International Journal of Analysis and Applications ISSN 2291-8639 Volume 5, Number 1 (2014), 56-67 http://www.etamaths.com INTEGRAL BOUNDARY VALUE PROBLEMS FOR FRACTIONAL IMPULSIVE INTEGRO DIFFERENTIAL EQUATIONS IN BANACH SPACES A. ANGURAJ1, M. KASTHURI2 AND P. KARTHIKEYAN 3,∗ Abstract. We study in this paper,the existence of solutions for fractional integro differential equations with impulsive and integral conditions by using fixed point method. We establish the Sufficient conditions and unique solution for given problem. An Example is also explained to the main results. 1. Introduction In the seventeenth century, Fractional calculus was originated and it has gained much attention in recent years by many researchers. Fractional differential equa- tions appears in a large number of fields of science and engineering, thermodynam- ics, elasticity, wave propagation, electric railway systems, telecommunication lines and also in chemistry, analysing kinetical reaction problems (see [1, 5, 6, 13, 15, 16]). Integral and anti-periodic boundary value conditions can be seen in models of a variety of physical, economic and biological processes, and they have been studied extensively in recent years (see [8, 9, 10, 11] ) and related references therein for boundary value problems with integral boundary conditions [1, 2, 3, 6]. In [14], the authors have studied the impulsive problems for fractional differential equations with boundary value conditions. J.R. Wang et al. in [7] discussed the ex- istence results for the boundary value problems for impulsive fractional differential equations. The authors in [17] proved the existence of solutions for multi-point non- linear differential equations of fractional orders with integral boundary conditions without impulsive conditions. Inspired by the above works, we consider the existence and uniqueness of solu- tions for impulsive fractional differential equations with integral boundary condi- tions Dα0+u(t) = f(t,u(t),Bu(s)), 1 < α ≤ 2,(1.1) t ∈ J ′ = J\{t1, ..., tm} ,J := [0,T], T > 0, u(t+k ) = u(t − k ) + yk, k = 1, 2, ...,m yk ∈ X,(1.2) I2−α0+ u(t)|t=0 = 0, D α−2 0+ u(T) = m∑ i=1 aiI α−1 0+ u(ξi),(1.3) 2000 Mathematics Subject Classification. 26A33, 34A37, 34K05, 34B15. Key words and phrases. Fractional integro-differential equations, Boundary value problem, Impulsive condition, fixed point theorem. c©2014 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 56 INTEGRAL BOUNDARY VALUE PROBLEMS 57 where Bu(s) = ∫ t 0 k(t,s,u(s))ds, 0 < ξi < T , T > 0, ai ∈ X, m ≥ 2, Dα0+ and Iα 0+ are the standard Riemann-Liouville fractional derivative and fractional integral respectively, f : J ×J ×X → X, k : J ×J ×X → X are jointly continuous and tk satisfy 0 = t0 < t1 < ... < tm < tm+1 = T, u(t + k ) = lim�→0+ u(tk + �) and u(t−k ) = lim�→0− u(tk + �) represent the right and left limits of u(t) at t = tk. In Section 2, we give definitions of fractional integral and derivative operators, lemma and some fixed point theorems. The main results discussed in section 3. Finally, in section 4, the example is also illustrated. 2. Preliminaries Let E = PC(J,X) = {u : J −→ X : u ∈ C((tk, tk+1],X)}k = 0, ...m,be a Ba- nach space with norm ‖u‖PC = supt∈J‖u(t)‖. and there exist u(t+k ) and u(t − k ),k = 1, 2, .....,m with u(t+k ) = u(t − k ), Set J ′ = [0,T]\{t1, t2, ....tm}. Theorem 2.1 ([12]). (Schaefer’s fixed point theorem) Let X be a Banach space. Assume that T : X → X is a completely continuous operator and the set V = {u ∈ X|u = µTu, 0 < µ < 1} is bounded. Then T has a fixed point in X. Theorem 2.2. (PC-Type Ascoli-Arzela Theorem) Let X be a Banach space and W ⊂ PC(J,X). If the following conditions are satisfied: (i): W is uniformly bounded subset of PC(J,X) (ii): W is equicontinuous in (tk, tk+1), k = 0, 1, 2, ...,m where t0 = 0, tm+1 = T; (iii): W(t) = {u(t)|u ∈ W, t ∈ J\{t1, ..., tm}},W(t+k = { u(t+k )|u ∈ W } and W(t−k = { u(t−k )|u ∈ W } is a relatively compact subsets of X. Then W is a relatively compact subsets of PC(J,X). Definition 2.3. The fractional integral of order α > 0 of a function y : (0,∞) → R is given by Iα0+y(t) = 1 Γ(α) ∫ t 0 (t−s)α−1y(s)ds, provided the right side is pointwise defined on (0,∞), where Γ(·) is the Gamma function. Definition 2.4. The fractional derivative of order α > 0 of a function y : (0,∞) → R is given by Dα0+y(t) = 1 Γ(n−α) ( d dt )n ∫ t 0 y(s) (t−s)α−n+1 ds, where n = [α] + 1, provided the right side is pointwise defined on (0,∞). Lemma 2.5. Let α > 0 and u ∈ C(0, 1) ∩ L1(0, 1).Then fractional differential equation Dα0+u(t) = 0 has u(t) = c1t α−1 + c2t α−2 + · · · + cNtα−N, ci ∈ R, N = [α] + 1, as unique solution. Lemma 2.6. Assume that u ∈ C(0, 1) ∩ L1(0, 1) with a fractional derivative of order α > 0 that belongs to C(0, 1) ∩L1(0, 1). Then Iα0+D α 0+u(t) = u(t) + c1t α−1 + c2t α−2 + · · · + cNtα−N, for some ci ∈ R, i = 1, 2, . . . ,N, where N is the smallest integer grater than or equal to α. 58 A. ANGURAJ1, M. KASTHURI2 AND P. KARTHIKEYAN 3,∗ Lemma 2.7 ([8]). Let α > 0, n = [α] + 1. Assume that u ∈ L1(0, 1) with a fractional integration of order n−α that belongs to ACn[0, 1]. Then the equality (Iα0+D α 0+u)(t) = u(t) − n∑ i=1 ((In−α0+ u)(t)) n−i|t=0 Γ(α− i + 1) tα−i holds almost everywhere on [0, 1]. Lemma 2.8 ([8]). (i) Let k ∈ N,α > 0. If Dαa+y(t) and (D α+k a+ y)(t) exist, then (DkDαa+)y(t) = (D α+k a+ y)(t); (ii) If α > 0,β > 0,α + β > 1, then (Iαa+I α a+)y(t) = (I α+β a+ y)(t) satisfies at any point on [a,b] for y ∈ Lp(a,b) and 1 ≤ p ≤∞; (iii) Let α > 0 and y ∈ C[a,b]. Then (Dαa+Iαa+)y(t) = y(t) holds on [a,b]; (iv) Note that for λ > −1,λ 6= α− 1,α− 2, . . . ,α−n, we have Dαtλ = Γ(λ + 1) Γ(λ−α + 1) tλ−α, Dαtα−i = 0, i = 1, 2, . . . ,n Lemma 2.9. For any y(t) ∈ PC(J,X), the linear impulsive fractional boundary- value problem (2.1) Dα0+u(t) = y(t), 1 < α ≤ 2, t ∈ [0,T], u(t+k ) = u(t − k ) + yk, k = 1, 2, ...,m yk ∈ X I2−α0+ u(t)|t=0 = 0, D α−2 0+ u(T) = m∑ i=1 aiI α−1 0+ u(ξi), has unique solution u(t) =   ∫ t 0 (t−s)α−1 Γ(α) y(s)ds + t α−1 Γ(α)(T−A) [ ∑m i=1 ai Γ(2α−1) ∫ ξi 0 (ξi −s)2α−2y(s)ds− ∫T 0 (T −s)y(s)ds ] , for t ∈ [0, t1) y1 + ∫ t 0 (t−s)α−1 Γ(α) y(s)ds + t α−1 Γ(α)(T−A) [ ∑m i=1 ai Γ(2α−1) ∫ ξi 0 (ξi −s)2α−2y(s)ds− ∫T 0 (T −s)y(s)ds ] , for t ∈ (t1, t2) y1 + y2 + ∫ t 0 (t−s)α−1 Γ(α) y(s)ds + t α−1 Γ(α)(T−A) [ ∑m i=1 ai Γ(2α−1) ∫ ξi 0 (ξi −s)2α−2y(s)ds− ∫T 0 (T −s)y(s)ds ] , for t ∈ (t2, t3) ... m∑ i=0 yi + ∫ t 0 (t−s)α−1 Γ(α) y(s)ds + t α−1 Γ(α)(T−A) [ ∑m i=1 ai Γ(2α−1) ∫ ξi 0 (ξi −s)2α−2y(s)ds− ∫T 0 (T −s)y(s)ds ] , for t ∈ (tm,T] where A = ∑m i=1 aiξ 2α−2 i /Γ(2α− 1) and T 6= A. INTEGRAL BOUNDARY VALUE PROBLEMS 59 Step:1 For t ∈ [0, t1] we have By Lemma 2.6. the solution of (2.1) can be written as u(t) = c1t α−1 + c2t α−2 + 1 Γ(α) ∫ t 0 (t−s)α−1y(s)ds. From I2−α0+ u(t)|t=0 = 0, and by Lemmas 2.7 and 2.8, we know that c2 = 0, and Dα−20+ u(t) = c1tΓ(α) + I 2 0+y(t), Iα−10+ u(t) = c1 Γ(α) Γ(2α− 1) t2α−2 + Iα−10+ I α 0+y(t), from Dα−20+ u(T) = ∑m i=1 aiI α−1 0+ u(ξi), we have c1 = 1 Γ(α)(T −A) [ ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2y(s)ds− ∫ T 0 (T −s)y(s)ds ] , where A = ∑m i=1 aiξ 2α−2 i /Γ(2α− 1) and T 6= A, so u(t) = ∫ t 0 (t−s)α−1 Γ(α) y(s)ds + tα−1 Γ(α)(T −A) [ ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2y(s)ds− ∫ T 0 (T −s)y(s)ds ] . Step:2 If t ∈ (t1, t2],with u(t+1 = u(t − 1 ) + y1 then we have u(t) = c1t α−1 + c2t α−2 + u(t+1 ) − 1 Γ(α) ∫ t1 0 (t1 −s)α−1y(s)ds + 1 Γ(α) ∫ t 0 (t−s)α−1y(s)ds, = c1t α−1 + c2t α−2 + u(t−1 ) + y1 − 1 Γ(α) ∫ t1 0 (t1 −s)α−1y(s)ds + 1 Γ(α) ∫ t 0 (t−s)α−1y(s)ds, = c1t α−1 + c2t α−2 + y1 + 1 Γ(α) ∫ t 0 (t−s)α−1y(s)ds. Then, u(t) = y1 + ∫ t 0 (t−s)α−1 Γ(α) y(s)ds + tα−1 Γ(α)(T −A) [ ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2y(s)ds− ∫ T 0 (T −s)y(s)ds ] . Preceding in this way, Step:3 For t ∈ (tm,T], we have u(t) = c1t α−1 + c2t α−2 + m∑ i=1 yi + 1 Γ(α) ∫ t 0 (t−s)α−1y(s)ds. Then, u(t) = m∑ i=1 yi + ∫ t 0 (t−s)α−1 Γ(α) y(s)ds 60 A. ANGURAJ1, M. KASTHURI2 AND P. KARTHIKEYAN 3,∗ + tα−1 Γ(α)(T −A) [ ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2y(s)ds− ∫ T 0 (T −s)y(s)ds ] . The proof is complete. � 3. Main Results In this section, we prove the existence and uniqueness results of the problem (1.1)-(1.3) by using the following assumptions: (H1) There exist positive functions L , such that |f(t,x,u) −f(t,y,v)| ≤ L[|x−y| + |u−v|], ∀t ∈ [0,T], x,y,u,v ∈ X, (H2) The function L satisfies 2L ≤ [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )]−1 + m∑ i=1 yi. (H3) There exists a positive constant L1 such that |f(t,u,v)| ≤ L1 for t ∈ [0,T], u,v ∈ X. Theorem 3.1. Assume that (H1), (H2)are satisfied, then the problem (1.1)–(1.3) has a unique solution. Proof: Choose r ≥ 2M1 [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] + m∑ i=1 yi Then we show that θBr ⊂ Br, where Br = {u ∈ E : ‖u‖ ≤ r}. Let us set supt∈[0,T] |f(t,s, 0)| = M1, Step :1 For t ∈ [0, t1], we have ‖(θu)(t)‖ = ∣∣∣∫ t 0 (t−s)α−1 Γ(α) f(s,u(s),Bu(s))ds + tα−1 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2f(s,u(s),Bu(s))ds − ∫ T 0 (T −s)f(s,u(s),Bu(s))ds )∣∣∣ ≤ [∫ t 0 (t−s)α−1 Γ(α) |f(s,u(s),Bu(s))|ds + tα−1 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2|f(s,u(s),Bu(s))|ds − ∫ T 0 (T −s)|f(s,u(s),Bu(s))|ds )] ≤ [∫ t 0 (t−s)α−1 Γ(α) (|f(s,u(s),Bu(s) −f(σ,s, 0)| + |f(σ,s, 0)|)ds INTEGRAL BOUNDARY VALUE PROBLEMS 61 + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2(|f(s,u(s),Bu(s) −f(σ,s, 0)| + |f(σ,s, 0)|)ds − ∫ T 0 (T −s)(|f(s,u(s),Bu(s) −f(σ,s, 0)| + |f(σ,s, 0)|)ds )] ≤ [ (2Lr + M1) (∫ t 0 (t−s)α−1 Γ(α) ds + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2ds− ∫ T 0 (T −s)ds ))] ≤ (2Lr + M1) [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] ≤ r Taking the maximum over the interval [0, t1], we obtain ‖θ(u)(t)‖≤ r. In view of (H1), for every t ∈ [0, t1], we have ‖(θx)(t) − (θy)(t)‖ = ∣∣∣∫ t 0 (t−s)α−1 Γ(α) (f(t,x) −f(t,y)ds + tα−1 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2(f(t,x,u) −f(t,y,v))ds − ∫ T 0 (T −s)(f(t,x,u) −f(t,y,v))ds )∣∣∣ ≤ [∫ t 0 (t−s)α−1 Γ(α) |(f(t,x,u) −f(t,y,v))|ds + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2|(f(t,x,u) −f(t,y,v))|ds − ∫ T 0 (T −s)|(f(t,x,u) −f(t,y,v))|ds )] ≤ [ L[‖x−y‖ + ‖u−v‖] (∫ t 0 (t−s)α−1 Γ(α) ds + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2ds− ∫ T 0 (T −s)ds ))] ≤ L[‖x−y‖ + ‖u−v‖] [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] = A[‖x−y‖ + ‖u−v‖], where A = L [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] , which depends only on the parameters involved in the problem. As A < 1, θ is contraction mapping for the interval t ∈ [0, t1]. Step :2 For t ∈ (t1, t2], we have ‖(θu)(t)‖ = ∣∣∣∫ t 0 (t−s)α−1 Γ(α) f(s,u(s),Bu(s))ds 62 A. ANGURAJ1, M. KASTHURI2 AND P. KARTHIKEYAN 3,∗ + tα−1 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2f(s,u(s),Bu(s))ds − ∫ T 0 (T −s)f(s,u(s),Bu(s))ds ) + y1 ∣∣∣ ≤ [ (2Lr + M1) (∫ t 0 (t−s)α−1 Γ(α) ds + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2ds− ∫ T 0 (T −s)ds )) + y1 ] ≤ (2Lr + M1) [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] + y1 ≤ r Taking the maximum over the interval (t1, t2], we obtain ‖θ(u)(t)‖≤ r. In view of (H1), for every t ∈ (t1, t2], we have ‖(θx)(t) − (θy)(t)‖ = ∣∣∣∫ t 0 (t−s)α−1 Γ(α) (f(t,x) −f(t,y)ds + tα−1 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2(f(t,x,u) −f(t,y,v))ds − ∫ T 0 (T −s)(f(t,x,u) −f(t,y,v))ds ) + y1 ∣∣∣ ≤ [ L[‖x−y‖ + ‖u−v‖] (∫ t 0 (t−s)α−1 Γ(α) ds + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2ds− ∫ T 0 (T −s)ds )) + y1 ] ≤ L[‖x−y‖ + ‖u−v‖] [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] + y1 = A[‖x−y‖ + ‖u−v‖], where A = L [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] + y1, As A < 1, θ is therefore a contraction in the interval t ∈ (t1, t2]. Preceding in this way, we got Step:3 For t ∈ (tm,T], we have ‖(θu)(t)‖ = ∣∣∣∫ t 0 (t−s)α−1 Γ(α) f(s,u(s),Bu(s))ds + tα−1 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2f(s,u(s),Bu(s))ds INTEGRAL BOUNDARY VALUE PROBLEMS 63 − ∫ T 0 (T −s)f(s,u(s),Bu(s))ds ) + m∑ i=1 yi ∣∣∣ ≤ [ (2Lr + M1) (∫ t 0 (t−s)α−1 Γ(α) ds + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2ds− ∫ T 0 (T −s)ds )) + m∑ i=1 yi ] ≤ (2Lr + M1) [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] + m∑ i=1 yi ≤ r Taking the maximum over the interval (tm,T], we obtain ‖θ(u)(t)‖≤ r. In view of (H1), for every t ∈ (tm,T], we have ‖(θx)(t) − (θy)(t)‖ = ∣∣∣∫ t 0 (t−s)α−1 Γ(α) (f(t,x) −f(t,y)ds + tα−1 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2(f(t,x,u) −f(t,y,v))ds − ∫ T 0 (T −s)(f(t,x,u) −f(t,y,v))ds ) + m∑ i=1 yi ∣∣∣ ≤ [ L[‖x−y‖ + ‖u−v‖] (∫ t 0 (t−s)α−1 Γ(α) ds + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2ds− ∫ T 0 (T −s)ds )) + m∑ i=1 yi ] ≤ L[‖x−y‖ + ‖u−v‖] [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] + m∑ i=1 yi = A[‖x−y‖ + ‖u−v‖], where A = L [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] + m∑ i=1 yi, which depends only on the parameters involved in the problem. Then by Banach fixed point theorem, the operator θ has fixed point in the interval t ∈ (tm,T]. � Theorem 3.2. Assume that (H1)-(H3) are satisfied. Then (1.1)-(1.3) has at least one solution. Proof: 64 A. ANGURAJ1, M. KASTHURI2 AND P. KARTHIKEYAN 3,∗ We define an operator P : PC(E) → PC(E), as (Pu)(t) = ∫ t 0 (t−s)α−1 Γ(α) f(t,u(s),v(s))ds + tα−1 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2f(t,u(s),v(s))ds − ∫ T 0 (T −s)f(t,u(s),v(s))ds ) + m∑ i=1 yi. (3.1) To show that the operator P is completely continuous. Clearly, continuity of the operator P follows from the continuity of f. Let Ω ⊂ E be bounded. Then, ∀u,v ∈ Ω together with (H3) we obtain (Pu)(t) ≤ ∫ t 0 (t−s)α−1 Γ(α) |f(t,u(s),v(s))|ds + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2|f(t,u(s),v(s))|ds − ∫ T 0 (T −s)|f(t,u(s),v(s))|ds ) + m∑ i=1 yi ≤ L1 [∫ t 0 (t−s)α−1 Γ(α) ds + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2ds− ∫ T 0 (T −s)ds )] + m∑ i=1 yi ≤ L1 [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 Γ(2α) − T 2 2 )] + m∑ i=1 yi, which implies ‖Pu‖≤ L1 [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 Γ(2α) − T 2 2 )] + m∑ i=1 yi < ∞. Hence, P(Ω) is uniformly bounded. For any s1,s2 ∈ [0, t1],u ∈ Ω, we have |(Pu)(s1) − (Pu)(s2)| = ∣∣∣∫ s1 0 (s1 −s)α−1 Γ(α) f(s,u(s),v(s))ds + sα−11 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2f(s,u(s),v(s))ds − ∫ T 0 (T −s)f (s,u(s))ds ) − ∫ s2 0 (s2 −s)α−1 Γ(α) f(s,u(s),v(s))ds− tα−12 Γ(α)(T −A) × ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2f(s,u(s),v(s))ds− ∫ T 0 (T −s)f(s,u(s),v(s))ds )∣∣∣ ≤ L1 ∣∣∣∫ s1 0 (s1 −s)α−1 − (s2 −s)α−1 Γ(α) ds INTEGRAL BOUNDARY VALUE PROBLEMS 65 + tα−11 −s α−1 2 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2ds − ∫ T 0 (T −s)ds ) − ∫ s2 s1 (s2 −s)α−1 Γ(α) ds ∣∣∣ ≤ L1 [∣∣∣∫ s1 0 (s1 −s)α−1 − (s2 −s)α−1 Γ(α) ds− ∫ s2 s1 (s2 −s)α−1 Γ(α) ds ∣∣∣ + ∣∣∣sα−11 −sα−12 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2ds− ∫ T 0 (T −s)ds )∣∣∣] → 0 as s1 → s2. Thus, by the PC-type Arzela-Ascoli theorem, P(Ω) is equicontinuous. Consequent- ly, the operator P is compact. Next, we consider the set S = {u ∈ E : u = µPu, 0 < µ < 1}, and show that it is bounded. Let u ∈ S; then u = µPu, 0 < µ < 1. For any t ∈ [0,T], we have u(t) = ∫ t 0 (t−s)α−1 Γ(α) f(t,u(s),v(s))ds + tα−1 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2f(t,u(s),v(s))ds − ∫ T 0 (T −s)f(t,u(s),v(s))ds ) + m∑ i=1 yi, and |u(t)| = µ|Pu| ≤ ∫ t 0 (t−s)α−1 Γ(α) |f(t,u(s),v(s))|ds + tα−1 Γ(α)(T −A) ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2|f(t,u(s),v(s))|ds − ∫ T 0 (T −s)|f(t,u(s),v(s))|ds ) + m∑ i=1 |yi| ≤ L1 [∫ t 0 (t−s)α−1 Γ(α) ds + tα−1 Γ(α)|T −A| ( ∑m i=1 ai Γ(2α− 1) ∫ ξi 0 (ξi −s)2α−2ds− ∫ T 0 (T −s)ds )] + m∑ i=1 |yi| ≤ max t∈[0,T] { L1 [ |tα| Γ(α + 1) + |tα−1| Γ(α)|T −A| (∑m i=1 aiξ 2α−1 Γ(2α) − T 2 2 )] + m∑ i=1 |yi| } = M∗. Thus, ‖u‖ ≤ M∗. So, the set S is bounded. Thus, by the conclusion of Theorem 3.1, the operator P has at least one fixed point, which implies that (1.1)-(1.2) has at least one solution. � 66 A. ANGURAJ1, M. KASTHURI2 AND P. KARTHIKEYAN 3,∗ 4. Example Consider the impulsive fractional integro differential equation cD 3 2 u(t) = et|u(t)| (9 + et)(1 + |u(t)|) + ∫ t 0 e−(s−t) 10 |u(s)|ds,(4.1) t ∈ J = [0, 2], t 6= 1 2 ,(4.2) y( 1 2 + ) = |u( 1 2 − )| 3 + |u( 1 2 − )| ,(4.3) I2−α0+ u(t)|t=0 = 0, D α−2 0+ u(T) = m∑ i=1 aiI α−1 0+ u(ξi),(4.4) where f(t,u,Bu) = e tu (9+et)(1+u) + Bu(t),a1 = 2,a2 = 3,ξ1 = 1/2,ξ2 = 1/3,T = 2 we have A = ∑m i=1 aiξ 2α−2 i /Γ(2α− 1) = 1 6= T = 1. Clearly, L = 1/10 as |f(t,x,u) −f(t,y,v)| ≤ 1 10 [|x−y| + |u−v|]. Further, L [ Tα Γ(α + 1) + Tα−1 Γ(α)|T −A| (∑m i=1 aiξ 2α−1 i Γ(2α) − T 2 2 )] + m∑ i=1 yi ≈ 0.01058612753 < 1. Thus, all the assumptions of Theorem 3.1 are satisfied and hence the problem (4.1)-(4.4) has unique solution. References [1] A. Anguraj, P. Karthikeyan, and G. M. NGuérékata; Nonlocal Cauchy problem for some fractional abstract integrodifferential equations in Banach space, Communications in Math- ematical Analysis , vol.55, no. 6, pp. 1?, 2009. [2] A. Anguraj, P. Karthikeyan and J.J. Trujillo; Existence of Solutions to Fractional Mixed Integrodifferential Equations with Nonlocal Initial Condition, Advances in Difference Equa- tions,Volume 2011, Article ID 690653,12pages, doi:10.1155/2011/690653 [3] B. Ahmad, J. J. Nieto; Existence Results for Nonlinear Boundary Value Problems of Frac- tional Integrodifferential Equations with Integral Boundary Conditions, Bound. Value Prob- l.(2009) Art. ID 708576, 11 pp.. [4] B. Ahmad, A. Alsaedi; Existence of approximate solutions of the forced Duffing equation with discontinuous type integral boundary conditions, Nonlinear Analysis, 10 (2009) 358-367. [5] C. Bai; Positive solutions for nonlinear fractional differential equations with coefficient that changes sign Nonlinear Analysis: Theory, Methods and Applications, 64 (2006) 677-685. [6] Z. Hu, W. Liu; Solvability for fractional order boundary value problem at resonance, Bound- ary value problem, 20(2011)1-10. [7] J. R Wang, Y. Z. and M. Feckan; On recent developments in the theory of boundary value problems for impulsive fractional differential equations, Computers and mathematics with Applications, 64(2012) 3008-3020. [8] A. A. Kilbas, H. M. Srivastava, J. J. Trujillo; Theory and Applications of Fractional Differen- tial Equations, North-Holland Mathematics Studies, 204. Elsevier Science B.V., Amsterdam, 2006. [9] V. Lakshmikantham, S. Leela, J. Vasundhara Devi; Theory of Fractional Dynamic Systems, Cambridge Academic Publishers, Cambridge, 2009. [10] J. Sabatier, O. P. Agrawal, J. A. T. Machado (Eds.); Advances in Fractional Calculus: Theoretical Developments and Applications in Physics and Engineering, Springer, Dordrecht, 2007. INTEGRAL BOUNDARY VALUE PROBLEMS 67 [11] S. G. Samko, A. A. Kilbas, O. I. Marichev; Fractional Integrals and Derivatives: Theory and Applications, Gordon and Breach, New York, NY, USA, 1993. [12] D. R. Smart; Fixed Point Theorems, Cambridge University Press, 1980. [13] X. Su; Boundary value problem for a coupled system of nonlinear fractional differential equations, Applied Mathematics Letters, 22 (2009) 64-69. [14] T.L. Guo and W. Jiang, Impulsive problems for fractional differential equations with bound- ary value conditions, Computers and mathematics with Applications, 64(2012) 3281-3291. [15] G. Wang, W. Liu; The existence of solutions for a fractional 2m-point boundary value prob- lems, Journal of Applied Mathematics. [16] G. Wang, W. Liu; Existence results for a coupled system of nonlinear fractional 2m-point boundary value problems at resonance, Advances in difference equations,doi:10.1186/1687- 1847-2011-44. [17] G.Wang, W. Liu, C. Ren, Existence Of Solutions For Multi-Point Nonlinear Differential Equations Of Fractional Orders With Integral Boundary Conditions , Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 54, pp. 1?0. 1Department of Mathematics, PSG College of Arts and Science, Coimbatore, TN, India 2Department of Mathematics, P.K.R. Arts College for Women, Gobichettipalayam, TN, India 3Department of Mathematics, KSR College of Arts and Science, Tiruchengode, TN, India ∗Corresponding author