Hindawi Publishing Corporation Boundary Value Problems Volume 2010, Article ID 236560, 15 pages doi:10.1155/2010/236560 Research Article Unbounded Solutions of Second-Order Multipoint Boundary Value Problem on the Half-Line Lishan Liu,1, Xinan Hao,1 and Yonghong Wu2 School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, China Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia Correspondence should be addressed to Lishan Liu, lls@mail.qfnu.edu.cn Received 14 May 2010; Revised September 2010; Accepted 11 October 2010 Academic Editor: Vicentiu Radulescu Copyright q 2010 Lishan Liu et al 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 This paper investigates the second-order multipoint boundary value problem on the half-line u t f t, u t ,u t 0, t ∈ R , αu − βu − n ki u ξi a ≥ 0, limt → ∞ u t b > 0, i where α > 0, β > 0, ki ≥ 0, ≤ ξi < ∞ i 1, 2, , n , and f : R × R × R → R is continuous We establish sufficient conditions to guarantee the existence of unbounded solution in a special function space by using nonlinear alternative of Leray-Schauder type Under the condition that f is nonnegative, the existence and uniqueness of unbounded positive solution are obtained based upon the fixed point index theory and Banach contraction mapping principle Examples are also given to illustrate the main results Introduction In this paper, we consider the following second-order multipoint boundary value problem on the half-line u t αu − βu − f t, u t , u t n ki u ξi i a ≥ 0, 0, t∈R , lim u t t→ ∞ b > 0, 1.1 where α > 0, β > 0, ki ≥ 0, < ξ1 < ξ2 < · · · < ξn < ∞, and f : R × R × R → R is continuous, 0, ∞ , R −∞, ∞ in which R The study of multipoint boundary value problems BVPs for second-order differential equations was initiated by Bicadze and Samarsk˘ and later continued by II’in and ı Boundary Value Problems Moiseev 2, and Gupta Since then, great efforts have been devoted to nonlinear multipoint BVPs due to their theoretical challenge and great application potential Many results on the existence of positive solutions for multi-point BVPs have been obtained, and for more details the reader is referred to 5–10 and the references therein The BVPs on the half-line arise naturally in the study of radial solutions of nonlinear elliptic equations and models of gas pressure in a semi-infinite porous medium 11–13 and have been also widely studied 14–27 When n 1, β 0, a b 0, BVP 1.1 reduces to the following three-point BVP on the half-line: u t f t, u t , u t u αu η , t ∈ 0, ∞ , 0, lim u t t→ ∞ 1.2 0, where α / 1, η ∈ 0, ∞ Lian and Ge 16 only studied the solvability of BVP 1.2 by the Leray-Schauder continuation theorem When ki 0, i 1, 2, , n, and nonlinearity f is variable separable, BVP 1.1 reduces to the second order two-point BVP on the half-line u Φ t f t, u, u au − bu 0, u0 ≥ 0, t ∈ 0, ∞ , lim u t 1.3 k > t→ ∞ Yan et al 17 established the results of existence and multiplicity of positive solutions to the BVP 1.3 by using lower and upper solutions technique Motivated by the above works, we will study the existence results of unbounded positive solution for second order multi-point BVP 1.1 Our main features are as follows Firstly, BVP 1.1 depends on derivative, and the boundary conditions are more general Secondly, we will study multi-point BVP on infinite intervals Thirdly, we will obtain the unbounded positive solution to BVP 1.1 Obviously, with the boundary condition in 1.1 , if the solution exists, it is unbounded Hence, we extend and generalize the results of 16, 17 to some degree The main tools used in this paper are Leray-Schauder nonlinear alternative and the fixed point index theory The rest of the paper is organized as follows In Section 2, we give some preliminaries and lemmas In Section 3, the existence of unbounded solution is established In Section 4, the existence and uniqueness of positive solution are obtained Finally, we formulate two examples to illustrate the main results Preliminaries and Lemmas Denote v0 t E t a/b C∞ R , R δ /Δ, where Δ α− x ∈ C1 R , R : lim t→ ∞1 n i ki / 0, δ β n i ki ξi Let x t exists, lim x t exists t→ ∞ v0 t 2.1 Boundary Value Problems For any x ∈ E, define x max sup ∞ t∈R x t , sup x t v0 t t∈R , 2.2 C∞ R , R is a Banach space with the norm · ∞ see 17 The Arzela-Ascoli theorem fails to work in the Banach space E due to the fact that the infinite interval 0, ∞ is noncompact The following compactness criterion will help us to resolve this problem then E C∞ R , R Then, M is relatively compact in E if the following Lemma 2.1 see 17 Let M ⊂ E conditions hold: a M is bounded in E; b the functions belonging to {y : y t x t / v0 t , x ∈ M} and {z : z t M} are locally equicontinuous on R ; c the functions from {y : y t are equiconvergent, at ∞ v0 t , x ∈ M} and {z : z t x t / x t ,x ∈ x t , x ∈ M} Throughout the paper we assume the following H1 Suppose that f t, 0, / 0, t ∈ R , and there exist nonnegative functions ≡ p t , q t , r t ∈ L1 0, ∞ with t , tq t , tr t ∈ L1 0, ∞ such that f t, ≤ p t |u| v0 t u, v n i H2 Δ α− H3 P1 q t |v| a.e t, u, v ∈ R × R × R r t , 2.3 Q1 < 1, where ki > ∞ ∞ p t dt, P1 Q1 2.4 q t dt 0 Denote ∞ P2 ∞ v0 t p t dt, Q2 v0 t q t dt, ∞ r t dt, R1 2.5 ∞ R2 v0 t r t dt Lemma 2.2 Supposing that σ t ∈ L1 0, ∞ with tσ t ∈ L1 0, ∞ , then BVP u t αu − βu − σ t n ki u ξi i 0, a ≥ 0, t∈R , lim u t t→ ∞ b>0 2.6 Boundary Value Problems has a unique solution ∞ a G t, s σ s ds ut bδ Δ bt, t∈R , 2.7 where ⎧ j ⎪ β Σi ki ξi Σn j ki s ⎪ i ⎪ ⎪ t, ⎪ ⎪ Δ ⎪ ⎪ ⎪ ⎪ t ∈ R , max t, ξ ≤ s ≤ ξ , j ⎨ j j G t, s in which ξ0 0, ξn ⎪β ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ j s, Δ t ∈ R , ξj ≤ s ≤ t, ξj m2 i m1 2.8 Σ n j ki s i Σi ki ξi ∞, and 0, 1, 2, , n, , j 0, 1, 2, , n, for m2 < m1 f i Proof Integrating the differential equation from t to ∞, one has ∞ u t t∈R σ s ds, b 2.9 t Then, integrating the above integral equation from to t, noticing that σ t ∈ L1 0, ∞ and tσ t ∈ L1 0, ∞ , we have ∞ t u t u0 bt n i Since αu − βu − ut a Δ β Δ β Σn k i i σ s ds ∞ Σn k i i σ s ds t 0 bt a ∞ t σ τ dτds sσ s ds tσ s ds ∞ ξi ξi ∞ sσ s ds s a, it holds that ∞ bδ t ki u ξi 2.10 σ τ dτ ds bt s Σn k i i σ τ dτds s ∞ ξi σ s ds ξi bδ Δ 2.11 By using arguments similar to those used to prove Lemma 2.2 in , we conclude that 2.7 holds This completes the proof Boundary Value Problems Now, BVP 1.1 is equivalent to ∞ G t, s f s, u s , u s ds u t a Letting v t u t − bt − bt, t∈R 2.12 bδ /Δ , t ∈ R , 2.12 becomes a ∞ v t bδ Δ a G t, s f s, v s bδ Δ bs, v s t∈R b ds, 2.13 For v ∈ E, define operator A : E → E by ∞ Av t a G t, s f s, v s bδ Δ bs, v s t∈R b ds, 2.14 Then, ∞ Av a f s, v s t t bδ Δ bs, v s b ds, t∈R 2.15 Set ⎧ ⎪t ⎨ γ t ⎪ ⎩1 δ , Δ δ , Δ t ∈ 0, , 2.16 t ∈ 1, ∞ Remark 2.3 G t, s is the Green function for the following associated homogeneous BVP on the half-line: u t f t, u t , u t αu − βu − 0, t∈R , n ki u ξi 0, i lim u t t→ ∞ 2.17 It is not difficult to testify that G t, s G τ, s ≥ , γ t v0 τ G t, s ≤ G s, s , ∀t, s, τ ∈ R , G t, s ≤ 1, v0 t 2.18 ∀t, s ∈ R Let us first give the following result of completely continuous operator Lemma 2.4 Supposing that H1 and H2 hold, then A : E → E is completely continuous 6 Boundary Value Problems Proof First, we show that A : E → E is well defined For any v ∈ E, there exists d1 > such that v ∞ ≤ d1 Then, ∞ |Av t | ≤ v0 t ∞ ≤ a G t, s f s, v s v0 t p s ≤ d1 Q1 bs, v s ds b ∞ |v s | v0 s b P1 bδ Δ b ds ∞ v s q s b ds r s ds 2.19 t∈R , R1 , so sup t∈R |Av t | ≤ d1 v0 t b P1 Q1 R1 2.20 Similarly, ∞ Av a f s, v s t bδ Δ t ∞ b q s v s b r s ds, p s |v s | v0 s b q s v s b r s ds ∞ ≤ t∈R 2.21 |v s | v0 s t t b ds p s ≤ sup Av bs, v s ≤ d1 b P1 Q1 t∈R , 2.22 R1 Further, ∞ |Av t | ≤ a G t, s f s, v s ∞ ≤ v0 s p s ≤ d1 Av t b P2 f s, v s a ∞ ≤ p s ≤ d1 b P1 |v s | v0 s Q1 bs, v s |v s | v0 s R2 < ∞, Q2 ∞ ≤ bδ Δ bδ Δ b ds q s v s b b r s ds 2.23 t∈R , bs, v s b q s v s R1 < ∞ b ds b r s ds 2.24 Boundary Value Problems On the other hand, for any t1 , t2 ∈ R and s ∈ R , by Remark 2.3, we have a |G t1 , s − G t2 , s | f s, v s ≤2 v0 s ≤2 v0 s p s p s bδ Δ |v s | v0 s q s bs, v s b q s v s b v b 2.25 r s b r s Hence, by H1 , the Lebesgue dominated convergence theorem, and the continuity of G t, s , for any t1 , t2 ∈ R , we have ∞ | Av t1 − Av t2 | ≤ a |G t1 , s − G t2 , s | f s, v s −→ 0, Av t1 − Av bs, v s b ds as t1 −→ t2 , t2 a f s, v s t2 bδ Δ t1 bδ Δ bs, v s b ds −→ 0, as t1 −→ t2 2.26 So, Av ∈ C1 R , R for any v ∈ E We can show that Av ∈ E In fact, by 2.23 and 2.24 , we obtain |Av t | v0 t lim t→ ∞1 0, then lim t→ ∞1 Av t v0 t ∞ lim Av t t→ ∞ f s, v s lim t→ ∞ a t 0, 2.27 bδ Δ bs, v s b ds Hence, A : E → E is well defined We show that A is continuous Suppose {vm } ⊆ E, v ∈ E, and limm → ∞ vm v Then, vm t → v t , vm t → v t as m → ∞, t ∈ R , and there exists r0 > such that vm ∞ ≤ r0 , m 1, 2, , v ∞ ≤ r0 The continuity of f implies that f t, vm t bt a bδ , vm t Δ b − f t, v t bt a bδ ,v t Δ −→ b 2.28 as m → ∞, t ∈ R Moreover, since f t, vm t ≤2 p t bt a q t bδ , vm t Δ r0 b b − f t, v t r t , t∈R , bt a bδ ,v t Δ b 2.29 Boundary Value Problems we have from the Lebesgue dominated convergence theorem that Avm − Av0 ∞ max sup t∈R | Avm t − Av t | , sup Avm v0 t t∈R ∞ ≤ f s, vm s a bs bδ , vm s Δ −→ t − Av b − f s, v s t bs a bδ ,v s Δ b ds m −→ ∞ 2.30 Thus, A : E → E is continuous We show that A : E → E is relatively compact a Let B ⊂ E be a bounded subset Then, there exists M > such that v v ∈ B By the similar proof of 2.20 and 2.22 , if v ∈ B, one has ∞≤ Av M b P1 Q1 ∞ R1 , ≤ M for all 2.31 which implies that A B is uniformly bounded b For any T > 0, if t1 , t2 ∈ 0, T , v ∈ B, we have Av t1 Av t2 − v0 t1 v0 t2 ∞ ≤ G t1 , s G t2 , s − v0 t1 v0 t2 ∞ ≤2 f s, v s a bs ≤2 M b P1 t1 − Av Av ≤ Q1 f s, v s b a bδ ,v s Δ b ds ds 2.32 R1 , bs a t1 b P1 bδ ,v s Δ bs t2 t2 ≤ M f s, v s Q1 bδ ,v s Δ b ds R1 Thus, for any ε > 0, there exists δ > such that if t1 , t2 ∈ 0, T , |t1 − t2 | < δ, v ∈ B, Boundary Value Problems then Av t1 Av t2 − v0 t1 v0 t2 Av t1 − Av Since T is arbitrary, then { AB t / continuous on R < ε, 2.33 < ε t2 v0 t } and { AB t } are locally equi- c For v ∈ B, from 2.27 , we have lim t→ ∞ lim t→ ∞ Av t Av s − lim s→ ∞1 v0 t v0 s Av t − lim Av s→ ∞ s lim Av t v0 t t→ ∞ lim Av t→ ∞ t 0, 2.34 0, which means that { AB t / v0 t } and { AB t } are equiconvergent at Lemma 2.1, A : E → E is relatively compact Therefore, A : E → E is completely continuous The proof is complete ∞ By Lemma 2.5 see 28, 29 Let E be Banach space, Ω be a bounded open subset of E, θ ∈ Ω, and A : Ω → E be a completely continuous operator Then either there exist x ∈ ∂Ω, λ > such that F x λx, or there exists a fixed point x∗ ∈ Ω Lemma 2.6 see 28, 29 Let Ω be a bounded open set in real Banach space E, let P be a cone of E, θ ∈ Ω, and let A : Ω ∩ P → P be completely continuous Suppose that λAx / x, ∀x ∈ ∂Ω ∩ P, λ ∈ 0, 2.35 Then, i A, Ω ∩ P, P 2.36 Existence Result In this section, we present the existence of an unbounded solution for BVP 1.1 by using the Leray-Schauder nonlinear alternative Theorem 3.1 Suppose that conditions H1 – H3 hold Then BVP 1.1 has at least one unbounded solution 10 Boundary Value Problems Proof Since f t, 0, / 0, by H1 , we have r t ≥ |f t, 0, |, a.e t ∈ R , which implies that ≡ R1 > Set b P1 Q1 R1 , − P1 − Q1 R ΩR {v ∈ E : v ∞ 3.1 < R} From Lemmas 2.2 and 2.4, BVP 1.1 has a solution v v t if and only if v is a fixed point of A in E So, we only need to seek a fixed point of A in E Suppose v ∈ ∂ΩR , λ > such that Av λv Then λR λ v Av ∞ ∞ max sup t∈R ∞ ≤ f s, v s bs a bδ ,v s Δ ≤ P1 P1 Q1 v ∞ Q1 R P1 P1 | Av t | , sup Av v0 t t∈R Q1 b Q1 b b t ds 3.2 1, 3.3 R1 R1 Therefore, λ ≤ P1 Q1 P1 Q1 b R R1 which contradicts λ > By Lemma 2.5, A has a fixed point v∗ ∈ ΩR Letting u∗ t v∗ t ∗ bt a bδ /Δ , t ∈ R , boundary conditions imply that u is an unbounded solution of BVP 1.1 Existence and Uniqueness of Positive Solution In this section, we restrict the nonlinearity f ≥ and discuss the existence and uniqueness of positive solution for BVP 1.1 Define the cone P ⊂ E as follows: P u ∈ E : u t ≥ γ t sup s∈R u s u , t∈R , ≥ v0 s v0 δ β sup u s Δ a/b s∈R 4.1 Lemma 4.1 Suppose that H1 and H2 hold Then, A : P → P is completely continuous Proof Lemma 2.4 shows that A : P → E is completely continuous, so we only need to prove A P ⊂ P Since f ∈ C R × R × R, R , Av t ≥ 0, t ∈ R , and from Remark 2.3, Boundary Value Problems 11 we have ∞ a G t, s f s, v s Av t bδ Δ ∞ ≥γ t ∞ γ t γ t G τ, s f s, v s v0 τ G τ, s f s, v s a Av τ , v0 τ b ds bδ Δ bs, v s b ds bδ /Δ a 1 bs, v s bs, v s b ds 4.2 v0 τ ∀t, τ ∈ R Then, Av t ≥ γ t sup τ∈R ∞ Av v0 0 Av τ , v0 τ t∈R , G 0, s f s, v s a ≥ ≥ ∞ β/Δ f s, v s b ds bs, v s b ds 4.3 bδ /Δ a/b β sup Av Δ a/b t∈R bs, v s v0 a δ bδ /Δ δ /Δ t Therefore, A P ⊂ P Theorem 4.2 Suppose that conditions H2 and H3 hold and the following condition holds: ≡ H1 suppose that f t, 0, , tf t, 0, ∈ L1 0, ∞ , f t, 0, / and there exist nonnegative functions p t , q t ∈ L1 0, ∞ with t , tq t ∈ L1 0, ∞ such that f t, v0 t u1 , v1 − f t, ≤ p t |u1 − u2 | v0 t u2 , v2 q t |v1 − v2 |, a.e t, ui , vi ∈ R × R × R, i 4.4 1, Then, BVP 1.1 has a unique unbounded positive solution Proof We first show that H1 implies H1 By 4.4 , we have f t, v0 t u, v ≤ p t |u| q t |v| f t, 0, , a.e t, u, v ∈ R × R × R By Lemma 4.1, A : P → P is completely continuous Let R R> b P1 Q1 R , − P1 − Q1 Ω {v ∈ E : v 4.5 ∞ f t, 0, dt Then, R > Set ∞ < R} 4.6 12 Boundary Value Problems For any v ∈ P ∩ ∂Ω, by 4.5 , we have ∞ | Av t | v0 t ≤ R G t, s f s, v s v0 t b P1 Q1 f s, v s a bs bδ ,v s Δ bδ ,v s Δ b t ∞ ≤ ≤ R b P1 Q1 b ds b ds a bs f s, v s t bδ ,v s Δ t∈R , R < R, ∞ Av a bs 4.7 ds t∈R R < R, Therefore, Av ∞ < v ∞ , for all v ∈ P ∩ ∂Ω, that is, λAv / v for any λ ∈ 0, , v ∈ P ∩ ∂Ω Then, Lemma 2.6 yields i A, P ∩ Ω, P 1, which implies that A has a fixed point v∗ ∈ P ∩ Ω ∗ ∗ v t bt a bδ /Δ , t ∈ R Then, u∗ is an unbounded positive solution of Let u t BVP 1.1 Next, we show the uniqueness of positive solution for BVP 1.1 We will show that A is a contraction In fact, by 4.4 , we have Av1 − Av2 ∞ max sup t∈R | Av1 t − Av2 t | , sup Av1 v0 t t∈R ∞ ≤ f s, v1 s bs a ∞ ≤ p s ≤ P1 bδ , v1 s Δ |v1 s − v2 s | v0 s Q1 v1 − v2 t − Av2 b − f s, v2 s q s v1 s − v2 s t a bs bδ , v2 s Δ b ds ds ∞ 4.8 So, A is indeed a contraction The Banach contraction mapping principle yields the uniqueness of positive solution to BVP 1.1 Examples Example 5.1 Consider the following BVP: u t 2e−4t u2 t u2 t 89u − 3u − 2e−3t iu i u t u t i 4 − arctan t t3 0, t∈R , 5.1 2, lim u t t→ ∞ 1, Boundary Value Problems 13 We have Δ 89 − i 61, δ i i i i f t, x2 x2 2e−3t t x, y 2e−4t f t, x, y 59, v0 t t 59 61 t 1, y3 arctan t − ∈ C R × R × R, R t3 y ≤ t e−4t |x| 5.2 arctan t t3 e−3t y Let p t t e−4t , e−3t , q t arctan t t3 r t 5.3 Then, p t , q t , r t ∈ L1 0, ∞ , t , tq t , tr t ∈ L1 0, ∞ , and it is easy to prove that H1 is satisfied By direct calculations, we can obtain that P1 9/16, Q1 1/3, P1 Q1 < By Theorem 3.1, BVP 5.1 has an unbounded solution Example 5.2 Consider the following BVP: u t t −3t |u t |3 e u t u2 t u2 t e−4t 89u − 3u − i iu 0, t∈R , 5.4 2, i arctan t t3 lim u t t→ ∞ In this case, we have Δ f t, x, y f t, 2 t e 61, δ x2 x2 −3t y e y4 −4t t x1 , y1 − f t, 59, v0 t t x2 , y2 t 1, arctan t ∈ C R × R × R, R , t3 ≤ 2e−4t |x1 − x2 | 5.5 −3t e y1 − y2 Let p t Then, P1 1/2, Q1 positive solution 1/4, P1 2e−4t , q t −3t e 5.6 Q1 < By Theorem 4.2, BVP 5.4 has a unique unbounded 14 Boundary Value Problems Acknowledgments The authors are grateful to the referees for valuable suggestions and comments The first and second authors were supported financially by the National Natural Science Foundation of China 11071141, 10771117 and the Natural Science Foundation of Shandong Province of China Y2007A23, Y2008A24 The third author was supported 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