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On the existence of a mild solution for impulsive hybrid fractional differential equations

Abstract

This paper is motivated by some recent contributions on the existence of solution of impulsive fractional differential equations and the theory of fractional hybrid differential equations by Agarwal, Ahmad, Baleanu, Benchohra, Fečkan, Nieto, Sun, Bai, Zhou, Zhang and Wang. Here, we derive new existence results of a mild solution of impulsive hybrid fractional differential equations. Finally, an example is given to illustrate the result.

1 Introduction

The study of differential equations of fractional order is motivated by the intensive development of the theory of fractional calculus itself (see [17]) and the application of fractional differential equations in the modeling of many physical phenomena. There have been many works on the theory of fractional calculus and applications of it. Fractional differential equations, including Riemann-Liouville fractional derivative or Caputo fractional derivative, have received more and more attention (see [818]).

In recent years, hybrid differential equations have attracted much attention. The theory of hybrid differential equations has been developed, and we refer the readers to the articles [1922]. The authors [23] discussed the following fractional hybrid differential equations involving Riemann-Liouville differential operators:

{ D 0 , t q L [ u ( t ) f ( t , u ( t ) ) ] = g ( t , u ( t ) ) , a.e.  t J : = [ 0 , T ] , u ( 0 ) = 0 ,
(1)

where D 0 , t q L is the Riemann-Liouville fractional derivative of order q(0,1) with the lower limit zero, fC(J×R,R{0}) and gC(J×R,R). They developed existence of solutions under mixed Lipschitz and Carathéodory conditions. Moreover, they have established some fundamental fractional differential inequalities and the comparison principle. Some recent papers have treated the problem of the existence of solutions for impulsive fractional differential equations.

The authors [24] considered the following basic impulsive Cauchy problems:

{ D 0 , t q c u ( t ) = f ( t , u ( t ) ) , t J : = J { t 1 , , t m } , Δ u ( t k ) = I k ( u ( t k ) ) , k = 1 , 2 , , m , u ( 0 ) = u 0 ,
(2)

where D 0 , t q c is the generalized Caputo fractional derivative of order q(0,1) with the lower limit zero and I k :RR and t k satisfy 0= t 0 < t 1 << t m < t m + 1 =T, u( t k + )= lim ε 0 + u( t k +ε) and u( t k )= lim ε 0 u( t k +ε) represent the right and left limits of u(t) at t= t k . They established some sufficient conditions for the existence of solutions.

In the recent paper [25], Herzallah obtained the existence of a mild solution for the fractional order hybrid differential equations:

{ D 0 , t q c [ u ( t ) f ( t , u ( t ) ) ] = g ( t , u ( t ) ) , a.e.  t J , u ( 0 ) = u 0 R .
(3)

In the present paper, we study the following impulsive hybrid fractional differential equations (IHFDE):

{ D 0 , t q c [ u ( t ) f ( t , u ( t ) ) ] = g ( t , u ( t ) ) , t J : = J { t 1 , , t m } , u ( t k + ) = u ( t k ) + I k ( u ( t k ) ) , k = 1 , 2 , , m , u ( 0 ) = u 0 ,
(4)

where D 0 , t q c is the generalized Caputo fractional derivative of order q(0,1) with the lower limit zero, fC(J×R,R{0}) and gC(J×R,R).

This paper is arranged as follows. In Section 2, we recall some concepts and some fractional calculation law and establish preparation results. In Section 3, we give the main results based on the Dhage fixed point theorem. In Section 4, we give an example to demonstrate the application of our main result.

2 Preliminaries

In this section, we recall some basic definitions and properties of the fractional calculus theory and preparation results. Throughout this paper, J 1 denotes the interval [ t 0 , t 1 ] and J k + 1 denotes the interval ( t k , t k + 1 ], k=1,2,,m. Let C(J,R) be the Banach algebra of all continuous functions from J into with the norm u c =sup{|u(t)|:tJ} for uC(J,R) and with multiplication (uv)(s)=u(s)v(s) for u,vC(J,R). Define PC(J,R) = {u:JR:uC(( t k , t k + 1 ],R), k=0,1,,m and there exist u( t k ) and u( t k + ), k=1,2,,m, with u( t k )=u( t k )} with the norm u P C :=sup{u(t):tJ} that is a Banach space.

We introduce the following known definitions. For more details, one can see [2].

Definition 2.1 The fractional integral of order γ with the lower limit zero for a function f:[0,)R is defined as I t γ f(t)= 1 Γ ( γ ) 0 t f ( s ) ( t s ) 1 γ dx, t>0, γ>0 provided the right-hand side is point-wise defined on [0,), where Γ() is the gamma function.

Definition 2.2 The Riemann-Liouville derivative of order γ with the lower limit zero for a function f:[0,)R can be written as D 0 , t γ L f(t)= 1 Γ ( n γ ) d n d t n 0 t f ( s ) ( t s ) γ + 1 n ds, t>0, n1<γ<n.

Definition 2.3 The Caputo derivative of order γ for a function f:[0,)R can be written as

D 0 , t γ c f(t)= L D 0 , t γ [ f ( t ) k = 0 n 1 t k k ! f ( k ) ( 0 ) ] ,t>0,n1<γ<n.

Remark (i) We have to explain that we follow the ideas from the recent contributions on impulsive fractional differential equations by Fečkan et al. [26, 27].

  1. (ii)

    In Definition 2.3, the integrable function f can be discontinuous.

  2. (iii)

    For more details and explanations on such interesting problems, one can refer to [[26], Discussions I-V, p.4214] and [[27], Remark 2.21(iii)].

Lemma 2.4 ([28])

Let S be a non-empty, closed convex and bounded subset of the Banach algebra X, and let A:XX and B:SX be two operators such that

  1. (a)

    A is Lipschitzian with a Lipschitz constant α;

  2. (b)

    B is completely continuous;

  3. (c)

    u 1 =A u 1 B u 2 u 1 S for all u 2 S;

  4. (d)

    αM<1, where M=B(S)=sup{B(u):uS}.

Then the operator equation AuBu=u has a solution in S.

Lemma 2.5 ([24])

Let q(0,1) and h:[0, T 0 ]R be continuous. A function uC([0, T 0 ],R) is a solution of the fractional integral equation

u(t)= u 0 1 Γ ( q ) 0 a ( a s ) q 1 h(s)ds+ 1 Γ ( q ) 0 t ( t s ) q 1 h(s)ds,

if and only if u is a solution of the following fractional Cauchy problems:

{ D 0 , t q c u ( t ) = h ( t ) , t [ 0 , T 0 ] , u ( a ) = u 0 , a > 0 .

We introduce the following hypotheses in what follows.

(H1) The function u u f ( t , u ) is increasing in for every t J 1 .

(H2) There exists a constant L k >0 such that |f(t, u 1 )f(t, u 2 )| L k | u 1 u 2 | for all t J k (k=1,2,,m+1) and u 1 , u 2 R.

(H3) There exists a function h k L ( J k ) (k=1,2,,m+1) such that |g(t,u(t))| h k (t), t J k (k=1,2,,m+1) for all u(t)R.

(H4) f(t,u(t)) is continuous on J k (k=1,2,,m+1) for every uPC(J,R).

Lemma 2.6 Assume that hypotheses (H1) and (H4) hold. Let q(0,1) and h:JR be continuous. A function u is a solution of the fractional integral equation

u(t)= { f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ] , t J 1 , f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + I 1 ( u ( t 1 ) ) f ( t 1 , u ( t 1 ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ] , t J 2 , f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + I 1 ( u ( t 1 ) ) f ( t 1 , u ( t 1 ) ) + I 2 ( u ( t 2 ) ) f ( t 2 , u ( t 2 ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ] , t J 3 , f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + i = 1 m I i ( u ( t i ) ) f ( t i , u ( t i ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ] , t J m + 1 ,
(5)

if and only if u is a solution of the following impulsive problem:

{ D 0 , t q c [ u ( t ) f ( t , u ( t ) ) ] = h ( t ) , t J , u ( t k + ) = u ( t k ) + I k ( u ( t k ) ) , k = 1 , 2 , , m , u ( 0 ) = u 0 .
(6)

Proof Assume that u satisfies (6). If t J 1 , then

D 0 , t q c [ u ( t ) f ( t , u ( t ) ) ] =h(t),t J 1 ,
(7)
u(0)= u 0 .
(8)

Operating by I 0 , t q on both sides of (7), one can obtain

u ( t ) f ( t , u ( t ) ) = u ( 0 ) f ( 0 , u ( 0 ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s = u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ,

i.e.,

u(t)=f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ] .

If t J 2 , then

D 0 , t q c [ u ( t ) f ( t , u ( t ) ) ] =h(t),t J 2 ,
(9)
u ( t 1 + ) =u ( t 1 ) + I 1 ( u ( t 1 ) ) .
(10)

According to Lemma 2.5 and the continuity of f(t,u(t)), we have

u ( t ) f ( t , u ( t ) ) = u ( t 1 + ) f ( t 1 , u ( t 1 ) ) 1 Γ ( q ) 0 t 1 ( t 1 s ) q 1 h ( s ) d s + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s = [ u ( t 1 ) + I 1 ( u ( t 1 ) ) ] f ( t 1 , u ( t 1 ) ) 1 Γ ( q ) 0 t 1 ( t 1 s ) q 1 h ( s ) d s + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s .

Since

u ( t 1 ) =f ( t 1 , u ( t 1 ) ) [ u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t 1 ( t 1 s ) q 1 h ( s ) d s ] ,

there exists

u ( t ) f ( t , u ( t ) ) = [ u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t 1 ( t 1 s ) q 1 h ( s ) d s ] + I 1 ( u ( t 1 ) ) f ( t 1 , u ( t 1 ) ) 1 Γ ( q ) 0 t 1 ( t 1 s ) q 1 h ( s ) d s + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s = u 0 f ( 0 , u 0 ) + I 1 ( u ( t 1 ) ) f ( t 1 , u ( t 1 ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ,

i.e.,

u(t)=f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + I 1 ( u ( t 1 ) ) f ( t , u ( t 1 ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ] .

If t J 3 , we have

u ( t ) f ( t , u ( t ) ) = u ( t 2 + ) f ( t 2 , u ( t 2 ) ) 1 Γ ( q ) 0 t 2 ( t 2 s ) q 1 h ( s ) d s + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s = u ( t 2 ) + I 2 ( u ( t 2 ) ) f ( t 2 , u ( t 2 ) ) 1 Γ ( q ) 0 t 2 ( t 2 s ) q 1 h ( s ) d s + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s .

For

u ( t 2 ) =f ( t 2 , u ( t 2 ) ) [ u 0 f ( 0 , u 0 ) + I 1 ( u ( t 1 ) ) f ( t 1 , u ( t 1 ) ) + 1 Γ ( q ) 0 t 2 ( t 2 s ) q 1 h ( s ) d s ] ,

we have that

u ( t ) f ( t , u ( t ) ) = u 0 f ( 0 , u 0 ) + I 1 ( u ( t 1 ) ) f ( t 1 , u ( t 1 ) ) + 1 Γ ( q ) 0 t 2 ( t 2 s ) q 1 h ( s ) d s + I 2 ( u ( t 2 ) ) f ( t 2 , u ( t 2 ) ) 1 Γ ( q ) 0 t 2 ( t 2 s ) q 1 h ( s ) d s + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s = u 0 f ( 0 , u 0 ) + I 1 ( u ( t 1 ) ) f ( t 1 , u ( t 1 ) ) + I 2 ( u ( t 2 ) ) f ( t 2 , u ( t 2 ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ,

i.e.,

u(t)=f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + i = 1 2 I i ( u ( t i ) ) f ( t i , u ( t i ) ) + I 2 ( u ( t 2 ) ) f ( t 2 , u ( t 2 ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ] .

If t J k + 1 (k=3,4,,m), using the same method, we get

u(t)=f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + i = 1 k I i ( u ( t i ) ) f ( t i , u ( t i ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d s ] .
(11)

Conversely, assume that u satisfies (5). If t J 1 , then we have

u(t)=f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d x ] .
(12)

Then, dividing by f(t,u(t)) and applying D 0 , t q c on both sides of (12), (7) is satisfied. Again, substituting t=0 in (12), we have u ( 0 ) f ( 0 , u ( 0 ) ) = u 0 f ( 0 , u 0 ) . Since u u f ( t , u ) is increasing in for t J 1 , the map u u f ( 0 , u ) is injective in . Then we get (8).

If t J 2 , then we have

u(t)=f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + I 1 ( u ( t 1 ) ) f ( t 1 , u ( t 1 ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 h ( s ) d x ] .
(13)

Then, dividing by f(t,u(t)) and applying D t q c on both sides of (13), (9) is satisfied. Again, by (H4), substituting t= t 1 in (12) and taking the limit of (13), then (13) minus (12) gives (10).

If t J k + 1 (k=2,4,,m), similarly we get

{ D 0 , t q c [ u ( t ) f ( t , u ( t ) ) ] = h ( t ) , t ( t k , t k + 1 ] , u ( t k + ) = u ( t k ) + I k ( u ( t k ) ) .
(14)

This completes the proof. □

Now we give the following definition.

Definition 2.7 If a function uPC(J,R) satisfies the fractional integral equation

u(t)= { f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t ( t s ) q 1 g ( s , u ( s ) ) d s ] , t J 1 , f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + I 1 ( u ( t 1 ) ) f ( t 1 , u ( t 1 ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 g ( s , u ( s ) ) d s ] , t J 2 , f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + I 1 ( u ( t 1 ) ) f ( t 1 , u ( t 1 ) ) + I 2 ( u ( t 2 ) ) f ( t 2 , u ( t 2 ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 g ( s , u ( s ) ) d s ] , t J 3 , f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + i = 1 m I i ( u ( t i ) ) f ( t i , u ( t i ) ) + 1 Γ ( q ) 0 t ( t s ) q 1 g ( s , u ( s ) ) d s ] , t J m + 1 ,
(15)

it is said to be a mild solution of (4).

3 Main results

In this section, we prove the existence of a mild solution for IHFDE (4) by Lemma 2.4 and Lemma 2.6.

Theorem 3.1 Assume that hypotheses (H1)-(H4) hold. Further, if

L k [ | u 0 f ( 0 , u 0 ) + i = 1 k 1 I i ( u i ( t i ) ) f ( t i , u i ( t i ) ) | + 1 Γ ( q + 1 ) j = 0 k ( t k t j ) q h j + 1 L ] < 1 , k = 1 , 2 , , m , m + 1

for all u i (t)C( J i ,R) and u i ( t i ) is bounded, then IHFDE (4) has a mild solution defined on J.

Proof By Lemma 2.6, IHFDE (4) is equivalent to the fractional integral equation (15).

If t J 1 , we have

u(t)=f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t ( t s ) q 1 g ( s , u ( s ) ) d s ] .
(16)

Set X 1 =C( J 1 ,R) and

S 1 = { u X 1 u N 1 } ,
(17)

where

N 1 = F 0 ( | u 0 | Γ ( q + 1 ) + t 1 q | f ( 0 , t 0 ) | h 1 L ) Γ ( q + 1 ) | f ( 0 , u 0 ) | L 1 ( | u 0 | Γ ( q + 1 ) + t 1 q | f ( 0 , t 0 ) | h 1 L ) ,
(18)

and F 0 = sup t J |f(t,0)|.

Define two operators A 1 : X 1 X 1 and B 1 : S 1 X 1 by

A 1 u(t)=f ( t , u ( t ) ) ,t J 1
(19)

and

B 1 u(t)= u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t ( t s ) q 1 g ( s , u ( s ) ) ds,t J 1 .
(20)

We will show that the operators A 1 and B 1 satisfy all the conditions of Lemma 2.4.

First, we will show that A 1 is a Lipschitz operator on X 1 with the Lipschitz constant L 1 . Set u 1 , u 2 X 1 . Then, by hypotheses (H2) and (H3), we have

| A 1 u 1 ( t ) A 1 u 2 ( t ) | = | f ( t , u 1 ( t ) ) f ( t , u 2 ( t ) ) | L 1 | u 1 ( t ) u 2 ( t ) | L 1 u 1 u 2
(21)

for all t J 1 . Taking supremum over t, we have

A 1 u 1 A 1 u 2 L 1 u 1 u 2
(22)

for all u 1 , u 2 X 1 .

Next, we will show that B 1 is a compact and continuous operator on S 1 into X 1 . We show that B 1 is continuous on S 1 . Let { u n } S 1 converge to a point u S 1 . Then, by the Lebesgue dominated convergence theorem, we have

lim n B 1 u n ( t ) = u 0 f ( 0 , u 0 ) + lim n 1 Γ ( q ) 0 t ( t s ) q 1 g ( s , u n ( s ) ) d s = u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t ( t s ) q 1 lim n g ( s , u n ( s ) ) d s = u 0 f ( 0 , u 0 ) + 1 Γ ( q ) 0 t ( t s ) q 1 g ( s , u ( s ) ) d s = B 1 u ( t )
(23)

for all t J 1 . So, we have obtained that B 1 is continuous on S 1 .

Next we will show that B 1 is a compact operator on S 1 . We show that B 1 ( S 1 ) is a uniformly bounded and equicontinuous set in S 1 . Let u S 1 be arbitrary. Then, by hypothesis (H3), we have

| B 1 u ( t ) | | u 0 f ( 0 , u 0 ) | + 1 Γ ( q ) 0 t ( t s ) q 1 | g ( s , u ( s ) ) | d s | u 0 f ( 0 , u 0 ) | + 1 Γ ( q ) ( 0 t ( t s ) q 1 h 1 ( s ) d s ) | u 0 f ( 0 , u 0 ) | + t 1 q Γ ( q + 1 ) h 1 L
(24)

for all t J 1 . Taking supremum over t, we have

B 1 u| u 0 f ( 0 , u 0 ) |+ t 1 q Γ ( q + 1 ) h 1 L

for all u S 1 . We have obtained that B 1 ( S 1 ) is uniformly bounded.

Let t 11 , t 12 J 1 with t 11 < t 12 . Then, for any u S 1 , we have

| B 1 u ( t 11 ) B 1 u ( t 12 ) | | 1 Γ ( q ) 0 t 11 ( t 11 s ) q 1 g ( s , u ( s ) ) d s 1 Γ ( q ) 0 t 11 ( t 12 s ) q 1 g ( s , u ( s ) ) d s | + | 1 Γ ( q ) 0 t 11 ( t 12 s ) q 1 g ( s , u ( s ) ) d s 1 Γ ( q ) 0 t 12 ( t 12 s ) q 1 g ( s , u ( s ) ) d s | h 1 L Γ ( q + 1 ) [ | t 12 q t 11 q ( t 12 t 11 ) q | + ( t 12 t 11 ) q ] .
(25)

Hence, for ε>0, there exists a constant δ>0 such that

| t 11 t 12 |<δ| B 1 u( t 11 ) B 1 u( t 12 )|<ε
(26)

for all t 11 , t 12 [0, t 1 ] and for all u S 1 . This obtains that B 1 ( S 1 ) is an equicontinuous set in  X 1 . By the Arzela-Ascoli theorem, we know that B 1 is compact. As a result B 1 is a complete continuous operator on S 1 .

Next, we show that hypothesis (c) of Lemma 2.4 is satisfied. Let u 11 X 1 and any u 12 S 1 such that u 11 = A 1 u 11 B 1 u 12 . Then, by assumption (H2), we have

| u 11 ( t ) | [ | f ( t , u 11 ( t ) ) f ( t , 0 ) | + | f ( t , 0 ) | ] [ | u 0 f ( 0 , u 0 ) | + 1 Γ ( q ) 0 t ( t s ) q 1 | g ( s , u 12 ( s ) ) | d s ] [ L 1 | u 11 ( t ) | + F 0 ] [ | u 0 f ( 0 , u 0 ) | + 1 Γ ( q ) 0 t ( t s ) q 1 h 1 ( s ) d s ] [ L 1 | u 11 ( t ) | + F 0 ] [ | u 0 f ( 0 , u 0 ) | + t 1 q Γ ( q + 1 ) h 1 L ] .
(27)

Thus,

| u 11 (t)| F 0 ( | u 0 | Γ ( q + 1 ) + t 1 q | f ( 0 , u 0 ) | h 1 L ) Γ ( q + 1 ) | f ( 0 , u 0 ) | L 1 ( | u 0 | Γ ( q + 1 ) + t 1 q | f ( 0 , t 0 ) | h 1 L ) .
(28)

Taking supremum over t, we have

u 11 F 0 ( | u 0 | Γ ( q + 1 ) + t 1 q | f ( 0 , t 0 ) | h 1 L ) Γ ( q + 1 ) | f ( 0 , u 0 ) | L 1 ( | u 0 | Γ ( q + 1 ) + t 1 q | f ( 0 , t 0 ) | h 1 L ) = N 1 .
(29)

Finally, we have

M 1 = B 1 ( S 1 ) =sup { B 1 u : u S 1 } | u 0 f ( 0 , u 0 ) |+ t 1 q Γ ( q + 1 ) h 1 L ,

and so

L 1 M 1 L 1 [ | u 0 f ( 0 , u 0 ) | + t 1 q Γ ( q + 1 ) h 1 L ] <1.
(30)

So, the operator equation A 1 u B 1 u=u has a solution denoted by u 1 (t) in S 1 .

If t J 2 , we set

u ( t ) = f ( t , u ( t ) ) [ u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) + 1 Γ ( q ) 0 t 1 ( t s ) q 1 g ( s , u 1 ( s ) ) d s ] + f ( t , u ( t ) ) t 1 t ( t s ) q 1 g ( s , u ( s ) ) d s .

Set X 2 =C( J 2 ,R) and

S 2 = { u X 2 u N 2 } ,
(31)

where

N 2 = F 0 H 2 1 L 2 H 2 ,
(32)
H 2 =| u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) |+ t 2 q Γ ( q + 1 ) h 1 L + ( t 2 t 1 ) q Γ ( q + 1 ) h 2 L .
(33)

Define two operators A 2 : X 2 X 2 and B 2 : S 2 X 2 by

A 2 u(t)=f ( t , u ( t ) ) ,t J 2
(34)

and

B 2 u ( t ) = u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) + 1 Γ ( q ) 0 t 1 ( t s ) q 1 g ( s , u 1 ( s ) ) d s + 1 Γ ( q ) t 1 t ( t s ) q 1 g ( s , u ( s ) ) d s , t J 2 .
(35)

The operators A 2 and B 2 satisfy all the conditions of Lemma 2.4. First, we prove that A 2 is a Lipschitz operator on X 2 with the Lipschitz constant L 2 . Set u 21 , u 22 X 2 . Then, by hypothesis (H2), we have

| A 2 u 21 ( t ) A 2 u 22 ( t ) | L 2 | u 21 ( t ) u 22 ( t ) | L 2 u 21 u 22
(36)

for all t J 2 . Taking supremum over t, we have

A 2 u 21 A 2 u 22 L 2 u 21 u 22
(37)

for all u 21 , u 22 X 2 .

Next, we prove that B 2 is a compact and continuous operator on S 2 into X 2 . We show that B 2 is continuous on S 2 . Let { u n } S 2 converge to a point u B 2 . Then, by the Lebesgue dominated convergence theorem, we have

lim n B 2 u n ( t ) = u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) + 1 Γ ( q ) 0 t 1 ( t s ) q 1 g ( s , u 1 ( s ) ) d s + 1 Γ ( q ) t 1 t ( t s ) q 1 lim n g ( s , u n ( s ) ) d s = u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) + 1 Γ ( q ) 0 t 1 ( t s ) q 1 g ( s , u 1 ( s ) ) d s + 1 Γ ( q ) t 1 t ( t s ) q 1 g ( s , u ( s ) ) d s = B 2 u ( t )
(38)

for all t J 2 . So, B 2 is continuous on S 2 .

We will prove that B 2 is a compact operator on S 2 . We show that B 2 ( S 2 ) is a uniformly bounded and equicontinuous set in S 2 . Let u S 2 be arbitrary. Then, by hypothesis (H3), we have

| B 2 u ( t ) | | u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) | + | 1 Γ ( q ) 0 t 1 ( t s ) q 1 g ( s , u 1 ( s ) ) d s | + | 1 Γ ( q ) t 1 t ( t s ) q 1 g ( s , u ( s ) ) d s | | u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) | + 1 Γ ( q ) 0 t 1 ( t s ) q 1 | g ( s , u 1 ( s ) ) | d s + 1 Γ ( q ) t 1 t ( t s ) q 1 | g ( s , u ( s ) ) | d s | u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) | + 1 Γ ( q ) 0 t 1 ( t s ) q 1 h 1 ( s ) d s + 1 Γ ( q ) t 1 t ( t s ) q 1 h 2 ( s ) d s | u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) | + t 2 q Γ ( q + 1 ) h 1 L + ( t 2 t 1 ) q Γ ( q + 1 ) h 2 L
(39)

for all t J 2 . Taking supremum over t, we have

B 2 u| u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) |+ t 2 q Γ ( q + 1 ) h 1 L + ( t 2 t 1 ) q Γ ( q + 1 ) h 2 L

for all u S 2 . So, B 2 ( S 2 ) is uniformly bounded.

Let t 21 , t 22 J 2 with t 21 < t 22 . Then, for any u S 2 , we have

| B 2 u ( t 21 ) B 2 u ( t 22 ) | | 1 Γ ( q ) 0 t 1 ( t 21 s ) q 1 g ( s , u 1 ( s ) ) d s 1 Γ ( q ) 0 t 1 ( t 22 s ) q 1 g ( s , u 1 ( s ) ) d s | + | 1 Γ ( q ) t 1 t 21 ( t 21 s ) q 1 g ( s , u ( s ) ) d s 1 Γ ( q ) t 1 t 21 ( t 22 s ) q 1 g ( s , u ( s ) ) d s | + | 1 Γ ( q ) t 1 t 21 ( t 22 s ) q 1 g ( s , u ( s ) ) d s 1 Γ ( q ) t 1 t 22 ( t 22 s ) q 1 g ( s , u ( s ) ) d s | h 1 L Γ ( q + 1 ) | t 22 q t 21 q + ( t 22 t 1 ) q ( t 21 t 1 ) q | + h 2 L Γ ( q + 1 ) [ | ( t 22 t 1 ) q ( t 21 t 1 ) q + ( t 22 t 21 ) q | + ( t 22 t 21 ) q ] .

Hence, for ε>0, there exists a constant δ>0 such that

| t 21 t 22 |<δ| B 2 u( t 21 ) B 2 u( t 22 )|<ε
(40)

for all t 21 , t 22 J 2 and for all u S 2 . So, B 2 ( S 2 ) is an equicontinuous set in X 2 . By the Arzela-Ascoli theorem, we know that B 2 is compact. As a result, B 2 is a complete continuous operator on S 2 .

Next, we show that hypothesis (c) of Lemma 2.4 is satisfied. Let u 21 X 2 and any u 22 S 2 such that u 21 = A 2 u 21 B 2 u 22 . Then, by assumptions (H2) and (H3), we have

| u 21 ( t ) | [ | f ( t , u 21 ( t ) ) f ( t , 0 ) | + | f ( t , 0 ) | ] [ | u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) | + 1 Γ ( q ) 0 t 1 ( t s ) q 1 | g ( s , u 1 ( s ) ) | d s + 1 Γ ( q ) t 1 t ( t s ) q 1 | g ( s , u 22 ( s ) ) | d s ] [ L 2 | u 21 ( t ) | + F 0 ] [ | u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) | + 1 Γ ( q ) 0 t 1 ( t s ) q 1 h 1 ( s ) d s + 1 Γ ( q ) t 1 t ( t s ) q 1 h 2 ( s ) d s ] [ L 2 | u 21 ( t ) | + F 0 ] [ | u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) | + t 2 q Γ ( q + 1 ) h 1 L + ( t 2 t 1 ) q Γ ( q + 1 ) h 2 L ] .

Thus,

| u 21 (t)| F 0 H 2 1 L 2 H 2 .
(41)

Taking supremum over t, we have

u 21 F 0 H 2 1 L 2 H 2 = N 2 .
(42)

Finally, we have

M 2 | u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) |+ t 2 q Γ ( q + 1 ) h 1 L + ( t 2 t 1 ) q Γ ( q + 1 ) h 2 L
(43)

and

L 2 M 2 L 2 [ | u 0 f ( 0 , u 0 ) + I 1 ( u 1 ( t 1 ) ) f ( t 1 , u 1 ( t 1 ) ) | + t 2 q Γ ( q + 1 ) h 1 L + ( t 2 t 1 ) q Γ ( q + 1 ) h 2 L ] < 1 .
(44)

So, the operator equation A 2 u B 2 u=u has a solution denoted by u 2 (t) in S 2 .

If t j k + 1 (k=2,3,,m), repeating the same process, we obtain u ( k + 1 ) (t)C( J k + 1 ,R) (k=2,3,,m). So, we get a mild solution of IHFDE (4). The proof is completed. □

4 Example

In this section we give a simple example to illustrate the usefulness of our main result.

Example 4.1 Let us consider the impulsive hybrid fractional differential equations

{ D 0 , t q c [ u ( t ) e υ 1 t + | u ( t ) | ] = e υ 2 t | u ( t ) | ( 1 + e t ) ( 1 + | u ( t ) | ) , t [ 0 , 1 ] t 1 , u ( t 1 + ) = u ( t 1 ) + ( 2 u ( t 1 ) ) , t 1 0 , 1 u ( 0 ) = 0 ,
(45)

where υ 1 >0, υ 2 >0 are constants and 0<q<1.

Set

f ( t , u ( t ) ) = e υ 1 t +|u(t)|,g ( t , u ( t ) ) = e υ 2 t | u ( t ) | ( 1 + e t ) ( 1 + | u ( t ) | )

for all (t,u)[0,1]×R.

Obviously,

f ( t , u ( t ) ) 0andf ( t 1 + , u ( t 1 + ) ) =f ( t 1 , u ( t 1 ) )

for all (t,u)[0,1]×R. There exist constants L k =1 (k=1,2) such that

|f(t, u 1 )f(t, u 2 )| L k | u 1 u 2 |

for all t J k (k=1,2) and u 1 , u 2 R.

By

u ( t ) f ( t , u ( t ) ) = { 1 e υ 1 t e υ 1 t + u ( t ) , u ( t ) 0 , 1 + e υ 1 t e υ 1 t u ( t ) , u ( t ) < 0 ,
(46)

we have that the function u u f ( t , u ( t ) ) is increasing in for [0, t 1 ].

For all uR and each t[0,1], we have

g ( t , u ( t ) ) = e υ t | u ( t ) | ( 1 + e t ) ( 1 + | u ( t ) | ) e υ t 1 + e t e υ t 2 .
(47)

So, choosing some υ 1 , υ 2 >0 large enough, we have

L 2 [ | 2 u ( t 1 ) e υ 1 t 1 + | u ( t 1 ) | | + 1 Γ ( q + 1 ) ( ( 1 t 1 ) q h 1 L 1 + 1 ) h 2 L 1 ] <1.

Thus all the assumptions in Theorem 3.1 are satisfied, our results can be applied to problem (45).

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Acknowledgements

The authors thank the referees for their careful reading of the manuscript and insightful comments, which helped to improve the quality of the paper. We would also like to acknowledge the valuable comments and suggestions from the editors, which vastly contributed to improving the presentation of the paper. This work was supported by the NSF of China (Nos. 11371183, 11271050) and the NSF of Shandong Province (No. ZR2013AM004).

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Ge, H., Xin, J. On the existence of a mild solution for impulsive hybrid fractional differential equations. Adv Differ Equ 2014, 211 (2014). https://doi.org/10.1186/1687-1847-2014-211

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Keywords

  • Caputo fractional derivative
  • impulsive hybrid differential equations
  • existence