# Existence of Solutions to Anti-Periodic Boundary Value Problem for Nonlinear Fractional Differential Equations with ImpulsesAcademic research paper on "Mathematics"

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## Academic research paper on topic "Existence of Solutions to Anti-Periodic Boundary Value Problem for Nonlinear Fractional Differential Equations with Impulses"

﻿Hindawi Publishing Corporation Advances in Difference Equations Volume 2011, Article ID 915689,17 pages doi:10.1155/2011/915689

Research Article

Existence of Solutions to Anti-Periodic Boundary Value Problem for Nonlinear Fractional Differential Equations with Impulses

Anping Chen1,2 and Yi Chen2

1 Department of Mathematics, Xiangnan University, Chenzhou, Hunan 423000, China

2 School of Mathematics and Computational Science, Xiangtan University, Xiangtan, Hunan 411005, China

Correspondence should be addressed to Anping Chen, chenap@263.net Received 20 October 2010; Revised 25 December 2010; Accepted 20 January 2011 Academic Editor: Dumitru Baleanu

Copyright © 2011 A. Chen and Y. Chen. 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.

This paper discusses the existence of solutions to antiperiodic boundary value problem for nonlinear impulsive fractional differential equations. By using Banach fixed point theorem, Schaefer fixed point theorem, and nonlinear alternative of Leray-Schauder type theorem, some existence results of solutions are obtained. An example is given to illustrate the main result.

1. Introduction

In this paper, we consider an antiperiodic boundary value problem for nonlinear fractional differential equations with impulses

CDau(t) = f (t,u(t)), t e [0,T], t = tk, k = 1,2,...,p, Au\t=tk = Ik (u(tk)), Au'\t=tk = Jk (u(tk)), k = 1,2,...,p, (1.1)

u(0) + u(T) = 0, u'(0) + u'(T) = 0,

where T is a positive constant, 1 < a < 2, CDa denotes the Caputo fractional derivative of order a, f e C([0,T] x R,R), Ik, Jk : R ^ R and {tk} satisfy that 0 = t0 < h < ti < ••• <tp < tp+1 = T, Au\t=tk = u(t+) - u(t-), Au'\t=tk = u'(t+) - u'(tk), u(t+) and u(tk) represent the right and left limits of u(t) at t = tk.

Fractional differential equations have proved to be an excellent tool in the mathematic modeling of many systems and processes in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control,

electromagnetic, porous media, and so forth. In consequence, the subject of fractional differential equations is gaining much importance and attention (see [1-6] and the references therein).

The theory of impulsive differential equations has found its extensive applications in realistic mathematic modeling of a wide variety of practical situations and has emerged as an important area of investigation in recent years. For the general theory of impulsive differential equations, we refer the reader to [7,8]. Recently, many authors are devoted to the study of boundary value problems for impulsive differential equations of integer order, see

Very recently, there are only a few papers about the nonlinear impulsive differential equations and delayed differential equations of fractional order.

Agarwal et al. in [13] have established some sufficient conditions for the existence of solutions for a class of initial value problems for impulsive fractional differential equations involving the Caputo farctional derivative. Ahmad et al. in [14] have discussed some existence results for the two-point boundary value problem involving nonlinear impulsive hybrid differential equation of fractional order by means of contraction mapping principle and Krasnoselskii's fixed point theorem. By the similar way, they have also obtained the existence results for integral boundary value problem of nonlinear impulsive fractional differential equations (see [15]). Tian et al. in [16] have obtained some existence results for the three-point impulsive boundary value problem involving fractional differential equations by the means of fixed points method. Maraaba et al. in [17,18] have established the existence and uniqueness theorem for the delay differential equations with Caputo fractional derivatives. Wang et al. in [19] have studied the existence and uniqueness of the mild solution for a class of impulsive fractional differential equations with time-varying generating operators and nonlocal conditions.

To the best of our knowledge, few papers exist in the literature devoted to the antiperiodic boundary value problem for fractional differential equations with impulses. This paper studies the existence of solutions of antiperiodic boundary value problem for fractional differential equations with impulses.

The organization of this paper is as follows. In Section 2, we recall some definitions of fractional integral and derivative and preliminary results which will be used in this paper. In Section 3, we will consider the existence results for problem (1.1). We give three results, the first one is based on Banach fixed theorem, the second one is based on Schaefer fixed point theorem, and the third one is based on the nonlinear alternative of Leray-Schauder type. In Section 4, we will give an example to illustrate the main result.

2. Preliminaries

In this section, we present some basic notations, definitions, and preliminary results which will be used throughout this paper.

Definition 2.1 (see [4]). The Caputo fractional derivative of order a of a function f : [0, to) ^ R is defined as

[9-12].

n - 1 < a <n, n = [a] + 1, (2.1)

where [a] denotes the integer part of the real number a.

Definition 2.2 (see [4]). The Riemann-Liouville fractional integral of order a> 0 of a function f (t), t > 0, is defined as

provided that the right side is pointwise defined on (0, to).

Definition 2.3 (see [4]). The Riemann-Liouville fractional derivative of order a > 0 of a continuous function f : (0, to) ^ R is given by

where n = [a] + 1 and [a] denotes the integer part of real number a, provided that the right side is pointwise defined on (0, to).

For the sake of convenience, we introduce the following notation.

Let J = [0,T],J0 = [0,t1],Jt = (ti,ti+1],i = 1,2.....p-1J = (tp,T]. J' = J\{tx,t2.....tp).

We define PC (J) = {u : [0,T] ^ R | u e C(J'),u(t+) and u(t-) exists, and u(tk) = u(tk), 1 < k < p}. Obviously, PC(J) is a Banach space with the norm ||u|| = supteJ|u(t)|.

Definition 2.4. A function u e PC(J) is said to be a solution of (1.1) if u satisfies the equation cDau(t) = f (t,u(t)) for t e J', the equations Au|t=ik = Ik(u(tk)), Au'|t=ik = Jk(u(tk)),k = 1,2,...,p, and the condition u(0) + u(T) = 0, u'(0) + u'(T) = 0.

Lemma 2.5 (see [20]). Let a > 0; then

for some c e R, i = 0,1,2,...,n- 1, n = [a]+ 1.

Lemma 2.6 (nonlinear alternative of Leray-Schauder type [21]). Let E be a Banach space with C c E closed and convex. Assume that U is a relatively open subset of C with 0 e U and A : U ^ C is continuous, compact map. Then either

(1) A has a fixed point in U, or

(2) there exists u e dU and 1 e (0,1) with u = XAu.

Lemma 2.7 (Schaefer fixed point theorem [22]). Let S be a convex subset of a normed linear space Q. and 0 e S.Let F : S ^ S be a completely continuous operator, and let

IaCDau(t) = u(t) + C0 + C1t + C2t2 + ••• + Cn-1tn-1,

Z(F) = {u e S : u = 1Fu, for some 0 < 1 < 1}.

Then either Z(F) is unbounded or F has a fixed point.

Lemma 2.8. Assume that y e C([0,T],R), T > 0, 1 < a < 2. A function u e PC(J) is a solution of the antiperiodic boundary value problem

CDau(t)= y(t), t e [0,T], t/tk, k = 1,2,...,p,

Au\t=tk = Ik(u(tk)), Au'\t=tk = Jk(u(tk)), k = 1,2.....p,

u(0) + u(T) = 0, u'(0) + u'(T) = 0,

if and only if u is a solution of the integral equation

u(t) =

rO)J c

(t - s)a-ly(s)ds -1

1 p+1 t

1 v-i I 1

r(a)J (,v ' 2T(a) t

fti fti i=\J ti-1

- s)a-ly(s)ds

2r(a - 1)£

p fti Z(T - ti)\ (

i=1 J ti-1

(ti - s)a-2y(s)ds

p+1 ti 1 2 (ti - s)a-2y(s)ds - (T - ti)Ji(u(ti)) i=1J ti-1 2 i=1

4r(a - 1) ^J u T 2t p 1 p

+—7—YiJi(u(ti)) - ^Ii(u(t^)), t e [0,t1],

t 1 k ti (t - s)a-1y(s)ds + (tr - s)a-1 y(s)ds

J tk r (a) i=\J ti-1

p+1 ti 1 k gt (ti - s)a-1y(s)ds + ^--1)g(t - ti)

x I (ti - s)a-2y(s)ds

2Y(a -1)

p ti Z(T - ti)\ (

i=1 J ti-1

(ti - s)a-2y(s)ds

p+1 ti k ^ (ti - s)a-2y(s)ds + £(t - ti)Ji(u(ti)) i=\J ti-1 i=1

4r(a - 1) tV ti-1 i=1

1 p T 2t p

- (T - ti)Ji(u(ti)) + --Jufr))

■ i=1

+JIi(u(ti)) - Ii(u(ti)), t e (tk,tk+1], 1 < k < p.

■ i=1

Proof. Assume that y satisfies (2.6). Using Lemma 2.5, for some constants c0, c\ e R, we have

u(t)= Iay(t) - C0 - c\t = —■ \ (t - s)a-1y(s)ds - C0 - c\t, t e [0,h]. (2.8)

r(a) J 0

Advances in Difference Equations Then, we obtain

(2.10)

u'(t) = ™-TT (t - s)a-2y(s)ds - c\, t e [0,t1].

r (a - i) j0

If t e (ti, t2], then we have

u(t) = r(a) J (t - s)a-1y(s)ds - d0 - d1(t - t1), u'(t) = nahrjit (t - s)a-2y(s)ds - d1,

where d0, d1 e R are arbitrary constants. Thus, we find that

u(t-) = t^t (t1 - s)a-1y(s)ds - C0 - C1t1, r(a) J 0

u(t+) = -do, 1 rf1

u'(t-) = r(a - 1J (t1 - s)a-2y(s)ds - C1,

u (t+) = -d1.

In view of Au^ = u(t+) - u(t-) = I1 (u(t1)) and Au'|i=i1 = u'(t+) - u'(t-) = J1(u(t1)), we have

-d0 = ft^t (t1 - s)a-1y(s)ds - C0 - C1t1 + h(u(h)), r(a) J 0

1 Ch _?

-d1 = w-TT (t1 - s)a 2y(s)ds - C1 + J1(u(t1)).

i(a - 1) J0

Hence, we obtain

(2.11)

(2.12)

1 1 1 P

u(t) = f(a) J t (t - s)a-1y(s)ds + ra Jo (t1 - s)a-1y(s)ds

f1 (t1 - s)a-2y(s)ds + (t - t1)J1(u(t1))

+ f (, - „y^.d .(, , )T )) (2.13)

r(a -1) J0

+ I1(u(t1)) - C0 - C1t, t e (t1,t2].

Repeating the process in this way, the solution u(t) for t e (tk, tk+i] can be written as

1 r 1 k r'

u(t) = r^y Jt (t - s)a-1y(s)ds + — g Jt (ti - s)a-1y(s)ds

1 k A' k

+ -- U)\ (ti - s)a-2y(s)ds + £(t - ti)Ji(u(U)) (2.14)

r (a 1) i=i J ti-i i=i

+^ Ii(u(ti)) - Co - cit, t e (tk,tk+i], k = 1,2,...,p.

On the other hand, by (2.14), we have

1 (T 1 P Ai

u(T) = ray K (T - s)a-1y(s)ds + m § J^- s)a-1y(s)ds

1 P ti P

+ rK-nZT - ti)\ (ti - s)a-2y(s)ds + £(T - ti )Ji(u(ti)) i(a 1) i=1 J ti-1 i=1

+ XIi(u(ti)) - Co - C1T, (2.15)

1 T 1 P ti

w(T ) = na-T) I(T - s)a-2y(s)ds+ra-T) g L- s)a-2y(s)ds

+ XJi(u(ti)) - C1.

By the boundary conditions u(0) + u(T) = 0, u'(0) + u'(T) = 0, we obtain

1 P+1 Ai 1 p fti

Co = 2m hl (ti- s)a-1y(s)ds + mo-T) g(T - ti)\ hi (ti- s)a-2y(s)ds

T P+1 fti t P

X (ti - s)a-2y(s)ds - Ji(u(ti))

i=1 J ti-1 4 i=1

4r(a - 1) t« ' 4

1 P 1 P

+ (T - ti)Ji(u(ti)) + -XIi(u(ti)), i=1 i=1

1 P+1 ti 1 P C1 = mO-T) s)'-1y(s)ds +1Z

(2.16)

Substituting the values of C0 and C1 into (2.8), (2.14), respectively, we obtain (2.7).

Conversely, we assume that u is a solution of the integral equation (2.7). By a direct computation, it follows that the solution given by (2.7) satisfies (2.6). The proof is completed. □

3. Main Result

In this section, our aim is to discuss the existence and uniqueness of solutions to the problem

Theorem 3.1. Assume that

(HI) there exists a constant L1 > 0 such that \f (t,u) - f (t,v)\ < L1\u - v\,for each t e J and all u, v e R;

(H2) there exist constants L2, L3 > 0 such that Ik(u) - Ik(v) < L2\u - v\, Jk(u) - Jk(v) < L3\u - v\,foreach t e J and all u, v e R, k = 1,2,...,p.

then problem (1.1) has a unique solution on J.

Proof. We transform the problem (1.1) into a fixed point problem. Define an operator T :

(1.1).

PC(J) ^ PC(J) by

(Tu)(t) = (tk - s)a-1f (s,u(s))ds

- 2fc0) ^ (t' - s)a-1f (s'u(s))ds

1 P+1 çti

+ V (t - tk ) Jk (u(tk )) - - > (T - t,-) J, (u(t,)) +->>(u(fO)

+ £ Ik(u(tk)) - WO), 2

0<tk<t ¿=1

where PC(/) is with the norm ||u|| = suptej\u(t)\. Let u, v e PC (J); then for each t e J, we have

\(Tu)(t) - (Tv)(t)\

I" (t - s)a-1\f (s,u(s)) - f (s,v(s)) \ds

r( ) £ \ (tk - s)a-1\f (s,u(s)) - f (s,v(s))\ds

1 (a) 0<tk<tJtk-i

p+1 fti

E (ti - s)a-1\f (s,u(s)) - f (s,v(s))\ds

i=1 ti-1

2r(a) ^J u

1 ^ .. .. \ a-2

1 tk \ \

(t - tk) (tk - s)a-2\f (s,u(s)) - f (s,v(s))\ds

i(a 1) 0<tk<t Jtk-1

-£(T - ti)\ (ti - s)a-2\f (s,u(s)) - f (s,v(s))\ds i=1 ti-1

, I I - Tj I I I Tj - S

2r(a - 1) i=

\T - 2t\ P^1 (ti

p ti \ \ E (ti - s)a-2\f (s,u(s)) - f (s,v(s))\ds

i=1 J ti-1

4r(a - 1) ^J ti.

1 p (3.3)

+ X (t - tk)\Jk(u(tk)) - Jk(v(tk))\ + (T - ti)\Ji(u(ti)) - Ji(v(ti))\

0<tk<t 2 i=1

+ £\Ji(u(ti)) - Ji(v(ti))\ + £ \!k(u(tk)) - Ik(v(tk))\

4 i=1 0<tk<t 1p

\Ii(u(ti)) - Ii(v(ti))\

i; ((_ s)«-1ds+^ ££ (t - s)«-1dS

7TLd|u - v|| P^1 ^^ .a-lA 3p 7Tp

4I(a - 1) §J^ (ti - s) ds + YL2|u - V| + "/J

TaL1 3(p + 1) TaL1 7(p + 1) TaL1

* f(0+1)|u - v| + 2T(a + 1) |u - V| + V) |u - v|

+ 3pL2|u - v|| + 7TPL3|u - v||

(3p + 5)Ta + 7( p + ^ Ta\ + / 3 T + 7Tt L^ 2r(a + 1) + 4r(a) )+ ^ 2L2 + TL3

||u - v||.

\\Tu - Tv\\ <

(3p + 5)Ta + 7( p + V Ta\ + J 3 L 7TT L1 ( 2r(a + 1) + 4T(a) + p{2L + TL

\u - v\\. (3.4)

L-itO+T-+7J4mF) +<2^, < 1, (3.5)

consequently T is a contraction; as a consequence of Banach fixed point theorem, we deduce that T has a fixed point which is a solution of the problem (1.1). □

Theorem 3.2. Assume that

(H3) the function f : J x R ^ R is continuous and there exists a constant Ni > 0 such that \f (t,u)\ < N for each t e J and all u e R;

(H4) the functions Ik, Jk : R ^ R are continuous and there exist constants N2, N3 > 0 such that \Ik (u)\ < N2, \Jk (u)\ < N3, for all u e R, k = 1,2,...,p.

Then the problem (1.1) has at least one solution on J.

Proof. We will use Schaefer fixed-point theorem to prove T has a fixed point. The proof will be given in several steps.

Step 1. T is continuous.

Let {un} be a sequence such that un ^ u in PC(J); we have

\(Tun)(t) - (Tu)(t)\

< (t - s)a-1 If (s,un(s)) - f (s,u(s))\ds

+ 77^% (tk - s)a-1\f(s,un(s)) - f(s,u(s))\ds

r (a) 0<tk<^ tk-1

1 p+1 ti \ \ + 5 )t iu - s)a-1\f (s,un (s)) - f (s,u(s))\ds

1 tk \ \ + r, n X (t - tk) (tk - s)a-2\f (s,un(s)) - f (s,u(s))\ds 1 (a 1)« Jtk-1

2r(a -1)

p ti \ \ %(T - ti)\ (ti - s)a-2\f (s,un(s)) - f (s,u(s))\ds

i=1 J ti-1

+ 4T(--1) ££ (ti - s)a-2\f (s,un(s)) - f (s,u(s))\ds

+ £ (t - tk)\Jk(un(tk)) - Jk(u(tk))\ + 2£(T - ti)\Ji(un(ti)) - Ji(u(ti))\

0<tk<t i=1

T-2\£\Ji(un(ti)) - Ji(u(ti))\ + £ \Ik(un(tk)) - Ik(u(tk))\

i=1 0<tk<t

\Ii(un(ti)) - Ii(u(ti))\

2 i=1 1t

(t - s)a-1\f (s,un(s)) - f (s,u(s))\ds

r(a) Jik

p+1 fti

3 pi1 ^

2r(a) £

p+1 ti \ \ £ (ti - s)a-1\f (s,un(s)) - f (s,u(s))\ds

i=1 J ti-1

7T p+1 ti \ \

£ (ti - s)a-2\f (s,un(s)) - f (s,u(s))\ds

i=1 J ti-1

4r(a - 1) 1=1 j ti_

3 p 7T p

+ ¿£\Ii(un(ti)) - I(u(ti))\ + 7^£\Ji(un(ti)) - Ji(u(h))\. i=1 i=1

Since f, I, J are continuous functions, then we have

||Tun - Tu|| —> 0, n —> to. (3.7)

Step 2. T maps bounded sets into bounded sets in PC(J).

Indeed, it is enough to show that for any r > 0, there exists a positive constant L such that, for each u e ^r = {u e PC(J) : ||u|| < r}, we have ||Tu|| < L. By (h3) and (H4), for each t e J, we can obtain

1 a-1 \\

\(Tu)(t)\ < f^) 1 (t - s)a-1\f (s,u(s))\ds

£ f (tk - s)a-1\f (s,u(s))\ds

<tk<t tk-1

r(a) 0<tk<t 1 p+1 t

1 V (t- - ^a-1

2r(a) &J t

p+1 ti \ \ £ (ti - s)a-1\f (s,u(s))\ds

i=1 J ti-1

^ £ (t - tk)t (tk - s)a-2\f (s,u(s))\ds

1) 0<tk<t Jtk-1

2r(a - 1)

p ti \ \ £(T - ti)\ (ti - s)a-2\f (s,u(s))\ds

i=1 J ti-1

Therefore,

p+1 ti \ \ X (ti - s)a-2\f (s,u(s))\d

i=1 J ti-1

\T - 2t\ 4r(a - 1) ^J tt

+ X (t - tk)\Jk(u(tk))\ + 2X(T - ti)\Ji(u(ti))\

0<tk<t 2

■ i=1

T 2t p 1 p \—r-\BJ>(u(ti))\ + X \Ik(u(tk))\ + ^\Ii(u(ti))\

0<tk<t

(t - s)a-1ds + 3N1

p+1 ti 2r(a) i^ J

(ti - s)a-1ds

maa-H)

p+1 ti Xf t

i=1 J ti-1

- s)a-2ds + ^N2 + 7TpN3

3p + 5 Ta 7 p + 1 Ta 3 7T

N zr^a+TT + Jir(a) )+ K3N2 + Tn

\\Tu\\ <

3p + 5 Ta 7 p + 1 Ta

2r(a +1) 41(a)

3 7T + 2 N2 + — N3

Step 3. T maps bounded sets into equicontinuous sets in PC(J).

Let Q.r be a bounded set of PC(J) as in Step 2, and let u e Q.r. For each t e J, we can estimate the derivative (Tu)'(t):

\(Tu)'(t)\ < fa^l) \t (t - s)a-2\f(s,u(s))\ds

1 tk \ \ fT—^Z (tk - s)a-2\f (s,u(s))\ds

i(a 1) 0<fk<f Jtk-1

1 p+1 ti \ \ X (ti - s)a-2\f (s,u(s))\ds

i=1 J ti-1

2T(a -1)

X \Jk(u(tk))\ + ^\Ji(u(tr))\

0<tk<t

(t- s)a-2ds +7Y I (ti - s)a-2ds + N3

N1 f^ \a-2 1 3N1 pX1 i^* \a-2 1 3p .

KO-T) I <'- s) ds + Z 1-1(i-- s) ds + T

N1Ta-1 3(p + 1) N1Ta-1 3p

< —1-+ —---+ — N3

" r(a) + 2r(a) + 2 N 3

(3p + 5)Ta-1\ 3p VK y 'N1 + N3:= M.

2r(a) / 1 2

(3.10)

Hence, let t, t" e J, t' < t"; we have

\ (Tu) (t") - (Tu)(t') \ = f\(Tu)'(s)\ds < M(t"-t;). (3.11)

So T(Qr) is equicontinuous in PC(J). As a consequence of Steps 1 to 3 together with the Arzela-Ascoli theorem, we can conclude that T : PC(J) ^ PC(J) is completely continuous.

Step 4. A priori bounds.

Now it remains to show that the set

Z(T) = {u e PC(J) : u = 1Tu for some 0 < 1 < 1} (3.12)

is bounded. Let u = 1Tu for some 0 < 1 < 1. Thus, for each t e J, we have

1 ft 1 ftk

u(t) = -n~)\ (t - s)a-1f (s,u(s))ds + f7-)X (tk - s)a-1f (s,u(s))ds

1 (a) Jtk 1 (a) 0<tk<tJtk-1

1 p+1 fti

- 2f(a) gL (ti - s)a-1f (s,u(s))ds

1 ftk. TT—^Z (t - tk) (tk - s)a-2f (s, u(s))ds

i(a 1) 0<tk<t Jtk-1

1 p Ck

2r(a - 1) g(T - ti)Jt (U - s)a-2f (s,u(s))ds (3.13)

'¿J ti-1

£ f' (ti - sr-2f (s,u(s))ds

1 p 1(T — 2t) p

+ 1 2 (t- tk)Jk(u(tk)) - (T- ti)Ji(u(ti)) + ( 4 ^Jr(u(ti))

0<tk<t 2 i=1 4 i=1 1p

+ 1 2 Ik(u(tk)) - Ii(u(ti)).

0<tk<t 2 i=1

For each t e J, by (H3) and (H4), we have

+ ^ <¡N2 + ?N3> (3.14)

This shows that the set Z(T) is bounded. As a consequence of Schaefer fixed-point theorem, we deduce that T has a fixed point which is a solution of the problem (1.1). □

In the following theorem we give an existence result for the problem (1.1) by applying the nonlinear alternative of Leray-Schauder type and by which the conditions (H3) and (H4) are weakened.

Theorem 3.3. Assume that (H2) and the following conditions hold.

(H5) There exists \$ e C(J) and y : [0, <) ^ (0, <) continuous and nondecreasing such that

\f(t,u)\< \$(t)y(\u\), t e J, u e R. (3.15)

(H6) There exist y*,y* : [0, <) ^ (0, <) continuous and nondecreasing such that

\Ik(u)\< y*(\u\), \Jk(u)\< y (\u\),u e R. (3.16)

(H7) There exists a number M* > 0 such that

\$*y (M*)((3p + 5)Ta/2F(a + 1)+7(p + 1)Ta/4F(a)) +p( (3/2)y*(M*) + (7T/4)y*(M*)) > 1,

(3.17)

where = sup{\$(t) : t e J}.

Then (1.1) has at least one solution on J.

Proof. Consider the operator T defined in Theorem 3.1. It can be easily shown that T is continuous and completely continuous. For X e (0,1) and each t e J, let u = XTu. Then from (H5) and (H6), and we have

\u(t)\< r-jjt (t - s)a-1\f (s,u(s))\ds

1 tk \ \ + TTa)Z (tk - s)a-l\f(s,u(s))\ds

L (a) 0<tk <^ tk-1

1 p+1 ti \ \ + Z\ti {ti - s)a-1\f (s,u(s))\ds

1 tk \ \

+ r(-1) 2 (t - tk) (tk - s)a-2\f (s,u(s))\ds

i(a 7) 0<tk<t Jtk-1

1 p ti \ \ + 2x0-7) - u ) J t {ti - s)a-2\f (s,u(s))\ds

p+1 ti \ \ -X (ti - s)a-2\f (s,u(s))\ds ' i=\J t-1

\T - 2t\ Y I " (t s)a-2 \

4Y(a - 1) £ J,

■ X (t - tk)\Jk(u(tk))\ + 2X(T - ti)\Ji(u(ti))\

0<tk<t 2 i=1 T 2t p 1 p

-—T-^ZJ(u(ti))\ + x \Ik(u(tk))\ + tx\Ii(u(ti))\

4 i=1 C<-k<- 2 i=1

< 1 I U „\a-1

r(a) \t (t - s)a-1 \$(s)y(\u(s)\)ds

+—x fk

r(a) ^k^ Jtk-1 p+1 ti

X (ti - s)a-1\$(s)y(\u(s)\)ds i=1J ti-1

(tk - s)a-1\$(s)y(\u(s)\)ds

21(a) %

1 tk f7-—r)Z(t - tk) (tk - s)a-2\$(s)y(\u(s)\)ds

i(a 7) C<-k<- •;-kkl

1 p ti Wo-T) - u) Jt (ti - s)a-2\$(s)y(\u(s)\)ds

p+1 ti

wO^T)g \t t (ti - s)a-2\$(s)y(\u(s)\)ds

\T - 2t\ v | (t s)a-2

+ X (t - tk)y*(\u(tk)\) + TX(T - ti)y*(\u(ti)\)

0<tk<t 2 i=1

it - 2tl p__1 p

+ ^Xy*(\u(ti)\) + x y*(\u(tk)\) + 7Xy*(\u(ti)\)

4 i=1 C<-k<- 2 i=1

Ta 3(p + 1)Ta

r(a + 1) r TV" 2r(a + 1) 7(p + 1)Ta 7pT__3p

"(Wu\\)^T(^- + JTy*(\\u\\) + 2 y

/(3p + 5) Ta 7(p + 1) Ta\ /3 7T_

=(-TfiaTTT++K 2 y*(M)+T^

(3.18)

^(||u||)((3p + 5)Ta/2r(a + 1)+7(p+1)Ta/4r(a))+ p( (3/2)^*(||u||) + (7T/4ty * (||u||)) < 1.

(3.19)

Then by (H7), there exists M* such that ||u|| = M*. Let

U = {u e PC(J) : ||u|| <M*}. (3.20)

The operator T : U ^ PC(J) is a continuous and completely continuous. From the choice of U, there is no u e dU such that u = 1Tu for some 0 < 1 < 1. As a consequence of the nonlinear alternative of Leray-Schauder type, we deduce that T has a fixed point u in U which is a solution of the problem (1.1). This completes the proof. □

4. Example

Let a = 3/2, T = 2n, p = 1. We consider the following boundary value problem:

Au\t=1/2 = ^ u( 2)), Au'\t=1/2 = j(u(2

CD3/2u(t) = f (t, u(t)), 0 < t < 2n, t

u(0) + u(2n) = 0, u'(0) + u'(2n) = 0,

cos tu

f (t,u) = -——^--, (t,u) e J x [0, to),

(t + 20)2 (1 + u)

I (u) = J (u) = u

10 + u 25 + u

Obviously L1 = 1/400, L2 = 1/10, L3 = 1/25. Further,

T (3p + 5)Ta + 7(p + ^Ta\ + /3T + 7TT 2F(a + 1) + 4F(a) ) + p\ 2L2 + TL

1 /32v^ „„ ^ \ 3 7n „

n+14v^; + 20 + 50 <1.

Thus, all the assumptions of Theorem 3.1 are satisfied. Hence, by the conclusion of Theorem 3.1, the impulsive fractional antiperiodic boundary value problem has a unique solution on [0,2n].

Acknowledgments

This work was supported by the Natural Science Foundation of China (10971173), the Natural Science Foundation of Hunan Province (10JJ3096), the Aid Program for Science and Technology Innovative Research Team in Higher Educational Institutions of Hunan Province, and the Construct Program of the Key Discipline in Hunan Province.

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