) and 𝑢 ( 0 ) = ℎ , ∑ 𝑢 ( 1 ) = 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝑢 ( 𝜂 𝑖 ∑ ) + ( 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ) ℎ . Lemma 2.2. Assume that conditions ( C 1 ) and ( C 2 ) are satisfied. Then, for each fixed constant 𝑢 0 > 0 , l i m 𝑡 → 0 + 𝑡  𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0  𝑑 𝑠 = 0 , ( 2 . 2 ) l i m 𝑡 → 1 − (  1 − 𝑡 ) 𝑡 𝜂 𝑚 − 2 𝑓  𝑠 , 𝑢 0  𝑑 𝑠 = 0 . ( 2 . 3 ) Proof. We only prove ( 2.2 ). And ( 2.3 ) can be proved similarly. For each fixed constant 𝑢 0 > 0 , let  𝑣 ( 𝑡 ) = 𝑡 𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0  𝑑 𝑠 f o r  𝑡 ∈ 0 , 𝜂 1  . ( 2 . 4 ) Then from the conditions ( C 1 ) and ( C 2 ) , we have  0 ≤ 𝑣 ( 𝑡 ) ≤ 𝜂 1 𝑡  𝑠 𝑓 𝑠 , 𝑢 0   𝑑 𝑠 ≤ 𝜂 1 0  𝑠 𝑓 𝑠 , 𝑢 0  𝑑 𝑠 < + ∞ f o r  𝑡 ∈ 0 , 𝜂 1  , 𝑣   ( 𝑡 ) = 𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0   𝑑 𝑠 − 𝑡 𝑓 𝑡 , 𝑢 0  f o r  𝑡 ∈ 0 , 𝜂 1  . ( 2 . 5 ) Hence from the conditions ( C 1 ) and ( C 2 ) , we have  𝜂 1 0 | | 𝑣  | |  ( 𝑡 ) 𝑑 𝑡 ≤ 𝜂 1 0  𝑑 𝑡 𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0   𝑑 𝑠 + 𝜂 1 0  𝑡 𝑓 𝑡 , 𝑢 0   𝑑 𝑡 = 2 𝜂 1 0  𝑡 𝑓 𝑡 , 𝑢 0  𝑑 𝑡 < + ∞ . ( 2 . 6 ) This implies that 𝑣  ( 𝑡 ) ∈ 𝐿 1 ( 0 , 𝜂 1 ) , and hence for each 𝑡 ∈ [ 0 , 𝜂 1 ] ,  𝑡 0 𝑣  (  𝜏 ) 𝑑 𝜏 = 𝑡 0  𝑑 𝜏 𝜂 1 𝜏 𝑓  𝑠 , 𝑢 0   𝑑 𝑠 − 𝑡 0  𝜏 𝑓 𝜏 , 𝑢 0   𝑑 𝜏 = 𝑡 𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0  𝑑 𝑠 = 𝑣 ( 𝑡 ) . ( 2 . 7 ) Thus, it follows from the absolute continuity of integral that l i m 𝑡 → 0 + 𝑣 ( 𝑡 ) = 0 , that is, l i m 𝑡 → 0 + 𝑡  𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0  𝑑 𝑠 = 0 . ( 2 . 8 ) This completes the proof of the lemma. In the following discussion 𝐺 ( 𝑡 , 𝑠 ) denotes Green’s function for Dirichlet problem: − 𝑢   [ ] , ( 𝑡 ) = 0 , 𝑡 ∈ 0 , 1 𝑢 ( 0 ) = 𝑢 ( 1 ) = 0 . ( 2 . 9 ) Then Green's function 𝐺 ( 𝑡 , 𝑠 ) can be expressed as follows:  𝐺 ( 𝑡 , 𝑠 ) = ( 1 − 𝑡 ) 𝑠 , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , ( 1 − 𝑠 ) 𝑡 , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 . ( 2 . 1 0 ) It is easy to see that Green’s function 𝐺 ( 𝑡 , 𝑠 ) has the following simple properties: (i) 0 ≤ 𝑡 ( 1 − 𝑡 ) 𝑠 ( 1 − 𝑠 ) ≤ 𝐺 ( 𝑡 , 𝑠 ) ≤ 𝑠 ( 1 − 𝑠 ) for ( 𝑡 , 𝑠 ) ∈ [ 0 , 1 ] × [ 0 , 1 ] ; (ii) 𝐺 ( 𝑡 , 𝑠 ) > 0 for ( 𝑡 , 𝑠 ) ∈ ( 0 , 1 ) × ( 0 , 1 ) ; (iii) 𝐺 ( 0 , 𝑠 ) = 𝐺 ( 1 , 𝑠 ) = 0 for 𝑠 ∈ [ 0 , 1 ] . By direct calculation, we can easily obtain the following result. Lemma 2.3. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Then, 𝑢 ( 𝑡 ) is a positive solution of BVP( 2 . 1 ℎ ) ( ℎ > 0 ) if and only if 𝑢 ∈ 𝐶 [ 0 , 1 ] is a solution of the following integral equation:  𝑢 ( 𝑡 ) = 1 0 𝑡 𝐺 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ℎ , ( ( 2 . 1 1 ) ℎ ) such that 𝑢 ( 𝑡 ) > ℎ > 0 on ( 0 , 1 ] . Lemma 2.4. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Suppose also that 𝑢 ∈ 𝐶 [ 0 , 1 ] is a solution of the following integral equation:  𝑢 ( 𝑡 ) = 1 0 𝑡 𝐺 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 , ( 2 . 1 2 ) such that 𝑢 ( 𝑡 ) > 0 on ( 0 , 1 ] . Then, 𝑢 ( 𝑡 ) is a positive solution of S B V P ( 1.1 ), ( 1.2 ). Proof. Since 𝑢 ∈ 𝐶 [ 0 , 1 ] is a solution of ( 2.12 ) with 𝑢 ( 𝑡 ) > 0 on ( 0 , 1 ] , then for each 𝑡 ∈ ( 0 , 1 ) ,  𝑡 0  𝑠 ( 1 − 𝑡 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 < + ∞ , 1 𝑡 𝑡 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 < + ∞ . ( 2 . 1 3 ) So for each 𝑡 ∈ ( 0 , 1 ) , we have  𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 < + ∞ , 1 𝑡 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 < + ∞ . ( 2 . 1 4 ) For convenience, let ∑ 𝑐 = ∶ ( 1 / ( 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 ∑ ) ) 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∫ 1 0 𝐺 ( 𝜂 𝑖 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 . Take 𝑡 ∈ ( 0 , 1 ) and Δ 𝑡 such that 𝑡 + Δ 𝑡 ∈ ( 0 , 1 ) , then from the definition of derivative, the mean value theorem of integral, and the absolute continuity of integral, we have l i m Δ 𝑡 → 0 𝑢 ( 𝑡 + Δ 𝑡 ) − 𝑢 ( 𝑡 ) Δ 𝑡 = l i m Δ 𝑡 → 0 1   Δ 𝑡 0 𝑡 + Δ 𝑡  𝑠 ( 1 − 𝑡 − Δ 𝑡 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 𝑡 + Δ 𝑡 ( −  1 − 𝑠 ) ( 𝑡 + Δ 𝑡 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 𝑡 0  𝑠 ( 1 − 𝑡 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 − 1 𝑡  𝑡 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑐 = l i m Δ 𝑡 → 0 1  −  Δ 𝑡 𝑡 0  𝑠 Δ 𝑡 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑡 𝑡 + Δ 𝑡 +  𝑠 ( 1 − 𝑡 − Δ 𝑡 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 1 𝑡 + Δ 𝑡  ( 1 − 𝑠 ) Δ 𝑡 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 − 𝑡 𝑡 + Δ 𝑡 𝑡   ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 + 𝑐 = − 𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑡 ( 1 − 𝑡 ) 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) + 1 𝑡 (  1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 − 𝑡 ( 1 − 𝑡 ) 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) + 𝑐 = − 𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 𝑡 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑐 . ( 2 . 1 5 ) Hence 𝑢   ( 𝑡 ) = − 𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 𝑡 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑐 f o r 𝑡 ∈ ( 0 , 1 ) . ( 2 . 1 6 ) Consequently 𝑢  ∈ 𝐶 ( 0 , 1 ) . Again, from the definition of derivative and the mean value theorem of integrals, we have l i m Δ 𝑡 → 0 𝑢  ( 𝑡 + Δ 𝑡 ) − 𝑢  ( 𝑡 ) Δ 𝑡 = l i m Δ 𝑡 → 0 1  −  Δ 𝑡 0 𝑡 + Δ 𝑡  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 𝑡 + Δ 𝑡 +  ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 − 1 𝑡  ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 = l i m Δ 𝑡 → 0 1  −  Δ 𝑡 𝑡 𝑡 + 1  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 − 𝑡 𝑡 + Δ 𝑡  ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 = l i m Δ 𝑡 → 0 1  −  Δ 𝑡 𝑡 𝑡 + Δ 𝑡  𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 = − 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) f o r 𝑡 ∈ ( 0 , 1 ) . ( 2 . 1 7 ) Hence 𝑢   ( 𝑡 ) = − 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) for 𝑡 ∈ ( 0 , 1 ) . In particular, 𝑢   ∈ 𝐶 ( 0 , 1 ) . On the other hand, from ( 2.12 ), we have 𝑢 ( 0 ) = 0 and 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  = 𝑚 − 2  𝑖 = 1 𝛼 𝑖   1 0 𝐺  𝜂 𝑖  𝜂 , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖   = , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  𝑓 ∑ , 𝑠 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  𝑓 = 1 , 𝑠 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 m − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 = 𝑢 ( 1 ) . ( 2 . 1 8 ) In summary, 𝑢 ( 𝑡 ) is a positive solution of SBVP( 1.1 ), ( 1.2 ). This completes the proof of the lemma. Remark 2.5. Assume that all conditions in Lemma 2.4 hold. Then (1) if 𝑓 ∈ 𝐶 ( [ 0 , 1 ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , we have [ ] 𝑢 ∈ 𝐶 0 , 1 ∩ 𝐶 1 [ 0 , 1 ) ∩ 𝐶 2 ( 0 , 1 ) ; ( 2 . 1 9 ) (2) if 𝑓 ∈ 𝐶 ( ( 0 , 1 ] × ( 0 , + ∞ ) , [ 0 , + ∞ ) ) , we get [ ] 𝑢 ∈ 𝐶 0 , 1 ∩ 𝐶 1 ] ( 0 , 1 ∩ 𝐶 2 ( 0 , 1 ) . ( 2 . 2 0 ) Lemma 2.6. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Then, for each constant ℎ > 0 , BVP( 2 . 1 ℎ ) has a unique solution 𝑢 ( 𝑡 ; ℎ ) with 𝑢 ( 𝑡 ; ℎ ) ≥ ℎ on [ 0 , 1 ] . Proof. We begin by defining an operator 𝑇 in 𝐷 ℎ by  ( 𝑇 𝑢 ) ( 𝑡 ) = 1 0 𝑡 𝐺 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ℎ , ( 2 . 2 1 ) where 𝐷 ℎ ∶ = { 𝑢 ∈ 𝐶 [ 0 , 1 ] ∶ 𝑢 ( 𝑡 ) ≥ ℎ o n [ 0 , 1 ] } is a convex closed set. Then from Lemma 2.2 and the condition ( C 2 ) , we have 𝑇 𝑢 ∈ 𝐶 [ 0 , 1 ] and 𝑇 𝑢 satisfies ( 𝑇 𝑢 )   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 , 𝑡 ∈ ( 0 , 1 ) , ( 𝑇 𝑢 ) ( 0 ) = ℎ , ( 𝑇 𝑢 ) ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖  𝜂 ( 𝑇 𝑢 ) 𝑖  +  1 − 𝑚 − 2  𝑖 = 1 𝛼 𝑖  ℎ . ( 2 . 2 2 ) We now apply Schauder fixed point theorem [ 30 ] to obtain the existence of a fixed point for 𝑇 . To do this, it suffices to verify that 𝑇 is continuous in 𝐷 ℎ and 𝑇 ( 𝐷 ℎ ) is a compact set. Take 𝑢 0 ∈ 𝐷 ℎ , and let { 𝑢 𝑘 } ∞ 𝑘 = 1 ⊂ 𝐷 ℎ such that ‖ ‖ 𝑢 𝑘 − 𝑢 0 ‖ ‖ 𝐶 [ 0 , 1 ] ⟶ 0 a s 𝑘 ⟶ ∞ . ( 2 . 2 3 ) Then for each 𝑡 ∈ ( 0 , 1 ) , 𝑓  𝑡 , 𝑢 𝑘   ( 𝑡 ) ⟶ 𝑓 𝑡 , 𝑢 0  ( 𝑡 ) a s 𝑘 ⟶ ∞ . ( 2 . 2 4 ) From the definition of 𝑇 , we have  𝑇 𝑢 𝑘   ( 𝑡 ) = 1 0  𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝑘  𝑡 ( 𝑠 ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  𝑓  , 𝑠 𝑠 , 𝑢 𝑘  ( 𝑠 ) 𝑑 𝑠 + ℎ . ( 2 . 2 5 ) Also, from the conditions ( C 1 ) and ( C 2 ) , we have 𝑓  𝑡 , 𝑢 0   ( 𝑡 ) + 𝑓 𝑡 , 𝑢 𝑘  ( 𝑡 ) ≤ 2 𝑓 ( 𝑡 , ℎ ) f o r  𝑡 ∈ ( 0 , 1 ) , 1 0 𝑠 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 < + ∞ . ( 2 . 2 6 ) Thus by Lebesgue-dominated convergence theorem, we have m a x 𝑡 ∈ [ 0 , 1 ] | |  𝑇 𝑢 𝑘   ( 𝑡 ) − 𝑇 𝑢 0  | | ≤  ( 𝑡 ) 1 0 | | 𝑓  𝐺 ( 𝑠 , 𝑠 ) 𝑠 , 𝑢 𝑘   ( 𝑠 ) − 𝑓 𝑠 , 𝑢 0  | | + ∑ ( 𝑠 ) 𝑑 𝑠 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖  1 0 | | 𝑓  𝐺 ( 𝑠 , 𝑠 ) 𝑠 , 𝑢 𝑘   ( 𝑠 ) − 𝑓 𝑠 , 𝑢 0  | | =  ∑ ( 𝑠 ) 𝑑 𝑠 1 + 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖   1 0 | | 𝑓  𝑠 ( 1 − 𝑠 ) 𝑠 , 𝑢 𝑘   ( 𝑠 ) − 𝑓 𝑠 , 𝑢 0  | | ( 𝑠 ) 𝑑 𝑠 ⟶ 0 a s 𝑘 ⟶ ∞ . ( 2 . 2 7 ) Therefore, 𝑇 ∶ 𝐷 ℎ → 𝐷 ℎ is continuous. Next we need to show that 𝑇 ( 𝐷 ℎ ) is a relatively compact subset of 𝐶 [ 0 , 1 ] . ( 1 ) From the definition of 𝑇 and the conditions ( C 1 ) and ( C 2 ) , for each 𝑢 ∈ 𝐷 ℎ we have 0 < ℎ ≤ ( 𝑇 𝑢 ) ( 𝑡 ) ≤ ( 𝑇 ℎ ) ( 𝑡 ) f o r [ ] . 𝑡 ∈ 0 , 1 ( 2 . 2 8 ) This implies that 𝑇 ( 𝐷 ℎ ) is uniformly bounded. ( 2 ) For each 𝑢 ∈ 𝐷 ℎ , since ( 𝑇 𝑢 )   ( 𝑡 ) = − 𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 𝑡 + 1 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 f o r [ ] , 𝑡 ∈ 0 , 1 ( 2 . 2 9 ) then | | ( 𝑇 𝑢 )  | | ≤  ( 𝑡 ) 𝑡 0  𝑠 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 + 1 𝑡 + 1 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 = ∶ 𝑀 ( 𝑡 ) f o r [ ] . 𝑡 ∈ 0 , 1 ( 2 . 3 0 ) Obviously 𝑀 ( 𝑡 ) ≥ 0 on [ 0 , 1 ] , and  1 0  𝑀 ( 𝑡 ) 𝑑 𝑡 = 2 1 0 1 𝑠 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖   , 𝑠 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 ≤ 2 1 0 𝑠 1 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝑠 =  ∑ ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 2 + 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖   1 0 𝑠 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 < + ∞ . ( 2 . 3 1 ) Thus 𝑀 ∈ 𝐿 1 ( 0 , 1 ) . From the absolute continuity of integral, we have that for each number 𝜀 > 0 , there is a positive number 𝛿 > 0 such that for all 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ] , if | 𝑡 1 − 𝑡 2 | < 𝛿 , then | ∫ 𝑡 2 𝑡 1 𝑀 ( 𝑡 ) 𝑑 𝑡 | < 𝜀 . It follows that for all 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ] with | 𝑡 1 − 𝑡 2 | < 𝛿 , we have | |  𝑡 ( 𝑇 𝑢 ) 2  −  𝑡 ( 𝑇 𝑢 ) 1  | | = | | | |  𝑡 2 𝑡 1 | | | | ≤ | | | |  ( 𝑇 𝑢 ) ′ ( 𝑡 ) 𝑑 𝑡 𝑡 2 𝑡 1 | | ( 𝑇 𝑢 )  | | | | | | ≤ | | | |  ( 𝑡 ) 𝑑 𝑡 𝑡 2 𝑡 1 𝑀 | | | | ( 𝑡 ) 𝑑 𝑡 < 𝜀 . ( 2 . 3 2 ) Therefore 𝑇 ( 𝐷 ℎ ) is equicontinuous on [ 0 , 1 ] . It follows from Ascoli-Arzela theorem that 𝑇 ( 𝐷 ℎ ) is a relatively compact subset of 𝐶 [ 0 , 1 ] . Consequently, by Schauder fixed point theorem [ 30 ], 𝑇 has a fixed point 𝑢 ( 𝑡 ; ℎ ) ∈ 𝐷 ℎ . Obviously, 𝑢 ( 𝑡 ; ℎ ) > ℎ > 0 on ( 0 , 1 ] . Hence from Lemma 2.3 , 𝑢 ( 𝑡 ; ℎ ) is a solution of BVP( 2 . 1 ℎ ). Next, we will show the uniqueness of solution. Let us suppose that 𝑢 1 ( 𝑡 ; ℎ ) , 𝑢 2 ( 𝑡 ; ℎ ) are two different solutions of BVP( 2 . 1 ℎ ). Then there exists 𝑡 0 ∈ ( 0 , 1 ] such that 𝑢 1 ( 𝑡 0 ; ℎ ) ≠ 𝑢 2 ( 𝑡 0 ; ℎ ) . Without loss of generality, assume that 𝑢 1 ( 𝑡 0 ; ℎ ) > 𝑢 2 ( 𝑡 0 ; ℎ ) . Let 𝑤 ( 𝑡 ) ∶ = 𝑢 1 ( 𝑡 ; ℎ ) − 𝑢 2 ( 𝑡 ; ℎ ) , then 𝑤 ( 0 ) = 0 , 𝑤 ( 𝑡 0 ) > 0 , and hence there exists 𝑡 1 ∈ [ 0 , 𝑡 0 ) such that 𝑤  𝑡 1  = 0 , 𝑤 ( 𝑡 ) > 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 0  . ( 2 . 3 3 ) Further we have 𝑤 ( 𝑡 ) > 0 on ( 𝑡 1 , 1 ] . In fact, assume to the contrary that the conclusion is false. Then there exists 𝑡 2 ∈ ( 𝑡 0 , 1 ] such that 𝑤 ( 𝑡 2 ) ≤ 0 . Thus there exists 𝑡 3 ∈ ( 𝑡 0 , 𝑡 2 ] such that 𝑤  𝑡 3  = 0 , 𝑤 ( 𝑡 ) > 0 f o r  𝑡 𝑡 ∈ 0 , 𝑡 3  . ( 2 . 3 4 ) Since 𝑤 ( 𝑡 1 ) = 0 , 𝑤 ( 𝑡 ) > 0 on ( 𝑡 1 , 𝑡 0 ] , then 𝑤    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ; ℎ ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ; ℎ ) ≥ 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 3  . ( 2 . 3 5 ) It follows from 𝑤 ( 𝑡 1 ) = 𝑤 ( 𝑡 3 ) = 0 that 𝑤 ( 𝑡 ) ≤ 0 on [ 𝑡 1 , 𝑡 3 ] . This is a contradiction to 𝑤 ( 𝑡 ) > 0 on ( 𝑡 1 , 𝑡 3 ) . Now we prove that 𝑤 ( 𝑡 ) ≥ 0 on [ 0 , 𝑡 1 ] . In fact, assume to the contrary that the conclusion is false. Then there exists 𝑡 4 ∈ ( 0 , 𝑡 1 ) such that 𝑤 ( 𝑡 4 ) < 0 . Since 𝑤 ( 0 ) = 𝑤 ( 𝑡 1 ) = 0 , then there exist 𝑡 5 , 𝑡 6 with 0 ≤ 𝑡 5 < 𝑡 4 < 𝑡 6 ≤ 𝑡 1 such that 𝑤  𝑡 5   𝑡 = 𝑤 6  = 0 , 𝑤 ( 𝑡 ) < 0 f o r  𝑡 𝑡 ∈ 5 , 𝑡 6  . ( 2 . 3 6 ) Thus, 𝑤    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ; ℎ ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ; ℎ ) ≤ 0 f o r  𝑡 𝑡 ∈ 5 , 𝑡 6  . ( 2 . 3 7 ) It follows from 𝑤 ( 𝑡 5 ) = 𝑤 ( 𝑡 6 ) that 𝑤 ( 𝑡 ) ≥ 0 on [ 𝑡 5 , 𝑡 6 ] . This is a contradiction to 𝑤 ( 𝑡 ) < 0 on ( 𝑡 5 , 𝑡 6 ) . In summary, we have 𝑤 ( 𝑡 ) ≥ 0 on [ 0 , 𝑡 1 ] and 𝑤 ( 𝑡 ) > 0 on ( 𝑡 1 , 1 ] . Thus  𝑤 ( 𝑡 ) = 1 0  𝑓  𝐺 ( 𝑡 , 𝑠 ) 𝑠 , 𝑢 1   ( 𝑠 ; ℎ ) − 𝑓 𝑠 , 𝑢 2 + 𝑡 ( 𝑠 ; ℎ )   𝑑 𝑠 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖 𝑓  , 𝑠   𝑠 , 𝑢 1   ( 𝑠 ; ℎ ) − 𝑓 𝑠 , 𝑢 2 ( 𝑠 ; ℎ )   𝑑 𝑠 ≤ 0 f o r ] . 𝑡 ∈ ( 0 , 1 ( 2 . 3 8 ) This is a contradiction to 𝑤 ( 𝑡 ) > 0 on ( 𝑡 1 , 1 ] . This completes the proof of the lemma. Lemma 2.7. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Then, the unique solution 𝑢 ( 𝑡 ; ℎ ) of BVP( 2 . 1 ℎ ) is nondecreasing in ℎ . Proof. Let 0 < ℎ 2 < ℎ 1 , and let 𝑢 ( 𝑡 ; ℎ 1 ) , 𝑢 ( 𝑡 ; ℎ 2 ) be the solutions of BVP ( 2 . 1 ) ℎ 1 and BVP ( 2 . 1 ) ℎ 2 , respectively. We will show 𝑢  𝑡 ; ℎ 1   ≥ 𝑢 𝑡 ; ℎ 2  f o r [ ] . 𝑡 ∈ 0 , 1 ( 2 . 3 9 ) Assume to the contrary that the above inequality is false. Then there exists 𝑡 0 ∈ ( 0 , 1 ] such that 𝑢 ( 𝑡 0 ; ℎ 1 ) < 𝑢 ( 𝑡 0 ; ℎ 2 ) . Since 𝑢 ( 0 ; ℎ 1 ) = ℎ 1 > ℎ 2 = 𝑢 ( 0 ; ℎ 2 ) , we have that there exists 𝑡 1 ∈ ( 0 , 𝑡 0 ) such that 𝑢  𝑡 1 ; ℎ 1   𝑡 = 𝑢 1 ; ℎ 2   , 𝑢 𝑡 ; ℎ 1   < 𝑢 𝑡 ; ℎ 2  f o r  𝑡 𝑡 ∈ 1 , 𝑡 0  . ( 2 . 4 0 ) Next we prove 𝑢 ( 𝑡 ; ℎ 1 ) < 𝑢 ( 𝑡 ; ℎ 2 ) on ( 𝑡 0 , 1 ] . In fact, assume to the contrary that the conclusion is false. Then there exists 𝑡 2 ∈ ( 𝑡 0 , 1 ] such that 𝑢  𝑡 2 ; ℎ 1   𝑡 = 𝑢 2 ; ℎ 2   , 𝑢 𝑡 ; ℎ 1   < 𝑢 𝑡 ; ℎ 2  f o r  𝑡 𝑡 ∈ 0 , 𝑡 2  . ( 2 . 4 1 ) Hence 𝑢    𝑡 ; ℎ 1  − 𝑢    𝑡 ; ℎ 2    = − 𝑓 𝑡 , 𝑢 𝑡 ; ℎ 1     + 𝑓 𝑡 , 𝑢 𝑡 ; ℎ 2   ≤ 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 2  . ( 2 . 4 2 ) It follows from 𝑢 ( 𝑡 𝑖 ; ℎ 1 ) = 𝑢 ( 𝑡 𝑖 ; ℎ 2 ) , 𝑖 = 1 , 2 that 𝑢 ( 𝑡 ; ℎ 1 ) ≥ 𝑢 ( 𝑡 ; ℎ 2 ) on [ 𝑡 1 , 𝑡 2 ] . This is a contradiction to 𝑢 ( 𝑡 ; ℎ 1 ) < 𝑢 ( 𝑡 ; ℎ 2 ) on ( 𝑡 1 , 𝑡 2 ) . Thus 𝑢 ( 𝑡 ; ℎ 1 ) < 𝑢 ( 𝑡 ; ℎ 2 ) on ( 𝑡 1 , 1 ] . This implies that 𝑢    𝑡 ; ℎ 1  − 𝑢    𝑡 ; ℎ 2    = − 𝑓 𝑡 , 𝑢 𝑡 ; ℎ 1     + 𝑓 𝑡 , 𝑢 𝑡 ; ℎ 2   ≤ 0 f o r  𝑡 𝑡 ∈ 1  . , 1 ( 2 . 4 3 ) It follows from 𝑢  ( 𝑡 1 ; ℎ 1 ) − 𝑢  ( 𝑡 1 ; ℎ 2 ) ≤ 0 that 𝑢  ( 𝑡 ; ℎ 1 ) − 𝑢  ( 𝑡 ; ℎ 2 ) ≤ 0 on [ 𝑡 1 , 1 ] . Hence, from 𝑢 ( 𝑡 ; ℎ 1 ) < 𝑢 ( 𝑡 ; ℎ 2 ) on ( 𝑡 1 , 1 ] , we have 𝑢  ( 1 ; ℎ 1 ) − 𝑢  ( 1 ; ℎ 2 ) < 0 . Thus 𝑢  1 ; ℎ 1   − 𝑢 1 ; ℎ 2   𝜂 < 𝑢 𝑚 − 2 ; ℎ 1   𝜂 − 𝑢 𝑚 − 2 ; ℎ 2  . ( 2 . 4 4 ) There are two cases to consider. Case 1 (see [ 𝑡 1 ≥ 𝜂 𝑚 − 2 ]). In this case, we have 𝑢  𝜂 𝑖 ; ℎ 1   𝜂 − 𝑢 𝑖 ; ℎ 2  ≥ 0 , 𝑖 = 1 , 2 , … , 𝑚 − 2 . ( 2 . 4 5 ) Hence from the boundary conditions of BVP( 2 . 1 ℎ ), we have 𝑢  1 ; ℎ 1   − 𝑢 1 ; ℎ 2  = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖 ; ℎ 1  +  1 − 𝑚 − 2  𝑖 = 1 𝛼 𝑖  ℎ 1 − 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖 ; ℎ 2  −  1 − 𝑚 − 2  𝑖 = 1 𝛼 𝑖  ℎ 2 ≥ 𝑚 − 2  𝑖 = 1 𝛼 𝑖  𝑢  𝜂 𝑖 ; ℎ 1   𝜂 − 𝑢 𝑖 ; ℎ 2   ≥ 0 . ( 2 . 4 6 ) This is a contradiction to 𝑢 ( 1 ; ℎ 1 ) − 𝑢 ( 1 ; ℎ 2 ) < 0 . Case 2 (see [ 𝑡 1 < 𝜂 𝑚 − 2 ]). In this case, we have 𝑢  1 ; ℎ 1   − 𝑢 1 ; ℎ 2   𝜂 < 𝑢 𝑚 − 2 ; ℎ 1   𝜂 − 𝑢 𝑚 − 2 ; ℎ 2  𝑢  𝜂 < 0 , 𝑚 − 2 ; ℎ 1   𝜂 − 𝑢 𝑚 − 2 ; ℎ 2   𝜂 ≤ 𝑢 𝑖 ; ℎ 1   𝜂 − 𝑢 𝑖 ; ℎ 2  , 𝑖 = 1 , 2 , … , 𝑚 − 3 . ( 2 . 4 7 ) It follows from ( C 0 ) that 𝑢  1 ; ℎ 1   − 𝑢 1 ; ℎ 2  < 𝑚 − 2  𝑖 = 1 𝛼 𝑖  𝑢  𝜂 𝑚 − 2 ; ℎ 1   𝜂 − 𝑢 𝑚 − 2 ; ℎ 2 ≤   𝑚 − 2  𝑖 = 1 𝛼 𝑖  𝑢  𝜂 𝑖 ; ℎ 1   𝜂 − 𝑢 𝑖 ; ℎ 2 .   ( 2 . 4 8 ) This is a contradiction to the boundary conditions of BVP( 2 . 1 ℎ ). In summary, we have 𝑢 ( 𝑡 ; ℎ 1 ) ≥ 𝑢 ( 𝑡 ; ℎ 2 ) on [ 0 , 1 ] . This completes the proof of the lemma. 3. Main Results We now state and prove our main results for singular second-order 𝑚 -point boundary value problem ( 1.1 ), ( 1.2 ). Theorem 3.1. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Then, SBVP( 1.1 ), ( 1.2 ) has at most one positive solution. Proof. Suppose that 𝑢 1 ( 𝑡 ) and 𝑢 2 ( 𝑡 ) are any two positive solutions of SBVP( 1.1 ), ( 1.2 ). We now prove that 𝑢 1 ( 𝑡 ) ≡ 𝑢 2 ( 𝑡 ) on [ 0 , 1 ] . To do this, let 𝑣 ( 𝑡 ) = 𝑢 1 ( 𝑡 ) − 𝑢 2 ( 𝑡 ) on [ 0 , 1 ] . We will show that 𝑣 ( 𝑡 ) ≡ 0 on [ 0 , 1 ] . There are three cases to consider. Case 1 (see [ 𝑣 ( 1 ) > 0 ]). In this case, we have that 𝑣 ( 𝑡 ) ≥ 0 on [ 0 , 1 ] . In fact, assume to the contrary that the conclusion is false. Then, there exists 𝑡 0 ∈ ( 0 , 1 ) such that 𝑣 ( 𝑡 0 ) < 0 . Since 𝑣 ( 0 ) = 0 and 𝑣 ( 1 ) > 0 , then there exist 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ) with 𝑡 1 < 𝑡 0 < 𝑡 2 such that 𝑣 ( 𝑡 ) < 0 o n  𝑡 1 , 𝑡 2   𝑡 , 𝑣 1   𝑡 = 𝑣 2  = 0 . ( 3 . 1 ) Thus 𝑣   ( 𝑡 ) = 𝑢 1   ( 𝑡 ) − 𝑢 2    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ) ≤ 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 2  . ( 3 . 2 ) Hence 𝑣 ( 𝑡 ) ≥ 0 on [ 𝑡 1 , 𝑡 2 ] , which is a contradiction to 𝑣 ( 𝑡 ) < 0 on ( 𝑡 1 , 𝑡 2 ) . Therefore 𝑣 ( 𝑡 ) ≥ 0 on [ 0 , 1 ] . Consequently 𝑣    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ) ≥ 0 f o r 𝑡 ∈ ( 0 , 1 ) . ( 3 . 3 ) Thus 𝑣 ( 𝑡 ) is convex on [ 0 , 1 ] . Since 𝑣 ( 1 ) > 0 and 𝑣 ( 1 ) = 𝑢 1 ( 1 ) − 𝑢 2 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢 1  𝜂 𝑖  − 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢 2  𝜂 𝑖  = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑣  𝜂 𝑖  , ( 3 . 4 ) then there exists 𝑖 0 ∈ { 1 , 2 , … , 𝑚 − 2 } such that 𝑣  𝜂 𝑖 0   𝑣  𝜂 = m a x 𝑖   ∶ 𝑖 = 1 , 2 , … , 𝑚 − 2 > 0 , ( 3 . 5 ) and hence from ( C 0 ) and 0 < 𝜂 𝑖 0 < 1 , we have 𝑣 ( 1 ) ≤ 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑣  𝜂 𝑖 0   𝜂 ≤ 𝑣 𝑖 0  < 1 𝜂 𝑖 0 𝑣  𝜂 𝑖 0  , ( 3 . 6 ) which is a contradiction to that 𝑣 ( 𝑡 ) is convex on [ 0 , 1 ] . Case 2 (see [ 𝑣 ( 1 ) = 0 ]). In this case, we have that 𝑣 ( 𝑡 ) ≡ 0 on [ 0 , 1 ] . In fact, assume to the contrary that the conclusion is false. Then, there exists 𝑡 0 ∈ ( 0 , 1 ) such that 𝑣 ( 𝑡 0 ) ≠ 0 . We may assume without loss of generality that 𝑣 ( 𝑡 0 ) > 0 . Then from 𝑣 ( 0 ) = 𝑣 ( 1 ) = 0 , there exist 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ] with 𝑡 1 < 𝑡 0 < 𝑡 2 such that 𝑣 ( 𝑡 ) > 0 o n  𝑡 1 , 𝑡 2   𝑡 , 𝑣 1   𝑡 = 𝑣 2  = 0 . ( 3 . 7 ) Thus 𝑣    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ) ≥ 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 2  . ( 3 . 8 ) Since 𝑣 ( 𝑡 1 ) = 𝑣 ( 𝑡 2 ) = 0 , then 𝑣 ( 𝑡 ) ≤ 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 2  , ( 3 . 9 ) which is a contradiction to that 𝑣 ( 𝑡 ) > 0 on ( 𝑡 1 , 𝑡 2 ) . Case 3 (see [ 𝑣 ( 1 ) < 0 ]). In this case, similar to the proof of Case 1 we can easily show that 𝑣 ( 𝑡 ) ≤ 0 on [ 0 , 1 ] . Consequently 𝑣    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ) ≤ 0 f o r 𝑡 ∈ ( 0 , 1 ) . ( 3 . 1 0 ) Thus 𝑣 ( 𝑡 ) is concave on [ 0 , 1 ] . Since ∑ 𝑣 ( 1 ) = 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝑣 ( 𝜂 𝑖 ) < 0 , then there exists 𝑖 1 ∈ { 1 , 2 , … , 𝑚 − 2 } such that 𝑣 ( 𝜂 𝑖 1 ) = m i n { 𝑣 ( 𝜂 𝑖 ) ∶ 𝑖 = 1 , 2 , … , 𝑚 − 2 } < 0 , and hence from 0 < 𝜂 𝑖 1 < 1 , we have 𝑣 ( 1 ) ≥ 𝑚 − 2  i = 1 𝛼 𝑖 𝑣  𝜂 𝑖 1   𝜂 ≥ 𝑣 𝑖 1  > 1 𝜂 𝑖 1 𝑣  𝜂 𝑖 1  , ( 3 . 1 1 ) which is a contradiction to that 𝑣 ( 𝑡 ) is concave on [ 0 , 1 ] . In summary, 𝑣 ( 𝑡 ) ≡ 0 on [ 0 , 1 ] , that is, 𝑢 1 ( 𝑡 ) ≡ 𝑢 2 ( 𝑡 ) on [ 0 , 1 ] . This completes the proof of the theorem. Theorem 3.2. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Then SBVP( 1.1 ), ( 1.2 ) has exactly one positive solution. Proof. The uniqueness of positive solution to SBVP( 1.1 ), ( 1.2 ) follows from Theorem 3.1 immediately. Thus we only need to show the existence. Let { ℎ 𝑗 } ∞ 𝑗 = 1 be a decreasing sequence that converges to the number 0 . Then from Lemma 2.6 , BVP ( 2 . 1 ) ℎ 𝑗 has a unique solution 𝑢 ( 𝑡 ; ℎ 𝑗 ) ∶ = 𝑢 𝑗 ( 𝑡 ) . From Lemma 2.7 and ( 2 . 1 1 ℎ ), we have that for each 𝑗 < 𝑘 , 0 ≤ 𝑢 𝑗 ( 𝑡 ) − 𝑢 𝑘 ( 𝑡 ) ≤ ℎ 𝑗 − ℎ 𝑘 f o r [ ] . 𝑡 ∈ 0 , 1 ( 3 . 1 2 ) Thus there exists 𝑢 ∈ 𝐶 [ 0 , 1 ] such that l i m 𝑗 → ∞ 𝑢 𝑗 ( 𝑡 ) = 𝑢 ( 𝑡 ) ≥ 0 , u n i f o r m l y o n [ ] . 0 , 1 ( 3 . 1 3 ) It is easy to see that 𝑢 ( 𝑡 ) satisfies boundary conditions ( 1.2 ). Now we prove that 𝑢 ( 𝑡 ) > 0 f o r ] . 𝑡 ∈ ( 0 , 1 ( 3 . 1 4 ) At first, we prove that 𝑢  𝜂 𝑖 0   𝑢  𝜂 = m a x 𝑖   ∶ 𝑖 = 1 , 2 , … , 𝑚 − 2 > 0 , ( 3 . 1 5 ) where 𝑖 0 ∈ { 1 , 2 , … , 𝑚 − 2 } . In fact, assume to the contrary that the conclusion is false. Then 𝑢 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  = 0 . ( 3 . 1 6 ) From the fact that each function in the sequence { 𝑢 𝑗 } ∞ 𝑗 = 1 is concave, we have that 𝑢 ( 𝑡 ) is concave. It follows from 𝑢 ( 0 ) = 𝑢 ( 𝜂 𝑖 0 ) = 𝑢 ( 1 ) = 0 that 𝑢 ( 𝑡 ) ≡ 0 on [ 0 , 1 ] . Thus when 𝑗 is large enough, 𝑢 𝑗 ( 𝑡 ) is small enough such that 𝑢 𝑗 ( 𝑡 ) ≤ ℎ 1 on [ 0 , 1 ] . Hence from condition ( C 1 ), we have 𝑢 𝑗  𝜂 𝑖 0  =  1 0 𝐺  𝜂 𝑖 0  𝑓  , 𝑠 𝑠 , 𝑢 𝑗  + 𝜂 ( 𝑠 ) 𝑑 𝑠 𝑖 0 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  𝑓  , 𝑠 𝑠 , 𝑢 𝑗  ( 𝑠 ) 𝑑 𝑠 + ℎ 𝑗 >  1 0 𝐺  𝜂 𝑖 0  𝑓  , 𝑠 𝑠 , ℎ 1  𝑑 𝑠 > 0 . ( 3 . 1 7 ) Let 𝑗 → ∞ , we have 𝑢  𝜂 𝑖 0  ≥  1 0 𝐺  𝜂 𝑖 0  𝑓  , 𝑠 𝑠 , ℎ 1  𝑑 𝑠 > 0 . ( 3 . 1 8 ) This is a contradiction to 𝑢 ( 𝜂 𝑖 0 ) = 0 . Thus 𝑢 ( 𝜂 𝑖 0 ) > 0 , and hence 𝑢 ( 1 ) > 0 . Since 𝑢 ( 𝑡 ) is concave, then 𝑢 ( 𝑡 ) > 0 on ( 0 , 1 ] . Since 𝑢 𝑗  ( 𝑡 ) = 1 0  𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝑗  𝑡 ( 𝑠 ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 i 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  𝑓  , 𝑠 𝑠 , 𝑢 𝑗  ( 𝑠 ) 𝑑 𝑠 + ℎ 𝑗 , ( 3 . 1 9 ) then passing to the limit, by Monotone convergence theorem [ 31 ], we have  𝑢 ( 𝑡 ) = 1 0 𝑡 𝐺 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 . ( 3 . 2 0 ) Therefore by Lemma 2.4 , 𝑢 ( 𝑡 ) is a positive solution of SBVP( 1.1 ), ( 1.2 ). This completes the proof of the theorem. Finally, we give an example to which our results can be applicable. Example 3.3. Consider the singular nonlinear second-order 𝑚 -point boundary value problem: 𝑢   + 1 𝑡 𝛽 1 ( 1 − 𝑡 ) 𝛽 2 𝑢 2 − 𝛽 1 𝑢 = 0 , 𝑡 ∈ ( 0 , 1 ) , ( 0 ) = 0 , 𝑢 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  , ( 3 . 2 1 ) where 𝑚 ≥ 3 , 0 < 𝜂 1 < 𝜂 2 < ⋯ < 𝜂 𝑚 − 2 < 1 , 𝛼 𝑖 > 0 ( 𝑖 = 1 , 2 , … , 𝑚 − 2 ) , ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ≤ 1 , and 𝛽 1 , 𝛽 2 ∈ ( 0 , 2 ) . Let 1 𝑓 ( 𝑡 , 𝑢 ) = 𝑡 𝛽 1 ( 1 − 𝑡 ) 𝛽 2 𝑢 2 − 𝛽 1 f o r ( 𝑡 , 𝑢 ) ∈ ( 0 , 1 ) × ( 0 , + ∞ ) . ( 3 . 2 2 ) Obviously, the function 𝑓 ( 𝑡 , 𝑢 ) is singular at 𝑡 = 0 , 1 and 𝑢 = 0 . It is easy to verify that 𝑓 ( 𝑡 , 𝑢 ) satisfies conditions ( C 1 ) and ( C 2 ) . So from Theorem 3.2 , SBVP( 3.21 ) has exactly one positive solution. However, we note that Theorem 2 in [ 7 ] cannot guarantee that SBVP( 3.21 ) has a unique positive solution, since  1 0 𝑡 ( 1 − 𝑡 ) 𝑓 ( 𝑡 , 𝜆 𝑡 ( 1 − 𝑡 ) ) 𝑑 𝑡 = + ∞ f o r 𝜆 > 0 . 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Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order m -Point Boundary Value Problem

Xuezhe, Lv; Minghe, Pei
Boundary Value Problems , Volume 2010 (2010) – Mar 31, 2010
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Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order m-Point Boundary Value Problem <meta name="citation_title" content="Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order m -Point Boundary Value Problem" /> 0 (i=1,2,…,m-2), 0<η1<η2<⋯<ηm-2<1 are constants, and f(t,u) can have singularities for t=0 and/or t=1 and for u=0. The main tool is the perturbation technique and Schauder fixed point theorem." /> //// 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 2010 (2010), Article ID 254928, 16 pages doi:10.1155/2010/254928 Research Article Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order 𝑚 -Point Boundary Value Problem Xuezhe Lv and Minghe Pei Department of Mathematics, Beihua University, JiLin City 132013, China Received 25 November 2009; Accepted 10 March 2010 Academic Editor: Ivan T. Kiguradze Copyright © 2010 Xuezhe Lv and Minghe Pei. 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 The existence and uniqueness of positive solution is obtained for the singular second-order 𝑚 -point boundary value problem 𝑢   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 for 𝑡 ∈ ( 0 , 1 ) , 𝑢 ( 0 ) = 0 , ∑ 𝑢 ( 1 ) = 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝑢 ( 𝜂 𝑖 ) , where 𝑚 ≥ 3 , 𝛼 𝑖 > 0 ( 𝑖 = 1 , 2 , … , 𝑚 − 2 ) , 0 < 𝜂 1 < 𝜂 2 < ⋯ < 𝜂 𝑚 − 2 < 1 are constants, and 𝑓 ( 𝑡 , 𝑢 ) can have singularities for 𝑡 = 0 and/or 𝑡 = 1 and for 𝑢 = 0 . The main tool is the perturbation technique and Schauder fixed point theorem. 1. Introduction In this paper, we investigate the existence and uniqueness of positive solution for the singular second-order differential equation 𝑢   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 , 𝑡 ∈ ( 0 , 1 ) ( 1 . 1 ) with the 𝑚 -point boundary conditions 𝑢 ( 0 ) = 0 , 𝑢 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  , ( 1 . 2 ) where 𝑚 ≥ 3 , 𝛼 𝑖 > 0 ( 𝑖 = 1 , 2 , … , 𝑚 − 2 ) , 0 < 𝜂 1 < 𝜂 2 < ⋯ < 𝜂 𝑚 − 2 < 1 are constants, and 𝑓 ( 𝑡 , 𝑢 ) can have singularities for 𝑡 = 0 and/or 𝑡 = 1 and for 𝑢 = 0 . Multipoint boundary value problems for second-order ordinary differential equations arise in many areas of applied mathematics and physics; see [ 1 – 3 ] and references therein. The study of three-point boundary value problems for nonlinear second-order ordinary differential equations was initiated by Lomtatidze [ 4 , 5 ]. Since then, the nonlinear second-order multipoint boundary value problems have been studied by many authors; see [ 1 – 3 , 6 – 29 ] and references therein. Most of all the works in the above mentioned references are nonsingular multipoint boundary value problems; see [ 1 – 3 , 10 – 17 , 20 – 23 , 25 , 26 , 28 , 29 ], but the works on the singularities have been quite rarely seen; see [ 4 – 8 , 18 , 19 , 24 , 27 ]. Recently, Du and Zhao [ 7 ], by constructing lower and upper solutions and together with the maximal principle, proved the existence and uniqueness of positive solutions for the following singular second-order 𝑚 -point boundary value problem: 𝑢   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 , 𝑡 ∈ ( 0 , 1 ) , 𝑢 ( 0 ) = 0 , 𝑢 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  , ( 1 . 3 ) where 𝑚 ≥ 3 , 0 < 𝛼 𝑖 < 1 ( 𝑖 = 1 , 2 , … , 𝑚 − 2 ) , 0 < 𝜂 1 < 𝜂 2 < ⋯ < 𝜂 𝑚 − 2 < 1 are constants, ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 < 1 , 𝑓 ( 𝑡 , 𝑢 ) is singular at 𝑡 = 0 , 𝑡 = 1 and 𝑢 = 0 , under conditions that ( H 1 ) 𝑓 ( 𝑡 , 𝑢 ) ∈ 𝐶 ( ( 0 , 1 ) × ( 0 , + ∞ ) , [ 0 , + ∞ ) ) , and 𝑓 ( 𝑡 , 𝑢 ) is decreasing in 𝑢 ; ( H 2 ) 𝑓 ( 𝑡 , 𝜆 ) ≢ 0 , ∫ 1 0 𝑡 ( 1 − 𝑡 ) 𝑓 ( 𝑡 , 𝜆 𝑡 ( 1 − 𝑡 ) ) 𝑑 𝑡 < + ∞ , for all 𝜆 > 0 . The purpose of this paper is to establish existence and uniqueness result of positive solution to SBVP( 1.1 ), ( 1.2 ) under conditions that are weaker than conditions in [ 7 ] and hence improve the result in [ 7 ] by using perturbation technique and Schauder fixed point theorem [ 30 ]. Throughout this paper, we make the following assumptions: ( C 0 ) 𝛼 𝑖 > 0 , 𝑖 = 1 , 2 , … , 𝑚 − 2 and ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ≤ 1 ; ( C 1 ) 𝑓 ∶ ( 0 , 1 ) × ( 0 , + ∞ ) → [ 0 , + ∞ ) is continuous and nonincreasing in 𝑢 for each fixed 𝑡 ∈ ( 0 , 1 ) ; ( C 2 ) ∫ 0 < 1 0 𝑠 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 0 ) 𝑑 𝑠 < + ∞ for each constant 𝑢 0 ∈ ( 0 , + ∞ ) . 2. Preliminary We consider the perturbation problems that are given by 𝑢   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 , 𝑡 ∈ ( 0 , 1 ) , 𝑢 ( 0 ) = ℎ , 𝑢 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  +  1 − 𝑚 − 2  𝑖 = 1 𝛼 𝑖  ℎ , ( ( 2 . 1 ) ℎ ) where ℎ is any nonnegative constant. Definition 2.1. For each fixed constant ℎ ≥ 0 , a function 𝑢 ( 𝑡 ) is said to be a positive solution of BVP( 2 . 1 ℎ ) if 𝑢 ∈ 𝐶 [ 0 , 1 ] ∩ 𝐶 2 ( 0 , 1 ) with 𝑢 ( 𝑡 ) > 0 on ( 0 , 1 ] such that 𝑢   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 holds for all 𝑡 ∈ ( 0 , 1 ) and 𝑢 ( 0 ) = ℎ , ∑ 𝑢 ( 1 ) = 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝑢 ( 𝜂 𝑖 ∑ ) + ( 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ) ℎ . Lemma 2.2. Assume that conditions ( C 1 ) and ( C 2 ) are satisfied. Then, for each fixed constant 𝑢 0 > 0 , l i m 𝑡 → 0 + 𝑡  𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0  𝑑 𝑠 = 0 , ( 2 . 2 ) l i m 𝑡 → 1 − (  1 − 𝑡 ) 𝑡 𝜂 𝑚 − 2 𝑓  𝑠 , 𝑢 0  𝑑 𝑠 = 0 . ( 2 . 3 ) Proof. We only prove ( 2.2 ). And ( 2.3 ) can be proved similarly. For each fixed constant 𝑢 0 > 0 , let  𝑣 ( 𝑡 ) = 𝑡 𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0  𝑑 𝑠 f o r  𝑡 ∈ 0 , 𝜂 1  . ( 2 . 4 ) Then from the conditions ( C 1 ) and ( C 2 ) , we have  0 ≤ 𝑣 ( 𝑡 ) ≤ 𝜂 1 𝑡  𝑠 𝑓 𝑠 , 𝑢 0   𝑑 𝑠 ≤ 𝜂 1 0  𝑠 𝑓 𝑠 , 𝑢 0  𝑑 𝑠 < + ∞ f o r  𝑡 ∈ 0 , 𝜂 1  , 𝑣   ( 𝑡 ) = 𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0   𝑑 𝑠 − 𝑡 𝑓 𝑡 , 𝑢 0  f o r  𝑡 ∈ 0 , 𝜂 1  . ( 2 . 5 ) Hence from the conditions ( C 1 ) and ( C 2 ) , we have  𝜂 1 0 | | 𝑣  | |  ( 𝑡 ) 𝑑 𝑡 ≤ 𝜂 1 0  𝑑 𝑡 𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0   𝑑 𝑠 + 𝜂 1 0  𝑡 𝑓 𝑡 , 𝑢 0   𝑑 𝑡 = 2 𝜂 1 0  𝑡 𝑓 𝑡 , 𝑢 0  𝑑 𝑡 < + ∞ . ( 2 . 6 ) This implies that 𝑣  ( 𝑡 ) ∈ 𝐿 1 ( 0 , 𝜂 1 ) , and hence for each 𝑡 ∈ [ 0 , 𝜂 1 ] ,  𝑡 0 𝑣  (  𝜏 ) 𝑑 𝜏 = 𝑡 0  𝑑 𝜏 𝜂 1 𝜏 𝑓  𝑠 , 𝑢 0   𝑑 𝑠 − 𝑡 0  𝜏 𝑓 𝜏 , 𝑢 0   𝑑 𝜏 = 𝑡 𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0  𝑑 𝑠 = 𝑣 ( 𝑡 ) . ( 2 . 7 ) Thus, it follows from the absolute continuity of integral that l i m 𝑡 → 0 + 𝑣 ( 𝑡 ) = 0 , that is, l i m 𝑡 → 0 + 𝑡  𝜂 1 𝑡 𝑓  𝑠 , 𝑢 0  𝑑 𝑠 = 0 . ( 2 . 8 ) This completes the proof of the lemma. In the following discussion 𝐺 ( 𝑡 , 𝑠 ) denotes Green’s function for Dirichlet problem: − 𝑢   [ ] , ( 𝑡 ) = 0 , 𝑡 ∈ 0 , 1 𝑢 ( 0 ) = 𝑢 ( 1 ) = 0 . ( 2 . 9 ) Then Green's function 𝐺 ( 𝑡 , 𝑠 ) can be expressed as follows:  𝐺 ( 𝑡 , 𝑠 ) = ( 1 − 𝑡 ) 𝑠 , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1 , ( 1 − 𝑠 ) 𝑡 , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1 . ( 2 . 1 0 ) It is easy to see that Green’s function 𝐺 ( 𝑡 , 𝑠 ) has the following simple properties: (i) 0 ≤ 𝑡 ( 1 − 𝑡 ) 𝑠 ( 1 − 𝑠 ) ≤ 𝐺 ( 𝑡 , 𝑠 ) ≤ 𝑠 ( 1 − 𝑠 ) for ( 𝑡 , 𝑠 ) ∈ [ 0 , 1 ] × [ 0 , 1 ] ; (ii) 𝐺 ( 𝑡 , 𝑠 ) > 0 for ( 𝑡 , 𝑠 ) ∈ ( 0 , 1 ) × ( 0 , 1 ) ; (iii) 𝐺 ( 0 , 𝑠 ) = 𝐺 ( 1 , 𝑠 ) = 0 for 𝑠 ∈ [ 0 , 1 ] . By direct calculation, we can easily obtain the following result. Lemma 2.3. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Then, 𝑢 ( 𝑡 ) is a positive solution of BVP( 2 . 1 ℎ ) ( ℎ > 0 ) if and only if 𝑢 ∈ 𝐶 [ 0 , 1 ] is a solution of the following integral equation:  𝑢 ( 𝑡 ) = 1 0 𝑡 𝐺 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ℎ , ( ( 2 . 1 1 ) ℎ ) such that 𝑢 ( 𝑡 ) > ℎ > 0 on ( 0 , 1 ] . Lemma 2.4. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Suppose also that 𝑢 ∈ 𝐶 [ 0 , 1 ] is a solution of the following integral equation:  𝑢 ( 𝑡 ) = 1 0 𝑡 𝐺 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 , ( 2 . 1 2 ) such that 𝑢 ( 𝑡 ) > 0 on ( 0 , 1 ] . Then, 𝑢 ( 𝑡 ) is a positive solution of S B V P ( 1.1 ), ( 1.2 ). Proof. Since 𝑢 ∈ 𝐶 [ 0 , 1 ] is a solution of ( 2.12 ) with 𝑢 ( 𝑡 ) > 0 on ( 0 , 1 ] , then for each 𝑡 ∈ ( 0 , 1 ) ,  𝑡 0  𝑠 ( 1 − 𝑡 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 < + ∞ , 1 𝑡 𝑡 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 < + ∞ . ( 2 . 1 3 ) So for each 𝑡 ∈ ( 0 , 1 ) , we have  𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 < + ∞ , 1 𝑡 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 < + ∞ . ( 2 . 1 4 ) For convenience, let ∑ 𝑐 = ∶ ( 1 / ( 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 ∑ ) ) 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∫ 1 0 𝐺 ( 𝜂 𝑖 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 . Take 𝑡 ∈ ( 0 , 1 ) and Δ 𝑡 such that 𝑡 + Δ 𝑡 ∈ ( 0 , 1 ) , then from the definition of derivative, the mean value theorem of integral, and the absolute continuity of integral, we have l i m Δ 𝑡 → 0 𝑢 ( 𝑡 + Δ 𝑡 ) − 𝑢 ( 𝑡 ) Δ 𝑡 = l i m Δ 𝑡 → 0 1   Δ 𝑡 0 𝑡 + Δ 𝑡  𝑠 ( 1 − 𝑡 − Δ 𝑡 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 𝑡 + Δ 𝑡 ( −  1 − 𝑠 ) ( 𝑡 + Δ 𝑡 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 𝑡 0  𝑠 ( 1 − 𝑡 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 − 1 𝑡  𝑡 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑐 = l i m Δ 𝑡 → 0 1  −  Δ 𝑡 𝑡 0  𝑠 Δ 𝑡 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑡 𝑡 + Δ 𝑡 +  𝑠 ( 1 − 𝑡 − Δ 𝑡 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 1 𝑡 + Δ 𝑡  ( 1 − 𝑠 ) Δ 𝑡 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 − 𝑡 𝑡 + Δ 𝑡 𝑡   ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 + 𝑐 = − 𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑡 ( 1 − 𝑡 ) 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) + 1 𝑡 (  1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 0 𝑥 0 2 0 0 𝑑 − 𝑡 ( 1 − 𝑡 ) 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) + 𝑐 = − 𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 𝑡 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑐 . ( 2 . 1 5 ) Hence 𝑢   ( 𝑡 ) = − 𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 𝑡 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑐 f o r 𝑡 ∈ ( 0 , 1 ) . ( 2 . 1 6 ) Consequently 𝑢  ∈ 𝐶 ( 0 , 1 ) . Again, from the definition of derivative and the mean value theorem of integrals, we have l i m Δ 𝑡 → 0 𝑢  ( 𝑡 + Δ 𝑡 ) − 𝑢  ( 𝑡 ) Δ 𝑡 = l i m Δ 𝑡 → 0 1  −  Δ 𝑡 0 𝑡 + Δ 𝑡  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 𝑡 + Δ 𝑡 +  ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 − 1 𝑡  ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 = l i m Δ 𝑡 → 0 1  −  Δ 𝑡 𝑡 𝑡 + 1  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 − 𝑡 𝑡 + Δ 𝑡  ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 = l i m Δ 𝑡 → 0 1  −  Δ 𝑡 𝑡 𝑡 + Δ 𝑡  𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 = − 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) f o r 𝑡 ∈ ( 0 , 1 ) . ( 2 . 1 7 ) Hence 𝑢   ( 𝑡 ) = − 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) for 𝑡 ∈ ( 0 , 1 ) . In particular, 𝑢   ∈ 𝐶 ( 0 , 1 ) . On the other hand, from ( 2.12 ), we have 𝑢 ( 0 ) = 0 and 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  = 𝑚 − 2  𝑖 = 1 𝛼 𝑖   1 0 𝐺  𝜂 𝑖  𝜂 , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖   = , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  𝑓 ∑ , 𝑠 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  𝑓 = 1 , 𝑠 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 m − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 = 𝑢 ( 1 ) . ( 2 . 1 8 ) In summary, 𝑢 ( 𝑡 ) is a positive solution of SBVP( 1.1 ), ( 1.2 ). This completes the proof of the lemma. Remark 2.5. Assume that all conditions in Lemma 2.4 hold. Then (1) if 𝑓 ∈ 𝐶 ( [ 0 , 1 ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , we have [ ] 𝑢 ∈ 𝐶 0 , 1 ∩ 𝐶 1 [ 0 , 1 ) ∩ 𝐶 2 ( 0 , 1 ) ; ( 2 . 1 9 ) (2) if 𝑓 ∈ 𝐶 ( ( 0 , 1 ] × ( 0 , + ∞ ) , [ 0 , + ∞ ) ) , we get [ ] 𝑢 ∈ 𝐶 0 , 1 ∩ 𝐶 1 ] ( 0 , 1 ∩ 𝐶 2 ( 0 , 1 ) . ( 2 . 2 0 ) Lemma 2.6. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Then, for each constant ℎ > 0 , BVP( 2 . 1 ℎ ) has a unique solution 𝑢 ( 𝑡 ; ℎ ) with 𝑢 ( 𝑡 ; ℎ ) ≥ ℎ on [ 0 , 1 ] . Proof. We begin by defining an operator 𝑇 in 𝐷 ℎ by  ( 𝑇 𝑢 ) ( 𝑡 ) = 1 0 𝑡 𝐺 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ℎ , ( 2 . 2 1 ) where 𝐷 ℎ ∶ = { 𝑢 ∈ 𝐶 [ 0 , 1 ] ∶ 𝑢 ( 𝑡 ) ≥ ℎ o n [ 0 , 1 ] } is a convex closed set. Then from Lemma 2.2 and the condition ( C 2 ) , we have 𝑇 𝑢 ∈ 𝐶 [ 0 , 1 ] and 𝑇 𝑢 satisfies ( 𝑇 𝑢 )   ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 , 𝑡 ∈ ( 0 , 1 ) , ( 𝑇 𝑢 ) ( 0 ) = ℎ , ( 𝑇 𝑢 ) ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖  𝜂 ( 𝑇 𝑢 ) 𝑖  +  1 − 𝑚 − 2  𝑖 = 1 𝛼 𝑖  ℎ . ( 2 . 2 2 ) We now apply Schauder fixed point theorem [ 30 ] to obtain the existence of a fixed point for 𝑇 . To do this, it suffices to verify that 𝑇 is continuous in 𝐷 ℎ and 𝑇 ( 𝐷 ℎ ) is a compact set. Take 𝑢 0 ∈ 𝐷 ℎ , and let { 𝑢 𝑘 } ∞ 𝑘 = 1 ⊂ 𝐷 ℎ such that ‖ ‖ 𝑢 𝑘 − 𝑢 0 ‖ ‖ 𝐶 [ 0 , 1 ] ⟶ 0 a s 𝑘 ⟶ ∞ . ( 2 . 2 3 ) Then for each 𝑡 ∈ ( 0 , 1 ) , 𝑓  𝑡 , 𝑢 𝑘   ( 𝑡 ) ⟶ 𝑓 𝑡 , 𝑢 0  ( 𝑡 ) a s 𝑘 ⟶ ∞ . ( 2 . 2 4 ) From the definition of 𝑇 , we have  𝑇 𝑢 𝑘   ( 𝑡 ) = 1 0  𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝑘  𝑡 ( 𝑠 ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  𝑓  , 𝑠 𝑠 , 𝑢 𝑘  ( 𝑠 ) 𝑑 𝑠 + ℎ . ( 2 . 2 5 ) Also, from the conditions ( C 1 ) and ( C 2 ) , we have 𝑓  𝑡 , 𝑢 0   ( 𝑡 ) + 𝑓 𝑡 , 𝑢 𝑘  ( 𝑡 ) ≤ 2 𝑓 ( 𝑡 , ℎ ) f o r  𝑡 ∈ ( 0 , 1 ) , 1 0 𝑠 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 < + ∞ . ( 2 . 2 6 ) Thus by Lebesgue-dominated convergence theorem, we have m a x 𝑡 ∈ [ 0 , 1 ] | |  𝑇 𝑢 𝑘   ( 𝑡 ) − 𝑇 𝑢 0  | | ≤  ( 𝑡 ) 1 0 | | 𝑓  𝐺 ( 𝑠 , 𝑠 ) 𝑠 , 𝑢 𝑘   ( 𝑠 ) − 𝑓 𝑠 , 𝑢 0  | | + ∑ ( 𝑠 ) 𝑑 𝑠 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖  1 0 | | 𝑓  𝐺 ( 𝑠 , 𝑠 ) 𝑠 , 𝑢 𝑘   ( 𝑠 ) − 𝑓 𝑠 , 𝑢 0  | | =  ∑ ( 𝑠 ) 𝑑 𝑠 1 + 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖   1 0 | | 𝑓  𝑠 ( 1 − 𝑠 ) 𝑠 , 𝑢 𝑘   ( 𝑠 ) − 𝑓 𝑠 , 𝑢 0  | | ( 𝑠 ) 𝑑 𝑠 ⟶ 0 a s 𝑘 ⟶ ∞ . ( 2 . 2 7 ) Therefore, 𝑇 ∶ 𝐷 ℎ → 𝐷 ℎ is continuous. Next we need to show that 𝑇 ( 𝐷 ℎ ) is a relatively compact subset of 𝐶 [ 0 , 1 ] . ( 1 ) From the definition of 𝑇 and the conditions ( C 1 ) and ( C 2 ) , for each 𝑢 ∈ 𝐷 ℎ we have 0 < ℎ ≤ ( 𝑇 𝑢 ) ( 𝑡 ) ≤ ( 𝑇 ℎ ) ( 𝑡 ) f o r [ ] . 𝑡 ∈ 0 , 1 ( 2 . 2 8 ) This implies that 𝑇 ( 𝐷 ℎ ) is uniformly bounded. ( 2 ) For each 𝑢 ∈ 𝐷 ℎ , since ( 𝑇 𝑢 )   ( 𝑡 ) = − 𝑡 0  𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 𝑡 + 1 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 f o r [ ] , 𝑡 ∈ 0 , 1 ( 2 . 2 9 ) then | | ( 𝑇 𝑢 )  | | ≤  ( 𝑡 ) 𝑡 0  𝑠 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 + 1 𝑡 + 1 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 = ∶ 𝑀 ( 𝑡 ) f o r [ ] . 𝑡 ∈ 0 , 1 ( 2 . 3 0 ) Obviously 𝑀 ( 𝑡 ) ≥ 0 on [ 0 , 1 ] , and  1 0  𝑀 ( 𝑡 ) 𝑑 𝑡 = 2 1 0 1 𝑠 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖   , 𝑠 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 ≤ 2 1 0 𝑠 1 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝑠 =  ∑ ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 2 + 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖   1 0 𝑠 ( 1 − 𝑠 ) 𝑓 ( 𝑠 , ℎ ) 𝑑 𝑠 < + ∞ . ( 2 . 3 1 ) Thus 𝑀 ∈ 𝐿 1 ( 0 , 1 ) . From the absolute continuity of integral, we have that for each number 𝜀 > 0 , there is a positive number 𝛿 > 0 such that for all 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ] , if | 𝑡 1 − 𝑡 2 | < 𝛿 , then | ∫ 𝑡 2 𝑡 1 𝑀 ( 𝑡 ) 𝑑 𝑡 | < 𝜀 . It follows that for all 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ] with | 𝑡 1 − 𝑡 2 | < 𝛿 , we have | |  𝑡 ( 𝑇 𝑢 ) 2  −  𝑡 ( 𝑇 𝑢 ) 1  | | = | | | |  𝑡 2 𝑡 1 | | | | ≤ | | | |  ( 𝑇 𝑢 ) ′ ( 𝑡 ) 𝑑 𝑡 𝑡 2 𝑡 1 | | ( 𝑇 𝑢 )  | | | | | | ≤ | | | |  ( 𝑡 ) 𝑑 𝑡 𝑡 2 𝑡 1 𝑀 | | | | ( 𝑡 ) 𝑑 𝑡 < 𝜀 . ( 2 . 3 2 ) Therefore 𝑇 ( 𝐷 ℎ ) is equicontinuous on [ 0 , 1 ] . It follows from Ascoli-Arzela theorem that 𝑇 ( 𝐷 ℎ ) is a relatively compact subset of 𝐶 [ 0 , 1 ] . Consequently, by Schauder fixed point theorem [ 30 ], 𝑇 has a fixed point 𝑢 ( 𝑡 ; ℎ ) ∈ 𝐷 ℎ . Obviously, 𝑢 ( 𝑡 ; ℎ ) > ℎ > 0 on ( 0 , 1 ] . Hence from Lemma 2.3 , 𝑢 ( 𝑡 ; ℎ ) is a solution of BVP( 2 . 1 ℎ ). Next, we will show the uniqueness of solution. Let us suppose that 𝑢 1 ( 𝑡 ; ℎ ) , 𝑢 2 ( 𝑡 ; ℎ ) are two different solutions of BVP( 2 . 1 ℎ ). Then there exists 𝑡 0 ∈ ( 0 , 1 ] such that 𝑢 1 ( 𝑡 0 ; ℎ ) ≠ 𝑢 2 ( 𝑡 0 ; ℎ ) . Without loss of generality, assume that 𝑢 1 ( 𝑡 0 ; ℎ ) > 𝑢 2 ( 𝑡 0 ; ℎ ) . Let 𝑤 ( 𝑡 ) ∶ = 𝑢 1 ( 𝑡 ; ℎ ) − 𝑢 2 ( 𝑡 ; ℎ ) , then 𝑤 ( 0 ) = 0 , 𝑤 ( 𝑡 0 ) > 0 , and hence there exists 𝑡 1 ∈ [ 0 , 𝑡 0 ) such that 𝑤  𝑡 1  = 0 , 𝑤 ( 𝑡 ) > 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 0  . ( 2 . 3 3 ) Further we have 𝑤 ( 𝑡 ) > 0 on ( 𝑡 1 , 1 ] . In fact, assume to the contrary that the conclusion is false. Then there exists 𝑡 2 ∈ ( 𝑡 0 , 1 ] such that 𝑤 ( 𝑡 2 ) ≤ 0 . Thus there exists 𝑡 3 ∈ ( 𝑡 0 , 𝑡 2 ] such that 𝑤  𝑡 3  = 0 , 𝑤 ( 𝑡 ) > 0 f o r  𝑡 𝑡 ∈ 0 , 𝑡 3  . ( 2 . 3 4 ) Since 𝑤 ( 𝑡 1 ) = 0 , 𝑤 ( 𝑡 ) > 0 on ( 𝑡 1 , 𝑡 0 ] , then 𝑤    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ; ℎ ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ; ℎ ) ≥ 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 3  . ( 2 . 3 5 ) It follows from 𝑤 ( 𝑡 1 ) = 𝑤 ( 𝑡 3 ) = 0 that 𝑤 ( 𝑡 ) ≤ 0 on [ 𝑡 1 , 𝑡 3 ] . This is a contradiction to 𝑤 ( 𝑡 ) > 0 on ( 𝑡 1 , 𝑡 3 ) . Now we prove that 𝑤 ( 𝑡 ) ≥ 0 on [ 0 , 𝑡 1 ] . In fact, assume to the contrary that the conclusion is false. Then there exists 𝑡 4 ∈ ( 0 , 𝑡 1 ) such that 𝑤 ( 𝑡 4 ) < 0 . Since 𝑤 ( 0 ) = 𝑤 ( 𝑡 1 ) = 0 , then there exist 𝑡 5 , 𝑡 6 with 0 ≤ 𝑡 5 < 𝑡 4 < 𝑡 6 ≤ 𝑡 1 such that 𝑤  𝑡 5   𝑡 = 𝑤 6  = 0 , 𝑤 ( 𝑡 ) < 0 f o r  𝑡 𝑡 ∈ 5 , 𝑡 6  . ( 2 . 3 6 ) Thus, 𝑤    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ; ℎ ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ; ℎ ) ≤ 0 f o r  𝑡 𝑡 ∈ 5 , 𝑡 6  . ( 2 . 3 7 ) It follows from 𝑤 ( 𝑡 5 ) = 𝑤 ( 𝑡 6 ) that 𝑤 ( 𝑡 ) ≥ 0 on [ 𝑡 5 , 𝑡 6 ] . This is a contradiction to 𝑤 ( 𝑡 ) < 0 on ( 𝑡 5 , 𝑡 6 ) . In summary, we have 𝑤 ( 𝑡 ) ≥ 0 on [ 0 , 𝑡 1 ] and 𝑤 ( 𝑡 ) > 0 on ( 𝑡 1 , 1 ] . Thus  𝑤 ( 𝑡 ) = 1 0  𝑓  𝐺 ( 𝑡 , 𝑠 ) 𝑠 , 𝑢 1   ( 𝑠 ; ℎ ) − 𝑓 𝑠 , 𝑢 2 + 𝑡 ( 𝑠 ; ℎ )   𝑑 𝑠 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖 𝑓  , 𝑠   𝑠 , 𝑢 1   ( 𝑠 ; ℎ ) − 𝑓 𝑠 , 𝑢 2 ( 𝑠 ; ℎ )   𝑑 𝑠 ≤ 0 f o r ] . 𝑡 ∈ ( 0 , 1 ( 2 . 3 8 ) This is a contradiction to 𝑤 ( 𝑡 ) > 0 on ( 𝑡 1 , 1 ] . This completes the proof of the lemma. Lemma 2.7. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Then, the unique solution 𝑢 ( 𝑡 ; ℎ ) of BVP( 2 . 1 ℎ ) is nondecreasing in ℎ . Proof. Let 0 < ℎ 2 < ℎ 1 , and let 𝑢 ( 𝑡 ; ℎ 1 ) , 𝑢 ( 𝑡 ; ℎ 2 ) be the solutions of BVP ( 2 . 1 ) ℎ 1 and BVP ( 2 . 1 ) ℎ 2 , respectively. We will show 𝑢  𝑡 ; ℎ 1   ≥ 𝑢 𝑡 ; ℎ 2  f o r [ ] . 𝑡 ∈ 0 , 1 ( 2 . 3 9 ) Assume to the contrary that the above inequality is false. Then there exists 𝑡 0 ∈ ( 0 , 1 ] such that 𝑢 ( 𝑡 0 ; ℎ 1 ) < 𝑢 ( 𝑡 0 ; ℎ 2 ) . Since 𝑢 ( 0 ; ℎ 1 ) = ℎ 1 > ℎ 2 = 𝑢 ( 0 ; ℎ 2 ) , we have that there exists 𝑡 1 ∈ ( 0 , 𝑡 0 ) such that 𝑢  𝑡 1 ; ℎ 1   𝑡 = 𝑢 1 ; ℎ 2   , 𝑢 𝑡 ; ℎ 1   < 𝑢 𝑡 ; ℎ 2  f o r  𝑡 𝑡 ∈ 1 , 𝑡 0  . ( 2 . 4 0 ) Next we prove 𝑢 ( 𝑡 ; ℎ 1 ) < 𝑢 ( 𝑡 ; ℎ 2 ) on ( 𝑡 0 , 1 ] . In fact, assume to the contrary that the conclusion is false. Then there exists 𝑡 2 ∈ ( 𝑡 0 , 1 ] such that 𝑢  𝑡 2 ; ℎ 1   𝑡 = 𝑢 2 ; ℎ 2   , 𝑢 𝑡 ; ℎ 1   < 𝑢 𝑡 ; ℎ 2  f o r  𝑡 𝑡 ∈ 0 , 𝑡 2  . ( 2 . 4 1 ) Hence 𝑢    𝑡 ; ℎ 1  − 𝑢    𝑡 ; ℎ 2    = − 𝑓 𝑡 , 𝑢 𝑡 ; ℎ 1     + 𝑓 𝑡 , 𝑢 𝑡 ; ℎ 2   ≤ 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 2  . ( 2 . 4 2 ) It follows from 𝑢 ( 𝑡 𝑖 ; ℎ 1 ) = 𝑢 ( 𝑡 𝑖 ; ℎ 2 ) , 𝑖 = 1 , 2 that 𝑢 ( 𝑡 ; ℎ 1 ) ≥ 𝑢 ( 𝑡 ; ℎ 2 ) on [ 𝑡 1 , 𝑡 2 ] . This is a contradiction to 𝑢 ( 𝑡 ; ℎ 1 ) < 𝑢 ( 𝑡 ; ℎ 2 ) on ( 𝑡 1 , 𝑡 2 ) . Thus 𝑢 ( 𝑡 ; ℎ 1 ) < 𝑢 ( 𝑡 ; ℎ 2 ) on ( 𝑡 1 , 1 ] . This implies that 𝑢    𝑡 ; ℎ 1  − 𝑢    𝑡 ; ℎ 2    = − 𝑓 𝑡 , 𝑢 𝑡 ; ℎ 1     + 𝑓 𝑡 , 𝑢 𝑡 ; ℎ 2   ≤ 0 f o r  𝑡 𝑡 ∈ 1  . , 1 ( 2 . 4 3 ) It follows from 𝑢  ( 𝑡 1 ; ℎ 1 ) − 𝑢  ( 𝑡 1 ; ℎ 2 ) ≤ 0 that 𝑢  ( 𝑡 ; ℎ 1 ) − 𝑢  ( 𝑡 ; ℎ 2 ) ≤ 0 on [ 𝑡 1 , 1 ] . Hence, from 𝑢 ( 𝑡 ; ℎ 1 ) < 𝑢 ( 𝑡 ; ℎ 2 ) on ( 𝑡 1 , 1 ] , we have 𝑢  ( 1 ; ℎ 1 ) − 𝑢  ( 1 ; ℎ 2 ) < 0 . Thus 𝑢  1 ; ℎ 1   − 𝑢 1 ; ℎ 2   𝜂 < 𝑢 𝑚 − 2 ; ℎ 1   𝜂 − 𝑢 𝑚 − 2 ; ℎ 2  . ( 2 . 4 4 ) There are two cases to consider. Case 1 (see [ 𝑡 1 ≥ 𝜂 𝑚 − 2 ]). In this case, we have 𝑢  𝜂 𝑖 ; ℎ 1   𝜂 − 𝑢 𝑖 ; ℎ 2  ≥ 0 , 𝑖 = 1 , 2 , … , 𝑚 − 2 . ( 2 . 4 5 ) Hence from the boundary conditions of BVP( 2 . 1 ℎ ), we have 𝑢  1 ; ℎ 1   − 𝑢 1 ; ℎ 2  = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖 ; ℎ 1  +  1 − 𝑚 − 2  𝑖 = 1 𝛼 𝑖  ℎ 1 − 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖 ; ℎ 2  −  1 − 𝑚 − 2  𝑖 = 1 𝛼 𝑖  ℎ 2 ≥ 𝑚 − 2  𝑖 = 1 𝛼 𝑖  𝑢  𝜂 𝑖 ; ℎ 1   𝜂 − 𝑢 𝑖 ; ℎ 2   ≥ 0 . ( 2 . 4 6 ) This is a contradiction to 𝑢 ( 1 ; ℎ 1 ) − 𝑢 ( 1 ; ℎ 2 ) < 0 . Case 2 (see [ 𝑡 1 < 𝜂 𝑚 − 2 ]). In this case, we have 𝑢  1 ; ℎ 1   − 𝑢 1 ; ℎ 2   𝜂 < 𝑢 𝑚 − 2 ; ℎ 1   𝜂 − 𝑢 𝑚 − 2 ; ℎ 2  𝑢  𝜂 < 0 , 𝑚 − 2 ; ℎ 1   𝜂 − 𝑢 𝑚 − 2 ; ℎ 2   𝜂 ≤ 𝑢 𝑖 ; ℎ 1   𝜂 − 𝑢 𝑖 ; ℎ 2  , 𝑖 = 1 , 2 , … , 𝑚 − 3 . ( 2 . 4 7 ) It follows from ( C 0 ) that 𝑢  1 ; ℎ 1   − 𝑢 1 ; ℎ 2  < 𝑚 − 2  𝑖 = 1 𝛼 𝑖  𝑢  𝜂 𝑚 − 2 ; ℎ 1   𝜂 − 𝑢 𝑚 − 2 ; ℎ 2 ≤   𝑚 − 2  𝑖 = 1 𝛼 𝑖  𝑢  𝜂 𝑖 ; ℎ 1   𝜂 − 𝑢 𝑖 ; ℎ 2 .   ( 2 . 4 8 ) This is a contradiction to the boundary conditions of BVP( 2 . 1 ℎ ). In summary, we have 𝑢 ( 𝑡 ; ℎ 1 ) ≥ 𝑢 ( 𝑡 ; ℎ 2 ) on [ 0 , 1 ] . This completes the proof of the lemma. 3. Main Results We now state and prove our main results for singular second-order 𝑚 -point boundary value problem ( 1.1 ), ( 1.2 ). Theorem 3.1. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Then, SBVP( 1.1 ), ( 1.2 ) has at most one positive solution. Proof. Suppose that 𝑢 1 ( 𝑡 ) and 𝑢 2 ( 𝑡 ) are any two positive solutions of SBVP( 1.1 ), ( 1.2 ). We now prove that 𝑢 1 ( 𝑡 ) ≡ 𝑢 2 ( 𝑡 ) on [ 0 , 1 ] . To do this, let 𝑣 ( 𝑡 ) = 𝑢 1 ( 𝑡 ) − 𝑢 2 ( 𝑡 ) on [ 0 , 1 ] . We will show that 𝑣 ( 𝑡 ) ≡ 0 on [ 0 , 1 ] . There are three cases to consider. Case 1 (see [ 𝑣 ( 1 ) > 0 ]). In this case, we have that 𝑣 ( 𝑡 ) ≥ 0 on [ 0 , 1 ] . In fact, assume to the contrary that the conclusion is false. Then, there exists 𝑡 0 ∈ ( 0 , 1 ) such that 𝑣 ( 𝑡 0 ) < 0 . Since 𝑣 ( 0 ) = 0 and 𝑣 ( 1 ) > 0 , then there exist 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ) with 𝑡 1 < 𝑡 0 < 𝑡 2 such that 𝑣 ( 𝑡 ) < 0 o n  𝑡 1 , 𝑡 2   𝑡 , 𝑣 1   𝑡 = 𝑣 2  = 0 . ( 3 . 1 ) Thus 𝑣   ( 𝑡 ) = 𝑢 1   ( 𝑡 ) − 𝑢 2    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ) ≤ 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 2  . ( 3 . 2 ) Hence 𝑣 ( 𝑡 ) ≥ 0 on [ 𝑡 1 , 𝑡 2 ] , which is a contradiction to 𝑣 ( 𝑡 ) < 0 on ( 𝑡 1 , 𝑡 2 ) . Therefore 𝑣 ( 𝑡 ) ≥ 0 on [ 0 , 1 ] . Consequently 𝑣    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ) ≥ 0 f o r 𝑡 ∈ ( 0 , 1 ) . ( 3 . 3 ) Thus 𝑣 ( 𝑡 ) is convex on [ 0 , 1 ] . Since 𝑣 ( 1 ) > 0 and 𝑣 ( 1 ) = 𝑢 1 ( 1 ) − 𝑢 2 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢 1  𝜂 𝑖  − 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢 2  𝜂 𝑖  = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑣  𝜂 𝑖  , ( 3 . 4 ) then there exists 𝑖 0 ∈ { 1 , 2 , … , 𝑚 − 2 } such that 𝑣  𝜂 𝑖 0   𝑣  𝜂 = m a x 𝑖   ∶ 𝑖 = 1 , 2 , … , 𝑚 − 2 > 0 , ( 3 . 5 ) and hence from ( C 0 ) and 0 < 𝜂 𝑖 0 < 1 , we have 𝑣 ( 1 ) ≤ 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑣  𝜂 𝑖 0   𝜂 ≤ 𝑣 𝑖 0  < 1 𝜂 𝑖 0 𝑣  𝜂 𝑖 0  , ( 3 . 6 ) which is a contradiction to that 𝑣 ( 𝑡 ) is convex on [ 0 , 1 ] . Case 2 (see [ 𝑣 ( 1 ) = 0 ]). In this case, we have that 𝑣 ( 𝑡 ) ≡ 0 on [ 0 , 1 ] . In fact, assume to the contrary that the conclusion is false. Then, there exists 𝑡 0 ∈ ( 0 , 1 ) such that 𝑣 ( 𝑡 0 ) ≠ 0 . We may assume without loss of generality that 𝑣 ( 𝑡 0 ) > 0 . Then from 𝑣 ( 0 ) = 𝑣 ( 1 ) = 0 , there exist 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ] with 𝑡 1 < 𝑡 0 < 𝑡 2 such that 𝑣 ( 𝑡 ) > 0 o n  𝑡 1 , 𝑡 2   𝑡 , 𝑣 1   𝑡 = 𝑣 2  = 0 . ( 3 . 7 ) Thus 𝑣    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ) ≥ 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 2  . ( 3 . 8 ) Since 𝑣 ( 𝑡 1 ) = 𝑣 ( 𝑡 2 ) = 0 , then 𝑣 ( 𝑡 ) ≤ 0 f o r  𝑡 𝑡 ∈ 1 , 𝑡 2  , ( 3 . 9 ) which is a contradiction to that 𝑣 ( 𝑡 ) > 0 on ( 𝑡 1 , 𝑡 2 ) . Case 3 (see [ 𝑣 ( 1 ) < 0 ]). In this case, similar to the proof of Case 1 we can easily show that 𝑣 ( 𝑡 ) ≤ 0 on [ 0 , 1 ] . Consequently 𝑣    ( 𝑡 ) = − 𝑓 𝑡 , 𝑢 1   ( 𝑡 ) + 𝑓 𝑡 , 𝑢 2  ( 𝑡 ) ≤ 0 f o r 𝑡 ∈ ( 0 , 1 ) . ( 3 . 1 0 ) Thus 𝑣 ( 𝑡 ) is concave on [ 0 , 1 ] . Since ∑ 𝑣 ( 1 ) = 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝑣 ( 𝜂 𝑖 ) < 0 , then there exists 𝑖 1 ∈ { 1 , 2 , … , 𝑚 − 2 } such that 𝑣 ( 𝜂 𝑖 1 ) = m i n { 𝑣 ( 𝜂 𝑖 ) ∶ 𝑖 = 1 , 2 , … , 𝑚 − 2 } < 0 , and hence from 0 < 𝜂 𝑖 1 < 1 , we have 𝑣 ( 1 ) ≥ 𝑚 − 2  i = 1 𝛼 𝑖 𝑣  𝜂 𝑖 1   𝜂 ≥ 𝑣 𝑖 1  > 1 𝜂 𝑖 1 𝑣  𝜂 𝑖 1  , ( 3 . 1 1 ) which is a contradiction to that 𝑣 ( 𝑡 ) is concave on [ 0 , 1 ] . In summary, 𝑣 ( 𝑡 ) ≡ 0 on [ 0 , 1 ] , that is, 𝑢 1 ( 𝑡 ) ≡ 𝑢 2 ( 𝑡 ) on [ 0 , 1 ] . This completes the proof of the theorem. Theorem 3.2. Assume that conditions ( C 0 ) , ( C 1 ) , and ( C 2 ) are satisfied. Then SBVP( 1.1 ), ( 1.2 ) has exactly one positive solution. Proof. The uniqueness of positive solution to SBVP( 1.1 ), ( 1.2 ) follows from Theorem 3.1 immediately. Thus we only need to show the existence. Let { ℎ 𝑗 } ∞ 𝑗 = 1 be a decreasing sequence that converges to the number 0 . Then from Lemma 2.6 , BVP ( 2 . 1 ) ℎ 𝑗 has a unique solution 𝑢 ( 𝑡 ; ℎ 𝑗 ) ∶ = 𝑢 𝑗 ( 𝑡 ) . From Lemma 2.7 and ( 2 . 1 1 ℎ ), we have that for each 𝑗 < 𝑘 , 0 ≤ 𝑢 𝑗 ( 𝑡 ) − 𝑢 𝑘 ( 𝑡 ) ≤ ℎ 𝑗 − ℎ 𝑘 f o r [ ] . 𝑡 ∈ 0 , 1 ( 3 . 1 2 ) Thus there exists 𝑢 ∈ 𝐶 [ 0 , 1 ] such that l i m 𝑗 → ∞ 𝑢 𝑗 ( 𝑡 ) = 𝑢 ( 𝑡 ) ≥ 0 , u n i f o r m l y o n [ ] . 0 , 1 ( 3 . 1 3 ) It is easy to see that 𝑢 ( 𝑡 ) satisfies boundary conditions ( 1.2 ). Now we prove that 𝑢 ( 𝑡 ) > 0 f o r ] . 𝑡 ∈ ( 0 , 1 ( 3 . 1 4 ) At first, we prove that 𝑢  𝜂 𝑖 0   𝑢  𝜂 = m a x 𝑖   ∶ 𝑖 = 1 , 2 , … , 𝑚 − 2 > 0 , ( 3 . 1 5 ) where 𝑖 0 ∈ { 1 , 2 , … , 𝑚 − 2 } . In fact, assume to the contrary that the conclusion is false. Then 𝑢 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  = 0 . ( 3 . 1 6 ) From the fact that each function in the sequence { 𝑢 𝑗 } ∞ 𝑗 = 1 is concave, we have that 𝑢 ( 𝑡 ) is concave. It follows from 𝑢 ( 0 ) = 𝑢 ( 𝜂 𝑖 0 ) = 𝑢 ( 1 ) = 0 that 𝑢 ( 𝑡 ) ≡ 0 on [ 0 , 1 ] . Thus when 𝑗 is large enough, 𝑢 𝑗 ( 𝑡 ) is small enough such that 𝑢 𝑗 ( 𝑡 ) ≤ ℎ 1 on [ 0 , 1 ] . Hence from condition ( C 1 ), we have 𝑢 𝑗  𝜂 𝑖 0  =  1 0 𝐺  𝜂 𝑖 0  𝑓  , 𝑠 𝑠 , 𝑢 𝑗  + 𝜂 ( 𝑠 ) 𝑑 𝑠 𝑖 0 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  𝑓  , 𝑠 𝑠 , 𝑢 𝑗  ( 𝑠 ) 𝑑 𝑠 + ℎ 𝑗 >  1 0 𝐺  𝜂 𝑖 0  𝑓  , 𝑠 𝑠 , ℎ 1  𝑑 𝑠 > 0 . ( 3 . 1 7 ) Let 𝑗 → ∞ , we have 𝑢  𝜂 𝑖 0  ≥  1 0 𝐺  𝜂 𝑖 0  𝑓  , 𝑠 𝑠 , ℎ 1  𝑑 𝑠 > 0 . ( 3 . 1 8 ) This is a contradiction to 𝑢 ( 𝜂 𝑖 0 ) = 0 . Thus 𝑢 ( 𝜂 𝑖 0 ) > 0 , and hence 𝑢 ( 1 ) > 0 . Since 𝑢 ( 𝑡 ) is concave, then 𝑢 ( 𝑡 ) > 0 on ( 0 , 1 ] . Since 𝑢 𝑗  ( 𝑡 ) = 1 0  𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝑗  𝑡 ( 𝑠 ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 i 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  𝑓  , 𝑠 𝑠 , 𝑢 𝑗  ( 𝑠 ) 𝑑 𝑠 + ℎ 𝑗 , ( 3 . 1 9 ) then passing to the limit, by Monotone convergence theorem [ 31 ], we have  𝑢 ( 𝑡 ) = 1 0 𝑡 𝐺 ( 𝑡 , 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝜂 𝑖 𝑚 − 2  𝑖 = 1 𝛼 𝑖  1 0 𝐺  𝜂 𝑖  , 𝑠 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 . ( 3 . 2 0 ) Therefore by Lemma 2.4 , 𝑢 ( 𝑡 ) is a positive solution of SBVP( 1.1 ), ( 1.2 ). This completes the proof of the theorem. Finally, we give an example to which our results can be applicable. Example 3.3. Consider the singular nonlinear second-order 𝑚 -point boundary value problem: 𝑢   + 1 𝑡 𝛽 1 ( 1 − 𝑡 ) 𝛽 2 𝑢 2 − 𝛽 1 𝑢 = 0 , 𝑡 ∈ ( 0 , 1 ) , ( 0 ) = 0 , 𝑢 ( 1 ) = 𝑚 − 2  𝑖 = 1 𝛼 𝑖 𝑢  𝜂 𝑖  , ( 3 . 2 1 ) where 𝑚 ≥ 3 , 0 < 𝜂 1 < 𝜂 2 < ⋯ < 𝜂 𝑚 − 2 < 1 , 𝛼 𝑖 > 0 ( 𝑖 = 1 , 2 , … , 𝑚 − 2 ) , ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ≤ 1 , and 𝛽 1 , 𝛽 2 ∈ ( 0 , 2 ) . Let 1 𝑓 ( 𝑡 , 𝑢 ) = 𝑡 𝛽 1 ( 1 − 𝑡 ) 𝛽 2 𝑢 2 − 𝛽 1 f o r ( 𝑡 , 𝑢 ) ∈ ( 0 , 1 ) × ( 0 , + ∞ ) . ( 3 . 2 2 ) Obviously, the function 𝑓 ( 𝑡 , 𝑢 ) is singular at 𝑡 = 0 , 1 and 𝑢 = 0 . It is easy to verify that 𝑓 ( 𝑡 , 𝑢 ) satisfies conditions ( C 1 ) and ( C 2 ) . So from Theorem 3.2 , SBVP( 3.21 ) has exactly one positive solution. However, we note that Theorem 2 in [ 7 ] cannot guarantee that SBVP( 3.21 ) has a unique positive solution, since  1 0 𝑡 ( 1 − 𝑡 ) 𝑓 ( 𝑡 , 𝜆 𝑡 ( 1 − 𝑡 ) ) 𝑑 𝑡 = + ∞ f o r 𝜆 > 0 . 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