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Hindawi Publishing Corporation Boundary Value Problems Volume 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 Correspondence should be addressed to Minghe Pei, peiminghe@ynu.ac.kr Received 25 November 2009; Accepted 10 March 2010 Academic Editor: Ivan T Kiguradze Copyright q 2010 X Lv and M 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 The existence and uniqueness of positive solution is obtained for the singular second-order mm−2 f t, u t for t ∈ 0, , u 0, u point boundary value problem u t i αi u ηi , where m ≥ 3, αi > i 1, 2, , m − , < η1 < η2 < · · · < ηm−2 < are constants, and f t, u can have singularities for t and/or t and for u The main tool is the perturbation technique and Schauder fixed point theorem 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 with the m-point boundary conditions m−2 u 0, u1 αi u ηi , 1.2 i where m ≥ 3, αi > i 1, 2, , m − , < η1 < η2 < · · · < ηm−2 < are constants, and f t, u can have singularities for t and/or t and for u 2 Boundary Value Problems 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, Since then, the nonlinear secondorder 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 , 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 u 0, t ∈ 0, , 0, 1.3 m−2 αi u ηi , u i where m ≥ 3, < αi < i 1, 2, , m − , < η1 < η2 < · · · < ηm−2 < are constants, m−2 0, t and u 0, under conditions that i αi < 1, f t, u is singular at t H1 f t, u ∈ C 0, × 0, ∞ , 0, ∞ , and f t, u is decreasing in u; H2 f t, λ / 0, ≡ t − t f t, λt − t dt < ∞, for all λ > 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 and hence improve the result in by using perturbation technique and Schauder fixed point theorem 30 Throughout this paper, we make the following assumptions: C0 αi > 0, i 1, 2, , m − and m−2 i αi ≤ 1; C1 f : 0, × 0, ∞ → 0, ∞ is continuous and nonincreasing in u for each fixed t ∈ 0, ; C2 < s − s f s, u0 ds < ∞ for each constant u0 ∈ 0, ∞ Preliminary We consider the perturbation problems that are given by u t f t, u t 0, m−2 u h, αi u ηi u i t ∈ 0, , 1− m−2 αi h, 2.1 h i 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 holds BVP 2.1 h if u ∈ C 0, ∩ C2 0, with u t > on 0, such that u t f t, u t m−2 αi u ηi − m−2 αi h for all t ∈ 0, and u h, u i i Boundary Value Problems Lemma 2.2 Assume that conditions C1 and C2 are satisfied Then, for each fixed constant u0 > 0, η1 lim t t→0 f s, u0 ds 2.2 0, t t lim− − t f s, u0 ds t→1 2.3 ηm−2 Proof We only prove 2.2 And 2.3 can be proved similarly For each fixed constant u0 > 0, let η1 v t t for t ∈ 0, η1 f s, u0 ds 2.4 t Then from the conditions C1 and C2 , we have η1 0≤v t ≤ η1 sf s, u0 ds ≤ t sf s, u0 ds < ∞ for t ∈ 0, η1 , η1 v t 2.5 f s, u0 ds − tf t, u0 for t ∈ 0, η1 t Hence from the conditions C1 and C2 , we have η1 η1 v t dt ≤ η1 dt η1 f s, u0 ds η1 tf t, u0 dt t tf t, u0 dt < ∞ 2.6 This implies that v t ∈ L1 0, η1 , and hence for each t ∈ 0, η1 , t t v τ dτ η1 dτ f s, u0 ds − t η1 τf τ, u0 dτ τ f s, u0 ds t v t 2.7 t Thus, it follows from the absolute continuity of integral that limt → v t 0, that is, η1 lim t t→0 f s, u0 ds 2.8 t This completes the proof of the lemma In the following discussion G t, s denotes Green’s function for Dirichlet problem: −u t u 0, t ∈ 0, , u 2.9 Boundary Value Problems Then Green’s function G t, s can be expressed as follows: G t, s ⎧ ⎨ − t s, ≤ s ≤ t ≤ 1, ⎩ − s t, ≤ t ≤ s ≤ 2.10 It is easy to see that Green’s function G t, s has the following simple properties: i ≤ t − t s − s ≤ G t, s ≤ s − s for t, s ∈ 0, × 0, ; ii G t, s > for t, s ∈ 0, × 0, ; iii G 0, s G 1, s for s ∈ 0, By direct calculation, we can easily obtain the following result Lemma 2.3 Assume that conditions C0 , C1 , and C2 are satisfied Then, u t is a positive solution of BVP 2.1 h h > if and only if u ∈ C 0, is a solution of the following integral equation: G t, s f s, u s ds ut 1− m−2 t m−2 i αi ηi αi G ηi , s f s, u s ds h, 2.11 i h such that u t > h > on 0, Lemma 2.4 Assume that conditions C0 , C1 , and C2 are satisfied Suppose also that u ∈ C 0, is a solution of the following integral equation: G t, s f s, u s ds u t m−2 t 1− m−2 i αi ηi αi i G ηi , s f s, u s ds, 2.12 such that u t > on 0, Then, u t is a positive solution of SBVP 1.1 , 1.2 Proof Since u ∈ C 0, is a solution of 2.12 with u t > on 0, , then for each t ∈ 0, , t s − t f s, u s ds < ∞, t − s f s, u s ds < ∞ 2.13 t So for each t ∈ 0, , we have t sf s, u s ds < ∞, 1 − s f s, u s ds < ∞ 2.14 t m−2 For convenience, let c : 1/ − m−2 αi ηi i i αi G ηi , s f s, u s ds Take t ∈ 0, and Δt such that t Δt ∈ 0, , then from the definition of derivative, the mean value theorem of Boundary Value Problems integral, and the absolute continuity of integral, we have lim Δt − u t Δt u t Δt → t Δt Δt → Δt lim t s − t f s, u s ds − Δt → Δt − t Δt t sΔtf s, u s ds Δt f s, u s ds t − s f s, u s ds c s − t − Δt f s, u s ds t t Δt t Δt − s Δtf s, u s ds − t − s f s, u s ds c t t t − t f t, u t sf s, u s ds − 1−s t t − t Δt − lim s − t − Δt f s, u s ds − s f s, u s ds − t − t f t, u t c t t sf s, u s ds − s f s, u s ds c t 2.15 Hence u t − t sf s, u s ds − s f s, u s ds c for t ∈ 0, 2.16 t Consequently u ∈ C 0, Again, from the definition of derivative and the mean value theorem of integrals, we have lim u t Δt → Δt − u t Δt Δt → Δt lim − t Δt sf s, u s ds t sf s, u s ds − lim Δt → Δt Δt → Δt lim −f t, u t Hence u t −f t, u t t Δt − t 1 − s f s, u s ds t sf s, u s ds − t − 1 − s f s, u s ds t Δt 2.17 − s f s, u s ds t t Δt f s, u s ds t for t ∈ 0, for t ∈ 0, In particular, u ∈ C 0, Boundary Value Problems On the other hand, from 2.12 , we have u m−2 m−2 αi u ηi i αi αi G ηi , s f s, u s ds 1− i m−2 1− m−2 i αi ηi αi ηi αi m−2 m−2 i αi ηi αi m−2 i αi ηi i G ηi , s f s, u s ds i 1 G ηi , s f s, u s ds αi i m−2 i 1− m−2 ηi G ηi , s f s, u s ds i m−2 and G ηi , s f s, u s ds u 2.18 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 if f ∈ C 0, × 0, ∞ , 0, ∞ , we have u ∈ C 0, ∩ C1 0, ∩ C2 0, ; 2.19 if f ∈ C 0, × 0, ∞ , 0, ∞ , we get u ∈ C 0, ∩ C1 0, ∩ C2 0, 2.20 Lemma 2.6 Assume that conditions C0 , C1 , and C2 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, Proof We begin by defining an operator T in Dh by 1− m−2 t G t, s f s, u s ds Tu t m−2 i αi ηi αi G ηi , s f s, u s ds h, 2.21 i where Dh : {u ∈ C 0, : u t ≥ h on 0, } is a convex closed set Then from Lemma 2.2 and the condition C2 , we have T u ∈ C 0, and T u satisfies Tu t f t, u t 0, t ∈ 0, , m−2 Tu h, αi T u ηi Tu i 1− m−2 2.22 αi h i We now apply Schauder fixed point theorem 30 to obtain the existence of a fixed point for T To this, it suffices to verify that T is continuous in Dh and T Dh is a compact set Boundary Value Problems Take u0 ∈ Dh , and let {uk }∞ ⊂ Dh such that k uk − u0 C 0,1 −→ as k −→ ∞ 2.23 Then for each t ∈ 0, , −→ f t, u0 t f t, uk t as k −→ ∞ 2.24 From the definition of T , we have m−2 t G t, s f s, uk s ds T uk t m−2 i 1− αi ηi αi i G ηi , s f s, uk s ds h 2.25 Also, from the conditions C1 and C2 , we have f t, u0 t f t, uk t ≤ 2f t, h for t ∈ 0, , 2.26 s − s f s, h ds < ∞ Thus by Lebesgue-dominated convergence theorem, we have max | T uk t − T u0 t | ≤ t∈ 0,1 G s, s f s, uk s − f s, u0 s ds 1− −→ m−2 i αi m−2 i αi ηi 1− G s, s f s, uk s − f s, u0 s ds m−2 i αi m−2 i αi ηi s − s f s, uk s − f s, u0 s ds as k −→ ∞ 2.27 Therefore, T : Dh → Dh is continuous Next we need to show that T Dh is a relatively compact subset of C 0, From the definition of T and the conditions C1 and C2 , for each u ∈ Dh we have < h ≤ Tu t ≤ Th t This implies that T Dh is uniformly bounded for t ∈ 0, 2.28 Boundary Value Problems For each u ∈ Dh , since Tu − t t 1 − s f s, u s ds sf s, u s ds t m−2 i 1− αi ηi 2.29 m−2 αi G ηi , s f s, u s ds for t ∈ 0, , i then Tu t ≤ t sf s, h ds − s f s, h ds t 1− m−2 m−2 i αi ηi αi G ηi , s f s, h ds 2.30 i for t ∈ 0, :M t Obviously M t ≥ on 0, , and 1 M t dt s − s f s, h ds ≤2 s − s f s, h ds 1− m−2 i αi m−2 i αi ηi m−2 1− m−2 i αi ηi 1− m−2 i αi ηi G ηi , s f s, h ds i m−2 1 αi αi i s − s f s, h ds 2.31 s − s f s, h ds < ∞ Thus M ∈ L1 0, From the absolute continuity of integral, we have that for each number ε > 0, there is a positive number δ > such that for all t1 , t2 ∈ 0, , if |t1 − t2 | < δ, then t | t2 M t dt| < ε It follows that for all t1 , t2 ∈ 0, with |t1 − t2 | < δ, we have | T u t2 − T u t1 | t2 t dt ≤ Tu t1 t2 Tu t dt ≤ t1 t2 M t dt < ε 2.32 t1 Therefore T Dh is equicontinuous on 0, It follows from Ascoli-Arzela theorem that T Dh is a relatively compact subset of C 0, Consequently, by Schauder fixed point theorem 30 , T has a fixed point u t; h ∈ Dh Obviously, u t; h > h > on 0, 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 u1 t; h , u2 t; h are two different solutions of BVP 2.1 h Then there exists t0 ∈ 0, such that u1 t0 ; h / u2 t0 ; h Without loss of generality, assume that u1 t0 ; h > u2 t0 ; h Let w t : u1 t; h − u2 t; h , then w 0, w t0 > 0, and hence there exists t1 ∈ 0, t0 such that w t1 0, w t >0 for t ∈ t1 , t0 2.33 Boundary Value Problems Further we have w t > on t1 , In fact, assume to the contrary that the conclusion is false Then there exists t2 ∈ t0 , such that w t2 ≤ Thus there exists t3 ∈ t0 , t2 such that w t3 Since w t1 0, w t >0 for t ∈ t0 , t3 2.34 0, w t > on t1 , t0 , then −f t, u1 t; h w t f t, u2 t; h ≥ for t ∈ t1 , t3 2.35 It follows from w t1 w t3 that w t ≤ on t1 , t3 This is a contradiction to w t > on t1 , t3 Now we prove that w t ≥ on 0, t1 In fact, assume to the contrary that the w t1 0, conclusion is false Then there exists t4 ∈ 0, t1 such that w t4 < Since w then there exist t5 , t6 with ≤ t5 < t4 < t6 ≤ t1 such that w t5 w t6 0, w t on t1 , Thus w t G t, s f s, u1 s; h − f s, u2 s; h ds m−2 t 1− m−2 i αi ηi αi i G ηi , s f s, u1 s; h − f s, u2 s; h ds 2.38 ≤ for t ∈ 0, This is a contradiction to w t > on t1 , This completes the proof of the lemma Lemma 2.7 Assume that conditions C0 , C1 , and C2 are satisfied Then, the unique solution u t; h of BVP 2.1 h is nondecreasing in h Proof Let < h2 < h1 , and let u t; h1 , u t; h2 be the solutions of BVP 2.1 respectively We will show u t; h1 ≥ u t; h2 for t ∈ 0, h1 and BVP 2.1 h2 , 2.39 10 Boundary Value Problems Assume to the contrary that the above inequality is false Then there exists t0 ∈ 0, such that h1 > h2 u 0; h2 , we have that there exists t1 ∈ 0, t0 u t0 ; h1 < u t0 ; h2 Since u 0; h1 such that u t1 ; h1 u t1 ; h2 , for t ∈ t1 , t0 u t; h1 < u t; h2 2.40 Next we prove u t; h1 < u t; h2 on t0 , In fact, assume to the contrary that the conclusion is false Then there exists t2 ∈ t0 , such that u t2 ; h1 u t2 ; h2 , for t ∈ t0 , t2 u t; h1 < u t; h2 2.41 Hence u t; h1 − u t; h2 −f t, u t; h1 ≤0 f t, u t; h2 for t ∈ t1 , t2 2.42 u ti ; h2 , i 1, that u t; h1 ≥ u t; h2 on t1 , t2 This is a It follows from u ti ; h1 contradiction to u t; h1 < u t; h2 on t1 , t2 Thus u t; h1 < u t; h2 on t1 , This implies that u t; h1 − u t; h2 −f t, u t; h1 ≤ for t ∈ t1 , f t, u t; h2 2.43 It follows from u t1 ; h1 − u t1 ; h2 ≤ that u t; h1 − u t; h2 ≤ on t1 , Hence, from u t; h1 < u t; h2 on t1 , , we have u 1; h1 − u 1; h2 < Thus u 1; h1 − u 1; h2 < u ηm−2 ; h1 − u ηm−2 ; h2 2.44 There are two cases to consider Case see t1 ≥ ηm−2 In this case, we have u ηi ; h1 − u ηi ; h2 ≥ 0, i 1, 2, , m − 2.45 Hence from the boundary conditions of BVP 2.1 h , we have m−2 u 1; h1 − u 1; h2 αi u ηi ; h1 i − m−2 αi h1 i m−2 αi u ηi ; h2 − i ≥ 1− m−2 1− m−2 αi h2 i αi u ηi ; h1 − u ηi ; h2 i This is a contradiction to u 1; h1 − u 1; h2 < ≥ 2.46 Boundary Value Problems 11 Case see t1 < ηm−2 In this case, we have u 1; h1 − u 1; h2 < u ηm−2 ; h1 − u ηm−2 ; h2 < 0, u ηm−2 ; h1 − u ηm−2 ; h2 ≤ u ηi ; h1 − u ηi ; h2 , i 2.47 1, 2, , m − It follows from C0 that u 1; h1 − u 1; h2 < m−2 αi u ηm−2 ; h1 − u ηm−2 ; h2 ≤ m−2 i αi u ηi ; h1 − u ηi ; h2 2.48 i This is a contradiction to the boundary conditions of BVP 2.1 h In summary, we have u t; h1 ≥ u t; h2 on 0, This completes the proof of the lemma 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 C0 , C1 , and C2 are satisfied Then, SBVP 1.1 , 1.2 has at most one positive solution Proof Suppose that u1 t and u2 t are any two positive solutions of SBVP 1.1 , 1.2 We now u1 t − u2 t on 0, We will show that prove that u1 t ≡ u2 t on 0, To this, let v t v t ≡ on 0, There are three cases to consider Case see v > In this case, we have that v t ≥ on 0, In fact, assume to the contrary that the conclusion is false Then, there exists t0 ∈ 0, such that v t0 < Since v 0 and v > 0, then there exist t1 , t2 ∈ 0, with t1 < t0 < t2 such that v t < on t1 , t2 , v t1 v t2 3.1 for t ∈ t1 , t2 3.2 Thus v t u1 t − u2 t −f t, u1 t f t, u2 t ≤0 Hence v t ≥ on t1 , t2 , which is a contradiction to v t < on t1 , t2 Therefore v t ≥ on 0, Consequently v t −f t, u1 t f t, u2 t ≥ for t ∈ 0, 3.3 Thus v t is convex on 0, Since v > and v u1 − u2 m−2 αi u1 ηi − i m−2 m−2 αi u2 ηi i αi v ηi , i 3.4 12 Boundary Value Problems then there exists i0 ∈ {1, 2, , m − 2} such that v ηi0 max v ηi : i 1, 2, , m − > 0, 3.5 and hence from C0 and < ηi0 < 1, we have v ≤ m−2 αi v ηi0 ≤ v ηi0 < i 1 v ηi0 , ηi0 3.6 which is a contradiction to that v t is convex on 0, Case see v In this case, we have that v t ≡ on 0, In fact, assume to the contrary that the conclusion is false Then, there exists t0 ∈ 0, such that v t0 / We may v 0, there exist assume without loss of generality that v t0 > Then from v t1 , t2 ∈ 0, with t1 < t0 < t2 such that v t >0 on t1 , t2 , v t1 v t2 3.7 Thus −f t, u1 t v t Since v t1 v t2 f t, u2 t ≥ for t ∈ t1 , t2 3.8 0, then v t ≤ for t ∈ t1 , t2 , 3.9 which is a contradiction to that v t > on t1 , t2 Case see v < In this case, similar to the proof of Case we can easily show that v t ≤ on 0, Consequently v t −f t, u1 t f t, u2 t ≤ for t ∈ 0, 3.10 m−2 Thus v t is concave on 0, Since v i αi v ηi < 0, then there exists i1 ∈ {1, 2, , m− min{v ηi : i 1, 2, , m − 2} < 0, and hence from < ηi1 < 1, we have 2} such that v ηi1 v ≥ m−2 αi v ηi1 ≥ v ηi1 > i 1 v ηi1 , ηi1 3.11 which is a contradiction to that v t is concave on 0, In summary, v t ≡ on 0, , that is, u1 t ≡ u2 t on 0, This completes the proof of the theorem Boundary Value Problems 13 Theorem 3.2 Assume that conditions C0 , C1 , and C2 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 {hj }∞ be a decreasing sequence that converges to the number Then from j Lemma 2.6, BVP 2.1 hj has a unique solution u t; hj : uj t From Lemma 2.7 and 2.11 h , we have that for each j < k, ≤ uj t − uk t ≤ hj − hk for t ∈ 0, 3.12 Thus there exists u ∈ C 0, such that u t ≥ 0, lim uj t j →∞ uniformly on 0, 3.13 It is easy to see that u t satisfies boundary conditions 1.2 Now we prove that u t >0 for t ∈ 0, 3.14 At first, we prove that max u ηi : i u ηi0 1, 2, , m − > 0, 3.15 where i0 ∈ {1, 2, , m − 2} In fact, assume to the contrary that the conclusion is false Then m−2 αi u ηi u 3.16 i From the fact that each function in the sequence {uj }∞ is concave, we have that u t is j u that u t ≡ on 0, Thus when j is concave It follows from u u ηi0 large enough, uj t is small enough such that uj t ≤ h1 on 0, Hence from condition C1 , we have G ηi0 , s f s, uj s ds uj ηi0 m−2 ηi0 1− m−2 i αi ηi αi i G ηi , s f s, uj s ds > G ηi0 , s f s, h1 ds > 0 hj 3.17 14 Boundary Value Problems Let j → ∞, we have u ηi0 ≥ 3.18 G ηi0 , s f s, h1 ds > 0 Thus u ηi0 > 0, and hence u > Since u t is concave, This is a contradiction to u ηi0 then u t > on 0, Since uj t m−2 i 1− m−2 t G t, s f s, uj s ds αi ηi αi G ηi , s f s, uj s ds hj , 3.19 i then passing to the limit, by Monotone convergence theorem 31 , we have 1− m−2 t G t, s f s, u s ds u t m−2 i αi ηi αi G ηi , s f s, u s ds 3.20 i 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 tβ β2 2−β1 u t ∈ 0, , 0, 3.21 m−2 u 0, u αi u ηi , i where m ≥ 3, < η1 < η2 < · · · < ηm−2 < 1, αi > β1 , β2 ∈ 0, Let f t, u tβ 1−t β2 2−β1 u i 1, 2, , m − , for t, u ∈ 0, × 0, ∞ m−2 i αi ≤ 1, 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