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Nonlinear second-order impulsive q-difference Langevin equation with boundary conditions

Abstract

In this paper, we discuss the existence and uniqueness of solutions for Langevinimpulsive q-difference equations with boundary conditions. Our studyrelies on Banach’s and Schaefer’s fixed point theorems. Illustrativeexamples are also presented.

MSC: 26A33, 39A13, 34A37.

1 Introduction and preliminaries

In recent years, the boundary value problems of fractional order differentialequations have emerged as an important area of research, since these problems haveapplications in various disciplines of science and engineering such as mechanics,electricity, chemistry, biology, economics, control theory, signal and imageprocessing, polymer rheology, regular variation in thermodynamics, biophysics,aerodynamics, viscoelasticity and damping, electro-dynamics of complex medium, wavepropagation, blood flow phenomena, etc.[15]. Many researchers have studied the existence theory for nonlinearfractional differential equations with a variety of boundary conditions, forinstance, see the papers [618], and the references therein.

The Langevin equation (first formulated by Langevin in 1908) is found to be aneffective tool to describe the evolution of physical phenomena in fluctuatingenvironments [19]. For some new developments on the fractional Langevin equation, see, forexample, [2027].

Nowadays there is a significant increase of activities in the area ofq-calculus due to its applications in various fields such as mathematics,mechanics, and physics. The book by Kac and Cheung [28] covers many of the fundamental aspects of the quantum calculus.A variety of new results can be found in the papers [2941] and the references cited therein.

Impulsive differential equations serve as basic models to study the dynamics ofprocesses that are subject to sudden changes in their states. Recent development inthis field has been motivated by many applied problems, such as control theory,population dynamics and medicine. For some recent works on the theory of impulsivedifferential equations, we refer the interested reader to the monographs [4244].

Recently in [45] the notions of q k -derivative and q k -integral on finite intervals were introduced. Let usrecall here these notions. For a fixed kN{0} let J k :=[ t k , t k + 1 ]R be an interval and 0< q k <1 be a constant. We define q k -derivative of a function f: J k R at a point t J k as follows.

Definition 1.1 Assume f: J k R is a continuous function and lett J k . Then the expression

D q k f ( t ) = f ( t ) f ( q k t + ( 1 q k ) t k ) ( 1 q k ) ( t t k ) , t t k , D q k f ( t k ) = lim t t k D q k f ( t ) ,
(1.1)

is called the q k -derivative of function f at t.

We say that f is q k -differentiable on J k provided D q k f(t) exists for all t J k . Note that if t k =0 and q k =q in (1.1), then D q k f= D q f, where D q is the well-known q-derivative of thefunction f(t) defined by

D q f(t)= f ( t ) f ( q t ) ( 1 q ) t .
(1.2)

In addition, we should define the higher q k -derivative of functions.

Definition 1.2 Let f: J k R be a continuous function, we call the second-order q k -derivative D q k 2 f provided D q k f is q k -differentiable on J k with D q k 2 f= D q k ( D q k f): J k R. Similarly, we define higher order q k -derivative D q k n : J k R.

The q k -integral is defined as follows.

Definition 1.3 Assume f: J k R is a continuous function. Then the q k -integral is defined by

t k t f(s) d q k s=(1 q k )(t t k ) n = 0 q k n f ( q k n t + ( 1 q k n ) t k )
(1.3)

for t J k . Moreover, if a( t k ,t) then the definite q k -integral is defined by

a t f ( s ) d q k s = t k t f ( s ) d q k s t k a f ( s ) d q k s = ( 1 q k ) ( t t k ) n = 0 q k n f ( q k n t + ( 1 q k n ) t k ) ( 1 q k ) ( a t k ) n = 0 q k n f ( q k n a + ( 1 q k n ) t k ) .

Note that if t k =0 and q k =q, then (1.3) reduces to q-integral of afunction f(t), defined by 0 t f(s) d q s=(1q)t n = 0 q n f( q n t) for t[0,).

For the basic properties of the q k -derivative and q k -integral we refer to [45].

In this paper we combine all the above subjects and investigate the nonlinearsecond-order impulsive q k -difference Langevin equation with boundary conditionsof the form

{ D q k ( D q k + λ ) x ( t ) = f ( t , x ( t ) ) , t J , t t k , Δ x ( t k ) = I k ( x ( t k ) ) , k = 1 , 2 , , m , D q k x ( t k + ) D q k 1 x ( t k ) = I k ( x ( t k ) ) , k = 1 , 2 , , m , α x ( 0 ) + β D q 0 x ( 0 ) = x ( T ) , γ x ( 0 ) + η D q 0 x ( 0 ) = D q m x ( T ) ,
(1.4)

where 0= t 0 < t 1 < t 2 << t k << t m < t m + 1 =T, f:J×RR is a continuous function, λ is a givenconstant, I k , I k C(R,R), Δx( t k )=x( t k + )x( t k ) for k=1,2,,m, x( t k + )= lim h 0 + x( t k +h), 0< q k <1 for k=0,1,2,,m, and α, β,γ, η are given constants.

The rest of this paper is organized as follows. In Section 2, we present apreliminary result which will be used in this paper. In Section 3, we willconsider the existence results for problem (1.4) while in Section 4, we willgive examples to illustrate our main results.

2 An auxiliary lemma

In this section, we present an auxiliary lemma which will be used throughout thispaper. Let J=[0,T], J 0 =[ t 0 , t 1 ], J k =( t k , t k + 1 ] for k=1,2,,m.

Lemma 2.1 LetλT(η+βλα)(α1)(η1)+γ(Tβ). The unique solution of problem (1.4) isgiven by

x ( t ) = δ 1 + δ 2 t Ω { k = 1 m ( t k 1 t k t k 1 s f ( r , x ( r ) ) d q k 1 r d q k 1 s λ t k 1 t k x ( s ) d q k 1 s + I k ( x ( t k ) ) ) + k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) ( T t k ) + t m T t m s f ( r , x ( r ) ) d q m r d q m s λ t m T x ( s ) d q m s } + δ 3 + δ 4 t Ω { k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) + t m T f ( s , x ( s ) ) d q m s } + 0 < t k < t ( t k 1 t k t k 1 s f ( r , x ( r ) ) d q k 1 r d q k 1 s λ t k 1 t k x ( s ) d q k 1 s + I k ( x ( t k ) ) ) + 0 < t k < t ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) ( t t k ) + t k t t k s f ( r , x ( r ) ) d q k r d q k s λ t k t x ( s ) d q k s ,
(2.1)

with 0 < 0 ()=0, where

Ω = ( α 1 ) ( η 1 ) λ T ( η + β λ α ) + γ ( T β ) , δ 1 = η 1 + β λ , δ 2 = λ ( η + β λ 1 α ) + 1 γ , δ 3 = T β , δ 4 = α 1 β λ .

Proof For t J 0 using q 0 -integral for the first equation of (1.4), we get

D q 0 x(t)= D q 0 x(0)+λx(0)+ 0 t f ( s , x ( s ) ) d q 0 sλx(t).

Setting x(0)=A and D q 0 x(0)=B, we have

D q 0 x(t)=λA+B+ 0 t f ( s , x ( s ) ) d q 0 sλx(t),
(2.2)

which leads to

D q 0 x( t 1 )=λA+B+ 0 t 1 f ( s , x ( s ) ) d q 0 sλx( t 1 ).
(2.3)

For t J 0 we obtain by q 0 -integrating (2.2),

x(t)=A+(λA+B)t+ 0 t 0 s f ( r , x ( r ) ) d q 0 r d q 0 sλ 0 t x(s) d q 0 s.

In particular, for t= t 1

x( t 1 )=A+(λA+B) t 1 + 0 t 1 0 s f ( r , x ( r ) ) d q 0 r d q 0 sλ 0 t 1 x(s) d q 0 s.
(2.4)

For t J 1 =( t 1 , t 2 ], q 1 -integrating (1.4), we have

D q 1 x(t)= D q 1 x ( t 1 + ) +λx ( t 1 + ) + t 1 t f ( s , x ( s ) ) d q 1 sλx(t).

From the second impulsive equations of (1.4), we have

D q 1 x ( t ) = λ A + B + 0 t 1 f ( s , x ( s ) ) d q 0 s + I 1 ( x ( t 1 ) ) + λ I 1 ( x ( t 1 ) ) + t 1 t f ( s , x ( s ) ) d q 1 s λ x ( t ) .
(2.5)

Applying q 1 -integral to (2.5) for t J 1 , we obtain

x ( t ) = x ( t 1 + ) + [ λ A + B + 0 t 1 f ( s , x ( s ) ) d q 0 s + I 1 ( x ( t 1 ) ) + λ I 1 ( x ( t 1 ) ) ] ( t t 1 ) + t 1 t t 1 s f ( r , x ( r ) ) d q 1 r d q 1 s λ t 1 t x ( s ) d q 1 s .
(2.6)

Using the second impulsive equation of (1.4) with (2.4) and (2.6), one has

x ( t ) = A + ( λ A + B ) t 1 + 0 t 1 0 s f ( r , x ( r ) ) d q 0 r d q 0 s λ 0 t 1 x ( s ) d q 0 s + I 1 ( x ( t 1 ) ) + [ λ A + B + 0 t 1 f ( s , x ( s ) ) d q 0 s + I 1 ( x ( t 1 ) ) + λ I 1 ( x ( t 1 ) ) ] ( t t 1 ) + t 1 t t 1 s f ( r , x ( r ) ) d q 1 r d q 1 s λ t 1 t x ( s ) d q 1 s = A + ( λ A + B ) t + 0 t 1 0 s f ( r , x ( r ) ) d q 0 r d q 0 s λ 0 t 1 x ( s ) d q 0 s + I 1 ( x ( t 1 ) ) + [ 0 t 1 f ( s , x ( s ) ) d q 0 s + I 1 ( x ( t 1 ) ) + λ I 1 ( x ( t 1 ) ) ] ( t t 1 ) + t 1 t t 1 s f ( r , x ( r ) ) d q 1 r d q 1 s λ t 1 t x ( s ) d q 1 s .

Repeating the above process, for tJ, we get

x ( t ) = A + ( λ A + B ) t + 0 < t k < t ( t k 1 t k t k 1 s f ( r , x ( r ) ) d q k 1 r d q k 1 s λ t k 1 t k x ( s ) d q k 1 s + I k ( x ( t k ) ) ) + 0 < t k < t ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) ( t t k ) + t k t t k s f ( r , x ( r ) ) d q k r d q k s λ t k t x ( s ) d q k s .
(2.7)

For t=T, we get

x ( T ) = ( 1 + λ T ) A + B T + k = 1 m ( t k 1 t k t k 1 s f ( r , x ( r ) ) d q k 1 r d q k 1 s λ t k 1 t k x ( s ) d q k 1 s + I k ( x ( t k ) ) ) + k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) ( T t k ) + t m T t m s f ( r , x ( r ) ) d q m r d q m s λ t m T x ( s ) d q m s .
(2.8)

It is easy to see that

D q k x ( t ) = λ A + B + 0 < t k < t ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) + t k t f ( s , x ( s ) ) d q k s λ x ( t ) .

For t=T and using x(T)=αA+βB, we have

D q m x ( T ) = λ A + B + k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) + t m T f ( s , x ( s ) ) d q m s λ x ( T ) = ( 1 α ) λ A + ( 1 λ β ) B + k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) + t m T f ( s , x ( s ) ) d q m s .
(2.9)

Applying the boundary conditions of (1.4) with (2.8) and (2.9), it follows that

A = η + λ β 1 Ω { k = 1 m ( t k 1 t k t k 1 s f ( r , x ( r ) ) d q k 1 r d q k 1 s λ t k 1 t k x ( s ) d q k 1 s + I k ( x ( t k ) ) ) + k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) ( T t k ) + t m T t m s f ( r , x ( r ) ) d q m r d q m s λ t m T x ( s ) d q m s } β T Ω { k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) + t m T f ( s , x ( s ) ) d q m s }

and

B = α 1 λ T Ω { k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) + t m T f ( s , x ( s ) ) d q m s } γ 1 + α λ Ω { k = 1 m ( t k 1 t k t k 1 s f ( r , x ( r ) ) d q k 1 r d q k 1 s λ t k 1 t k x ( s ) d q k 1 s + I k ( x ( t k ) ) ) + k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) ( T t k ) + t m T t m s f ( r , x ( r ) ) d q m r d q m s λ t m T x ( s ) d q m s } .

Substituting the values of A and B into (2.7), we get (2.1) asrequired. The proof is completed. □

3 Main results

Let PC(J,R) = {x:JR: x(t) is continuous everywhere except for some t k at which x( t k + ) and x( t k ) exist and x( t k )=x( t k ), k=1,2,,m}. PC(J,R) is a Banach space with the norm x P C =sup{|x(t)|;tJ}.

From Lemma 2.1, we define an operator S:PC(J,R)PC(J,R) by

( S x ) ( t ) = δ 1 + δ 2 t Ω { k = 1 m ( t k 1 t k t k 1 s f ( r , x ( r ) ) d q k 1 r d q k 1 s λ t k 1 t k x ( s ) d q k 1 s + I k ( x ( t k ) ) ) + k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) ( T t k ) + t m T t m s f ( r , x ( r ) ) d q m r d q m s λ t m T x ( s ) d q m s } + δ 3 + δ 4 t Ω { k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) + t m T f ( s , x ( s ) ) d q m s } + 0 < t k < t ( t k 1 t k t k 1 s f ( r , x ( r ) ) d q k 1 r d q k 1 s λ t k 1 t k x ( s ) d q k 1 s + I k ( x ( t k ) ) ) + 0 < t k < t ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) ( t t k ) + t k t t k s f ( r , x ( r ) ) d q k r d q k s λ t k t x ( s ) d q k s ,
(3.1)

where constants δ 1 , δ 2 , δ 3 , δ 4 , and Ω are defined as in Lemma 2.1. Itshould be noticed that problem (1.4) has solutions if and only if the operator has fixed points.

Our first result is an existence and uniqueness result for the impulsive boundaryvalue problem (1.4) by using the Banach contraction mapping principle.

For convenience, we set

Λ 1 = ( | δ 1 | + | δ 2 | T + | Ω | | Ω | ) [ L 1 k = 1 m + 1 ( t k t k 1 ) 2 1 + q k 1 + | λ | k = 1 m + 1 ( t k t k 1 ) + m L 2 + k = 1 m ( L 1 ( t k t k 1 ) + L 3 + | λ | L 2 ) ( T t k ) ] + ( | δ 3 | + | δ 4 | T | Ω | ) [ L 1 T + m L 3 + m | λ | L 2 ]
(3.2)

and

Λ 2 = ( | δ 1 | + | δ 2 | T + | Ω | | Ω | ) [ K 1 k = 1 m + 1 ( t k t k 1 ) 2 1 + q k 1 + m K 2 + k = 1 m ( K 1 ( t k t k 1 ) + K 3 + | λ | K 2 ) ( T t k ) ] + ( | δ 3 | + | δ 4 | T | Ω | ) [ K 1 T + m K 3 + m | λ | K 2 ] .
(3.3)

Theorem 3.1 Assume that the following conditions hold:

(H1) f:[0,T]×RRis a continuous function and there exists a constant L 1 >0such that

|f(t,x)f(t,y)| L 1 |xy|

for eachtJandx,yR.

(H2) The functions I k , I k :RRare continuous and there exist constants L 2 , L 3 >0such that

| I k (x) I k (y)| L 2 |xy|and| I k (x) I k (y)| L 3 |xy|

for eachx,yR, k=1,2,,m.

If

Λ 1 δ<1,
(3.4)

where Λ 1 is defined by (3.2), then the impulsive q k -difference Langevin boundary value problem(1.4) has a unique solution on J.

Proof Firstly, we transform the impulsive q k -difference Langevin boundary value problem (1.4) intoa fixed point problem, x=Sx, where the operator is defined by (3.1).Applying the Banach contraction mapping principle, we shall show that has a fixed point which is the unique solution of theboundary value problem (1.4).

Let K 1 , K 2 , and K 3 be nonnegative constants such that K 1 = sup t J |f(t,0)|, K 2 =sup{| I k (0)|:k=1,2,,m}, and K 3 =sup{| I k (0)|:k=1,2,,m}. We choose a suitable constant ρ by

ρ Λ 2 1 ε ,

where δε<1 and Λ 2 defined by (3.3). Now, we will show thatS B ρ B ρ , where a set B ρ is defined as B ρ ={xPC(J,R):xρ}. For x B ρ , we have

S x sup t J { | δ 1 | + | δ 2 | t | Ω | { k = 1 m ( t k 1 t k t k 1 s | f ( r , x ( r ) ) | d q k 1 r d q k 1 s + | λ | t k 1 t k | x ( s ) | d q k 1 s + | I k ( x ( t k ) ) | ) + k = 1 m ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | + | λ | | I k ( x ( t k ) ) | ) ( T t k ) + t m T t m s | f ( r , x ( r ) ) | d q m r d q m s + | λ | t m T | x ( s ) | d q m s } + | δ 3 | + | δ 4 | t | Ω | { k = 1 m ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | + | λ | | I k ( x ( t k ) ) | ) + t m T | f ( s , x ( s ) ) | d q m s } + 0 < t k < t ( t k 1 t k t k 1 s | f ( r , x ( r ) ) | d q k 1 r d q k 1 s + | λ | t k 1 t k | x ( s ) | d q k 1 s + | I k ( x ( t k ) ) | ) + 0 < t k < t ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | + | λ | | I k ( x ( t k ) ) | ) ( t t k ) + t k t t k s | f ( r , x ( r ) ) | d q k r d q k s + | λ | t k t | x ( s ) | d q k s } | δ 1 | + | δ 2 | T | Ω | { k = 1 m ( t k 1 t k t k 1 s ( | f ( r , x ( r ) ) f ( r , 0 ) | + | f ( r , 0 ) | ) d q k 1 r d q k 1 s + | λ | t k 1 t k x d q k 1 s + ( | I k ( x ( t k ) ) I k ( 0 ) | + | I k ( 0 ) | ) ) + k = 1 m ( t k 1 t k ( | f ( s , x ( s ) ) f ( s , 0 ) | + | f ( s , 0 ) | ) d q k 1 s + ( | I k ( x ( t k ) ) I k ( 0 ) | + | I k ( 0 ) | ) + | λ | ( | I k ( x ( t k ) ) I k ( x ( 0 ) ) | + | I k ( x ( 0 ) ) | ) ) ( T t k ) + t m T t m s ( | f ( r , x ( r ) ) f ( r , 0 ) | + | f ( r , 0 ) | ) d q m r d q m s + | λ | t m T x d q m s } + | δ 3 | + | δ 4 | T | Ω | { k = 1 m ( t k 1 t k ( | f ( s , x ( s ) ) f ( s , 0 ) | + | f ( s , 0 ) | ) d q k 1 s + ( | I k ( x ( t k ) ) I k ( 0 ) | + | I k ( 0 ) | ) + | λ | ( | I k ( x ( t k ) ) I k ( x ( 0 ) ) | + | I k ( x ( 0 ) ) | ) ) + t m T ( | f ( s , x ( s ) ) f ( s , 0 ) | + | f ( s , 0 ) | ) d q m s } + k = 1 m ( t k 1 t k t k 1 s ( | f ( r , x ( r ) ) f ( r , 0 ) | + | f ( r , 0 ) | ) d q k 1 r d q k 1 s + | λ | t k 1 t k x d q k 1 s + ( | I k ( x ( t k ) ) I k ( x ( 0 ) ) | + | I k ( x ( 0 ) ) | ) ) + k = 1 m ( t k 1 t k ( | f ( s , x ( s ) ) f ( s , 0 ) | + | f ( s , 0 ) | ) d q k 1 s + ( | I k ( x ( t k ) ) I k ( 0 ) | + | I k ( 0 ) | ) + | λ | ( | I k ( x ( t k ) ) I k ( x ( 0 ) ) | + | I k ( x ( 0 ) ) | ) ) ( T t k ) + t m T t m s ( | f ( r , x ( r ) ) f ( r , 0 ) | + | f ( r , 0 ) | ) d q m r d q m s + | λ | t m T x d q m s | δ 1 | + | δ 2 | T | Ω | { k = 1 m ( t k 1 t k t k 1 s ( L 1 ρ + K 1 ) d q k 1 r d q k 1 s + | λ | t k 1 t k ρ d q k 1 s + ( L 2 ρ + K 2 ) ) + k = 1 m ( t k 1 t k ( L 1 ρ + K 1 ) d q k 1 s + ( L 3 ρ + K 3 ) + | λ | ( L 2 ρ + K 2 ) ) ( T t k ) + t m T t m s ( L 1 ρ + K 1 ) d q m r d q m s + | λ | t m T ρ d q m s } + | δ 3 | + | δ 4 | T | Ω | { k = 1 m ( t k 1 t k ( L 1 ρ + K 1 ) d q k 1 s + ( L 3 ρ + K 3 ) + | λ | ( L 2 ρ + K 2 ) ) + t m T ( L 1 ρ + K 1 ) d q m s } + k = 1 m ( t k 1 t k t k 1 s ( L 1 ρ + K 1 ) d q k 1 r d q k 1 s + | λ | t k 1 t k ρ d q k 1 s + ( L 2 ρ + K 2 ) ) + k = 1 m ( t k 1 t k ( L 1 ρ + K 1 ) d q k 1 s + ( L 3 ρ + K 3 ) + | λ | ( L 2 ρ + K 2 ) ) ( T t k ) + t m T t m s ( L 1 ρ + K 1 ) d q m r d q m s + | λ | t m T ρ d q m s = Λ 1 ρ + Λ 2 ( δ + 1 ε ) ρ ρ ,

which implies that S B ρ B ρ .

For any x,yPC(J,R) and for each tJ, we have

| S x ( t ) S y ( t ) | | δ 1 | + | δ 2 | t | Ω | { k = 1 m ( t k 1 t k t k 1 s | f ( r , x ( r ) ) f ( r , y ( r ) ) | d q k 1 r d q k 1 s + | λ | t k 1 t k | x ( s ) y ( s ) | d q k 1 s + | I k ( x ( t k ) ) I k ( y ( t k ) ) | ) + k = 1 m ( t k 1 t k | f ( s , x ( s ) ) f ( s , y ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) I k ( y ( t k ) ) | + | λ | | I k ( x ( t k ) ) I k ( y ( t k ) ) | ) ( T t k ) + t m T t m s | f ( r , x ( r ) ) f ( r , y ( r ) ) | d q m r d q m s + | λ | t m T | x ( s ) y ( s ) | d q m s } + | δ 3 | + | δ 4 | t | Ω | { k = 1 m ( t k 1 t k | f ( s , x ( s ) ) f ( s , y ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) I k ( y ( t k ) ) | + | λ | | I k ( x ( t k ) ) I k ( y ( t k ) ) | ) + t m T | f ( s , x ( s ) ) f ( s , y ( s ) ) | d q m s } + 0 < t k < t ( t k 1 t k t k 1 s | f ( r , x ( r ) ) f ( r , y ( r ) ) | d q k 1 r d q k 1 s + | λ | t k 1 t k | x ( s ) y ( s ) | d q k 1 s + | I k ( x ( t k ) ) I k ( y ( t k ) ) | ) + 0 < t k < t ( t k 1 t k | f ( s , x ( s ) ) f ( s , y ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) I k ( y ( t k ) ) | + | λ | | I k ( x ( t k ) ) I k ( y ( t k ) ) | ) ( t t k ) + t k t t k s | f ( r , x ( r ) ) f ( r , y ( r ) ) | d q k r d q k s + | λ | t k t | x ( s ) y ( s ) | d q k s | δ 1 | + | δ 2 | T | Ω | x y { k = 1 m ( L 1 ( t k t k 1 ) 2 1 + q k 1 + | λ | ( t k t k 1 ) + L 2 ) + k = 1 m ( L 1 ( t k t k 1 ) + L 3 + | λ | L 2 ) ( T t k ) + L 1 ( T t m ) 2 1 + q m + | λ | ( T t m ) } + | δ 3 | + | δ 4 | T | Ω | x y { k = 1 m ( L 1 ( t k t k 1 ) + L 3 + | λ | L 2 ) + L 1 ( T t m ) } + x y k = 1 m ( L 1 ( t k t k 1 ) 2 1 + q k 1 + | λ | ( t k t k 1 ) + L 2 ) + x y k = 1 m ( L 1 ( t k t k 1 ) + L 3 + | λ | L 2 ) ( T t k ) + L 1 ( T t m ) 2 1 + q m x y + | λ | ( T t m ) x y = Λ 1 x y ,

which implies that SxSy Λ 1 xy. As Λ 1 <1, is a contraction.Therefore, by the Banach contraction mapping principle, we find that has a fixed point which is the unique solution of problem(1.4). □

The second existence result is based on Schaefer’s fixed point theorem.

Theorem 3.2 Assume that the following conditions hold:

(H3) f:J×RRis a continuous function and there exists a constant M 1 >0such that

|f(t,x)| M 1

for eachtJand allxR.

(H4) The functions I k , I k :RRare continuous and there exist constants M 2 , M 3 >0such that

| I k (x)| M 2 and| I k (x)| M 3

for allxR, k=1,2,,m.

If

| δ 1 | + | δ 2 | T + | Ω | | Ω | |λ|T<1,
(3.5)

then the impulsive q k -difference Langevin boundary value problem(1.4) has at least one solution on J.

Proof We shall use Schaefer’s fixed point theorem to prove that theoperator defined by (3.1) has a fixed point. We divide the proof intofour steps.

Step 1: Continuity of.

Let { x n } be a sequence such that x n x in PC(J,R). Since f is a continuous function onJ×R and I k , I k are continuous functions on fork=1,2,,m, we have

f ( t , x n ( t ) ) f ( t , x ( t ) ) , I k ( x n ( t k ) ) I k ( x ( t k ) ) and I k ( x n ( t k ) ) I k ( x ( t k ) )

for k=1,2,,m, as n.

Then, for each tJ, we get

| ( S x n ) ( t ) ( S x ) ( t ) | | δ 1 | + | δ 2 | t | Ω | { k = 1 m ( t k 1 t k t k 1 s | f ( r , x n ( r ) ) f ( r , x ( r ) ) | d q k 1 r d q k 1 s + λ t k 1 t k | x n ( s ) x ( s ) | d q k 1 s + | I k ( x n ( t k ) ) I k ( x ( t k ) ) | ) + k = 1 m ( t k 1 t k | f ( s , x n ( s ) ) f ( s , x ( s ) ) | d q k 1 s + | I k ( x n ( t k ) ) I k ( x ( t k ) ) | + | λ | | I k ( x n ( t k ) ) I k ( x ( t k ) ) | ) ( T t k ) + t m T t m s | f ( r , x n ( r ) ) f ( r , x ( r ) ) | d q m r d q m s + | λ | t m T | x n ( s ) x ( s ) | d q m s } + | δ 3 | + | δ 4 | t Ω { k = 1 m ( t k 1 t k | f ( s , x n ( s ) ) f ( s , x ( s ) ) | d q k 1 s + | I k ( x n ( t k ) ) I k ( x ( t k ) ) | + | λ | | I k ( x n ( t k ) ) I k ( x ( t k ) ) | ) + t m T | f ( s , x n ( s ) ) f ( s , x ( s ) ) | d q m s } + 0 < t k < t ( t k 1 t k t k 1 s | f ( r , x n ( r ) ) f ( r , x ( r ) ) | d q k 1 r d q k 1 s + | λ | t k 1 t k | x n ( s ) x ( s ) | d q k 1 s + | I k ( x n ( t k ) ) I k ( x ( t k ) ) | ) + 0 < t k < t ( t k 1 t k | f ( s , x n ( s ) ) f ( s , x ( s ) ) | d q k 1 s + | I k ( x n ( t k ) ) I k ( x ( t k ) ) | + | λ | | I k ( x n ( t k ) ) I k ( x ( t k ) ) | ) ( t t k ) + t k t t k s | f ( r , x n ( r ) ) f ( r , x ( r ) ) | d q k r d q k s + | λ | t k t | x n ( s ) x ( s ) | d q k s ,

which gives S x n Sx0 as n. This means that is continuous.

Step 2: maps bounded sets into bounded sets inPC(J,R).

Let us prove that for any ρ >0, there exists a positive constant σsuch that for each x B ρ ={xPC(J,R):x ρ }, we have Sxσ. For any x B ρ , we have

| ( S x ) ( t ) | | δ 1 | + | δ 2 | T | Ω | { k = 1 m ( t k 1 t k t k 1 s | f ( r , x ( r ) ) | d q k 1 r d q k 1 s + | λ | t k 1 t k | x ( s ) | d q k 1 s + | I k ( x ( t k ) ) | ) + k = 1 m ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | + | λ | | I k ( x ( t k ) ) | ) ( T t k ) + t m T t m s | f ( r , x ( r ) ) | d q m r d q m s + | λ | t m T | x ( s ) | d q m s } + | δ 3 | + | δ 4 | T | Ω | { k = 1 m ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | + | λ | | I k ( x ( t k ) ) | ) + t m T | f ( s , x ( s ) ) | d q m s } + k = 1 m ( t k 1 t k t k 1 s | f ( r , x ( r ) ) | d q k 1 r d q k 1 s + | λ | t k 1 t k | x ( s ) | d q k 1 s + | I k ( x ( t k ) ) | ) + k = 1 m ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | + | λ | | I k ( x ( t k ) ) | ) ( T t k ) + t m T t m s | f ( r , x ( r ) ) | d q m r d q m s + | λ | t m T | x ( s ) | d q m s | δ 1 | + | δ 2 | T | Ω | { k = 1 m ( M 1 ( t k t k 1 ) 2 1 + q k 1 + ρ | λ | ( t k t k 1 ) + M 2 ) + k = 1 m ( M 1 ( t k t k 1 ) + M 3 + | λ | M 2 ) ( T t k ) + M 1 ( T t m ) 2 1 + q m + ρ | λ | ( T t m ) } + | δ 3 | + | δ 4 | T | Ω | { k = 1 m ( M 1 ( t k t k 1 ) + M 3 + | λ | M 2 ) + M 1 ( T t m ) } + k = 1 m ( M 1 ( t k t k 1 ) 2 1 + q k 1 + ρ | λ | ( t k t k 1 ) + M 2 ) + ρ | λ | ( T t m ) + k = 1 m ( M 1 ( t k t k 1 ) + M 3 + | λ | M 2 ) ( T t k ) + M 1 ( T t m ) 2 1 + q m : = σ .

Hence, we deduce that Sxσ.

Step 3: maps bounded sets into equicontinuous sets ofPC(J,R).

Let τ 1 , τ 2 J i =( t i , t i + 1 ] for some i{0,1,2,,m}, τ 1 < τ 2 , B ρ be a bounded set of PC(J,R) as in Step 2, and let x B ρ . Then we have

| ( S x ) ( τ 2 ) ( S x ) ( τ 1 ) | | δ 2 | | τ 2 τ 1 | | Ω | { k = 1 m ( t k 1 t k t k 1 s | f ( r , x ( r ) ) | d q k 1 r d q k 1 s + | λ | t k 1 t k | x ( s ) | d q k 1 s + | I k ( x ( t k ) ) | ) + k = 1 m ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | + | λ | | I k ( x ( t k ) ) | ) ( T t k ) + t m T t m s | f ( r , x ( r ) ) | d q m r d q m s + | λ | t m T | x ( s ) | d q m s } + | δ 4 | | τ 2 τ 1 | | Ω | { k = 1 m ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | + | λ | | I k ( x ( t k ) ) | ) + t m T | f ( s , x ( s ) ) | d q m s } + | λ | | t i τ 2 x ( s ) d q i s t i τ 1 x ( s ) d q i s | + | τ 2 τ 1 | k = 1 i ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | + | λ | | I k ( x ( t k ) ) | ) + | t i τ 2 t i s f ( r , x ( r ) ) d q i r d q i s t i τ 1 t i s f ( r , x ( r ) ) d q i r d q i s | | δ 2 | | τ 2 τ 1 | | Ω | { k = 1 m ( M 1 ( t k t k 1 ) 2 1 + q k 1 + ρ | λ | ( t k t k 1 ) + M 2 ) + k = 1 m ( M 1 ( t k t k 1 ) + M 3 + | λ | M 2 ) ( T t k ) + M 1 ( T t m ) 2 1 + q m + ρ | λ | ( T t m ) } + | δ 4 | | τ 2 τ 1 | | Ω | { k = 1 m ( M 1 ( t k t k 1 ) + M 3 + | λ | M 2 ) + M 1 ( T t m ) } + | τ 2 τ 1 | ρ | λ | + | τ 2 τ 1 | k = 1 i ( M 1 ( t k t k 1 ) + M 3 + | λ | M 2 ) + | τ 2 τ 1 | M 1 ( τ 2 + τ 1 + 2 t i ) 1 + q i .

The right-hand side of the above inequality is independent of x and tendsto zero as τ 1 τ 2 . As a consequence of Steps 1 to 3, together with theArzelá-Ascoli theorem, we deduce that S:PC(J,R)PC(J,R) is completely continuous.

Step 4: We show that the set

E= { x P C ( J , R ) : x = κ S x  for some  0 < κ < 1 }

is bounded.

Let xE. Then x(t)=κ(Sx)(t) for some 0<κ<1. Thus, for each tJ, we have

x ( t ) = κ ( S x ) ( t ) = κ ( δ 1 + δ 2 t ) Ω { k = 1 m ( t k 1 t k t k 1 s f ( r , x ( r ) ) d q k 1 r d q k 1 s λ t k 1 t k x ( s ) d q k 1 s + I k ( x ( t k ) ) ) + k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) ( T t k ) + t m T t m s f ( r , x ( r ) ) d q m r d q m s λ t m T x ( s ) d q m s } + κ ( δ 3 + δ 4 t ) Ω { k = 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) + t m T f ( s , x ( s ) ) d q m s } + κ 0 < t k < t ( t k 1 t k t k 1 s f ( r , x ( r ) ) d q k 1 r d q k 1 s λ t k 1 t k x ( s ) d q k 1 s + I k ( x ( t k ) ) ) + κ 0 < t k < t ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) + λ I k ( x ( t k ) ) ) ( t t k ) + κ t k t t k s f ( r , x ( r ) ) d q k r d q k s κ λ t k t x ( s ) d q k s .

This implies by (H3) and (H4) that for eachtJ, we have

x | δ 1 | + | δ 2 | T | Ω | { k = 1 m ( t k 1 t k t k 1 s M 1 d q k 1 r d q k 1 s + | λ | t k 1 t k | x ( s ) | d q k 1 s + M 2 ) + k = 1 m ( t k 1 t k M 1 d q k 1 s + M 3 + | λ | M 2 ) ( T t k ) + t m T t m s M 1 d q m r d q m s + | λ | t m T | x ( s ) | d q m s } + | δ 3 | + | δ 4 | T | Ω | { k = 1 m ( t k 1 t k M 1 d q k 1 s + M 3 + | λ | M 2 ) + t m T M 1 d q m s } + k = 1 m ( t k 1 t k t k 1 s M 1 d q k 1 r d q k 1 s + | λ | t k 1 t k | x ( s ) | d q k 1 s + M 2 ) + k = 1 m ( t k 1 t k M 1 d q k 1 s + M 3 + | λ | M 2 ) ( T t k ) + t m T t m s M 1 d q m r d q m s + | λ | t m T | x ( s ) | d q m s | δ 1 | + | δ 2 | T + | Ω | | Ω | { M 1 k = 1 m + 1 ( t k t k 1 ) 2 1 + q k 1 + | λ | x T + m M 2 + k = 1 m ( M 1 ( t k t k 1 ) + M 3 + | λ | M 2 ) ( T t k ) } + | δ 3 | + | δ 4 | T | Ω | { M 1 T + m M 3 + m | λ | M 2 } .

Setting

Γ = | δ 1 | + | δ 2 | T + | Ω | | Ω | { M 1 k = 1 m + 1 ( t k t k 1 ) 2 1 + q k 1 + m M 2 + k = 1 m ( M 1 ( t k t k 1 ) + M 3 + | λ | M 2 ) ( T t k ) } + | δ 3 | + | δ 4 | T | Ω | { M 1 T + m M 3 + m | λ | M 2 } ,

we have

x | δ 1 | + | δ 2 | T + | Ω | | Ω | |λ|xT+Γ,

which yields

x Γ 1 | δ 1 | + | δ 2 | T + | Ω | | Ω | | λ | T :=M.

This shows that the set E is bounded. As a consequence of Schaefer’sfixed point theorem, we conclude that has a fixed point whichis a solution of the impulsive q k -difference Langevin boundary value problem(1.4). □

4 Examples

Example 4.1 Consider the following boundary value problem for the second-orderimpulsive q k -difference Langevin equation:

{ D ( 2 k + 1 5 k + 2 ) 1 2 ( D ( 2 k + 1 5 k + 2 ) 1 2 + 1 10 ) x ( t ) = t e 2 t ( 10 + t ) 2 | x ( t ) | ( 1 + | x ( t ) | ) , t J = [ 0 , 1 ] , t t k , Δ x ( t k ) = | x ( t k ) | 9 ( 9 + | x ( t k ) | ) , t k = k 10 , k = 1 , 2 , , 9 , D ( 2 k + 1 5 k + 2 ) 1 2 x ( t k + ) D ( 2 k 1 5 k 3 ) 1 2 x ( t k ) = 1 8 tan 1 ( 1 10 x ( t k ) ) , t k = k 10 , k = 1 , 2 , , 9 , 1 7 x ( 0 ) + 2 9 D 1 2 x ( 0 ) = x ( 1 ) , 2 7 x ( 0 ) + 1 9 D 1 2 x ( 0 ) = D 19 47 x ( 1 ) .
(4.1)

Here q k = ( 2 k + 1 ) / ( 5 k + 2 ) , k=0,1,2,,9, m=9, T=1, λ=1/10, α=1/7, β=2/9, γ=2/7, η=1/9, f(t,x)=(t|x(t)|)/( e 2 t ( 10 + t ) 2 (1+|x(t)|)), I k (x)=|x|/(9(9+|x|)), and I k (x)=(1/8) tan 1 (x/10). Since

| f ( t , x ) f ( t , y ) | ( 1 / 100 ) | x y | , | I k ( x ) I k ( y ) | ( 1 / 81 ) | x y | and | I k ( x ) I k ( y ) | ( 1 / 80 ) | x y | ,

then (H1) and (H2) are satisfied with L 1 =(1/100), L 2 =(1/81), L 3 =(1/80). We can find that Ω=3,103/3,150, δ 1 =(13)/15, δ 2 =46/75, δ 3 =7/9, δ 4 =(277)/315 and thus

Λ 1 0.920497882<1.

Hence, by Theorem 3.1, the boundary value problem (4.1) has a unique solutionon [0,1].

Example 4.2 Consider the following boundary value problem for the second-orderimpulsive q k -difference Langevin equation:

{ D ( k + 1 3 k + 4 ) 2 ( D ( k + 1 3 k + 4 ) 2 + 1 5 ) x ( t ) = 3 t 3 ( 4 + x 2 ) 1 2 , t J = [ 0 , 1 ] , t t k , Δ x ( t k ) = 2 k cos 2 π t k + t 2 | x ( t k ) | , t k = k 10 , k = 1 , 2 , , 9 , D ( k + 1 3 k + 4 ) 2 x ( t k + ) D ( k 3 k + 1 ) 2 x ( t k ) = 4 sin ( ( π t ) / 2 ) 3 k + | x ( t k ) | cos 2 2 t , t k = k 10 , k = 1 , 2 , , 9 , 1 4 x ( 0 ) + 1 5 D 1 16 x ( 0 ) = x ( 1 ) , 2 9 x ( 0 ) + 1 7 D 1 16 x ( 0 ) = D 100 961 x ( 1 ) .
(4.2)

Here q k = ( ( k + 1 ) / ( 3 k + 4 ) ) 2 , k=0,1,2,,9, m=9, T=1, λ=1/5, α=1/4, β=1/5, γ=2/9, η=1/7, f(t,x)=((3 t 3 )/ ( 4 + x 2 ) 1 / 2 ), I k (x)=((2k cos 2 πt)/(k+ t 2 |x|)), and I k (x)=((4sin((πt)/2))/(3k+|x| cos 2 2t)). Clearly,

|f(t,x)|=| 3 t 3 ( 4 + x 2 ) 1 2 | 3 2 ,| I k (x)|=| 2 k cos 2 π t k + t 2 | x | |2

and

| I k (x)|=| 4 sin ( ( π t ) / 2 )