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

Boundary Value Problems20142014:85

https://doi.org/10.1186/1687-2770-2014-85

Received: 11 November 2013

Accepted: 7 April 2014

Published: 2 May 2014

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.

Keywords

q k -derivative q k -integralimpulsive q k -difference equationexistenceuniquenessLangevin equation

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 k N { 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 let t 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 = ( 1 q ) 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 × R R is a continuous function, λ is a givenconstant, I k , I k C ( R , R ) , Δ x ( 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 Let λ T ( η + β λ α ) ( α 1 ) ( η 1 ) + γ ( T β ) . 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 ) + λ x ( 0 ) + 0 t f ( s , x ( s ) ) d q 0 s λ x ( t ) .
Setting x ( 0 ) = A and D q 0 x ( 0 ) = B , we have
D q 0 x ( t ) = λ A + B + 0 t f ( s , x ( s ) ) d q 0 s λ x ( t ) ,
(2.2)
which leads to
D q 0 x ( t 1 ) = λ A + B + 0 t 1 f ( s , x ( s ) ) d q 0 s λ x ( t 1 ) .
(2.3)
For t J 0 we obtain by q 0 -integrating (2.2),
x ( t ) = A + ( λ A + B ) t + 0 t 0 s f ( r , x ( r ) ) d q 0 r d q 0 s λ 0 t x ( s ) d q 0 s .
In particular, for t = t 1
x ( t 1 ) = 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 .
(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 + ) + λ x ( t 1 + ) + t 1 t f ( s , x ( s ) ) d q 1 s λ x ( 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 t J , 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 ) = α A + β B , 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 P C ( J , R ) = { x : J R : 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 }. P C ( J , R ) is a Banach space with the norm x P C = sup { | x ( t ) | ; t J } .

From Lemma 2.1, we define an operator S : P C ( J , R ) P C ( 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 ] × R R is a continuous function and there exists a constant L 1 > 0 such that
| f ( t , x ) f ( t , y ) | L 1 | x y |

for each t J and x , y R .

(H2) The functions I k , I k : R R are continuous and there exist constants L 2 , L 3 > 0 such that
| I k ( x ) I k ( y ) | L 2 | x y | and | I k ( x ) I k ( y ) | L 3 | x y |

for each x , y R , 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 = S x , 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 δ ε < 1 and Λ 2 defined by (3.3). Now, we will show that S B ρ B ρ , where a set B ρ is defined as B ρ = { x P C ( J , R ) : x ρ } . 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 , y P C ( J , R ) and for each t J , 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 S x S y Λ 1 x y . 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 × R R is a continuous function and there exists a constant M 1 > 0 such that
| f ( t , x ) | M 1

for each t J and all x R .

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

for all x R , 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 P C ( J , R ) . Since f is a continuous function on J × R and I k , I k are continuous functions on for k = 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 t J , 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 S x 0 as n . This means that is continuous.

Step 2: maps bounded sets into bounded sets in P C ( J , R ) .

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

Step 3: maps bounded sets into equicontinuous sets of P C ( 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 P C ( 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 : P C ( J , R ) P C ( 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 x E . Then x ( t ) = κ ( S x ) ( t ) for some 0 < κ < 1 . Thus, for each t J , 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 each t J , 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 | x T + 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 ) = ( ( 2 k cos 2 π t ) / ( k + t 2 | x | ) ) , and I k ( x ) = ( ( 4 sin ( ( π t ) / 2 ) ) / ( 3 k + | x | cos 2 2 t ) ) . 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 ) 3 k + | x | cos 2 2 t | 4 3 .
We can find that
| δ 1 | + | δ 2 | T + | Ω | | Ω | | λ | T = 13 , 958 26 , 273 < 1 ,

where Ω = ( α 1 ) ( η 1 ) λ T ( η + β λ α ) + γ ( T β ) = 26 , 273 / 31 , 500 , δ 1 = η 1 + β λ = ( 143 ) / 175 and δ 2 = λ ( η + β λ 1 α ) + 1 γ = 17 , 777 / 31 , 500 .

Hence, by Theorem 3.2, the boundary value problem (4.2) has at least onesolution on [ 0 , 1 ] .

Authors’ information

Sotiris K Ntouyas is a member of Nonlinear Analysis and Applied Mathematics (NAAM) -Research Group at King Abdulaziz University, Jeddah, Saudi Arabia.

Declarations

Acknowledgements

We would like to thank the reviewers for their valuable comments and suggestionson the manuscript. The research of J Tariboon is supported by KingMongkut’s University of Technology North Bangkok, Thailand.

Authors’ Affiliations

(1)
Nonlinear Dynamic Analysis Research Center, Department of Mathematics,Faculty of Applied Science, King Mongkut’s University of Technology, North Bangkok, Bangkok, Thailand
(2)
Department of Mathematics, University of Ioannina, Ioannina, Greece

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© Tariboon and Ntouyas; licensee Springer. 2014

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