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Hindawi Publishing Corporation Boundary Value Problems Volume 2007, Article ID 73176, 16 pages doi:10.1155/2007/73176 Research Article Extremal Solutions of Periodic Boundary Value Problems for First-Order Impulsive Integrodifferential Equations of Mixed-Type on Time Scales Yongkun Li and Hongtao Zhang Received 12 October 2006; Accepted 21 May 2007 Recommended by Ivan Kiguradze We consider the existence of minimal and maximal solutions of periodic boundary value problems for first-order impulsive integrodifferential equations of mixed-type on time scales by establishing a comparison result and using the monotone iterative technique. Copyright © 2007 Y. Li and H. Zhang. 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. 1. Introduction The theory of calculus on time scales (see [1, 2] a nd references cited therein) was ini- tiated by Stefan Hilger in his Ph.D. thesis in 1990 [3] in order to unify continuous and discrete analyses, and it has a tremendous potential for applications and has recently re- ceived much attention since his foundational work. In this paper, we will study the peri- odic boundary value problem for the first-order impulsive integrodifferential equations of mixed-type (PBVP): u Δ (t) = f  t,u(t),[Tu](t),[Su](t)  , t = t k , t ∈ J T , u  t + k  − u  t − k  = I k  u  t − k  , k = 1,2, , p, u(0) = u(T), (1.1) where T is a time scale which has the subspace topology inherited from the standard topology on R.ForeachintervalJ of R, we denote by J T = J ∩ T, f ∈ C[J T × R × R × R ,R], J = [0,T], I k ∈ C[R,R], where u(t + k )andu(t − k ) represent right and left limits of u(t)att = t k (k = 1,2, , p) in the sense of time scales, and in addition, if t k is right scattered, then y(t + k ) = y(t k ), whereas if t k is left scattered, then y(t − k ) = y(t k ), 2 Boundary Value Problems 0 <t 1 <t 2 < ··· <t k < ··· <t p <T, [Tu](t) =  t 0 k(t,s)u(s)Δs,[Su](t) =  T 0 h(t,s)u(s)Δs, (1.2) k(t,s) ∈ C[D,R + ], D ={(t,s) ∈ J T × J T : t ≥ s}, h(t,s) ∈ C[J T × J T ,R + ], R + = [0,+∞), k 0 = max{k(t,s):(t,s) ∈ D}, h 0 = max{h(t,s):(t,s) ∈ J T × J T }. The study of impulsive dynamic equations on time scales has been initiated by Hender- son [4], Benchohra et al. [5], and Atici and Biles [6]. Extremal solutions of PBVP for im- pulsive differential equations and difference equations has been studied by some authors (see [7, 8]). In this paper, we will obtain an inequality on time scales. And then, using this inequalit y, a comparison result is obtained. At last, we obtain an existence theorem of minimal and maximal solutions of PBVP (1.1) by using monotone iterative technique (see [7–9]). 2. Preliminaries and comparison principle In this section, we will first recall some basic definitions and lemmas, which are used in what follows. Let T be a nonempty closed subset (time scale) of R.Theforwardandbackwardjump operators σ,ρ : T → T, and the graininess μ : T → R + are defined, respectively, by σ(t) = inf{s ∈ T : s>t}, ρ(t) = sup{s ∈ T : s<t}, μ(t) = σ(t) − t. (2.1) Apointt ∈ T is called left dense if t>inf T and ρ(t) = t,leftscatteredifρ(t) <t, right dense if t<sup T and σ(t) = t, and right scattered if σ(t) >t.IfT has a left-scattered maximum m,then T k = T \{ m}; otherwise T k = T .IfT has a right-scattered minimum m,then T k = T \{ m}; otherwise T k = T . A function f : T → R is rig ht-dense continuous provided it is continuous at right- dense point in T and its left-side limits exist at left-dense points in T.If f is continuous at each right-dense point and each left-dense point, then f is said to be continuous function on T. For y : T → R and t ∈ T k , we define the delta derivative of y(t), y Δ (t)tobethenumber (if it exists) with the property that for a given ε>0, there exists a neighborhood U of t such that    y  σ(t)  − y(s)  − y Δ (t)  σ(t) − s    <ε   σ(t) − s   (2.2) for all s ∈ U. If y is continuous, then y is right-dense continuous, and if y is delta differentiable at t,theny is continuous at t. Lemma 2.1 (see [1]). Assume that f ,g : T → R are delta differentiable at t ∈ T k .Then, ( fg) Δ (t) = f Δ (t)g(t)+ f  σ(t)  g Δ (t) = f (t)g Δ (t)+ f Δ (t)g  σ(t)  . (2.3) Y. Li and H. Z hang 3 Let y be right-dense continuous. If Y Δ (t) = y(t), then we define the delta integral by  t a y(s)Δs = Y (t) − Y(a). (2.4) A function r : T → R is called regressive if 1+μ(t)r(t) = 0 (2.5) for all t ∈ T k . If r is regressive function, then the generalized exponential function e r is defined by e r (t,s) = exp   t s ξ μ(τ)  r(τ)  Δτ  for s,t ∈ T (2.6) with the cylinder transformation ξ h (z) = ⎧ ⎪ ⎨ ⎪ ⎩ Log(1 + hz) h if h = 0, z if h = 0. (2.7) Let p,q : T → R be two regressive functions, we define p ⊕ q := p + q + μpq, p :=− p 1+μp , p  q := p ⊕ (q). (2.8) Then, the generalized exponential function has the following properties. Lemma 2.2 (see [1]). Assume that p,q : T → R are two regressive funct ions, then (i) e 0 (t,s) ≡ 1 and e p (t,t) ≡ 1; (ii) e p (σ(t),s) = (1 + μ(t)p(t))e p (t,s); (iii) e p (t,σ(s)) = e p (t,s)/(1 + μ(s)p(s)); (iv) 1/e p (t,s) = e p (t,s); (v) e p (t,s) = 1/e p (s,t) = e p (s,t); (vi) e p (t,s)e p (s,r) = e p (t,r); (vii) e p (t,s)e q (t,s) = e p⊕q (t,s); (viii) e p (t,s)/e q (t,s) = e pq (t,s). Lemma 2.3 [1]. Let r : T → R be right-dense continuous and regressive, a ∈ T,andy a ∈ R. The unique solution of the initial value problem y Δ (t) = r(t)y(t)+h(t), y(a) = y a , (2.9) is given by y(t) = e r (t,a)y a +  t a e r  t,σ(s)  h(s)Δs. (2.10) Throughout this paper, we assume that, for each k = 1, , p, the points of impulse t k are right dense. For convenience, we introduce the notation PC[J T ,R] ={u : J T → R,u(t) 4 Boundary Value Problems is continuous everywhere except some t k at which u(t − k )andu(t + k ) exist and u(t − k ) = u(t k )}. Evidently, PC[J T ,R] is a Banach space with norm u PC = sup t∈J T |u(t)|.LetJ  T = J T \{t 1 ,t 2 , ,t p }, C 1 [J  T ,R] ={u Δ (t)iscontinuousonJ  T }, Ω = PC[J T ,R] ∩ C 1 [J  T ,R], T + = T ∩ R + , PC 1 [T + ,R] = PC[T + ,R] ∩ C 1 [T + ,R]. A function u ∈ Ω is called a solution of PBVP (1.1) if it satisfies (1.1). Next, we combine [10, 11] to obtain an inequality as fol lows. Lemma 2.4. Assume that (A 0 ) the sequence {t k } satisfies 0 ≤ t 0 <t 1 <t 2 < ···<t k < ··· with lim k→+∞ t k = +∞, (A 1 ) m ∈ PC 1 [T + ,R] is right-dense continuous at t k for k = 1,2, , (A 2 )inf t∈J T {μ(t)p(t)} > −1.Fork = 1,2, ,t ≥ t 0 , m Δ (t) ≥ p(t)m(t)+q(t), t = t k , m  t + k  ≥ d k m  t k  + b k , (2.11) where p,q ∈ C(T + ,R), d k ≥ 0,andb k are real constants. Then, m(t) ≥ m  t 0   t 0 <t k <t d k e p  t,t 0  +  t t 0  s<t k <t d k e p  t,σ(s)  q(s)Δs +  t 0 <t k <t  t k <t j <t d j e p  t,t k  b k . (2.12) Proof. By condition (A 2 ), we know that e p (σ(t),t 0 ) ≥ 0fort ∈ [t 0 ,+∞) T . For the follow- ing inequality: m Δ (t) ≥ p(t)m(t)+q(t), (2.13) on multiplying e p (σ(t),t 0 ) and arranging the terms, we obtain e p  σ(t),t 0  m Δ (t) − p(t)m(t)e p  σ(t),t 0  ≥ e p  σ(t),t 0  q(t), (2.14) which is the same as  e p  t,t 0  m(t)  Δ ≥ e p  σ(t),t 0  q(t). (2.15) Integrating (2.15)fromt 0 to t 1 ,then e p  t 1 ,t 0  m  t 1  ≥ m  t 0  +  t 1 t 0 e p  σ(s),t 0  q(s)Δs. (2.16) Again integrating (2.15)fromt 1 to t,wheret ∈ (t 1 ,t 2 ], then e p  t,t 0  m(t) ≥ e p  t 1 ,t 0  m  t + 1  +  t t 1 e p  σ(s),t 0  q(s)Δs ≥ e p  t 1 ,t 0  d 1 m  t 1  + b 1  +  t t 1 e p  σ(s),t 0  q(s)Δs ≥ d 1  m  t 0  +  t 1 t 0 e p  σ(s),t 0  q(s)Δs  + b 1 e p  t 1 ,t 0  +  t t 1 e p  σ(s),t 0  q(s)Δs, (2.17) Y. Li and H. Z hang 5 that is, m(t) ≥ m  t 0  d 1 e p  t,t 0  +  t t 0  s<t k <t d k e p  t,σ(s)  q(s)Δs + b 1 e p  t,t 1  . (2.18) Repeating the above procession for t ∈ [t 0 ,+∞) T ,wehave m(t) ≥ m  t 0   t 0 <t k <t d k e p  t,t 0  +  t t 0  s<t k <t d k e p  t,σ(s)  q(s)Δs +  t 0 <t k <t  t k <t j <t d j e p  t,t k  b k . (2.19) Thus the proof of Lemma 2.4 is complete.  The following comparison result plays an important role in this paper. Lemma 2.5. Let t 0 = 0, t p+1 = T. Assume that u ∈ Ω satisfies u Δ (t) ≥−a(t)u(t) − b(t)[Tu](t) − c(t)[Su](t), t = t k , t ∈ J T , u  t + k  − u  t k  ≥− L k u  t k  , k = 1,2, , p, u(0) ≥ u(T), (2.20) where a,b, c ∈ C[J T ,R + ], a is not identically vanishing, and sup t∈J T {μ(t)a(t)} < 1, 0 ≤ L k < 1(k = 1,2, , p).If  Bk 0 + Ch 0  e (−a) (T,0) ≤   0<t k <T  1 − L k   2  T 0  s<t k <T  1 − L k  Δs (2.21) with B = sup t∈J T {b(t)  t 0 e (−a) (σ(t),s)Δs} and C = sup t∈J T {c(t)  T 0 e (−a) (σ(t),s)Δs}, then u(t) ≥ 0 for t ∈ J T . Proof. Let p(t) = u(t)e (−a) (t,0) for t ∈ J T .Thenp ∈ Ω satisfies p Δ (t) ≥−b(t)  t 0 e (−a)  σ(t),s  k(t,s)p(s)Δs − c(t)  T 0 e (−a)  σ(t),s  h(t,s)p(s)Δs, t = t k , t ∈ J T , p  t + k  − p  t k  ≥− L k p  t k  , k = 1,2, , p, p(0) ≥ e (−a) (T,0)p(T). (2.22) We now prove p(t) ≥ 0fort ∈ J T . (2.23) 6 Boundary Value Problems Assume that (2.23) is not true. Then, there are two cases: (a) there exists t ∗ 1 ∈ J T such that p(t ∗ 1 ) < 0andp(t) ≤ 0fort ∈ J T ; (b) there exists t ∗ 1 ,t ∗ 2 ∈ J T such that p(t ∗ 1 ) < 0andp(t ∗ 2 ) > 0. In case (a), (2.22) implies that p Δ (t) ≥ 0, t = t k , t ∈ J T , p  t + k  − p  t k  ≥ 0, k = 1,2, , p. (2.24) This means that p(t) is nondecreasing in J T ; therefore, p(0) ≤ p  t ∗ 1  < 0, p(0) ≤ p(T) ≤ 0, (2.25) which contradicts p(T) ≤ e (−a) (T,0)p(0) < 0. In case (b) let sup t∈J T p(t) = λ.Then,λ>0 and there exists t i <t ∗ 0 ≤ t i+1 for some i such that p(t ∗ 0 ) = λ or p(t + i ) = λ. We may assume that p(t ∗ 0 ) = λ (since, in case of p(t + i ) = λ, the proof is similar). From (2.22), we have p Δ (t) ≥−λk 0 b(t)  t 0 e (−a)  σ(t),s  Δs − λh 0 c(t)  T 0 e (−a)  σ(t),s  Δs ≥−λ  Bk 0 + Ch 0  , t = t k , t ∈ J T . (2.26) For t ∈ [t ∗ 0 ,T] T , k = i +1,i +2, , p, p Δ (t) ≥−λ  Bk 0 + Ch 0  , t = t k , p  t + k  ≥  1 − L k  p  t k  . (2.27) By Lemma 2.4,wehave p(t) ≥ p  t ∗ 0   t ∗ 0 <t k <t  1 − L k  +  t t ∗ 0  s<t k <t  1 − L k  − λ  Bk 0 + Ch 0  Δs. (2.28) Let t = T in (2.28), then p(T) ≥ λ  t ∗ 0 <t k <T  1 − L k  − λ  Bk 0 + Ch 0   T t ∗ 0  s<t k <T (1 − L k )Δs. (2.29) If p(T) < 0, then (2.29)gives  Bk 0 + Ch 0  >  t ∗ 0 <t k <T  1 − L k   T t ∗ 0  s<t k <T  1 − L k  Δs ≥  0<t k <T  1 − L k   T 0  s<t k <T  1 − L k  Δs , (2.30) which contradicts (2.21), so, we have p(T) ≥ 0, and by (2.22), p(0) ≥ p(T)e −a (T,0) ≥ 0. Hence, 0 <t ∗ 1 <T.Lett j <t ∗ 1 ≤ t j+1 for some j.Wefirstassumethatt ∗ 0 <t ∗ 1 ,soi ≤ j.Let t = t ∗ 1 in (2.28), we have 0 >p  t ∗ 1  ≥ λ  t ∗ 0 <t k <t ∗ 1  1 − L k  +  t ∗ 1 t ∗ 0  s<t k <t ∗ 1  1 − L k  − λ  Bk 0 + Ch 0  Δs, (2.31) Y. Li and H. Z hang 7 which gives  Bk 0 + Ch 0  >  t ∗ 0 <t k <t ∗ 1  1 − L k   t ∗ 1 t ∗ 0  s<t k <t ∗ 1  1 − L k  Δs ≥  0<t k <T  1 − L k   T 0  s<t k <T  1 − L k  Δs , (2.32) which contradicts (2.21). Next we assume that t ∗ 1 <t ∗ 0 .Soj ≤ i.Fort ∈ J T , k = 1,2, , p, p Δ (t) ≥−λ  Bk 0 + Ch 0  , t = t k , p  t + k  ≥  1 − L k  p  t k  . (2.33) By Lemma 2.4,wehave p(t) ≥ p(0)  0<t k <t  1 − L k  +  t 0  s<t k <t  1 − L k  − λ  Bk 0 + Ch 0  Δs. (2.34) Let t = t ∗ 1 in (2.34), then 0 >p  t ∗ 1  ≥ p(0)  0<t k <t ∗ 1  1 − L k  − λ  Bk 0 + Ch 0   t ∗ 1 0  s<t k <t ∗ 1  1 − L k  Δs, (2.35) which implies p(0)  0<t k <t ∗ 1 (1 − L k ) <λ  Bk 0 + Ch 0   t ∗ 1 0  s<t k <t ∗ 1  1 − L k  Δs. (2.36) By (2.22), we obtain λ  Bk 0 + Ch 0   t ∗ 1 0  s<t k <t ∗ 1  1 − L k  Δs>e (−a) (T,0)p(T)  0<t k <t ∗ 1  1 − L k  . (2.37) From (2.29), (2.37), we have λ  Bk 0 + Ch 0   t ∗ 1 0  s<t k <t ∗ 1  1 − L k  Δs >e (−a) (T,0)  0<t k <t ∗ 1  1 − L k   λ  t ∗ 0 <t k <T  1 − L k  − λ  Bk 0 + Ch 0   T t ∗ 0  s<t k <T  1 − L k  Δs  (2.38) or  0<t k <t ∗ 1  1 − L k   t ∗ 0 <t k <T  1 − L k  <  Bk 0 + Ch 0   0<t k <t ∗ 1  1 − L k   T t ∗ 0  s<t k <T  1 − L k  Δs +  Bk 0 + Ch 0  e (−a) (T,0)  t ∗ 1 0  s<t k <t ∗ 1  1 − L k  Δs. (2.39) 8 Boundary Value Problems Hence   0<t k <T  1 − L k   2 ≤  0<t k <t ∗ 1  1 − L k   t ∗ 0 <t k <T  1 − L k   0<t k <T  1 − L k  <  Bk 0 + Ch 0   0<t k <t ∗ 1  1 − L k   0<t k <T  1 − L k   T t ∗ 0  s<t k <T  1 − L k  Δs +  Bk 0 + Ch 0  e (−a) (T,0)  0<t k <T  1 − L k   t ∗ 1 0  s<t k <t ∗ 1  1 − L k  Δs <  Bk 0 + Ch 0  e (−a) (T,0)  T 0  s<t k <T  1 − L k  Δs, (2.40) which contradicts (2.21). Thus the proof of Lemma 2.5 is complete.  For any δ(t) ∈ PC[J T ,R]andη ∈ Ω, a,b,c ∈ C[J T ,R + ], a is not identically vanishing, and 0 ≤ L k < 1(k = 1,2, , p), I k ∈ C[R,R](k = 1,2, , p), we consider the linear peri- odic boundary value problem for a linear impulsive integrodifferential equation(PBVP): u Δ (t)+a(t)u(t) =−b(t)[Tu](t) − c(t)[Su](t)+δ(t), t = t k , t ∈ J T , u  t + k  − u  t k  =− L k u  t k  + I k  η  t k  + L k η  t k  , k = 1,2 , p, u(0) = u(T). (2.41) Lemma 2.6. u ∈ Ω is a solution of PBVP (2.41)ifandonlyifu ∈ PC[J T ,R] is a solution of the following impulsive integral equation: u(t) =  T 0 G(t,s)  δ(s) − b(s)[Tu](s) − c(s)[Su](s)  Δs +  0<t k <T G  t,t k  e (−a)  σ  t k  ,t k  − L k u  t k  + I k  η  t k  + L k η  t k  , t ∈ J T , (2.42) where G(t,s) = 1 1 − e (−a) (T,0) ⎧ ⎪ ⎨ ⎪ ⎩ e (−a)  t,σ(s)  ,0≤ s<t≤ T, e (−a) (T,0)e (−a)  t,σ(s)  ,0≤ t ≤ s ≤ T. (2.43) Proof. Assume that u ∈ Ω is a solution of (2.41). For the first equation of (2.41), using Lemma 2.3 on t ∈ [0,t 1 ] T ,wehave u(t) = e (−a) (t,0)u(0) +  t 0 e (−a)  t,σ(s)  δ(s) − b(s)[Tu](s) − c(s)[Su](s)  Δs. (2.44) Y. Li and H. Z hang 9 Then u  t 1  = e (−a)  t 1 ,0  u(0) +  t 1 0 e (−a)  t 1 ,σ(s)  δ(s) − b(s)[Tu](s) − c(s)[Su](s)  Δs. (2.45) Again using Lemma 2.3 on t ∈ (t 1 ,t 2 ] T ,then u(t) = u  t + 1  e (−a)  t,t 1  +  t t 1 e (−a)  t,σ(s)  δ(s) − b(s)[Tu](s) − c(s)[Su](s)  Δs = u  t 1  e (−a)  t,t 1  +  t t 1 e (−a)  t,σ(s)  δ(s) − b(s)[Tu](s) − c(s)[Su](s)  Δs + e (−a)  t,t 1  − L 1 u  t 1  + I 1  η  t 1  + L 1 η  t 1  = e (−a) (t,0)u(0) +  t 0 e (−a)  t,σ(s)  δ(s) − b(s)[Tu](s) − c(s)[Su](s)  Δs + e (−a)  t,t 1  − L 1 u  t 1  + I 1  η  t 1  + L 1 η  t 1  . (2.46) Repeating the above procession for t ∈ J T ,wehave u(t) = u(0)e (−a) (t,0)+  t 0 e (−a)  t,σ(s)  δ(s) − b(s)[Tu](s) − c(s)[Su](s)  Δs +  0<t k <t e (−a)  t,t k  − L k u  t k  + I k  η  t k  + L k η  t k  . (2.47) Setting t = T in (2.47) and using the boundary condition u(0) = u(T), we obtain u(0) = 1 1 − e (−a) (T,0)   T 0 e (−a)  T,σ(s)  δ(s) − b(s)[Tu](s) − c(s)[Su](s)  Δs +  0<t k <T e (−a)  T,t k  − L k u  t k  + I k  η  t k  + L k η  t k   . (2.48) Substituting (2.48)into(2.47), we see that u ∈ PC[J T ,R] satisfies (2.42). If u ∈ PC[J T ,R]isasolutionof(2.42), then u ∈ C 1 (J  T ,R)and u Δ (t)+a(t)u(t) =−b(t)[Tu](t) − c(t)[Su](t)+δ(t), t = t k , t ∈ J T , u  t + k  − u  t k  =− L k u  t k  + I k  η  t k  + L k η  t k  , k = 1,2 , p. (2.49) Setting t = 0,T in (2.42), respectively, we have u(T) = 1 1 − e (−a) (T,0)   T 0 e (−a)  T,σ(s)  δ(s) − b(s)[Tu](s) − c(s)[Su](s)  Δs +  0<t k <T e (−a)  T,t k  − L k u  t k  + I k  η  t k  + L k η  t k   = u(0). (2.50) Therefore, u ∈ Ω is a solution of (2.41). Thus Lemma 2.6 is proved.  10 Boundary Value Problems Lemma 2.7. Assume that a,b,c ∈ C[J T ,R + ] and 0 ≤ L k < 1(k = 1,2, , p), I k ∈ C[R,R] (k = 1,2 , p), δ ∈ PC[J T ,R], η ∈ Ω, and the following inequality holds: 1 1 − e (−a) (T,0)   T 0  k 0 sb(s)+Th 0 c(s)  Δs + p  k=1 L k  < 1. (2.51) Then PBVP (2.41)possessesauniquesolutioninΩ. Proof. For any u ∈ Ω, consider the operator F defined by the formula (Fu)(t) =  T 0 G(t,s)  δ(s) − b(s)[Tu](s) − c(s)[Su](s)  Δs +  0<t k <T G  t,t k  e (−a)  σ  t k  ,t k  − L k u  t k  + I k  η  t k  + L k η  t k  , t ∈ J T . (2.52) Then Fu ∈ Ω, that is, FΩ ⊂ Ω. For every u,v ∈ Ω,t ∈ J T ,wehave   (Fu)(t) − (Fv)(t)   ≤  T 0 G(t,s)  b(s)   [Tu](s) − [Tv](s)   + c(s)   [Su](s) − [Sv](s)    Δs +  0<t k <T G  t,t k  e (−a)  σ  t k  ,t k  L k   u  t k  − v  t k    < 1 1 − e (−a) (T,0)   T 0  k 0 sb(s)+Th 0 c(s)  Δs + p  k=1 L k   u − v PC . (2.53) Hence Fu− Fv PC = sup t∈J T   (Fu)(t) − (Fv)(t)   ≤ αu − v PC , (2.54) where α = 1 1 − e (−a) (T,0)   T 0  k 0 sb(s)+Th 0 c(s)  Δs + p  k=1 L k  < 1. (2.55) Thus the operator F is a contraction on Ω. That is, there is a unique element u ∈ Ω such that u = Fu. Therefore, u is the unique solution of PBVP (2.41). The proof of Lemma 2.7 is complete.  Lemma 2.8. u ∈ Ω is a solution of PBVP (1.1)ifandonlyifu ∈ PC[J T ,R] is solution of the following integral equation: u(t) =  T 0 G(t,s)  f  s,u(s),[Tu](s),[Su](s)  + a(s)u(s)  Δs +  0<t k <1 G  t,t k  e (−a)  σ  t k  ,t k  I k  u  t k  , (2.56) [...]... 2, pp 119–132, 2005 [7] X Liu and D Guo, Periodic boundary value problems for impulsive integro-differential equations of mixed type in Banach spaces,” Chinese Annals of Mathematics Series B, vol 19, no 4, pp 517–528, 1998 [8] G S Ladde and S Sathananthan, Periodic boundary value problem for impulsive integrodifferential equations of Volterra type,” Journal of Mathematical and Physical Sciences, vol... [9] L H Erbe and D J Guo, Periodic boundary value problems for second order integrodifferential equations of mixed type,” Applicable Analysis, vol 46, no 3-4, pp 249–258, 1992 [10] V Lakshmikantham, D D Ba˘nov, and P S Simeonov, Theory of Impulsive Differential Equaı tions, vol 6 of Series in Modern Applied Mathematics, World Scientific, Teaneck, NJ, USA, 1989 16 Boundary Value Problems [11] W G Kelley... Mathematics, vol 18, no 1-2, pp 18–56, 1990 [4] J Henderson, “Double solutions of impulsive dynamic boundary value problems on a time scale,” Journal of Difference Equations and Applications, vol 8, no 4, pp 345–356, 2002 [5] M Benchohra, J Henderson, S K Ntouyas, and A Ouahab, “On first order impulsive dynamic equations on time scales,” Journal of Difference Equations and Applications, vol 10, no 6, pp 541–548,... (3.17) that is, u∗ and v∗ are the minimal and maximal solutions of the PBVP (1.1) in the interval [u0 ,v0 ] The proof of Theorem 3.1 is complete Acknowledgments The authors thank the referees for their many thoughtful suggestions that lead to an improved exposition of manuscript This work is supported by the National Natural Sciences Foundation of China References [1] M Bohner and A Peterson, Dynamic... ≤ T, 0 ≤ t ≤ s ≤ T (2.57) The proof of Lemma 2.8 is similar to that of Lemma 2.6 and we will omit it here 3 Main results In this section, we will use the monotone iterative technique to prove the existence of minimal and maximal solutions of the PBVP (1.1) Theorem 3.1 Assume that the following conditions hold (H1 ) There exist functions u0 ,v0 ∈ Ω, u0 (t) ≤ v0 (t) for all t ∈ JT such that uΔ (t) ≤... 0 . Corporation Boundary Value Problems Volume 2007, Article ID 73176, 16 pages doi:10.1155/2007/73176 Research Article Extremal Solutions of Periodic Boundary Value Problems for First-Order Impulsive. consider the existence of minimal and maximal solutions of periodic boundary value problems for first-order impulsive integrodifferential equations of mixed-type on time scales by establishing a comparison. 119–132, 2005. [7] X. Liu and D. Guo, Periodic boundary value problems for impulsive integro-differential equa- tions of mixed type in Banach spaces,” Chinese Annals of Mathematics. Series B, vol. 19,