Bai, Chuanzhi

Liu, Yang

Wang, Chunli

Xie, Dapeng

Xie Dapeng

Bai Chuanzhi

Liu Yang

Wang Chunli

University of Szeged

Electronic Journal of Qualitative Theory of Differential Equations2008, No. 8, 1-10; http://www.math.u-szeged.hu/ejqtde/Existence of positive solutions fornth-orderboundary value problem with signchanging nonlinearityDapeng Xiea,b, Chuanzhi Baia,∗, Yang Liua,b, Chunli Wanga,ba Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsu 223300, P R Chinab Department of Mathematics, Yanbian University, Yanji, Jilin 133002, P R ChinaAbstractIn this paper, we investigate the existence of positive solutions for singularnth-order bound-ary value problemu(n)(t) + a(t) f (t, u(t)) = 0, 0 ≤ t ≤ 1,u(i)(0) = u(n−2)(1) = 0, 0 ≤ i ≤ n − 2,wheren ≥ 2, a ∈ C((0,1), [0,+∞)) may be singular att = 0 and (or)t = 1 and the nonlinearterm f is continuous and is allowed to change sign. Our proofs are bas d on the method of lowersolution and topology degree theorem.Keywords: Existence; Singular; Positive solution; Cone; Sign changing nonlinearity.1. IntroductionBoundary value problems for higher order differential equations play a very impor-tant role in both theories and applications. Existence of positive solutions for nonlinearhigher order has been studied in the literature by using the Krasnosel’skii and Guofixed point theorem, Leggett-Williams fixed point theorem, Lower- and upper- solu-tions method and so on. We refer the reader to [2-9] for some rec nt results. However,to the best of our knowledge, few papers can be found in the literature fornth-orderboundary value problem with sign changing nonlinearity, most papers are dealing withthe existence of positive solutions when the nonlinear termf is nonnegative. For ex-ample, in [3], by using the Krasnosel’skii and Guo fixed pointtheorem, Eloe and Hen-derson studied the existence of positive solutions for the following boundary valueproblem∗E-mail address: czbai8@sohu.comEJQTDE, 2008 No. 8, p. 1{u(n)(t) + a(t) f (u(t)) = 0, t ∈ (0, 1),u(i)(0) = u(n−2)(1) = 0, 0 ≤ i ≤ n − 2,(1.1)where(A1) f : [0,∞] → [0,∞) is continuous;(A2) a : (0, 1) → [0,∞) is continuous and does not vanish identically on anysubinterval;(A3) f is either superlinear or sublinear.Motivated by the above works, in this paper, we study the exist nce of positive so-lutions for singularnth-order boundary value problem with sign changing nonlinearityas follows{u(n)(t) + a(t) f (t, u(t)) = 0, t ∈ (0, 1),u(i)(0) = u(n−2)(1) = 0, 0 ≤ i ≤ n − 2.(1.2)Throughout this paper, we assume the following conditions hold.(C1) f : [0, 1] × [0,∞)→ (−∞,+∞) is continuous;(C2) a : (0, 1)→ [0,∞) is continuous, and 0<∫ 10a(t)dt < ∞, if n = 2;0 <∫ 10(1− t)a(t)dt < ∞, if n ≥ 3.The purpose of this paper is to establish the existence of positive olutions for BVP(1.2) by constructing available operator and combining themethod of lower solutionwith the method of topology degree.The rest of this paper is organized as follows: in section 2, we present some prelim-inaries and lemmas. Section 3 is devoted to prove the existence of positive solutionsfor BVP (1.2). An example is considered in section 4 to illustrate our main results.2. Preliminary LemmasLemma 2.1. Suppose thaty(t) ∈ C[0, 1], then boundary value problem{u(n)(t) + y(t) = 0, 0 ≤ t ≤ 1,u(i)(0) = u(n−2)(1) = 0, 0 ≤ i ≤ n − 2,(2.1)has a unique solutionu(t) =∫ 10G(t, s)y(s)ds,whereEJQTDE, 2008 No. 8, p. 2G(t, s) =1(n − 1)!(1− s)tn−1 − (t − s)n−1, 0 ≤ s ≤ t ≤ 1,(1− s)tn−1, 0 ≤ t < s ≤ 1.Proof. The proof follows by direct calculations.Lemma 2.2. G(t, s) has the following properties.(1) 0≤ G(t, s) ≤ k(s), t, s ∈ [0, 1], wherek(s) =1− s(n − 1)!;(2)∂i∂tiG(t, s) > 0 on (0, 1)× (0, 1), 0 ≤ i ≤ n − 2;(3) G′′tt (t, s) :=∂2∂t2 G(t, s) ≤ (n − 1)(n − 2)k(s), n ≥ 3.Proof. It is easy to check that (1) and (3) hold. The proof of (2), please see [1].Remark 2.1. By (C2) and Lemma 2.2, we have0 <∫ 10G(t, s)a(s)ds < ∞, n ≥ 2, and 0<∫ 10∫ 10G′′ττ(τ, s)a(s)dτds < ∞, n ≥ 3.By the definition of completely continuous operator, we can check that the follow-ing lemma holds.Lemma 2.3. Let P is a cone ofX = C[0, 1]. SupposeT : P → X is completelycontinuous. DefineA : T X → P by(Ay)(t) = max{y(t), 0}, y ∈ T X.Then,A ◦ T : P→ P is also a completely continuous operator.3. Main resultsLet X = C[0, 1], P = {u ∈ X : u(t) ≥ 0, t ∈ [0, 1]} with ‖u‖ = maxt∈[0,1]|u(t)|. Setf1(t, u) = max{0, f (t, u)}, f2(t, u) = max{0,− f (t, u)},andδ =∫ 10k(s)a(s)ds, w(t) =∫ 10G(t, s)a(s)ds.For convenience, we introduce the following notationsEJQTDE, 2008 No. 8, p. 3f∞ = lim supu→∞max0≤t≤1f (t, u)u, f 0 = lim supu→0+max0≤t≤1f (t, u)u,f 1∞ = lim supu→∞max0≤t≤1f1(t, u)u, f 10 = lim supu→0+max0≤t≤1f1(t, u)u,f 2∞ = lim supu→∞max0≤t≤1f2(t, u)u, f 20 = lim supu→0+max0≤t≤1f2(t, u)u.Theorem 3.1. Suppose that(C1) and(C2) hold, in addition assumef 1∞ = A < +∞,f 2∞ = B < +∞ (or f 10 = A < +∞, f 20 = B < +∞) with A + B <1δand there existr, λwith r > λ > 0 such thatλ ≤ mint∈[0,1]f (t, λw(t)), n = 2,λ ≥ maxt∈[0,1]f (t, λw(t)), n ≥ 3.(3.1)Then BVP(1.2)has a positive solutionu∗(t) satisfying0 < λw(t) ≤ u∗(t), 0 < t < 1 and ‖u∗‖ ≤ r.Proof. Letg(t, u(t)) ={f (t, u(t)), u(t) ≥ λw(t),f (t, λw(t)), u(t) ≤ λw(t).(3.2)Define the operatorT : P→ X by(Tu)(t) =∫ 10G(t, s)a(s)g(s, u(s))ds, 0 ≤ t ≤ 1. (3.3)Similar to the proof of Lemma 2.1 in [7], we can easily check thatT is a completelycontinuous operator.Define the operatorA : X → P by(Au)(t) = max{u(t), 0}. (3.4)By Lemma 2.3, we getA ◦ T : P→ P is also completely continuous.If f 1∞ = A < +∞, f 2∞ = B < +∞, then by hypothesisA + B <1δ, we may takeA0 > A, B0 > B such thatA0 + B0 < 1δ , f 1∞ < A0 and f 2∞ < B0. Thus, we chooseL > 0 such thatf1(t, u) < A0u, f2(t, u) < B0u, if u ≥ L, t ∈ [0, 1], (3.5)EJQTDE, 2008 No. 8, p. 4and there existsr > L such thatf1(t, u) < A0r, f2(t, u) < B0r, if λw(t) ≤ u(t) ≤ L, t ∈ [0, 1]. (3.6)LetΩ = {u ∈ P : ‖u‖ < r}. Then, foru ∈ ∂Ω, we have by (3.5) and (3.6) thatmaxt∈[0,1]u∈[λw(t),r]f1(t, u) = maxmaxt∈[0,1]u∈[λw(t),L]f1(t, u), maxt∈[0,1]u∈[L,r]f1(t, u)< A0r. (3.7)Similarly,maxt∈[0,1]u∈[λw(t),r]f2(t, u) < B0r. (3.8)Thus, for eachu ∈ ∂Ω, from (3.7) and (3.8), we have(A ◦ T )u(t) = max{∫ 10G(t, s)a(s)g(s, u(s))ds, 0}≤∫ 10G(t, s)a(s)|g(s, u(s))|ds≤ maxt∈[0,1]u∈[0,r]|g(t, u)|∫ 10k(s)a(s)ds= δ maxt∈[0,1]u∈[λw(t),r]| f (t, u)|= δ maxt∈[0,1]u∈[λw(t),r]( f1(t, u) + f2(t, u))< δ(A0 + B0)r < r = ‖u‖, (3.9)which implies‖(A ◦ T )u‖ < ‖u‖, ∀u ∈ ∂Ω.Thus, we havedegP{I − A ◦ T, Ω, 0} = 1,where degP means the degree on coneP. Hence,A ◦ T has a fixed pointu∗ in Ω, i.e.,(A ◦ T )(u∗) = u∗, u∗ ∈ Ω.If f 10 = A < +∞, f 20 = B < +∞, then we takeA0 > A, B0 > B such thatA0 + B0 < 1δ , f 10 < A0 and f 20 < B0. Taker > 0 such thatEJQTDE, 2008 No. 8, p. 5f1(t, u) < A0u, f2(t, u) < B0u, if 0 < u(t) ≤ r, t ∈ [0, 1].Then, foru ∈ ∂Ω, similar to the proof of (3.9), we have thatA ◦ T has a fixed pointu∗in Ω. Hence, in any case we always have thatA ◦ T has a fixed pointu∗ in Ω.In the following, we shall show the following relation holds(Tu∗)(t) ≥ λw(t), t ∈ [0, 1]. (3.10)Assume the contrary, then there existst0 ∈ [0, 1] such that(Tu∗)(t0) − λw(t0) = mint∈[0,1]{(Tu∗)(t) − λw(t)} = M < 0. (3.11)Obviously,t0 , 0, sot0 ∈ (0, 1] and(Tu∗)′(t0) − λw′(t0) = 0. (3.12)There are two cases to consider.Case 1. t0 = 1. It is obvious thatTu∗(t)−λw(t) on [0, 1] is continuous. From (3.11), wehave that there exists1 ∈ [0, 1) such that (Tu∗)(t1)−λw(t1) = 0 and (Tu∗)(t)−λw(t) < 0for t ∈ (t1, 1].If n = 2, then by (3.1), (3.3) and (3.12), one has(Tu∗)′(t) − λw′(t) = (Tu∗)′(1)− λw′(1)−∫ 1t[(Tu∗)′(s) − λw′(s)]′ds=∫ 1ta(s)[g(s, u∗(s)) − λ]ds=∫ 1ta(s)[ f (s, λw(s)) − λ]ds≥ [ mint∈[0,1]f (t, λw(t)) − λ]∫ 1ta(s)ds ≥ 0.Then, we have that (Tu∗)′(t) − λw′(t) ≥ 0.If n ≥ 3, from (3.1), (3.3) and (3.12), we get(Tu∗)′(t) − λw′(t) = (Tu∗)′(1)− λw′(1)−∫ 1t[(Tu∗)′(τ) − λw′(τ)]′dτ=∫ 1t∫ 10G′′ττ(τ, s)a(s)[λ − g(s, u∗(s))]dsdτ=∫ 1t∫ 10G′′ττ(τ, s)a(s)[λ − f (s, λw(s))]dsdτEJQTDE, 2008 No. 8, p. 6≥ [λ − maxt∈[0,1]f (t, λw(t))]∫ 1t∫ 10G′′ττ(τ, s)a(s)dsdτ ≥ 0.Then, we have (Tu∗)′(t) − λw′(t) ≥ 0. Therefore, in any case we always have that(Tu∗)′(t) − λw′(t) ≥ 0, which implies(Tu∗)(t0) − λw(t0) = (Tu∗)(1)− λw(1) ≥ (Tu∗)(t1) − λw(t1) = 0.It contradicts (3.11), so (3.10) holds.Case 2. t0 ∈ (0, 1). Obviously,Tu∗(t) − λw(t) on [0, 1] is continuous. By (3.11) andt0 ∈ (0, 1), we have that there existst2 ∈ [0, t0)∪ (t0, 1] such that (Tu∗)(t2)− λw(t2) = 0and (Tu∗)(t) − λw(t) < 0 for t ∈ (t2, t0] or t ∈ [t0, t2). Without loss of generality, weassume thatt ∈ (t2, t0].If n = 2, we have by (3.1), (3.3) and (3.12) that(Tu∗)′(t) − λw′(t) = (Tu∗)′(t0) − λw′(t0) −∫ t0t[(Tu∗)′(s) − λw′(s)]′ds=∫ t0ta(s)[g(s, u∗(s)) − λ]ds=∫ t0ta(s)[ f (s, λw(s)) − λ]ds≥ [ mint∈[0,1]f (t, λw(t)) − λ]∫ t0ta(s)ds ≥ 0.Thus, (Tu∗)′(t) − λw′(t) ≥ 0.If n ≥ 3, from (3.1), (3.3), (3.12) and Lemma 2.2 (2), we obtain(Tu∗)′(t) − λw′(t) = (Tu∗)′(t0) − λw′(t0) −∫ t0t[(Tu∗)′(τ) − λw′(τ)]′dτ=∫ t0t∫ 10G′′ττ(τ, s)a(s)[λ − g(s, u∗(s)]dsdτ=∫ t0t∫ 10G′′ττ(τ, s)a(s)[λ − f (s, λw(s))]dsdτ≥ [λ − maxt∈[0,1]f (t, λw(t))]∫ t0t∫ 10G′′ττ(τ, s)a(s)dsdτ ≥ 0.Thus, (Tu∗)′(t)−λw′(t) ≥ 0. Hence, in any case we always have that (Tu∗)′(t)−λw′(t) ≥0, which implies(Tu∗)(t0) − λw(t0) ≥ (Tu∗)(t2) − λw(t2) = 0.It contradicts (3.11), so (3.10) holds. Thus, (A ◦ T )u∗ = Tu∗ = u∗, u∗ ∈ Ω, i.e., BVPEJQTDE, 2008 No. 8, p. 7(1.2) has a positive solutionu∗(t) satisfying 0< λw(t) ≤ u∗(t), 0< t < 1, and‖u∗‖ ≤ r.Corollary 3.1 Suppose that(C1) and(C2) hold, in addition assumef 1∞ = 0, f 2∞ = 0 (or f 10 = 0, f 20 = 0 ), if there exists a constantλ > 0 such thatλ ≤ mint∈[0,1]f (t, λw(t)), n = 2,λ ≥ maxt∈[0,1]f (t, λw(t)), n ≥ 3.Then BVP(1.2)has a positive solution.Theorem 3.2. Suppose that(C2) holds, in addition assumef (t, 0) ≥ 0, a(t) f (t, 0) . 0and f 1∞ = A < +∞ , f 2∞ = B < +∞ ( or f 10 = A < +∞, f 20 = B < +∞ ) withA + B < 1δ. Then BVP(1.2)has a positive solution.Proof: Similar to the proof of Theorem 3.1, we can complete the proofof Theorem3.2, so we omit it here.Corollary 3.4. Suppose that(C2) holds, in addition assumef (t, 0) ≥ 0, a(t) f (t, 0) . 0,f 1∞ = 0 and f 2∞ = 0 ( or f 10 = 0 and f 20 = 0 ). Then BVP(1.2) has a positivesolution.Corollary 3.3. Suppose that(C2) holds, in addition assumef : [0, 1]×[0,∞)→ [0,∞),a(t) f (t, 0) . 0 and f∞ = A < +∞ ( or f 0 = A < +∞ ) with A <1δ. Then BVP(1.2)hasa positive solution.Remark 3.1. Suppose that (C2) holds, in addition assumef : [0, 1] × [0,∞)→ [0,∞),a(t) f (t, 0) . 0 and f∞ = 0 ( or f 0 = 0 ). Then BVP (1.2) has a positive solution.4. An exampleExample 4.1. Let a(t) =(n − 1)!2(1− t), n ≥ 3, andf (t, u) =[1 + ln(1+ (e − 1)t)](u + u12 + 1), (t, u) ∈ [0, 1] × [0, 1],[1 + ln(1+ (e − 1)t)][3 − (2e−2 + 6)(u − 1)], (t, u) ∈ [0, 1] × [1, 2],−[1 + ln(1+ (e − 1)t)](e−u + 32 + sinπu)u, (t, u) ∈ [0, 1] × [2,∞).By simple calculation, we have0 < δ =∫ 10k(s)a(s)ds =∫ 101−s(n−1)! ·(n−1)!2(1−s) ds =12 < ∞.EJQTDE, 2008 No. 8, p. 8It is easy to see thatf ∈ C([0, 1]× [0,∞),R) and f (t, 0) > 0, a(t) f (t, 0) . 0,∀t ∈ [0, 1].Owing to f (t, u) > 0, (t, u) ∈ [0, 1] × [0, 1], and f (t, u) < 0, (t, u) ∈ [0, 1] × [2,∞), wehavef1(t, u) = f (t, u), f2(t, u) = 0, (t, u) ∈ [0, 1] × [0, 1],andf1(t, u) = 0, f2(t, u) = − f (t, u), (t, u) ∈ [0, 1] × [2,∞).By calculating, we obtain thatA = f 1∞ =12 , B = f 2∞ =12 ,and soA + B = 1 < 2 = 1δ. Therefore, by Theorem 3.2, BVP (1.2) has a positivesolution.AcknowledgmentThe authors are very grateful to the referee for his/her helpful comments.References[1] P.W. Eloe, Sign properties of Green’s functions for two classes of boundary valueproblems, Canad. Math. Bull. 30(1987), 28-35.[2] P.W. Eloe and B. Ahmad, Positive solutions to a nonlinearnth-order boundaryvalue problem with nonlocal conditions, Applied Mathematics Letters, 18(2005)521-527.[3] P.W. Eloe and J. Henderson, Positive solutions for higher order ordinary differ-ential equations, Electronic Journal of Differential Equations, Vol. 1995(1995),No. 03, pp. 1-8.[4] J.R. Graef, C. Qian and B. Yang, Multiple symmetric positive solutions of a classof boundary value problem for higher order ordinary differential equations, Proc.Amer. Math. Soc, 131(2003) 577-585.[5] J.R. Graef and B. Yang, Positive solutions to a multi-point higher order boundaryvalue problem, J. Math. Anal. Appl, 316(2006) 409-421.[6] Y.P. Guo, Y.D. Ji and J.G. Zhang, Three positive solutions for a nonlinearnth-orderm-point boundary value problem, Nonlinear Anal. (2007), doi:10.1016/j.na.2007. 03. 041.EJQTDE, 2008 No. 8, p. 9[7] X.A. Hao, L.H. Liu and Y.H. Wu, Positive solutions for nonli earnth-order singu-lar nonlocal Boundary Value Problem, Boundary Value Problems, Volume 2007,Article ID 74517, 10 pages doi:10. 1155/2007/74517.[8] L. Liu, X.G. Zhang and Y.H. Wu, Existence of positive solutions for singularhigher-order differential equations, Nonlinear Anal. (2007), doi: 10. 1016/j. na.2007. 04.32.[9] Z.L. Yang, Existence and uniqueness of positive solutions for a higer order bound-ary value problem, Computers and Mathematics with Applications. 54(2007)220-228.(Received August 17, 2007)EJQTDE, 2008 No. 8, p. 10

Existence of positive solutions for nth-order boundary value problem with sign changing nonlinearity

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Existence of positive solutions for nth-order boundary value problem with sign changing nonlinearity

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