Song, Ruipeng

Zhai, Chengbo

Zhai Chengbo

Song Ruipeng

University of Szeged

Electronic Journal of Qualitative Theory of Differential Equations2010, No. 76, 1-9; http://www.math.u-szeged.hu/ejqtde/Existence and uniqueness of positive solutions forNeumann problems of second order impulsivedifferential equationsChengbo Zhai∗, Ruipeng SongSchool of Mathematical Sciences, Shanxi University, Taiyuan 030006 Shanxi, P.R. ChinaAbstract This work is concerned with the existence and uniqueness of positive solutions for Neu-mann boundary value problems of second order impulsive differential equations. The result isobtained by using a fixed point theorem of generalized concave operators.MSC: 34B18; 34B37Keywords: Existence and uniqueness; positive solution; fixed point theorem of generalized concaveoperators; Neumann boundary value problem; second order impulsive differential equation1 IntroductionImpulsive differential equations have been studied extensively in recent years. The theoryof impulsive differential equations describes processes which experience a sudden change of theirstate at certain moments. Processes with such a character arise naturally and often, especiallyin phenomena studied in physics, chemical technology, population dynamics, biotechnology andeconomics. Second order impulsive differential equations have been studied by many authors withmuch of the attention given to positive solutions. For a small sample of such work, we referthe reader to works [1-11]. The results of these papers are based on Schauder fixed point theorem,Leggett-Williams theorem, fixed point index theorems in cones, Krasnoselskii’s fixed point theorem,the method of upper-lower solutions, fixed point theorems in cones and so on. In this paper,weconsider the existence and uniqueness of positive solutions for the following Neumann boundaryvalue problems of second order impulsive differential equations:−u′′(t) + γ2u(t) = f(t, u(t)), t 6= tk, k = 1, 2, · · ·m,△u′ |t=tk= Ik(u(tk)), k = 1, 2, · · ·m,u′(0) = u′(1) = 0,(1.1)∗Corresponding author. E-mail address: cbzhai215@sohu.com. The research was supported by the Youth ScienceFoundation of Shanxi Province (2010021002-1).EJQTDE, 2010 No. 76, p. 1where γ is a positive constant,f ∈ C[J × R,R], J = [0, 1], 0 < t1 < t2 < · · · < tm < 1,△u′ |t=tk=u′(t+k ) − u′(t−k ), u′(t+k ), u′(t−k ) denote the right limit (left limit)of u′(t) at t = tk, respectively. Ik ∈C[R,R], k = 1, 2, · · ·m.Neumann boundary value problem for the ordinary differential equations and elliptic equationsis an important kind of boundary value problems. During the last two decades, Neumann boundaryvalue problems have deserved the attention of many researchers [12-21]. However, few papers can befound in the literature on the existence of positive solutions for Neumann boundary value problemsfor second-order impulsive differential equations. In this paper, we shall study the problem (1.1) andnot suppose the existence of upper-lower solutions. Different from the above works mentioned, inthis paper we will use a fixed point theorem of generalized concave operators to show the existenceand uniqueness of positive solutions for the problem (1.1).2 PreliminariesSuppose that E is a real Banach space which is partially ordered by a cone P ⊂ E, i.e., x ≤ yif and only if y − x ∈ P. By θ we denote the zero element of E. A non-empty closed convex setP ⊂ E is called a cone if it satisfies (i) x ∈ P, λ ≥ 0 ⇒ λx ∈ P ; (ii) x ∈ P,−x ∈ P ⇒ x = θ.Moreover, P is called normal if there exists a constant N > 0 such that, for x, y ∈ E, θ ≤ x ≤ yimplies ‖x‖ ≤ N‖y‖; in this case N is called the normality constant of P . We say that an operatorA : E → E is increasing(decreasing) if x ≤ y implies Ax ≤ Ay(Ax ≥ Ay).For x, y ∈ E, the notation x ∼ y means that there exist λ > 0 and µ > 0 such that λx ≤ y ≤ µx.Clearly, ∼ is an equivalence relation. Given h > θ(i.e., h ≥ θ and h 6= θ), we denote by Ph the setPh = {x ∈ E| x ∼ h}. Clearly, Ph ⊂ P is convex and λPh = Ph for λ > 0.We now present a fixed point theorem of generalized concave operators which will be used inthe latter proof. See [22] for further information.Theorem 2.1(from the Lemma 2.1 and Theorem 2.1 in [22]). Let h > θ and P be a normal cone.Assume that: (D1) A : P → P is increasing and Ah ∈ Ph; (D2) For any x ∈ P and t ∈ (0, 1), thereexists α(t) ∈ (t, 1] such that A(tx) ≥ α(t)Ax. Then (i) there are u0, v0 ∈ Ph and r ∈ (0, 1) suchthat rv0 ≤ u0 < v0, u0 ≤ Au0 ≤ Av0 ≤ v0; (ii) operator equation x = Ax has a unique solution inPh.Remark 2.2. An operator A is said to be generalized concave if A satisfies condition (D2).In what follows, for the sake of convenience, let J ′ = J\{t1, t2, · · · , tm}, C[J,R] = {u|u : J → Ris continuous}, PC1[J,R] = {u ∈ C[J,R]|u′(t) is continuous at t 6= tk and left continuous att = tk, u′(t+k ) exists, k = 1, 2, · · · ,m}. Evidently, C[J,R] is a Banach space with the norm ‖u‖C =sup{|u(t)| : t ∈ J} and PC1[J,R] is a Banach space with the norm ‖u‖PC1 = sup{‖u‖C , ‖u′‖C}.Definition 2.3. A function u ∈ PC1[J,R]⋂C2[J ′,R] is called a solution of the problem (1.1), ifit satisfies the problem (1.1).Lemma 2.4. u ∈ PC1[J,R] ∩ C2[J1,R] is a solution of the problem (1.1) if only and if u ∈EJQTDE, 2010 No. 76, p. 2PC1[J,R] is the solution of the following integral equation:u(t) =∫ 10G(t, s)f(s, u(s))ds −m∑k=1G(t, tk)Ik(u(tk)), (2.1)whereG(t, s) =1ρ{ψ(s)ψ(1 − t), 0 ≤ s ≤ t ≤ 1,ψ(t)ψ(1 − s), 0 ≤ t ≤ s ≤ 1,(2.2)withρ =12γ(eγ − e−γ), ψ(t) =12(eγt + e−γt).Proof. First suppose that u ∈ PC1[J,R] ∩C2[J1,R] is a solution of the problem (1.1). Then−u′′(t) + γ2u(t) = f(t, u(t)), t 6= tk.That is, (e−2γt(eγtu(t))′)′ = −f(t, u(t))e−γt, t 6= tk. Lety(t) = e−2γt(eγtu(t))′ = e−γt(u′(t) + γu(t)). (2.3)Theny′(t) = −f(t, u(t))e−γt, t 6= tk. (2.4)It is easy to see by integration of (2.4) thaty(t1) − y(0) = −∫ t10f(s, u(s))e−γsds,y(t) − y(t+1 ) = −∫ tt1f(s, u(s))e−γsds, t1 < t ≤ t2.Soy(t) = y(0) −∫ t0f(s, u(s))e−γsds+ y(t+1 ) − y(t1), t1 < t ≤ t2.In the same way,we can show thaty(t) = y(0) −∫ t0f(s, u(s))e−γsds+∑0<tk<te−γtkIk(u(tk)). (2.5)In view of (2.3),we have(eγtu(t))′ = e2γt(y(0) −∫ t0f(s, u(s))e−γsds+∑0<tk<te−γtkIk(u(tk))).Letz(t) = eγtu(t), l(t) = e2γt(y(0) −∫ t0f(s, u(s))e−γsds+∑0<tk<te−γtkIk(u(tk))). (2.6)EJQTDE, 2010 No. 76, p. 3Then z(t) = z(0) +∫ t0 l(s)ds, t ∈ J. Further,u(t) = z(t)e−γt = e−γt(u(0) +∫ t0l(s)ds), t ∈ J. (2.7)By calculation,we can get∫ t0l(s)ds =12γ{y(0)(e2γt − 1) +∫ t0eγsf(s, u(s))ds − e2γt∫ t0e−γsf(s, u(s))ds+∑0<tk<t(e2γt − e2γtk )e−γtkIk(u(tk)). (2.8)Substituting (2.8) into (2.7),we obtainu(t) =12γ{(γu(0) − u′(0))e−γt + (γu(0) + u′(0))eγt + e−γt∫ t0eγsf(s, u(s))ds− eγt∫ t0e−γsf(s, u(s))ds+∑0<tk<teγ(t−tk)Ik(u(tk))−∑0<tk<te−γ(t−tk)Ik(u(tk)), t ∈ J, (2.9)u′(t) =12{−(γu(0) − u′(0))e−γt + (γu(0) + u′(0))eγt − e−γt∫ t0eγsf(s, u(s))ds− eγt∫ t0e−γsf(s, u(s))ds+∑0<tk<t(eγ(t−tk) + e−γ(t−tk))Ik(u(tk)), t ∈ J. (2.10)In view of u′(0) = u′(1) = 0, we haveu(0) =1γ(eγ − e−γ){∫ 10(e−γ(1−s) + eγ(1−s))f(s, u(s))ds−m∑k=1(eγ(1−tk) + e−γ(1−tk))Ik(u(tk))}. (2.11)Substituting (2.11) into (2.9) and making use of the fact thatm∑k=1Ik(u(tk)) =∑0<tk<1Ik(u(tk)) =∑0<tk<tIk(u(tk)) +∑t≤tk<1Ik(u(tk)),we obtainu(t) =eγt + e−γt2γ(eγ − e−γ){∫ 10(e−γ(1−s) + eγ(1−s))f(s, u(s))ds −m∑k=1(eγ(1−tk) + e−γ(1−tk))Ik(u(tk))}+12γ∫ t0(eγ(s−t) − eγ(t−s))f(s, u(s))ds +12γ∑0<tk<t(eγ(t−tk) − e−γ(t−tk))Ik(u(tk))=1ρ∫ t0ψ(s)ψ(1 − t)f(s, u(s))ds +1ρ∫ 1tψ(t)ψ(1 − s)f(s, u(s))ds−m∑k=11ρψ(t)ψ(1 − tk)Ik(u(tk)) +12γ∑0<tk<t(eγ(t−tk) − e−γ(t−tk))Ik(u(tk))=∫ 10G(t, s)f(s, u(s))ds −m∑k=1G(t, tk)Ik(u(tk)).EJQTDE, 2010 No. 76, p. 4That is, u(t) is a solution of the equation (2.1).Conversely,assume that u ∈ C[J,R] is a solution of the equation (2.1). Direct differentiation of(2.1) implies, for t 6= tku′(t) =∫ 10Gt(t, s)f(s, u(s))ds −m∑k=1Gt(t, tk)Ik(u(tk))=1ρ[(ψ(1 − t))′∫ t0ψ(s)f(s, u(s))ds + ψ′(t)∫ 1tψ(1 − s)f(s, u(s))ds]−1ρψ′(t)m∑k=1ψ(1 − tk)Ik(u(tk)) +12∑0<tk<t(eγ(t−tk) + e−γ(t−tk))Ik(u(tk)),u′′(t) =1ρ[(ψ(1 − t))′′∫ t0ψ(s)f(s, u(s))ds + (ψ(1 − t))′ψ(t)f(t, u(t))+ ψ′′(t)∫ 1tψ(1 − s)f(s, u(s))ds− ψ′(t)ψ(1 − t)f(t, u(t))]−1ρψ′′(t)m∑k=1ψ(1 − tk)Ik(u(tk)) +γ2∑0<tk<t(eγ(t−tk) − e−γ(t−tk))Ik(u(tk)).Making use of the factsψ′′(1 − t) = γ2ψ(1 − t), ψ′′(t) = γ2ψ(t), (ψ(1 − t))′ψ(t) − ψ′(t)ψ(1 − t) = −ρ,we can easily obtain u′′(t) = γ2u(t) − f(t, u(t)), t 6= tk. Moreover,△u′ |t=tk= u′(t+k ) − u′(t−k ) = Ik(u(tk)).So u ∈ C2[J ′,R] and it is easy to verify that u′(0) = u′(1) = 0 and the lemma is proved. 2Remark 2.5. To the best of our knowledge, the expression (2.1) is new for the Neumann problem.Similar expressions have been obtained for periodic problems of first order and for higher orderordinary differential equations with impulses, see Theorem 2.2 in [23] and Lemma 2.1 in [24,25].Lemma 2.6. (i) 0 < G(t, s) ≤ G(t, t), G(t, s) ≤ G(s, s), 0 ≤ t, s ≤ 1;(ii) G(t, s) ≥ Cψ(t)ψ(1 − t)G(t0, s), t, t0, s ∈ [0, 1], where C = 1/ψ2(1).(See [20])3 Existence and uniqueness of positive solutions for problem (1.1)In this section, we apply Theorem 2.1 to study the problem (1.1) and we obtain a new result onthe existence and uniqueness of positive solutions. The method used here is new to the literatureand so is the existence and uniqueness result to the second-order impulsive differential equations.Set P̃ = {x ∈ C[J,R]|x(t) ≥ 0, t ∈ J}, the standard cone. It is clear that P̃ is a normal cone inC[J,R] and the normality constant is 1. Our main result is summarized in the following theorem.Theorem 3.1. Assume that(H1) f(t, 0) ≥ 0, f(t, a) > 0, t ∈ [0, 1] and f(t, x) is increasing in x ∈ [0,∞) for each t ∈ [0, 1], wherea = 14(eγ + e−γ + 2);EJQTDE, 2010 No. 76, p. 5(H2) Ik(0) ≤ 0 and Ik(x) is decreasing in x ∈ [0,∞), k = 1, 2, · · · ,m;(H3) for any λ ∈ (0, 1) and x ≥ 0, there exist α1(λ), α2(λ) ∈ (λ, 1] such thatf(t, λx) ≥ α1(λ)f(t, x), Ik(λx) ≤ α2(λ)Ik(x), k = 1, 2, · · · ,m.(H4)m∑k=1G(tk, tk)Ik(b) < 0, where b =12(eγ + e−γ).Then (i) there exist u0, v0 ∈ P̃h such thatu0(t) ≤∫ 10G(t, s)f(s, u0(s))ds −m∑k=1G(t, tk)Ik(u0(tk)), t ∈ J,v0(t) ≥∫ 10G(t, s)f(s, v0(s))ds −m∑k=1G(t, tk)Ik(v0(tk)), t ∈ J ;(ii) the problem (1.1) has a unique positive solution x∗ in P̃h⋂PC1[J,R], whereh(t) = ψ(t)ψ(1 − t), t ∈ [0, 1].Remark 3.2. Some examples of αi(λ), i = 1, 2 which satisfy the condition (H3) are:(1) αi(λ) = λδi , i = 1, 2,∀ λ ∈ (0, 1), where δi ∈ (0, 1).(2) αi(λ) = λ(1 + ηi(λ)) with 0 < ηi(λ) ≤1λ− 1,∀ λ ∈ (0, 1), i = 1, 2.Remark 3.3. It is easy to check that a = min{h(t) : t ∈ [0, 1]}, b = max{h(t) : t ∈ [0, 1]}, wherea, b are given as in (H1), (H4).Proof of Theorem 3.1 Define an operator A : C[J,R] → C[J,R] byAu(t) =∫ 10G(t, s)f(s, u(s))ds −m∑k=1G(t, tk)Ik(u(tk)).Then from Lemma 2.4, u ∈ PC1[J,R]⋂C2[J ′,R] is a solution of the problem (1.1) if and only ifu ∈ PC1[J,R] is a fixed point of the operator A. Firstly, we show that A : P̃ → P̃ is increasing,generalized concave. For any u ∈ P̃ , from (H1), (H2), we obtain Au(t) ≥ 0, t ∈ [0, 1]. Further, alsofrom (H1), (H2), we can easily prove that A : P̃ → P̃ is increasing. For any λ ∈ (0, 1) and u ∈ P̃ ,from (H1) − (H3), we haveA(λu)(t) =∫ 10G(t, s)f(s, λu(s))ds −m∑k=1G(t, tk)Ik(λu(tk))≥∫ 10G(t, s)α1(λ)f(s, u(s))ds −m∑k=1G(t, tk)α2(λ)Ik(u(tk)).Set α(λ) = min{α1(λ), α2(λ)}, λ ∈ (0, 1). Then α(λ) ∈ (λ, 1]. We haveA(λu)(t) ≥ α(λ)[∫ 10G(t, s)f(s, u(s))ds −m∑k=1G(t, tk)Ik(u(tk))]= α(λ)Au(t).That is, A(λu) ≥ α(λ)Au, u ∈ P̃ , λ ∈ (0, 1). So A : P̃ → P̃ is generalized concave.EJQTDE, 2010 No. 76, p. 6Secondly, we prove Ah ∈ P̃h. To illuminate this, setr1 = mint∈[0,1]f(t, a), r2 = maxt∈[0,1]f(t, b).Then from (H1), we have r2 ≥ r1 > 0. Further, from (H1), (H2) and Lemma 2.6,Ah(t) =∫ 10G(t, s)f(s, h(s))ds −m∑k=1G(t, tk)Ik(h(tk))≥∫ 10G(t, s)f(s, h(s))ds ≥∫ 10Cψ(t)ψ(1 − t)G(t0, s)f(s, a)ds≥ Cr1∫ 10G(t0, s)ds · h(t).Note that∫ 10G(t0, s)ds =1ρ∫ t00ψ(s)ψ(1 − t0)ds+1ρ∫ 1t0ψ(t0)ψ(1 − s)ds =1γ2,we have Ah(t) ≥ Cr1γ2h(t). It follows from Lemma 2.6 and (H4) thatAh(t) =∫ 10G(t, s)f(s, h(s))ds −m∑k=1G(t, tk)Ik(h(tk))≤∫ 10G(t, t)f(s, b)ds −m∑k=1G(tk, tk)Ik(b)≤r2ρh(t) −4eγ + e−γ + 2h(t)m∑k=1G(tk, tk)Ik(b)=[r2ρ−4eγ + e−γ + 2m∑k=1G(tk, tk)Ik(b)]h(t).Hence,Cr1γ2h ≤ Ah ≤[r2ρ−4eγ + e−γ + 2m∑k=1G(tk, tk)Ik(b)]h.That is, Ah ∈ P̃h. Finally, an application of Theorem 2.1 implies that (i) there are u0, v0 ∈ P̃h suchthat u0 ≤ Au0, Av0 ≤ v0; (ii) operator equation u = Au has a unique solution x∗ in P̃h. That is,u0(t) ≤∫ 10G(t, s)f(s, u0(s))ds −m∑k=1G(t, tk)Ik(u0(tk)), t ∈ J,v0(t) ≥∫ 10G(t, s)f(s, v0(s))ds −m∑k=1G(t, tk)Ik(v0(tk)), t ∈ J ;and the problem (1.1) has a unique solution x∗ in P̃h. Moreover, from Lemma 2.4 we know thatx∗ ∈ PC1[J,R]. Evidently, x∗ is a positive solution of the problem (1.1).2Remark 3.4. For the case of Ik = 0, k = 1, 2, . . . ,m, the problem(1.1) reduces to the followingNeumann boundary value problem for ordinary differential equations:{−u′′(t) + γ2u(t) = f(t, u(t)), 0 < t < 1,u′(0) = u′(1) = 0,(3.1)EJQTDE, 2010 No. 76, p. 7We can establish the existence and uniqueness of positive solutions for the problem (3.1) by usingthe same method used in this paper, which is new to the literature. So the method used in thispaper is different from previous ones in literature and the result obtained in this paper is new.4 An exampleTo illustrate how our main result can be used in practice we present an example.Example 4.1. Consider the following boundary value problem−u′′(t) + (ln 2)2u(t) = uβ(t) + q(t), t ∈ J, t 6= 12 ,∆u′|t= 12= − 4√u(12),u′(0) = u′(1) = 0,(4.1)where β ∈ (0, 1) and q : [0, 1] → [0,+∞) is a continuous function.Conclusion. The impulsive problem (4.1) has a unique positive solution in P̃h⋂PC1[J,R], whereh(t) =58+14(21−2t + 22t−1), t ∈ [0, 1].Proof. The problem (4.1) can be regarded as a boundary value problem of the form (1.1), whereγ = ln 2, t1 =12 , f(t, x) := xβ + q(t), I1(x) := −x14 . After a simple calculation, we get a = 98 , b =54andf(t, a) = (98)β + q(t) > 0, G(t1, t1)I1(b) = G(12,12)I1(54) = −e2 + 14(e2 − 1)(54)14 < 0.It is not difficult to see that the conditions (H1), (H2) and (H4) hold. In addition, let α1(λ) =λβ, α2(λ) = λ14 . Then, the condition (H3) of Theorem 3.1 holds. Hence, by Theorem 3.1, theconclusion follows, and the proof is complete.Remark 4.1. Example 4.1 implies that there is a large number of functions that satisfy theconditions of Theorem 3.1. 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Existence and uniqueness of positive solutions for Neumann problems of second order impulsive differential equations

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Existence and uniqueness of positive solutions for Neumann problems of second order impulsive differential equations

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