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Advances in Difference Equations This Provisional PDF corresponds to the article as it appeared upon acceptance Fully formatted PDF and full text (HTML) versions will be made available soon Periodic boundary value problems for nonlinear first-order impulsive dynamic equations on time scales Advances in Difference Equations 2012, 2012:12 doi:10.1186/1687-1847-2012-12 Da-Bin Wang (wangdb@lut.cn) ISSN Article type 1687-1847 Research Submission date 23 August 2011 Acceptance date 15 February 2012 Publication date 15 February 2012 Article URL http://www.advancesindifferenceequations.com/content/2012/1/12 This peer-reviewed article was published immediately upon acceptance It can be downloaded, printed and distributed freely for any purposes (see copyright notice below) For information about publishing your research in Advances in Difference Equations go to http://www.advancesindifferenceequations.com/authors/instructions/ For information about other SpringerOpen publications go to http://www.springeropen.com © 2012 Wang ; licensee Springer This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited Periodic boundary value problems for nonlinear first-order impulsive dynamic equations on time scales Da-Bin Wang Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu 730050, People’s Republic of China Email address: wangdb@lut.cn Abstract By using the classical fixed point theorem for operators on cone, in this article, some results of one and two positive solutions to a class of nonlinear first-order periodic boundary value problems of impulsive dynamic equations on time scales are obtained Two examples are given to illustrate the main results in this article Keywords: time scale; periodic boundary value problem; positive solution; fixed point; impulsive dynamic equation Mathematics Subject Classification: 39A10; 34B15 1 Introduction Let T be a time scale, i.e., T is a nonempty closed subset of R Let 0, T be points in T, an interval (0, T )T denoting time scales interval, that is, (0, T )T := (0, T ) ∩ T Other types of intervals are defined similarly The theory of impulsive differential equations is emerging as an important area of investigation, since it is a lot richer than the corresponding theory of differential equations without impulse effects Moreover, such equations may exhibit several real world phenomena in physics, biology, engineering, etc (see [1–3]) At the same time, the boundary value problems for impulsive differential equations and impulsive difference equations have received much attention [4–18] On the other hand, recently, the theory of dynamic equations on time scales has become a new important branch (see, for example, [19–21]) Naturally, some authors have focused their attention on the boundary value problems of impulsive dynamic equations on time scales [22–36] However, to the best of our knowledge, few papers concerning PBVPs of impulsive dynamic equations on time scales with semi-position condition In this article, we are concerned with the existence of positive solutions for the following PBVPs of impulsive dynamic equations on time scales with semi-position condition   △  x (t) + f (t, x(σ(t))) = 0,      + − −  x(tk ) − x(tk ) = Ik (x(tk )),      x(0) = x(σ(T )),  t ∈ J := [0, T ]T , k = 1, 2, , m, t = tk , k = 1, 2, , m, (1.1) where T is an arbitrary time scale, T > is fixed, 0, T ∈ T, f ∈ C (J × [0, ∞) , (−∞, ∞)), Ik ∈ C ([0, ∞) , [0, ∞)) , tk ∈ (0, T )T , < t1 < · · · < tm < T, and for each k = 1, 2, , m, x(t+ ) = limh→0+ x(tk + h) and x(t− ) = limh→0− x(tk + h) represent the right and left limits of k k x(t) at t = tk We always assume the following hypothesis holds (semi-position condition): (H) There exists a positive number M such that M x − f (t, x) ≥ for x ∈ [0, ∞) , t ∈ [0, T ]T By using a fixed point theorem for operators on cone [37], some existence criteria of positive solution to the problem (1.1) are established We note that for the case T = R and Ik (x) ≡ 0, k = 1, 2, , m, the problem (1.1) reduces to the problem studied by [38] and for the case Ik (x) ≡ 0, k = 1, 2, , m, the problem (1.1) reduces to the problem (in the one-dimension case) studied by [39] In the remainder of this section, we state the following fixed point theorem [37] Theorem 1.1 Let X be a Banach space and K ⊂ X be a cone in X Assume Ω1 , Ω2 are bounded open subsets of X with ∈ Ω1 ⊂ Ω1 ⊂ Ω2 and Φ : K ∩ (Ω2 \Ω1 ) → K is a completely continuous operator If (i) There exists u0 ∈ K\{0} such that u − Φu = λu0 , u ∈ K ∩ ∂Ω2 , λ ≥ 0; Φu = τ u, u ∈ K ∩ ∂Ω1 , τ ≥ 1, or (ii) There exists u0 ∈ K\{0} such that u − Φu = λu0 , u ∈ K ∩ ∂Ω1 , λ ≥ 0; Φu = τ u, u ∈ K ∩ ∂Ω2 , τ ≥ Then Φ has at least one fixed point in K ∩ (Ω2 \Ω1 ) Preliminaries Throughout the rest of this article, we always assume that the points of impulse tk are right-dense for each k = 1, 2, , m We define P C = {x ∈ [0, σ(T )]T → R : xk ∈ C(Jk , R), k = 0, 1, 2, , m and there exist x(t+ ) and x(t− ) with x(t− ) = x(tk ), k = 1, 2, , m , k k k where xk is the restriction of x to Jk = (tk , tk+1 ]T ⊂ (0, σ(T )]T , k = 1, 2, , m and J0 = [0, t1 ]T , tm+1 = σ(T ) Let X = {x : x ∈ P C, x(0) = x(σ(T ))} with the norm x = sup t∈[0,σ(T )]T |x(t)| , then X is a Banach space Lemma 2.1 Suppose M > and h : [0, T ]T → R is rd-continuous, then x is a solution of σ(T ) x(t) = G(t, s)h(s)△s + where G(t, s) =        m G(t, tk )Ik (x(tk )), k=1 eM (s,t)eM (σ(T ),0) , eM (σ(T ),0)−1 ≤ s ≤ t ≤ σ(T ), eM (s,t) , eM (σ(T ),0)−1 ≤ t < s ≤ σ(T ), if and only if x is a solution of the boundary value problem t ∈ [0, σ(T )]T ,   △  x (t) + M x(σ(t)) = h(t),      + − −  x(tk ) − x(tk ) = Ik (x(tk )),      x(0) = x(σ(T ))  t ∈ J := [0, T ]T , t = tk , k = 1, 2, , m, k = 1, 2, , m, Proof Since the proof similar to that of [34, Lemma 3.1], we omit it here Lemma 2.2 Let G(t, s) be defined as in Lemma 2.1, then eM (σ(T ), 0) ≤ G(t, s) ≤ eM (σ(T ), 0) − eM (σ(T ), 0) − for all t, s ∈ [0, σ(T )]T Proof It is obviously, so we omit it here Remark 2.1 Let G(t, s) be defined as in Lemma 2.1, then σ(T ) G(t, s)△s = M For u ∈ X, we consider the following problem:   △  x (t) + M x(σ(t)) = M u(σ(t)) − f (t, u(σ(t))),      + − −  x(tk ) − x(tk ) = Ik (x(tk )),       x(0) = x(σ(T )) t ∈ [0, T ]T , t = tk , k = 1, 2, , m, k = 1, 2, , m, (2.1) It follows from Lemma 2.1 that the problem (2.1) has a unique solution: σ(T ) m G(t, tk )Ik (x(tk )), G(t, s)hu (s)△s + x(t) = k=1 where hu (s) = M u(σ(s)) − f (s, u(σ(s))), s ∈ [0, T ]T We define an operator Φ : X → X by t ∈ [0, σ(T )]T , σ(T ) m G(t, s)hu (s)△s + Φ(u)(t) = G(t, tk )Ik (u(tk )), t ∈ [0, σ(T )]T k=1 It is obvious that fixed points of Φ are solutions of the problem (1.1) Lemma 2.3 Φ : X → X is completely continuous Proof The proof is divided into three steps Step 1: To show that Φ : X → X is continuous Let {un }∞ be a sequence such that un → u (n → ∞) in X Since f (t, u) and Ik (u) are n=1 continuous in x, we have |hun (t) − hu (t)| = |M (un − u) − (f (t, un ) − f (t, u))| → (n → ∞), |Ik (un (tk )) − Ik (u(tk ))| → (n → ∞) So |Φ(un )(t) − Φ(u)(t)| σ(T ) m G(t, s) [hun (s) − hu (s)] △s + = k=1 ≤ G(t, tk ) [Ik (un (tk )) − Ik (u(tk ))]  σ(T ) eM (σ(T ), 0)  eM (σ(T ), 0) − m |hun (s) − hu (s)| △s + k=1  |Ik (un (tk )) − Ik (u(tk ))| → (n → ∞), which leads to Φun − Φu → (n → ∞) That is, Φ : X → X is continuous Step 2: To show that Φ maps bounded sets into bounded sets in X Let B ⊂ X be a bounded set, that is, ∃ r > such that ∀ u ∈ B we have u ≤ r Then, for any u ∈ B, in virtue of the continuities of f (t, u) and Ik (u), there exist c > 0, ck > such that |f (t, u)| ≤ c, |Ik (u)| ≤ ck , k = 1, 2, , m We get σ(T ) m G(t, tk )Ik (u(tk )) G(t, s)hu (s)△s + |Φ(u)(t)| = k=1 σ(T ) ≤ m G(t, tk ) |Ik (u(tk ))| G(t, s) |hu (s)| △s + k=1 m eM (σ(T ), 0) σ(T ) (M r + c) + ck ≤ eM (σ(T ), 0) − k=1 Then we can conclude that Φu is bounded uniformly, and so Φ(B) is a bounded set Step 3: To show that Φ maps bounded sets into equicontinuous sets of X Let t1 , t2 ∈ (tk , tk+1 ]T ∩ [0, σ(T )]T , u ∈ B, then |Φ(u)(t1 ) − Φ(u)(t2 )| σ(T ) m |G(t1 , s) − G(t2 , s)| |hu (s)| △s + ≤ |G(t1 , tk ) − G(t2 , tk )| |Ik (u(tk ))| k=1 The right-hand side tends to uniformly zero as |t1 − t2 | → Consequently, Steps 1–3 together with the Arzela–Ascoli Theorem shows that Φ : X → X is completely continuous Let K = {u ∈ X : u(t) ≥ δ u , where δ = eM (σ(T ), 0) t ∈ [0, σ(T )]T } , ∈ (0, 1) It is not difficult to verify that K is a cone in X From condition (H) and Lemma 2.2, it is easy to obtain following result: Lemma 2.4 Φ maps K into K Main results For convenience, we denote f = lim+ sup max f (t, u) , u f ∞ = lim sup max f0 = lim+ inf f (t, u) , u f∞ = lim inf u→0 u→0 t∈[0,T ]T t∈[0,T ]T u→∞ u→∞ t∈[0,T ]T t∈[0,T ]T and I0 = lim+ u→0 Ik (u) u→∞ u Ik (u) , u I∞ = lim Now we state our main results Theorem 3.1 Suppose that (H1 ) f0 > 0, f ∞ < 0, I0 = for any k; or (H2 ) f∞ > 0, f < 0, I∞ = for any k Then the problem (1.1) has at least one positive solutions f (t, u) , u f (t, u) u Proof Firstly, we assume (H1 ) holds Then there exist ε > and β > α > such that f (t, u) ≥ εu, Ik (u) ≤ t ∈ [0, T ]T , u ∈ (0, α] , [eM (σ(T ), 0) − 1]ε u, u ∈ (0, α] , 2M meM (σ(T ), 0) for any k, (3.1) (3.2) and f (t, u) ≤ −εu, t ∈ [0, T ]T , u ∈ [β, ∞) (3.3) Let Ω1 = {u ∈ X : u < r1 } , where r1 = α Then u ∈ K ∩ ∂Ω1 , < δα = δ u ≤ u(t) ≤ α, in view of (3.1) and (3.2) we have σ(T ) Φ(u)(t) = m G(t, s)hu (s)△s + G(t, tk )Ik (u(tk )) k=1 σ(T ) m G(t, s)(M − ε)u(σ(s))△s + ≤ G(t, tk ) k=1 (M − ε) eM (σ(T ), 0) ≤ u + M eM (σ(T ), 0) − ε = M− M < u , t ∈ [0, σ(T )]T , u which yields Φ(u) < u Therefore m k=1 [eM (σ(T ), 0) − 1]ε u(tk ) 2M meM (σ(T ), 0) [eM (σ(T ), 0) − 1]ε u 2M meM (σ(T ), 0) Φu = τ u, u ∈ K ∩ ∂Ω1 , τ ≥ (3.4) On the other hand, let Ω2 = {u ∈ X : u < r2 } , where r2 = β δ Choose u0 = 1, then u0 ∈ K\{0}.We assert that u − Φu = λu0 , u ∈ K ∩ ∂Ω2 , λ ≥ Suppose on the contrary that there exist u ∈ K ∩ ∂Ω2 and λ ≥ such that u − Φu = λu0 Let ζ = mint∈[0,σ(T )]T u(t), then ζ ≥ δ u = δr2 = β, we have from (3.3) that u(t) = Φ(u)(t) + λ σ(T ) m G(t, tk )Ik (u(tk )) + λ G(t, s)hu (s)△s + = k=1 σ(T ) G(t, s)hu (s)△s + λ ≥ (M + ε) ≥ ζ + λ, M t ∈ [0, σ(T )]T Therefore, ζ= t∈[0,σ(T )]T u(t) ≥ (M + ε) ζ + λ > ζ, M which is a contradiction 10 (3.5) It follows from (3.4), (3.5) and Theorem 1.1 that Φ has a fixed point u∗ ∈ K ∩ (Ω2 \Ω1 ), and u∗ is a desired positive solution of the problem (1.1) Next, suppose that (H2 ) holds Then we can choose ε′ > and β ′ > α′ > such that f (t, u) ≥ ε′ u, Ik (u) ≤ t ∈ [0, T ]T , [eM (σ(T ), 0) − 1]ε′ u, 2M meM (σ(T ), 0) u ∈ [β ′ , ∞) , (3.6) u ∈ [β ′ , ∞) for any k, (3.7) u ∈ (0, α′ ] (3.8) and f (t, u) ≤ −ε′ u, t ∈ [0, T ]T , Let Ω3 = {u ∈ X : u < r3 } , where r3 = α′ Then for any u ∈ K ∩ ∂Ω3 , < δ u ≤ u(t) ≤ u = α′ It is similar to the proof of (3.5), we have u − Φu = λu0 , u ∈ K ∩ ∂Ω3 , Let Ω4 = {u ∈ X : u < r4 } , where r4 = β′ δ λ ≥ (3.9) Then for any u ∈ K ∩ ∂Ω4 , u(t) ≥ δ u = δr4 = β ′ , by (3.6) and (3.7), it is easy to obtain Φu = τ u, u ∈ K ∩ ∂Ω4 , τ ≥ (3.10) It follows from (3.9), (3.10) and Theorem 1.1 that Φ has a fixed point u∗ ∈ K ∩ (Ω4 \Ω3 ), and u∗ is a desired positive solution of the problem (1.1) 11 Theorem 3.2 Suppose that (H3 ) f < 0, f ∞ < 0; (H4 ) there exists ρ > such that min{f (t, u) − u|t ∈ [0, T ]T , [eM (σ(T ), 0) − 1] u, M meM (σ(T ), 0) Ik (u) ≤ δρ ≤ u ≤ ρ} > 0; δρ ≤ u ≤ ρ, (3.11) for any k (3.12) Then the problem (1.1) has at least two positive solutions Proof By (H3 ), from the proof of Theorem 3.1, we should know that there exist β ′′ > ρ > α′′ > such that u − Φu = λu0 , u ∈ K ∩ ∂Ω5 , λ ≥ 0, (3.13) u − Φu = λu0 , u ∈ K ∩ ∂Ω6 , λ ≥ 0, (3.14) where Ω5 = {u ∈ X : u < r5 } , Ω6 = {u ∈ X : u < r6 } , r5 = α′′ , r6 = β ′′ δ By (3.11) of (H4 ), we can choose ε > such that f (t, u) ≥ (1 + ε)u, t ∈ [0, T ]T , δρ ≤ u ≤ ρ (3.15) Let Ω7 = {u ∈ X : u < ρ} , for any u ∈ K ∩ ∂Ω7 , δρ = δ u ≤ u(t) ≤ u = ρ, from (3.12) and (3.15), it is similar to the proof of (3.4), we have 12 Φu = τ u, u ∈ K ∩ ∂Ω7 , τ ≥ (3.16) By Theorem 1.1, we conclude that Φ has two fixed points u∗∗ ∈ K ∩ (Ω6 \Ω7 ) and u∗∗∗ ∈ K ∩ (Ω7 \Ω5 ), and u∗∗ and u∗∗∗ are two positive solution of the problem (1.1) Similar to Theorem 3.2, we have: Theorem 3.3 Suppose that (H4 ) f0 > 0, f∞ > 0, I0 = 0, I∞ = 0; (H5 ) there exists ρ > such that max{f (t, u)|t ∈ [0, T ]T , δρ ≤ u ≤ ρ} < Then the problem (1.1) has at least two positive solutions Examples Example 4.1 Let T = [0, 1] ∪ [2, 3] We consider the following problem on T   △  x (t) + f (t, x(σ(t))) = 0, t ∈ [0, 3] ,   T    − 1+ −x =I x ,  x      x(0) = x(3),  where T = 3, f (t, x) = x − (t + 1)x2 , and I(x) = x2 Let M = 1,then, it is easy to see that 13 t = 2, (4.1) M x − f (t, x) = (t + 1)x2 ≥ for x ∈ [0, ∞) , t ∈ [0, 3]T , and f0 ≥ 1, f ∞ = −∞, and I0 = Therefore, by Theorem 3.1, it follows that the problem (4.1) has at least one positive solution Example 4.2 Let T = [0, 1] ∪ [2, 3] We consider the following problem on T   △  x (t) + f (t, x(σ(t))) = 0,      + −x  x      x(0) = x(3),  1− =I x t ∈ [0, 3]T , t = 2, (4.2) , where T = 3, f (t, x) = 4e1−4e x − (t + 1)x2 e−x , and I(x) = x2 e−x Choose M = 1, ρ = 4e2 , then δ = , 2e2 it is easy to see that M x − f (t, x) = x(1 − 4e1−4e ) + (t + 1)x2 e−x ≥ for x ∈ [0, ∞) , f0 ≥ 4e1−4e > 0, f∞ ≥ 4e1−4e > 0, and 14 t ∈ [0, 3]T , I0 = , I∞ = 0, max{f (t, u)|t ∈ [0, T ]T , δρ ≤ u ≤ ρ} = max f (t, u)|t ∈ [0, 3]T , ≤ u ≤ 4e2 = 16e3−4e (1−e) < Therefore, together with Theorem 3.3, it follows that the problem (4.2) has at least two positive solutions Competing interests The authors have no competing interests to declare Acknowledgment The author thankful to the anonymous referee for his/her 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Periodic boundary value problems for first-order impulsive dynamic equations on time scales Nonlinear Anal 69, 4074–4087 (2008) [28] Graef, JR, Ouahab, A: Extremal solutions for nonresonance impulsive. .. theorem for operators on cone, in this article, some results of one and two positive solutions to a class of nonlinear first-order periodic boundary value problems of impulsive dynamic equations on time