Existence results for impulsive nonlinear fractional differential equation with mixed boundary conditions

Abstract

In this paper, the existence and uniqueness of solutions for an impulsive mixed boundary value problem of nonlinear differential equations of fractional order are obtained. Our results are based on some fixed point theorems. Some examples are also presented to illustrate the main results.

Bai et al. Boundary Value Problems (2016) 2016:63 DOI 10.1186/s13661-016-0573-z

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Web End = Existence results for impulsive nonlinear fractional differential equation with mixed boundary conditions

Zhanbing Bai1*, Xiaoyu Dong1 and Chun Yin2

*Correspondence: mailto:[email protected]

Web End [email protected]

1College of Mathematics and System Science, Shandong University of Science and Technology, Qianwangang Road, Qingdao, 266590, P.R. ChinaFull list of author information is available at the end of the article

Abstract

In this paper, the existence and uniqueness of solutions for an impulsive mixed boundary value problem of nonlinear dierential equations of fractional order are obtained. Our results are based on some xed point theorems. Some examples are also presented to illustrate the main results.

MSC: 34B15; 34A08

Keywords: fractional dierential equations; impulse; mixed boundary value problem; xed point theorem

1 Introduction

Recently, boundary value problems of nonlinear fractional dierential equations have been addressed by several researchers. Fractional dierential equations arise in many engineering and scientic disciplines as the mathematical modeling of systems and processes in the elds of physics, chemistry, control theory, biology, economics, blood ow phenomena, signal and image processing, biophysics, aerodynamics, tting of experimental data, etc. For details, see [] and the references therein.

Impulsive dierential equations, which provide a natural description of observed evolution processes, are regarded as important mathematical tools for a better understanding of several real world problems in the applied sciences. Recently, the boundary value problems of impulsive dierential equations of integer order have been studied extensively in the literature (see [, , ]). In [, ], Wang et al. gave a new concept of some impulsive dierential equations with fractional derivative, which is a correction of that of piecewise continuous solutions used in [, , ].

This paper is strongly motivated by the above research papers. We investigate the existence and uniqueness of solutions for a mixed boundary value problem of nonlinear impulsive dierential equations of fractional order given by

CDq+ u(t) = f (t, u(t)), t J ,

u(tk) = Ik(u(tk)), u (tk) = Jk(u(tk)), k = , , . . . , p, u() + u () = , u() + u () = ,

(.)

2016 Bai et al. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License

(http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro

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Bai et al. Boundary Value Problems (2016) 2016:63 Page 2 of 11

where CDq+ is the Caputo fractional derivative of order q (, ), f C(J R, R). Ik, Jk C(R, R), J = [, ], J = J \ {t, t, . . . , tp}, the {tk} satisfy = t < t < t < < tp < tp+ = , p N, u(tk) = u(t+k) u(tk), u (tk) = u (t+k) u (tk), where u(t+k) and u(tk) represent the right and left limits of u(t) at t = tk.

A function u PC(J, R) is said to be a solution of problem (.) if u(t) = uk(t) for t (tk, tk+) and uk C([, tk+], R) satises CDq

+ u(t) = f (t, u(t)) a.e. on (, tk+) with the restriction that uk(t) on [, tk) is just uk(t) and the conditions u(tk) = Ik(u(tk)), u (tk) = Jk(u(tk)), k = , , . . . , p with u() + u () = , u() + u () = .

The rest of this paper is organized as follows. In Section , we give some notations, recall some concepts and preparation results. In Section , we give the main results, the rst result based on Banach contraction principle, the second result based on Krasnoselskiis xed point theorem. Two examples are given in Section to demonstrate the application of our main results.

2 Preliminaries

In this section, we introduce preliminary facts which are used throughout this paper.Let J = [, t], J = (t, t], . . . , Jp = (tp, tp], Jp = (tp, ]. We have

PC(J) = u : [, ] R | u C J , and u t+k , u tk exist, and

u t

k = u(tk), k p .

Obviously, PC(J) is a Banach space with the norm

u PC = sup

t

u(t) .

Denition . The fractional integral of order q of a function f : [, ) R is dened as

Iq+f (t) =

(q)

t

k! f (k)()(t s)qn+ ds, t > , n = [q], (.)

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

Remark . In the case f (t) Cn[, +), there is CDq

+ f (t) = Inq+f (n)(t). That is to say that

Denition . is just the usual Caputos fractional derivative. In this paper, we consider an impulsive problem, so Denition . is appropriate.

Lemma . ([]) Let M be a closed, convex, and nonempty subset of a Banach space X, and A, B the operators such that

() Ax + By M whenever x, y M;

f (s)(t s)q ds, t > , q > , (.)

provided the right side is point-wise dened on (, ), where () is the gamma function.

Denition . The Caputo derivative of fractional order q for a function f : [, ) R is dened as

CDq+ f (t) =

(n q)

dn dtn

sk

t

f (s) n

k=

Bai et al. Boundary Value Problems (2016) 2016:63 Page 3 of 11

() A is compact and continuous;() B is a contraction mapping.

Then there exists z M such that z = Az + Bz.

Lemma . ([]) The set F PC([, T], Rn) is relatively compact if and only if:(i) F is bounded, that is, x C for each x F and some C > ;(ii) F is quasi-equicontinuous in [, T]. That is to say that for any > there exists > such that if x F; k N; , (tk, tk], and | | < , we have |x() x()| < .

Lemma . ([]) For q > , the general solution of the fractional dierential equation

CDq+ u(t) = is given byu(t) = c + ct + ct + + cntn,

where ci R, i = , , , . . . , n , n = [q].

In view of Lemma ., it follows that

Iq+ CDq+ u (t) = u(t) + c + ct + ct + + cntn,

where ci R, i = , , , . . . , n , n = [q].

Lemma . Let q (, ) and h : J R be continuous. A function u given by

u(t) =

(t s)qh(s) ds + t (q)

( s)qh(s) ds

+ t

t

(q)

( s)qh(s) ds, t [, t];

(t s)qh(s) ds + t (q)

( s)qh(s) ds

+ t

t

( s)qh(s) ds + ( t) pj= Jj(u(tj))( tj)

+ ( t) p

j= Ij(u(tj)) (t tj) p

j=k+ Jj(u(tj)) pj=k+ Ij(u(tj)),

t (tk, tk+], k = , , . . . , p ;

(q)

(t s)qh(s) ds + t (q)

( s)qh(s) ds

+ t

t

( s)qh(s) ds + ( t) pj= Jj(u(tj))( tj)

+ ( t) p

j= Ij(u(tj)), t (tp, tp+],

(.)

(q)

is a unique solution of the following impulsive problem:

CDq+ u(t) = h(t), t J , u(tk) = Ik(u(tk)), u (tk) = Jk(u(tk)), k = , , . . . , p,

u() + u () = , u() + u () = .

(.)

Proof With Lemma ., a general solution u of the equation CDq+u(t) = h(t) on each interval (tk, tk+] (k = , , , . . . , p) is given by

u(t) =

(q)

t

(t s)qh(s) ds + ak + bkt, for t (tk, tk+], (.)

Bai et al. Boundary Value Problems (2016) 2016:63 Page 4 of 11

where t = and tp+ = . Then we have

u (t) =

(q )

t

(t s)qh(s) ds + bk, for t (tk, tk+]. (.)

We have

u() = a, u () = b,

u() =

(q)

( s)qh(s) ds + ap + bp, u () =

(q )

( s)qh(s) ds + bp.

So applying the boundary conditions (.), we have

a + b = , (.)

(q)

( s)qh(s) ds +

( s)qh(s) ds + ap + bp = . (.)

Furthermore, using the impulsive condition u (tk) = u (t+k) u (tk) = Jk(u(tk)), we derive

bk = bk + Jk u(tk) , (.)

bk = bp

p

j=k+

(q )

Jj u(tj) (k = , , . . . , p ). (.)

In the same way, using the impulsive condition u(tk) = u(t+k) u(tk) = Ik(u(tk)), we derive

ak + bktk = ak + bktk + Ik u(tk) , (.)

which by (.) implies that

ak = ak Jk u(tk) tk + Ik u(tk) . (.)

Thus

ak = ap +

p

j=k+

Ij u(tj) (k = , , , . . . , p ). (.)

Combining (.), (.), (.) with (.) yields

ap =

(q)

( s)qh(s) ds +

Jj u(tj) tj

p

j=k+

(q )

( s)qh(s) ds

p

j=

Jj u(tj) (tj ) +

p

j=

Ij u(tj) , (.)

Bai et al. Boundary Value Problems (2016) 2016:63 Page 5 of 11

bp =

(q)

( s)qh(s) ds

(q )

( s)qh(s) ds

Jj u(tj) (tj )

p

j=

Ij u(tj) . (.)

Furthermore, by (.), (.), (.), (.) we have

ak =

(q)

( s)qh(s) ds +

+

p

j=

(q )

( s)qh(s) ds

p

j=

Jj u(tj) (tj ) +

p

j=

Ij u(tj)

+

p

j=k+

Jj u(tj) tj

p

j=k+

Ij u(tj) (k = , , , . . . , p ), (.)

bk =

(q)

( s)qh(s) ds

(q )

( s)qh(s) ds

+

p

j=

Jj u(tj) (tj )

p

j=

Ij u(tj)

Jj u(tj) (k = , , , . . . , p ). (.)

Hence for k = , , , . . . , p , (.) and (.) imply

ak + bkt = t

(q)

( s)qh(s) ds +

p

j=k+

t (q )

( s)qh(s) ds

+ ( t)

p

j=

Jj u(tj) ( tj) + ( t)

p

j=

Ij u(tj)

(t tj)

p

j=k+

Jj u(tj)

p

j=k+

Ij u(tj) . (.)

For k = p, (.) and (.) imply

ak + bkt = t

(q)

( s)qh(s) ds +

t (q )

( s)qh(s) ds

Jj u(tj) ( tj) + ( t)

p

j=

Ij u(tj) . (.)

Now it is clear that (.), (.), (.) imply that (.) holds.

Conversely, assume that u satises (.). By a direct computation, it follows that the solution given by (.) satises (.).

3 Main results

This section deals with the existence and uniqueness of solutions to problem (.).

+ ( t)

p

j=

Bai et al. Boundary Value Problems (2016) 2016:63 Page 6 of 11

Theorem . Let f : J R R be a continuous function. Suppose there exist positive constants L, L, L, M, M such that

(A) |f (t, x) f (t, y)| L|x y|, for all t J, x, y R;(A) |Ik(x) Ik(y)| L|x y|, |Jk(x) Jk(y)| L|x y|, |Ik(x)| M, |Jk(x)| M, x, y R, k = , , . . . , p,with

L

(q + )( + q) , L

(q + ) +

(q)

+ p(L + L) < .

Then problem (.) has a unique solution on J.

Proof Dene an operator T : PC(J) PC(J)

(Tu)(t) :=

(q)

t

(t s)qf

s, u(s) ds + t

(q)

( s)qf

s, u(s) ds

+ t

(q )

( s)qf

s, u(s) ds + ( t)

p

j=

Jj u(tj) ( tj)

+ ( t)

p

j=

Ij u(tj) (t tj)

p

j=k+

Jj u(tj)

p

j=k+

Ij u(tj) ,

t (tk, tk+], k = , , , . . . , p.

Let suptJ |f (t, )| = M, and Br = {u PC(J, R) | u PC r}, where

r

+ q

(q + )M + p(M + M)

.

Step . We show that TBr Br. For u Br, t J, we have

(Tu)(t)

(q)

t

(t s)q

f s, u(s) ds +

(q)

( s)q

f s, u(s) ds

+

(q )

( s)q

f s, u(s) ds +

p

j=

Jj u(tj)

p

j=

+ Ij u(tj) +

p

j=k+

Jj u(tj) +

p

j=k+

Ij u(tj)

t

(t s)q

(q)

f s, u(s) f (s, ) ds +

t

(t s)q

f (s, ) ds

+

(q)

( s)q

f s, u(s) f (s, ) ds +

( s)q

f (s, ) ds

+

(q )

( s)q

f s, u(s) f (s, ) ds

Bai et al. Boundary Value Problems (2016) 2016:63 Page 7 of 11

+

(q )

( s)q

f (s, ) ds +

p

j=

Jj u(tj)

p

j=

+ Ij u(tj) +

p

j=k+

Jj u(tj) +

p

j=k+

Ij u(tj)

Lr

(q + ) +

M (q + ) +

Lr

(q + ) +

M (q + ) +

Lr

(q) +

M (q)

+ pM + pM + pM + pM

= L + p

(q + )r +

+ p (q + )M + p(M + M).

Since

L

(q + )( + q) , r

+ q

(q + )M + p(M + M)

,

we have

(Tu)(t) r, TBr Br.

Step . T is a contraction mapping. For x, y Br and t J, we have

(Tx)(t) (Ty)(t)

=

(q)

s, x(s) ds + t

(q)

( s)qf

t

(t s)qf

s, x(s) ds

+ t

(q )

( s)qf

s, x(s) ds + ( t)

p

j=

Jj x(tj) (tj )

+ ( t)

p

j=

Ij x(tj) (t tj)

p

j=k+

Jj x(tj)

p

j=k+

Ij x(tj)

(q)

t

(t s)qf

s, y(s) ds + t

(q)

( s)qf

s, y(s) ds

+ t

(q )

( s)qf

s, y(s) ds + ( t)

p

j=

Jj y(tj) (tj )

+ ( t)

p

j=

Ij y(tj) (t tj)

p

j=k+

Jj y(tj)

p

j=k+

Ij y(tj)

t

(t s)q

(q)

f s, x(s) f s, y(s) ds

+

(q)

( s)q

f s, x(s) f s, y(s) ds

+

(q )

( s)q

f s, x(s) f s, y(s) ds

Bai et al. Boundary Value Problems (2016) 2016:63 Page 8 of 11

p

j=

+ Jj x(tj) Jj y(tj) +

p

j=

Ij x(tj) Ij y(tj)

p

j=k+

+ Jj x(tj) Jj y(tj) +

p

j=k+

Ij x(tj) Ij y(tj)

L (q + ) x y PC +

L (q + ) x y PC +

L (q) x y PC

p

j=

+ L x y PC +

p

j=

L x y PC

p

j=k+

+ L x y PC +

p

k=j+

L x y PC

L (q) x y PC + pL x y PC

+ pL x y PC + pL x y PC + pL x y PC

=

L (q + ) x y PC +

L

(q + ) +

(q)

+ p(L + L)

x y PC.

Since

L

(q + ) +

(q)

+ p(L + L) < ,

T is a contraction mapping. Thus, the conclusion follows by the contraction mapping principle.

Theorem . Assume that |f (t, u)| (t) for all (t, u) J R where L/ (J, R) and (, q ), furthermore, there exist positive constants L, L, M, M such that |Ik(x)

Ik(y)| L|x y|, |Jk(x) Jk(y)| L|x y|, |Ik(x)| M, |Jk(x)| M, x, y R, k = , , . . . , p, with p(L + L) < . Then problem (.) has at least one solution on J.

Proof Choose

r L

+ p(M + M)

(q)(q )

+

(q )(q)

and denote

Br = u PC(J, R) | u PC r .

Dene the operators P and Q on Br as

(Pu)(t) =

(q)

t

(t s)qf

s, u(s) ds + t

(q)

( s)qf

s, u(s) ds

+ t

(q )

( s)qf

s, u(s) ds,

Bai et al. Boundary Value Problems (2016) 2016:63 Page 9 of 11

(Qu)(t) = ( t)

p

k=

Jk u(tk) ( tk) + ( t)

p

k=

Ik u(tk)

Ik u(tk) .

For any u, v Br and t J, using the condition that |f (t, u)| (t) and the Hlder inequality,

t

(t s)qf s, u(s) ds

t

(t s)

(t tk)

p

k=j+

Jk u(tk)

p

k=j+

q ds

t

(s)

ds

L (J)

(q )

,

t

( s)qf s, u(s) ds

t

( s)

q ds

t

(s)

ds

L (J)

(q )

,

t

( s)qf s, u(s) ds

t

( s)

q ds

t

(s)

ds

L (J)

(q)

.

Therefore,

Pu + Qv PC

L

(J)

(J)

(q )(q) + pM + pM + pM + pM

= L

(q)(q )

+

L

+ p(M + M).

Thus Pu + Qv Br. It is obvious that Q is a contraction mapping (the proof is just similar to Theorem .). On the other hand, the continuity of f implies that the operator P is continuous. Also, P is uniformly bounded on Br since

Pu PC

L

(q)(q )

+

(q )(q)

(J)

(q)(q )

+

L

(J)

(q )(q)

r.

Now we prove the quasi-equicontinuity of the operator P.

Let = J Br, fmax = sup(t,u) |f (t, u)|. For any tk < < tk+, we have

(Pu)() (Pu)()

=

(q)

(

s)qf s, u(s) ds +

(q)

( s)qf

s, u(s) ds

Bai et al. Boundary Value Problems (2016) 2016:63 Page 10 of 11

+

(q )

( s)qf

s, u(s) ds

(q)

(

s)qf s, u(s) ds

(q)

( s)qf

s, u(s) ds

(q )

( s)qf

s, u(s) ds

fmax

(q)

( s)q ( s)q ds +

(

s)q ds

+

( )fmax

(q)

( s)q ds

+

( )fmax

(q )

( s)q ds

fmax

( )q + q q + (q + ) +

(q)

,

which tends to zero as . This shows that P is quasi-equicontinuous on the interval (tk, tk+]. It is obvious that P is compact by Lemma ., so P is relatively compact on Br.

Thus all the assumptions of Lemma . are satised and problem (.) has at least one solution on J.

4 Example

Example . Consider the following impulsive fractional boundary value problem:

CD

sin u(t)

+u(t) , t [, ], t = , u( ) =

|u(

+u(t) =

)|

+|u(

)| ,

u ( ) =

|u(

)|

+|u(

)|

(.)

u() + u () = , u() + u () = .

Obviously, L = /, L = /, L = /, M = /, M = /, p = ,

(q + ) ( + q) =

, L <

(q + ) ( + q) ,

< .

Thus, all the assumptions in Theorem . are satised. Hence, the impulsive fractional boundary value problem (.) has a unique solution on [, ].

Example . Consider the following impulsive fractional boundary value problem:

CD

L

(q + ) +

(q)

+ p(L + L) =

+

+u(t) =

et

(t+) |

u(t)|

+|u(t)| , t [, ], t = , u( ) = +|u(

)|

u ( ) = |u(

)|

+|u(

)|

(.)

+|u(

)| ,

u() + u () = , u() + u () = .

Set

f (t, u) = et

(t + )

|u| + |u|

, (t, u) [, ] [, ).

Obviously,

f (t, u)

et(t + ) .

Bai et al. Boundary Value Problems (2016) 2016:63 Page 11 of 11

Set

[, ], R .

Thus, all the assumptions in Theorem . are satised. Hence, the impulsive fractional boundary value problem (.) has at least one solution on [, ].

Competing interests

The authors declare that they have no competing interests.

Authors contributions

All authors contributed equally to the writing of this paper. All authors read and approved the nal manuscript.

Author details

1College of Mathematics and System Science, Shandong University of Science and Technology, Qianwangang Road, Qingdao, 266590, P.R. China. 2School of Automation Engineering, University of Electronic Science and Technology of China, Chengdu, 611731, P.R. China.

Acknowledgements

The authors express their sincere thanks to the anonymous reviews for their valuable suggestions and corrections for improving the quality of the paper. This work is supported by NSFC (11571207, 61503064), the Taishan Scholar project.

Received: 26 October 2015 Accepted: 8 March 2016

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L = L = , M = , M = and (t) = et

(t + ) L