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Three-point boundary value problems for nonlinear second-order impulsive q-difference equations

Advances in Difference Equations20142014:31

https://doi.org/10.1186/1687-1847-2014-31

Received: 11 November 2013

Accepted: 7 January 2014

Published: 27 January 2014

Abstract

The quantum calculus on finite intervals was studied recently by the authors in Adv. Differ. Equ. 2013:282, 2013, where the concepts of q k -derivative and q k -integral of a function f : J k : = [ t k , t k + 1 ] R have been introduced. In this paper, we prove existence and uniqueness results for nonlinear second-order impulsive q k -difference three-point boundary value problems, by using Banach’s contraction mapping principle and Krasnoselskii’s fixed-point theorem.

MSC:26A33, 39A13, 34A37.

Keywords

  • q k -derivative
  • q k -integral
  • impulsive q k -difference equation
  • existence
  • uniqueness
  • three-point boundary conditions
  • fixed-point theorems

1 Introduction

In this article, we investigate the nonlinear second-order impulsive q k -difference equation with three-point boundary conditions
{ D q k 2 x ( t ) = f ( t , x ( t ) ) , t J : = [ 0 , T ] , 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 , x ( 0 ) = 0 , x ( T ) = x ( η ) ,
(1.1)

where 0 = t 0 < t 1 < t 2 < < t k < < t m < t m + 1 = T , f : J × R R is a continuous function, 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 ) , η ( t j , t j + 1 ) a constant for some j { 0 , 1 , 2 , , m } and 0 < q k < 1 for k = 0 , 1 , 2 , , m .

The theory of quantum calculus on finite intervals was developed recently by the authors in [1]. In [1] the concepts of q k -derivative and q k -integral of a function f : J k : = [ t k , t k + 1 ] R , are defined and their basic properties proved. As applications, existence and uniqueness results for initial value problems for first- and second-order impulsive q k -difference equations are proved.

The book by Kac and Cheung [2] covers many of the fundamental aspects of the quantum calculus. In recent years, the topic of q-calculus has attracted the attention of several researchers and a variety of new results can be found in the papers [315] and the references cited therein.

Impulsive differential equations, that is, differential equations involving an impulse effect, appear as a natural description of observed evolution phenomena of several real-world problems. For some monographs on impulsive differential equations we refer to [1618].

In the present paper we prove existence and uniqueness results for the impulsive boundary value problem (1.1) by using Banach’s contraction mapping principle and Krasnoselskii’s fixed-point theorem. The rest of this paper is organized as follows: In Section 2 we present the notions of q k -derivative and q k -integral on finite intervals and collect their properties. The main results are proved in Section 3, while examples illustrating the results are presented in Section 4.

2 Preliminaries

In this section we present the notions of q k -derivative and q k -integral on finite intervals. 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 2.1 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 ) ,
(2.1)

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 (2.1), then D q k f = D q f , where D q is the well-known q-derivative of the function f ( t ) defined by
D q f ( t ) = f ( t ) f ( q t ) ( 1 q ) t .
(2.2)

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

Definition 2.2 Let f : J k R is 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 the higher-order q k -derivative D q k n : J k R .

The properties of the q k -derivative are summarized in the following theorem.

Theorem 2.3 Assume f , g : J k R are q k -differentiable on J k . Then:
  1. (i)
    The sum f + g : J k R is q k -differentiable on J k with
    D q k ( f ( t ) + g ( t ) ) = D q k f ( t ) + D q k g ( t ) .
     
  2. (ii)
    For any constant α, α f : J k R is q k -differentiable on J k with
    D q k ( α f ) ( t ) = α D q k f ( t ) .
     
  3. (iii)
    The product f g : J k R is q k -differentiable on J k with
    D q k ( f g ) ( t ) = f ( t ) D q k g ( t ) + g ( q k t + ( 1 q k ) t k ) D q k f ( t ) = g ( t ) D q k f ( t ) + f ( q k t + ( 1 q k ) t k ) D q k g ( t ) .
     
  4. (iv)
    If g ( t ) g ( q k t + ( 1 q k ) t k ) 0 , then f g is q k -differentiable on J k with
    D q k ( f g ) ( t ) = g ( t ) D q k f ( t ) f ( t ) D q k g ( t ) g ( t ) g ( q k t + ( 1 q k ) t k ) .
     
Definition 2.4 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 ) ,
(2.3)
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 (2.3) reduces to the q-integral of a function 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 , ) .

Theorem 2.5 For t J k , the following formulas hold:
  1. (i)

    D q k t k t f ( s ) d q k s = f ( t ) ;

     
  2. (ii)

    t k t D q k f ( s ) d q k s = f ( t ) ;

     
  3. (iii)

    a t D q k f ( s ) d q k s = f ( t ) f ( a ) for a ( t k , t ) .

     

3 Main results

Let J = [ 0 , T ] , J 0 = [ t 0 , t 1 ] , J k = ( t k , t k + 1 ] for k = 1 , 2 , , m . 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 } .

Lemma 3.1 The unique solution of problem (1.1) is given by
x ( t ) = t k = 1 j ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) ) t T η k = j + 1 m ( t k 1 t k t k 1 s f ( σ , x ( σ ) ) d q k 1 σ d q k 1 s + I k ( x ( t k ) ) ) t T η k = j + 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) ) ( T t k ) + t T η t j η t j s f ( σ , x ( σ ) ) d q j σ d q j s t T η t m T t m s f ( σ , x ( σ ) ) d q m σ d q m s + 0 < t k < t ( t k 1 t k t k 1 s f ( σ , x ( σ ) ) d q k 1 σ 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 ) ) ) ( t t k ) + t k t t k s f ( σ , x ( σ ) ) d q k σ d q k s ,
(3.1)

with 0 < 0 ( ) = 0 .

Proof For t J 0 , taking the q 0 -integral for the first equation of (1.1), we get
D q 0 x ( t ) = D q 0 x ( 0 ) + 0 t f ( s , x ( s ) ) d q 0 s ,
(3.2)
which yields
D q 0 x ( t 1 ) = D q 0 x ( 0 ) + 0 t 1 f ( s , x ( s ) ) d q 0 s .
(3.3)
For t J 0 we obtain by q 0 -integrating (3.2),
x ( t ) = x ( 0 ) + D q 0 x ( 0 ) t + 0 t 0 s f ( σ , x ( σ ) ) d q 0 σ d q 0 s : = A + B t + 0 t 0 s f ( σ , x ( σ ) ) d q 0 σ d q 0 s ( x ( 0 ) = A , D q 0 x ( 0 ) = B ) .
In particular, for t = t 1
x ( t 1 ) = A + B t 1 + 0 t 1 0 s f ( σ , x ( σ ) ) d q 0 σ d q 0 s .
(3.4)
For t J 1 = ( t 1 , t 2 ] , q 1 -integrating (1.1), we have
D q 1 x ( t ) = D q 1 x ( t 1 + ) + t 1 t f ( s , x ( s ) ) d q 1 s .
Using the third condition of (1.1) with (3.3), it follows that
D q 1 x ( t ) = B + 0 t 1 f ( s , x ( s ) ) d q 0 s + I 1 ( x ( t 1 ) ) + t 1 t f ( s , x ( s ) ) d q 1 s .
(3.5)
Taking the q 1 -integral to (3.5) for t J 1 , we obtain
x ( t ) = x ( t 1 + ) + [ B + 0 t 1 f ( s , x ( s ) ) d q 0 s + I 1 ( x ( t 1 ) ) ] ( t t 1 ) + t 1 t t 1 s f ( σ , x ( σ ) ) d q 1 σ d q 1 s .
(3.6)
Applying the second equation of (1.1) with (3.4) and (3.6), we get
x ( t ) = A + B t 1 + 0 t 1 0 s f ( σ , x ( σ ) ) d q 0 σ d q 0 s + I 1 ( x ( t 1 ) ) + [ B + 0 t 1 f ( s , x ( s ) ) d q 0 s + I 1 ( x ( t 1 ) ) ] ( t t 1 ) + t 1 t t 1 s f ( σ , x ( σ ) ) d q 1 σ d q 1 s = A + B t + 0 t 1 0 s f ( σ , x ( σ ) ) d q 0 σ 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 ) ) ] ( t t 1 ) + t 1 t t 1 s f ( σ , x ( σ ) ) d q 1 σ d q 1 s .
Repeating the above process, for t J , we get
x ( t ) = A + B t + 0 < t k < t ( t k 1 t k t k 1 s f ( σ , x ( σ ) ) d q k 1 σ 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 ) ) ) ( t t k ) + t k t t k s f ( σ , x ( σ ) ) d q k σ d q k s .
(3.7)
The first boundary condition of (1.1) implies A = 0 . The second boundary condition of (1.1) yields
k = 1 m ( t k 1 t k t k 1 s f ( σ , x ( σ ) ) d q k 1 σ 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 ) ) ) ( T t k ) + t m T t m s f ( σ , x ( σ ) ) d q m σ d q m s + B T = k = 1 j ( t k 1 t k t k 1 s f ( σ , x ( σ ) ) d q k 1 σ d q k 1 s + I k ( x ( t k ) ) ) + k = 1 j ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) ) ( η t k ) + t j η t j s f ( σ , x ( σ ) ) d q j σ d q j s + B η ,
which implies
B = k = 1 j ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) ) 1 T η k = j + 1 m ( t k 1 t k t k 1 s f ( σ , x ( σ ) ) d q k 1 σ d q k 1 s + I k ( x ( t k ) ) ) 1 T η k = j + 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) ) ( T t k ) + 1 T η t j η t j s f ( σ , x ( σ ) ) d q j σ d q j s 1 T η t m T t m s f ( σ , x ( σ ) ) d q m σ d q m s .

Substituting the constant B into (3.7), we obtain (3.1) as required. □

In view of Lemma 3.1, we define an operator A : P C ( J , R ) P C ( J , R ) by
( A x ) ( t ) = t k = 1 j ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) ) t T η k = j + 1 m ( t k 1 t k t k 1 s f ( σ , x ( σ ) ) d q k 1 σ d q k 1 s + I k ( x ( t k ) ) ) t T η k = j + 1 m ( t k 1 t k f ( s , x ( s ) ) d q k 1 s + I k ( x ( t k ) ) ) ( T t k ) + t T η t j η t j s f ( σ , x ( σ ) ) d q j σ d q j s t T η t m T t m s f ( σ , x ( σ ) ) d q m σ d q m s + 0 < t k < t ( t k 1 t k t k 1 s f ( σ , x ( σ ) ) d q k 1 σ 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 ) ) ) ( t t k ) + t k t t k s f ( σ , x ( σ ) ) d q k σ d q k s .
(3.8)

It should be noticed that problem (1.1) has solutions if and only if the operator A has fixed points.

For convenience, we set
Φ k = [ ( t k t k 1 ) ( T t k ) + ( t k t k 1 ) 2 1 + q k 1 ] M 1 + M 2 + ( T t k ) M 3 ,
(3.9)
Ψ k = [ ( t k t k 1 ) ( T t k ) + ( t k t k 1 ) 2 1 + q k 1 ] L 1 + L 2 + ( T t k ) L 3 ,
(3.10)

for k = 1 , , m .

Theorem 3.2 Assume that:

(H1) The function f : [ 0 , T ] × R R is a continuous 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 : = T k = 1 j [ ( t k t k 1 ) L 1 + L 3 ] + T T η k = j + 1 m Ψ k + T L 1 T η ( ( η t j ) 2 1 + q j + ( T t m ) 2 1 + q m ) + k = 1 m Ψ k + ( T t m ) 2 1 + q m L 1 δ < 1 ,
(3.11)

then the impulsive q k -difference boundary value problem (1.1) has a unique solution on J.

Proof First, we transform the problem (1.1) into a fixed-point problem, x = A x , where the operator A is defined by (3.8). By using Banach’s contraction principle, we shall show that A has a fixed point which is the unique solution of problem (1.1).

Set sup t J | f ( t , 0 ) | = M 1 < , sup { | I k ( 0 ) | : k = 1 , 2 , , m } = M 2 < , sup { | I k ( 0 ) | : k = 1 , 2 , , m } = M 3 < and a constant
ρ = T k = 1 j [ ( t k t k 1 ) M 1 + M 3 ] + T T η k = j + 1 m Φ k + T M 1 T η ( ( η t j ) 2 1 + q j + ( T t m ) 2 1 + q m ) + k = 1 m Φ k + ( T t m ) 2 1 + q m M 1 .
(3.12)
Choosing r ρ 1 ε , where δ ε < 1 , we show that A B r B r , where B r = { x P C ( J , R ) : x r } . For x B r , we have
A x sup t J { t k = 1 j ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | ) + t T η k = j + 1 m ( t k 1 t k t k 1 s | f ( σ , x ( σ ) ) | d q k 1 σ d q k 1 s + | I k ( x ( t k ) ) | ) + t T η k = j + 1 m ( t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | I k ( x ( t k ) ) | ) ( T t k ) + t T η t j η t j s | f ( σ , x ( σ ) ) | d q j σ d q j s + t T η t m T t m s | f ( σ , x ( σ ) ) | d q m σ d q m s + 0 < t k < t ( t k 1 t k t k 1 s | f ( σ , x ( σ ) ) | d q k 1 σ 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 ) ) | ) ( t t k ) + t k t t k s | f ( σ , x ( σ ) ) | d q k σ d q k s } T k = 1 j ( 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 ) | ) + T T η k = j + 1 m ( t k 1 t k t k 1 s ( | f ( σ , x ( σ ) ) f ( σ , 0 ) | + | f ( σ , 0 ) | ) d q k 1 σ d q k 1 s + | I k ( x ( t k ) ) I k ( 0 ) | + | I k ( 0 ) | ) + T T η k = j + 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 ) | ) ( T t k ) + T T η t j η t j s ( | f ( σ , x ( σ ) ) f ( σ , 0 ) | + | f ( σ , 0 ) | ) d q j σ d q j s + T T η t m T t m s ( | f ( σ , x ( σ ) ) f ( σ , 0 ) | + | f ( σ , 0 ) | ) d q m σ d q m s + k = 1 m ( t k 1 t k t k 1 s ( | f ( σ , x ( σ ) ) f ( σ , 0 ) | + | f ( σ , 0 ) | ) d q k 1 σ 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 ) | ) ( T t k ) + t m T t m s ( | f ( σ , x ( σ ) ) f ( σ , 0 ) | + | f ( σ , 0 ) | ) d q m σ d q m s T k = 1 j ( ( t k t k 1 ) ( L 1 r + M 1 ) + L 3 r + M 3 ) + T T η k = j + 1 m ( ( t k t k 1 ) 2 1 + q k 1 ( L 1 r + M 1 ) + L 2 r + M 2 ) + T T η k = j + 1 m ( ( t k t k 1 ) ( L 1 r + M 1 ) + L 3 r + M 3 ) ( T t k ) + T T η ( ( η t j ) 2 1 + q j + ( T t m ) 2 1 + q m ) ( L 1 r + M 1 ) + k = 1 m ( ( t k t k 1 ) 2 1 + q k 1 ( L 1 r + M 1 ) + L 2 r + M 2 ) + k = 1 m ( ( t k t k 1 ) ( L 1 r + M 1 ) + L 3 r + M 3 ) ( T t k ) + ( T t m ) 2 1 + q m ( L 1 r + M 1 ) = r Λ + ρ ( δ + 1 ε ) r r .

It follows that A B r B r .

For x , y P C ( J , R ) and for each t J , we have
A x A y sup t J { t k = 1 j ( 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 ) ) | ) + t T η k = j + 1 m ( t k 1 t k t k 1 s | f ( σ , x ( σ ) ) f ( σ , y ( σ ) ) | d q k 1 σ d q k 1 s + | I k ( x ( t k ) ) I k ( y ( t k ) ) | ) + t T η k = j + 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 ) ) | ) ( T t k ) + t T η t j η t j s | f ( σ , x ( σ ) ) f ( σ , y ( σ ) ) | d q j σ d q j s + t T η t m T t m s | f ( σ , x ( σ ) ) f ( σ , y ( σ ) ) | d q m σ d q m s + 0 < t k < t ( t k 1 t k t k 1 s | f ( σ , x ( σ ) ) f ( σ , y ( σ ) ) | d q k 1 σ 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 ) ) | ) ( t t k ) + t k t t k s | f ( σ , x ( σ ) ) f ( σ , y ( σ ) ) | d q k σ d q k s } T x y k = 1 j [ ( t k t k 1 ) L 1 + L 3 ] + T x y T η k = j + 1 m ( ( t k t k 1 ) 2 1 + q k 1 L 1 + L 2 ) + T x y T η k = j + 1 m ( ( t k t k 1 ) L 1 + L 3 ) ( T t k ) + T x y T η ( ( η t j ) 2 1 + q j + ( T t m ) 2 1 + q m ) L 1 + k = 1 m ( ( t k t k 1 ) 2 1 + q k 1 L 1 + L 2 ) x y + k = 1 m ( ( t k t k 1 ) L 1 + L 3 ) ( T t k ) x y + ( T t m ) 2 1 + q m L 1 x y = Λ x y .

As Λ < 1 , A is a contraction. Hence, by Banach’s contraction mapping principle, we find that A has a fixed point which is the unique solution of problem (1.1). □

Our next result is based on Krasnoselskii’s fixed-point theorem.

Lemma 3.3 (Krasnoselskii’s fixed-point theorem) [19]

Let M be a closed, bounded, convex and nonempty subset of a Banach space X. Let A, B be the operators such that (a) A x + B y M whenever x , y M ; (b) A is compact and continuous; (c) B is a contraction mapping. Then there exists z M such that z = A z + B z .

Further, we use the notation
θ 1 = T k = 1 j ( t k t k 1 ) + T T η k = j + 1 m ( t k t k 1 ) 2 1 + q k 1 + T T η k = j + 1 m ( T t k ) ( t k t k 1 ) + T ( η t j ) 2 ( T η ) ( 1 + q j ) + T ( T t m ) 2 ( T η ) ( 1 + q m ) + k = 1 m + 1 ( t k t k 1 ) 2 1 + q k 1 + k = 1 m ( T t k ) ( t k t k 1 ) ,
(3.13)
and
θ 2 = j T N 2 + ( m j ) T N 1 T η + m N 1 + N 2 k = 1 m ( T t k ) + T N 2 T η j + 1 m ( T t k ) .
(3.14)

Theorem 3.4 Let f : J × R R be a continuous function. Assume that (H2) holds and in addition suppose that:

(H3) | f ( t , x ) | μ ( t ) , ( t , x ) J × R , and μ C ( J , R + ) .

(H4) There exist constants N 1 , N 2 > 0 such that | I k ( x ) | N 1 and | I k ( x ) | N 2 for all x R , for k = 1 , 2 , , m .

Then the impulsive q k -difference boundary value problem (1.1) has at least one solution on J provided that
j T L 3 + m L 2 + T ( m j ) L 2 T η + L 3 k = 1 m ( T t k ) < 1 .
(3.15)
Proof Firstly, we define sup t J | μ ( t ) | = μ . Choosing a suitable ball B R = { x P C ( J , R ) : x R } , where
R μ θ 1 + θ 2 ,
(3.16)
and θ 1 , θ 2 are defined by (3.13), (3.14), respectively, we define the operators S 1 and S 2 on B R by
( S 1 x ) ( t ) = t k = 1 j t k 1 t k f ( s , x ( s ) ) d q k 1 s t T η k = j + 1 m t k 1 t k t k 1 s f ( σ , x ( σ ) ) d q k 1 σ d q k 1 s t T η k = j + 1 m ( T t k ) t k 1 t k f ( s , x ( s ) ) d q k 1 s + t T η t j η t j s f ( σ , x ( σ ) ) d q j σ d q j s t T η t m T t m s f ( σ , x ( σ ) ) d q m σ d q m s + 0 < t k < t t k 1 t k t k 1 s f ( σ , x ( σ ) ) d q k 1 σ d q k 1 s + 0 < t k < t ( t t k ) t k 1 t k f ( s , x ( s ) ) d q k 1 s + t k t t k s f ( σ , x ( σ ) ) d q k σ d q k s , t [ 0 , T ] ,
and
( S 2 x ) ( t ) = t k = 1 j I k ( x ( t k ) ) t T η k = j + 1 m I k ( x ( t k ) ) t T η k = j + 1 m ( T t k ) I k ( x ( t k ) ) + 0 < t k < t I k ( x ( t k ) ) + 0 < t k < t ( t t k ) I k ( x ( t k ) ) , t [ 0 , T ] .
For any x , y B R , we have
S 1 x + S 2 y μ [ T k = 1 j ( t k t k 1 ) + T T η k = j + 1 m ( t k t k 1 ) 2 1 + q k 1 + T T η k = j + 1 m ( T t k ) ( t k t k 1 ) + T ( η t j ) 2 ( T η ) ( 1 + q j ) + T ( T t m ) 2 ( T η ) ( 1 + q m ) + k = 1 m + 1 ( t k t k 1 ) 2 1 + q k 1 + k = 1 m ( T t k ) ( t k t k 1 ) ] + j T N 2 + ( m j ) T N 1 T η + m N 1 + N 2 k = 1 m ( T t k ) + T N 2 T η j + 1 m ( T t k ) = μ θ 1 + θ 2 R .

Hence, S 1 x + S 2 y B R .

To show that S 2 is a contraction, for x , y P C ( J , R ) , we have
S 2 x S 2 y T k = 1 j | I k ( x ( t k ) ) I k ( y ( t k ) ) | + T T η k = j + 1 m | I k ( x ( t k ) ) I k ( y ( t k ) ) | + k = 1 m | I ( x ( t k ) ) I k ( y ( t k ) ) | + k = 1 m ( t t k ) | I k ( x ( t k ) ) I k ( y ( t k ) ) | [ j T L 3 + m L 2 + T ( m j ) L 2 T η + L 3 k = 1 m ( T t k ) ] x y .

From (3.15), it follows that S 2 is a contraction.

Next, the continuity of f implies that the operator S 1 is continuous. Further, S 1 is uniformly bounded on B R by
S 1 x μ θ 1 .
Now we shall prove the compactness of S 1 . Setting sup ( t , x ) J × B R | f ( t , x ) | = f < , then for each τ 1 , τ 2 ( t l , t l + 1 ) for some l { 0 , 1 , , m } with τ 2 > τ 1 , we have
| ( S 1 x ) ( τ 2 ) ( S 1 x ) ( τ 1 ) | | τ 2 τ 1 | k = 1 j t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | τ 2 τ 1 | T η k = j + 1 m t k 1 t k t k 1 s | f ( σ , x ( σ ) ) | d q k 1 σ d q k 1 s + | τ 2 τ 1 | T η k = j + 1 m ( T t k ) t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | τ 2 τ 1 | T η t j η t j s | f ( σ , x ( σ ) ) | d q j σ d q j s + | τ 2 τ 1 | T η t m T t m s | f ( σ , x ( σ ) ) | d q m σ d q m s + | τ 2 τ 1 | k = 1 l t k 1 t k | f ( s , x ( s ) ) | d q k 1 s + | t l τ 2 t l s | f ( σ , x ( σ ) ) | d q l σ d q l s t l τ 1 t l s | f ( σ , x ( σ ) ) | d q l σ d q l s | | τ 2 τ 1 | f [ k = 1 j ( t k t k 1 ) + 1 T η k = j + 1 m ( t k t k 1 ) 2 1 + q k 1 + ( η t j ) 2 ( T η ) ( 1 + q j ) + ( T t m ) 2 ( T η ) ( 1 + q m ) + 1 T η k = j + 1 m ( T t k ) ( t k t k 1 ) + k = 1 l ( t k t k 1 ) + ( τ 1 + τ 2 + 2 t l ) 1 + q l ] .

As τ 1 τ 2 , the right hand side above (which is independent of x) tends to zero. Therefore, the operator S 1 is equicontinuous. Since S 1 maps bounded subsets into relatively compact subsets, it follows that S 1 is relative compact on B R . Hence, by the Arzelá-Ascoli theorem, S 1 is compact on B R . Thus all the assumptions of Lemma 3.3 are satisfied. Hence, by the conclusion of Lemma 3.3, the impulsive q k -difference boundary value problem (1.1) has at least one solution on J. □

4 Examples

Example 4.1 Consider the following nonlinear second-order impulsive q k -difference equation with three-point boundary condition:
{ D 4 5 + k 2 x ( t ) = e cos 2 t | x ( t ) | ( 6 + t ) 2 ( 1 + | x ( t ) | ) , t J = [ 0 , 1 ] , t t k = k 10 , Δ x ( t k ) = | x ( t k ) | 8 ( 7 + | x ( t k ) | ) , k = 1 , 2 , , 9 , D 4 5 + k x ( t k + ) D 4 4 + k x ( t k ) = 1 6 tan 1 ( 1 8 x ( t k ) ) , k = 1 , 2 , , 9 , x ( 0 ) = 0 , x ( 1 ) = x ( 1 4 ) .
(4.1)
Here q k = 4 / ( 5 + k ) for k = 0 , 1 , 2 , , 9 , m = 9 , T = 1 , η = 1 / 4 , j = 2 , f ( t , x ) = ( e cos 2 t | x | ) / ( ( 6 + t ) 2 ( 1 + | x | ) ) , I k ( x ) = | x | / ( 8 ( 7 + | x | ) ) and I k ( x ) = ( 1 / 6 ) tan 1 ( x / 8 ) . Since
| f ( t , x ) f ( t , y ) | ( 1 / 36 ) | x y | , | I k ( x ) I k ( y ) | ( 1 / 56 ) | x y | and | I k ( x ) I k ( y ) | ( 1 / 48 ) | x y | ,
then (H1) and (H2) are satisfied with L 1 = ( 1 / 36 ) , L 2 = ( 1 / 56 ) , L 3 = ( 1 / 48 ) . We can show that
Λ 0.5730986482 < 1 .

Hence, by Theorem 3.2, the three-point impulsive q k -difference boundary value problem (4.1) has a unique solution on [ 0 , 1 ] .

Example 4.2 Consider the following nonlinear second-order impulsive q k -difference equation with three-point boundary condition:
{ D 3 6 + k 2 x ( t ) = sin 2 ( π t ) ( t + 4 ) 2 | x ( t ) | ( 1 + | x ( t ) | ) , t J = [ 0 , 1 ] , t t k = k 10 , Δ x ( t k ) = | x ( t k ) | 9 ( 7 + | x ( t k ) | ) , k = 1 , 2 , , 9 , D 3 6 + k x ( t k + ) D 3 5 + k x ( t k ) = | x ( t k ) | 4 ( 5 + | x ( t k ) | ) , k = 1 , 2 , , 9 , x ( 0 ) = 0 , x ( 1 ) = x ( 9 20 ) .
(4.2)
Set q k = 3 / ( 6 + k ) for k = 0 , 1 , 2 , , 9 , m = 9 , T = 1 , η = 9 / 20 , j = 4 , f ( t , x ) = ( sin 2 ( π t ) | x | ) / ( ( t + 4 ) 2 ( 1 + | x | ) ) , I k ( x ) = | x | / ( 9 ( 7 + | x | ) ) and I k ( x ) = | x | / ( 4 ( 5 + | x | ) ) . Since
| I k ( x ) I k ( y ) | ( 1 / 63 ) | x y | and | I k ( x ) I k ( y ) | ( 1 / 20 ) | x y | ,
then (H2) is satisfied with L 2 = ( 1 / 63 ) , L 3 = ( 1 / 20 ) . It is easy to verify that | f ( t , x ) | μ ( t ) 1 , I k ( x ) N 1 = 1 / 9 and I k ( x ) N 2 = 1 / 4 for all t [ 0 , 1 ] , x R , k = 1 , , m . Thus (H3) and (H4) are satisfied. We can show that
j T L 3 + m L 2 + T ( m j ) L 2 T η + L 3 k = 1 m ( T t k ) = 19741 27720 < 1 .

Hence, by Theorem 3.3, the three-point impulsive q k -difference boundary value problem (4.2) has at least one solution 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

This research of J Tariboon is supported by King Mongkut’s University of Technology North Bangkok, Thailand.

Authors’ Affiliations

(1)
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

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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