Research Article 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

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, [email protected]

Received 20 October 2010; Revised 25 December 2010; Accepted 20 January 2011

Academic Editor: Dumitru Baleanu

Copyright q 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

C α D ut ft, ut,t∈ 0,T,t/ tk,k 1, 2,...,p, Δu| I ut , Δu| J ut ,k , ,...,p, ttk k k ttk k k 1 2 1.1 u0 uT 0,u0 u T 0,

C α where T is a positive constant, 1 <α≤ 2, D denotes the Caputo fractional derivative of α f ∈ C ,T × R, R I J R → R {t } t

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 α of a function f : 0, ∞ → R is defined as

t CDαft 1 t − sn−α−1f nsds, n − <α

Deﬁnition 2.2 see 4. The Riemann-Liouville fractional integral of order α>0ofafunction ft, t>0, is deﬁned as

t Iαft 1 t − sα−1fsds, Γα 2.2 0 provided that the right side is pointwise deﬁned on 0, ∞.

Deﬁnition 2.3 see 4. The Riemann-Liouville fractional derivative of order α>0ofa continuous function f : 0, ∞ → R is given by

d n t Dαft 1 t − sn−α−1fsds, Γn − α dt 2.3 0 where n α 1andα denotes the integer part of real number α, provided that the right side is pointwise deﬁned on 0, ∞.

For the sake of convenience, we introduce the following notation. J ,T,J ,t ,J t ,t ,i , ,...,p− ,J t ,T J J\{t ,t ,...,t } Let 0 0 0 1 i i i 1 1 2 1 p p . 1 2 p . J{u ,T → R | u ∈ CJ,ut ut− ut−ut , ≤ We deﬁne PC : 0 k and k exists, and k k 1 k ≤ p}. Obviously, PCJ is a Banach space with the norm u supt∈J |ut|.

Deﬁnition 2.4. A function u ∈ PCJ is said to be a solution of 1.1 if u satisﬁes the equation cDαutft, ut t ∈ J Δu| I ut Δu| J ut ,k for , the equations ttk k k , ttk k k 1, 2,...,p, and the condition u0 uT0,u0 u T0.

Lemma 2.5 see 20. Let α>0;then

IαCDαu t u t c c t c t2 ··· c tn−1, 0 1 2 n−1 2.4 for some ci ∈ R, i 0, 1, 2,...,n− 1, n α 1.

Lemma 2.6 nonlinear alternative of Leray-Schauder type 21. Let E be a Banach space with C ⊆ E closed and convex. Assume that U is a relatively open subset of C with 0 ∈ U and A : U → C is continuous, compact map. Then either

1 A has a ﬁxed point in U,or 2 there exists u ∈ ∂U and λ ∈ 0, 1 with u λAu.

Lemma 2.7 Schaefer ﬁxed point theorem 22. Let S be a convex subset of a normed linear space Ω and 0 ∈ S.LetF : S → S be a completely continuous operator, and let

ζF {u ∈ S : u λFu, for some 0 <λ<1}. 2.5

Then either ζF is unbounded or F has a ﬁxed point. 4 Advances in Diﬀerence Equations

Lemma 2.8. Assume that y ∈ C0,T,R,T>0, 1 <α≤ 2. A function u ∈ PCJ is a solution of the antiperiodic boundary value problem

C α D ut yt,t∈ 0,T,t/ tk,k 1, 2,...,p, Δu| I ut , Δu| J ut ,k , ,...,p, ttk k k ttk k k 1 2 2.6 u0 uT 0,u0 u T 0, if and only if u is a solution of the integral equation

⎧ p ⎪ t 1 ti ⎪ 1 α−1 1 α−1 ⎪ t − s ysds − ti − s ysds ⎪Γα 2Γα t ⎪ 0 i1 i−1 ⎪ p ⎪ ti ⎪ 1 α−2 ⎪ − T − ti ti − s ysds ⎪ 2Γα − 1 t ⎪ i1 i−1 ⎪ ⎪ p t p ⎪ T − t 1 i ⎪ 2 t − sα−2ysds − 1 T − t J ut ⎪ Γα − i i i i ⎪ 4 1 i ti− 2 i ⎪ 1 1 1 ⎪ p p ⎪ T − t ⎪ 2 J ut − 1 I ut ,t∈ ,t , ⎪ i i i i 0 1 ⎪ 4 i 2 i ⎪ 1 1 ⎪ ⎪ t k ti ⎪ 1 t − sα−1ysds 1 t − sα−1ysds ⎪ i ⎪Γα t Γα t ⎪ k i1 i−1 ⎪ ⎨ p 1 ti k ut 1 α−1 1 ⎪ − ti − s ysds t − ti 2.7 ⎪ 2Γα t Γα − 1 ⎪ i1 i−1 i1 ⎪ ⎪ ti ⎪ α−2 ⎪ × ti − s ysds ⎪ ⎪ ti− ⎪ 1 ⎪ p ⎪ ti ⎪ − 1 T − t t − sα−2ysds ⎪ i i ⎪ 2Γα − 1 t ⎪ i1 i−1 ⎪ ⎪ p ⎪ T − t 1 ti k ⎪ 2 α−2 ⎪ ti − s ysds t − tiJiuti ⎪ Γα − t ⎪ 4 1 i i−1 i ⎪ 1 1 ⎪ p p ⎪ 1 T − 2t ⎪ − T − t J ut J ut ⎪ i i i i i ⎪ 2 i 4 i ⎪ 1 1 ⎪ k p ⎪ ⎪ I u t − 1 I u t ,t∈ t ,t , ≤ k ≤ p. ⎩ i i i i k k 1 1 i1 2 i1

y c ,c ∈ R Proof. Assume that satisﬁes 2.6 .UsingLemma 2.5, for some constants 0 1 , we have

t ut Iαyt − c − c t 1 t − sα−1ysds − c − c t, t ∈ ,t . 0 1 Γα 0 1 0 1 2.8 0 Advances in Diﬀerence Equations 5

Then, we obtain

t ut 1 t − sα−2ysds − c ,t∈ ,t . Γα − 1 0 1 2.9 1 0

t ∈ t ,t If 1 2 , then we have

t ut 1 t − sα−1ysds − d − d t − t , Γα 0 1 1 t1 2.10 t ut 1 t − sα−2ysds − d , Γα − 1 1 t1

d ,d ∈ R where 0 1 are arbitrary constants. Thus, we ﬁnd that

t 1 u t− 1 t − sα−1ysds − c − c t , 1 Γα 1 0 1 1 0

u t −d , 1 0 2.11 t 1 u t− 1 t − sα−2ysds − c , 1 Γα − 1 1 1 0

u t −d . 1 1

− − Δu|tt ut − ut I ut Δu |tt u t − u t J ut In view of 1 1 1 1 1 and 1 1 1 1 1 , we have

t1 −d 1 t − sα−1ysds − c − c t I ut , 0 Γα 1 0 1 1 1 1 0 2.12 t1 −d 1 t − sα−2ysds − c J ut . 1 Γα − 1 1 1 1 1 0

Hence, we obtain

t t1 ut 1 t − sα−1ysds 1 t − sα−1ysds Γα Γα 1 t1 0 t t − t 1 1 t − sα−2ysds t − t J ut 2.13 Γα − 1 1 1 1 1 0 I u t − c − c t, t ∈ t ,t . 1 1 0 1 1 2 6 Advances in Diﬀerence Equations

ut t ∈ t ,t Repeating the process in this way, the solution for k k 1 can be written as t k ti 1 α−1 1 α−1 ut t − s ysds ti − s ysds Γα t Γα t k i1 i−1 k ti k 1 α−2 t − ti ti − s ysds t − tiJiuti 2.14 Γα − 1 t i1 i−1 i1 k I u t − c − c t, t ∈ t ,t ,k , ,...,p. i i 0 1 k k 1 1 2 i1

On the other hand, by 2.14, we have

p T ti 1 α−1 1 α−1 uT T − s ysds ti − s ysds Γα t Γα t p i1 i−1 p p ti 1 α−2 T − ti ti − s ysds T − tiJiuti Γα − 1 t i1 i−1 i1 p I u t − c − c T, i i 0 1 2.15 i1 p T ti 1 α−2 1 α−2 u T T − s ysds ti − s ysds Γα − 1 t Γα − 1 t p i1 i−1 p J u t − c . i i 1 i1

By the boundary conditions u0 uT0,u0 u T0, we obtain

p p 1 ti ti c 1 t − s α−1y s ds 1 T − t t − s α−2y s ds 0 i i i 2Γα t 2Γα − 1 t i1 i−1 i1 i−1 p p T 1 ti T α−2 − ti − s ysds − Jiuti 4Γα − 1 t 4 i1 i−1 i1 2.16 p p 1 1 T − tiJiuti Iiuti, 2 i1 2 i1 p p 1 ti c 1 t − s α−2y s ds 1 J u t . 1 i i i 2Γα − 1 t 2 i1 i−1 i1

c c Substituting the values of 0 and 1 into 2.8 , 2.14 , respectively, we obtain 2.7 . Conversely, we assume that u is a solution of the integral equation 2.7.Bya direct computation, it follows that the solution given by 2.7 satisﬁes 2.6. The proof is completed. Advances in Diﬀerence Equations 7

3. Main Result

In this section, our aim is to discuss the existence and uniqueness of solutions to the problem 1.1.

Theorem 3.1. Assume that

H L > |ft, u − ft, v|≤L |u − v| t ∈ J 1 there exists a constant 1 0 such that 1 , for each and all u, v ∈ R;

H L ,L > I u − I v ≤ L |u − v| J u − J v ≤ 2 there exist constants 2 3 0 such that k k 2 , k k L |u − v| t ∈ J u, v ∈ R, k , ,...,p 3 , for each and all 1 2 .

α α 3p 5 T 7 p 1 T 3 7T L p L L < 1, 3.1 1 2Γα 1 4Γα 2 2 4 3

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 : PCJ → PCJ by

t tk 1 α−1 1 α−1 Tut t − s fs, usds tk − s fs, usds Γα t Γα t k 0

p 1 ti 1 α−1 − ti − s fs, usds 2Γα t i1 i−1 tk 1 α−2 t − tk tk − s fs, usds Γα − t 1 0

|Tut − Tvt|

t ≤ 1 t − sα−1fs, us − fs, vsds Γα tk tk 1 α−1 tk − s fs, us − fs, vs ds Γα t 0

p 1 ti 1 α−1 ti − s fs, us − fs, vs ds 2Γα t i1 i−1 tk 1 α−2 t − tk tk − s fs, us − fs, vs ds Γα − t 1 0

p ti 1 α−2 T − ti ti − s fs, us − fs, vs ds 2Γα − 1 t i1 i−1 p |T − t| 1 ti 2 α−2 ti − s fs, us − fs, vs ds 4Γα − 1 t i1 i−1 p 1 3.3 t − tk|Jkutk − Jkvtk| T − ti|Jiuti − Jivti| 0

T αL p T αL p T αL ≤ 1 u − v 3 1 1 u − v 7 1 1 u − v Γα 1 2Γα 1 4Γα p Tp 3 L u − v 7 L u − v 2 2 4 3 p T α p T α T L 3 5 7 1 p 3L 7 L u − v. 1 2Γα 1 4Γα 2 2 4 3 Advances in Diﬀerence Equations 9

Therefore,

α α 3p 5 T 7 p 1 T 3 7T Tu− Tv≤ L p L L u − v. 3.4 1 2Γα 1 4Γα 2 2 4 3

Since

α α 3p 5 T 7 p 1 T 3 7T L p L L < 1, 3.5 1 2Γα 1 4Γα 2 2 4 3 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

H f J × R → R N > 3 the function : is continuous and there exists a constant 1 0 such that |ft, u|≤N t ∈ J u ∈ R 1 for each and all ; H I ,J R → R N ,N > 4 the functions k k : are continuous and there exist constants 2 3 0 such |I u|≤N |J u|≤N u ∈ R k , ,...,p that k 2, k 3, for all , 1 2 .

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 PCJ; we have

|Tunt − Tut| t ≤ 1 t − sα−1fs, u s − fs, usds Γα n tk tk 1 α−1 tk − s fs, uns − fs, us ds Γα t 0

p 1 ti 1 α−1 ti − s fs, uns − fs, us ds 2Γα t i1 i−1 tk 1 α−2 t − tk tk − s fs, uns − fs, us ds Γα − t 1 0

p |T − t| 1 ti 2 α−2 ti − s fs, uns − fs, us ds 4Γα − 1 t i1 i−1 p 1 t − tk|Jkuntk − Jkutk| T − ti|Jiunti − Jiuti| 0

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

Tun − Tu−→0,n−→ ∞ . 3.7

Step 2. T maps bounded sets into bounded sets in PCJ. Indeed, it is enough to show that for any r>0, there exists a positive constant L such that, for each u ∈ Ωr {u ∈ PCJ : u≤r}, we have Tu≤L.ByH3 and H4, for each t ∈ J, we can obtain

t |Tut| ≤ 1 t − sα−1fs, usds Γα t k tk 1 α−1 tk − s fs, us ds Γα t 0

p |T − t| 1 ti 2 α−2 ti − s fs, us ds 4Γα − 1 t i1 i−1 p 1 t − tk|Jkutk| T − ti|Jiuti| 0

p N t N 1 ti 1 α−1 3 1 α−1 ≤ t − s ds ti − s ds Γα t 2Γα t k i1 i−1

p TN 1 ti p Tp 7 1 t − s α−2ds 3 N 7 N i 2 3 4Γα − 1 t 2 4 i1 i−1 p T α p T α T ≤ N 3 5 7 1 p 3N 7 N . 1 2Γα 1 4Γα 2 2 4 3 3.8

Therefore,

α α 3p 5 T 7 p 1 T 3 7T Tu≤ N p N N : L. 3.9 1 2Γα 1 4Γα 2 2 4 3

Step 3. T maps bounded sets into equicontinuous sets in PCJ. Let Ωr be a bounded set of PCJ as in Step 2,andletu ∈ Ωr . For each t ∈ J, we can estimate the derivative Tu t:

t Tut ≤ 1 t − sα−2fs, usds Γα − 1 tk tk 1 α−2 tk − s fs, us ds Γα − t 1 0

p 1 ti 1 α−2 ti − s fs, us ds 2Γα − 1 t i1 i−1 p 1 |Jkutk| |Jiuti| 0

p N t N 1 ti p ≤ 1 t − sα−2ds 3 1 t − s α−2ds 3 N i 3 Γα − 1 t 2Γα − 1 t 2 k i1 i−1

N T α−1 p N T α−1 p ≤ 1 3 1 1 3 N Γα 2Γα 2 3

α− 3p 5 T 1 3p N N : M. 2Γα 1 2 3 3.10

Hence, let t ,t ∈ J, t

So TΩr is equicontinuous in PCJ. As a consequence of Steps 1 to 3 together with the Arzela-Ascoli theorem, we can conclude that T :PCJ → PCJ is completely continuous.

Step 4. A priori bounds. Now it remains to show that the set

ζT {u ∈ PCJ : u λTu for some 0 <λ<1} 3.12 is bounded. Let u λTu for some 0 <λ<1. Thus, for each t ∈ J, we have λ t λ tk α−1 α−1 ut t − s fs, usds tk − s fs, usds Γα t Γα t k 0

p λ 1 ti α−1 − ti − s fs, usds 2Γα t i1 i−1 λ tk α−2 t − tk tk − s fs, usds Γα − t 1 0

For each t ∈ J,byH3 and H4, we have

α α 3p 5 T 7 p 1 T 3 7T u≤N p N N . 3.14 1 2Γα 1 4Γα 2 2 4 3

This shows that the set ζT 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 φ ∈ CJ and ψ : 0, ∞ → 0, ∞ continuous and nondecreasing such that ft, u ≤ φtψ|u|,t∈ J, u ∈ R. 3.15

∗ H6 There exist ψ , ψ∗ : 0, ∞ → 0, ∞ continuous and nondecreasing such that

∗ ∗ |Iku| ≤ ψ |u|, |Jku| ≤ ψ |u|,u∈ R. 3.16

∗ H7 There exists a number M > 0 such that

M∗ φ∗ψM∗ 3p 5 T α/2Γα 1 7 p 1 T α/4Γα p 3/2ψ∗M∗ 7T/4ψ∗M∗

> 1, 3.17

∗ where φ sup{φt : t ∈ 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 λ ∈ 0, 1 and each t ∈ J,letu λTu. Then from H5 and H6, and we have t |ut| ≤ 1 t − sα−1fs, usds Γα tk tk 1 α−1 tk − s fs, us ds Γα t 0

p 1 ti 1 α−1 ti − s φsψ|us|ds 2Γα t i1 i−1 tk 1 α−2 t − tk tk − s φsψ|us|ds Γα − t 1 0

T α p T α ≤ φ∗ψu φ∗ψu3 1 Γα 1 2Γα 1

p T α pT p φ∗ψu7 1 7 ψ∗u 3 ψ∗u 4Γα 4 2 p T α p T α T φ∗ψu 3 5 7 1 p 3ψ∗u 7 ψ∗u . 2Γα 1 4Γα 2 4

3.18 Advances in Diﬀerence Equations 15

Thus,

u φ∗ψu 3p 5 T α/2Γα 1 7 p 1 T α/4Γα p 3/2ψ∗u 7T/4ψ∗u

≤ 1. 3.19

∗ ∗ Then by H7, there exists M such that u / M .Let

∗ U {u ∈ PCJ : u

The operator T : U → PCJ is a continuous and completely continuous. From the choice of U, there is no u ∈ ∂U such that u λTu for some 0 <λ<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 α 3/2, T 2π, p 1. We consider the following boundary value problem:

C / 1 D3 2ut ft, ut, 0 ≤ t ≤ 2π, t / , 2 1 1 4.1 Δu|t / I u , Δu |t / J u , 1 2 2 1 2 2 u0 u2π 0,u0 u 2π 0, where

cos tu ft, u , t, u ∈ J × 0, ∞, t 2021 u 4.2 u u Iu ,Ju . 10 u 25 u

L / ,L / ,L / Obviously 1 1 400 2 1 10 3 1 25. Further,

p T α p T α T L 3 5 7 1 p 3L 7 L 1 2Γα 1 4Γα 2 2 4 3 √ 4.3 1 32 2 √ 3 7π π 14 2π < 1. 400 3 20 50 16 Advances in Diﬀerence Equations

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, 2π.

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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