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

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 $f:J_k:=[t_k,t_{k+1}]\to \mathbb{R}$ 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.

1 Introduction

In this article, we investigate the nonlinear second-order impulsive $q_k$-difference equation with three-point boundary conditions

$$\{\begin{array}{c}{D}_{q_k}^{2}x(t)=f(t,x(t)),t\in J:=[0,T],t\ne t_k,\hfill \\ \mathrm{\Delta }x(t_k)=I_k(x(t_k)),k=1,2,\dots ,m,\hfill \\ D_{q_k}x(t_k^{+})-D_{q_{k-1}}x(t_k)=I_k^{\ast }(x(t_k)),k=1,2,\dots ,m,\hfill \\ x(0)=0,x(T)=x(\eta ),\hfill \end{array}$$

(1.1)

where $0=t_0<t_1<t_2<\cdots <t_k<\cdots <t_m<t_{m+1}=T$, $f:J\times \mathbb{R}\to \mathbb{R}$ is a continuous function, $I_k,I_k^{\ast }\in C(\mathbb{R},\mathbb{R})$, $\mathrm{\Delta }x(t_k)=x(t_k^{+})-x(t_k)$ for $k=1,2,\dots ,m$, $x(t_k^{+})={lim}_{h\to 0}x(t_k+h)$, $\eta \in (t_j,t_{j+1})$ a constant for some $j\in \{0,1,2,\dots ,m\}$ and $0<q_k<1$ for $k=0,1,2,\dots ,m$.

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 $f:J_k:=[t_k,t_{k+1}]\to \mathbb{R}$, 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 [3–15] 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 [16–18].

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 $k\in \mathbb{N}\cup \{0\}$ let $J_k:=[t_k,t_{k+1}]\subset \mathbb{R}$ be an interval and $0<q_k<1$ be a constant. We define $q_k$-derivative of a function $f:J_k\to \mathbb{R}$ at a point $t\in J_k$ as follows.

Definition 2.1 Assume $f:J_k\to \mathbb{R}$ is a continuous function and let $t\in J_k$. Then the expression

$$D_{q_k}f(t)=\frac{f(t)-f(q_{k}t+(1-q_k)t_k)}{(1-q_k)(t-t_k)},t\ne t_k,D_{q_k}f(t_k)=\underset{t\to t_k}{lim}{D}_{q_k}f(t),$$

(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 $t\in J_k$. 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

$$D_{q}f(t)=\frac{f(t)-f(qt)}{(1-q)t}.$$

(2.2)

In addition, we should define the higher $q_k$-derivative of functions.

Definition 2.2 Let $f:J_k\to \mathbb{R}$ 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 $D_{q_k}^{2}f=D_{q_k}(D_{q_k}f):J_k\to \mathbb{R}$. Similarly, we define the higher-order $q_k$-derivative $D_{q_k}^n:J_k\to \mathbb{R}$.

The properties of the $q_k$-derivative are summarized in the following theorem.

Theorem 2.3 Assume $f,g:J_k\to \mathbb{R}$ are $q_k$-differentiable on $J_k$. Then:

• (i) The sum $f+g:J_k\to \mathbb{R}$ is $q_k$-differentiable on $J_k$ with

  $$D_{q_k}(f(t)+g(t))=D_{q_k}f(t)+D_{q_k}g(t).$$

• (ii) For any constant α, $\alpha f:J_k\to \mathbb{R}$ is $q_k$-differentiable on $J_k$ with

  $$D_{q_k}(\alpha f)(t)=\alpha D_{q_k}f(t).$$

• (iii) The product $fg:J_k\to \mathbb{R}$ is $q_k$-differentiable on $J_k$ with

  $$\begin{array}{rcl}{D}_{q_k}(fg)(t)& =& f(t)D_{q_k}g(t)+g(q_{k}t+(1-q_k)t_k)D_{q_k}f(t)\\ =& g(t)D_{q_k}f(t)+f(q_{k}t+(1-q_k)t_k)D_{q_k}g(t).\end{array}$$

• (iv) If $g(t)g(q_{k}t+(1-q_k)t_k)\ne 0$, then $\frac{f}{g}$ is $q_k$-differentiable on $J_k$ with

  $$D_{q_k}\left(\frac{f}{g}\right)(t)=\frac{g(t)D_{q_k}f(t)-f(t)D_{q_k}g(t)}{g(t)g(q_{k}t+(1-q_k)t_k)}.$$

Definition 2.4 Assume $f:J_k\to \mathbb{R}$ is a continuous function. Then the $q_k$-integral is defined by

$${\int }_{t_k}^{t}f(s)d_{q_k}s=(1-q_k)(t-t_k)\sum _{n=0}^{\mathrm{\infty }}{q}_k^{n}f(q_k^{n}t+(1-q_k^n)t_k),$$

(2.3)

for $t\in J_k$. Moreover, if $a\in (t_k,t)$ then the definite $q_k$-integral is defined by

$$\begin{array}{rcl}{\int }_a^{t}f(s)d_{q_k}s& =& {\int }_{t_k}^{t}f(s)d_{q_k}s-{\int }_{t_k}^{a}f(s)d_{q_k}s\\ =& (1-q_k)(t-t_k)\sum _{n=0}^{\mathrm{\infty }}{q}_k^{n}f(q_k^{n}t+(1-q_k^n)t_k)\\ -(1-q_k)(a-t_k)\sum _{n=0}^{\mathrm{\infty }}{q}_k^{n}f(q_k^{n}a+(1-q_k^n)t_k).\end{array}$$

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 ${\int }_0^{t}f(s)d_{q}s=(1-q)t{\sum }_{n=0}^{\mathrm{\infty }}{q}^{n}f(q^{n}t)$ for $t\in [0,\mathrm{\infty })$.

Theorem 2.5 For $t\in J_k$, the following formulas hold:

• (i) $D_{q_k}{\int }_{t_k}^{t}f(s)d_{q_k}s=f(t)$;

• (ii) ${\int }_{t_k}^t{D}_{q_k}f(s)d_{q_k}s=f(t)$;

• (iii) ${\int }_a^t{D}_{q_k}f(s)d_{q_k}s=f(t)-f(a)$ for $a\in (t_k,t)$.

3 Main results

Let $J=[0,T]$, $J_0=[t_0,t_1]$, $J_k=(t_k,t_{k+1}]$ for $k=1,2,\dots ,m$. Let $PC(J,\mathbb{R})$ = {$x:J\to \mathbb{R}:x(t)$ is continuous everywhere except for some $t_k$ at which $x(t_k^{+})$ and $x(t_k^{-})$ exist and $x(t_k^{-})=x(t_k)$, $k=1,2,\dots ,m$}. $PC(J,\mathbb{R})$ is a Banach space with the norm ${\parallel x\parallel }_{PC}=sup\{|x(t)|;t\in J\}$.

Lemma 3.1 The unique solution of problem (1.1) is given by

$$\begin{array}{rl}x(t)=& -t\sum _{k=1}^j({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))\\ -\frac{t}{T-\eta }\sum _{k=j+1}^m({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^{s}f(\sigma ,x(\sigma ))d_{q_{k-1}}\sigma d_{q_{k-1}}s+I_k(x(t_k)))\\ -\frac{t}{T-\eta }\sum _{k=j+1}^m({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))(T-t_k)\\ +\frac{t}{T-\eta }{\int }_{t_j}^{\eta }{\int }_{t_j}^{s}f(\sigma ,x(\sigma ))d_{q_j}\sigma d_{q_j}s\\ -\frac{t}{T-\eta }{\int }_{t_m}^T{\int }_{t_m}^{s}f(\sigma ,x(\sigma ))d_{q_m}\sigma d_{q_m}s\\ +\sum _{0<t_k<t}({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^{s}f(\sigma ,x(\sigma ))d_{q_{k-1}}\sigma d_{q_{k-1}}s+I_k(x(t_k)))\\ +\sum _{0<t_k<t}({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))(t-t_k)\\ +{\int }_{t_k}^t{\int }_{t_k}^{s}f(\sigma ,x(\sigma ))d_{q_k}\sigma d_{q_k}s,\end{array}$$

(3.1)

with ${\sum }_{0<0}(\cdot )=0$.

Proof For $t\in J_0$, taking the $q_0$-integral for the first equation of (1.1), we get

$$D_{q_0}x(t)=D_{q_0}x(0)+{\int }_0^{t}f(s,x(s))d_{q_0}s,$$

(3.2)

which yields

$$D_{q_0}x(t_1)=D_{q_0}x(0)+{\int }_0^{t_1}f(s,x(s))d_{q_0}s.$$

(3.3)

For $t\in J_0$ we obtain by $q_0$-integrating (3.2),

$$\begin{array}{rl}x(t)& =x(0)+D_{q_0}x(0)t+{\int }_0^t{\int }_0^{s}f(\sigma ,x(\sigma ))d_{q_0}\sigma d_{q_0}s\\ :=A+Bt+{\int }_0^t{\int }_0^{s}f(\sigma ,x(\sigma ))d_{q_0}\sigma d_{q_0}s(x(0)=A,D_{q_0}x(0)=B).\end{array}$$

In particular, for $t=t_1$

$$x(t_1)=A+B{t}_1+{\int }_0^{t_1}{\int }_0^{s}f(\sigma ,x(\sigma ))d_{q_0}\sigma d_{q_0}s.$$

(3.4)

For $t\in J_1=(t_1,t_2]$, $q_1$-integrating (1.1), we have

$$D_{q_1}x(t)=D_{q_1}x\left(t_1^{+}\right)+{\int }_{t_1}^{t}f(s,x(s))d_{q_1}s.$$

Using the third condition of (1.1) with (3.3), it follows that

$$D_{q_1}x(t)=B+{\int }_0^{t_1}f(s,x(s))d_{q_0}s+I_1^{\ast }(x(t_1))+{\int }_{t_1}^{t}f(s,x(s))d_{q_1}s.$$

(3.5)

Taking the $q_1$-integral to (3.5) for $t\in J_1$, we obtain

$$\begin{array}{rcl}x(t)& =& x\left(t_1^{+}\right)+[B+{\int }_0^{t_1}f(s,x(s))d_{q_0}s+I_1^{\ast }(x(t_1))](t-t_1)\\ +{\int }_{t_1}^t{\int }_{t_1}^{s}f(\sigma ,x(\sigma ))d_{q_1}\sigma d_{q_1}s.\end{array}$$

(3.6)

Applying the second equation of (1.1) with (3.4) and (3.6), we get

$$\begin{array}{rcl}x(t)& =& A+B{t}_1+{\int }_0^{t_1}{\int }_0^{s}f(\sigma ,x(\sigma ))d_{q_0}\sigma d_{q_0}s+I_1(x(t_1))\\ +[B+{\int }_0^{t_1}f(s,x(s))d_{q_0}s+I_1^{\ast }(x(t_1))](t-t_1)\\ +{\int }_{t_1}^t{\int }_{t_1}^{s}f(\sigma ,x(\sigma ))d_{q_1}\sigma d_{q_1}s\\ =& A+Bt+{\int }_0^{t_1}{\int }_0^{s}f(\sigma ,x(\sigma ))d_{q_0}\sigma d_{q_0}s+I_1(x(t_1))\\ +[{\int }_0^{t_1}f(s,x(s))d_{q_0}s+I_1^{\ast }(x(t_1))](t-t_1)\\ +{\int }_{t_1}^t{\int }_{t_1}^{s}f(\sigma ,x(\sigma ))d_{q_1}\sigma d_{q_1}s.\end{array}$$

Repeating the above process, for $t\in J$, we get

$$\begin{array}{rl}x(t)=& A+Bt\\ +\sum _{0<t_k<t}({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^{s}f(\sigma ,x(\sigma ))d_{q_{k-1}}\sigma d_{q_{k-1}}s+I_k(x(t_k)))\\ +\sum _{0<t_k<t}({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))(t-t_k)\\ +{\int }_{t_k}^t{\int }_{t_k}^{s}f(\sigma ,x(\sigma ))d_{q_k}\sigma d_{q_k}s.\end{array}$$

(3.7)

The first boundary condition of (1.1) implies $A=0$. The second boundary condition of (1.1) yields

$$\begin{array}{r}\sum _{k=1}^m({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^{s}f(\sigma ,x(\sigma ))d_{q_{k-1}}\sigma d_{q_{k-1}}s+I_k(x(t_k)))\\ +\sum _{k=1}^m({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))(T-t_k)\\ +{\int }_{t_m}^T{\int }_{t_m}^{s}f(\sigma ,x(\sigma ))d_{q_m}\sigma d_{q_m}s+BT\\ =\sum _{k=1}^j({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^{s}f(\sigma ,x(\sigma ))d_{q_{k-1}}\sigma d_{q_{k-1}}s+I_k(x(t_k)))\\ +\sum _{k=1}^j({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))(\eta -t_k)\\ +{\int }_{t_j}^{\eta }{\int }_{t_j}^{s}f(\sigma ,x(\sigma ))d_{q_j}\sigma d_{q_j}s+B\eta ,\end{array}$$

which implies

$$\begin{array}{rl}B=& -\sum _{k=1}^j({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))\\ -\frac{1}{T-\eta }\sum _{k=j+1}^m({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^{s}f(\sigma ,x(\sigma ))d_{q_{k-1}}\sigma d_{q_{k-1}}s+I_k(x(t_k)))\\ -\frac{1}{T-\eta }\sum _{k=j+1}^m({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))(T-t_k)\\ +\frac{1}{T-\eta }{\int }_{t_j}^{\eta }{\int }_{t_j}^{s}f(\sigma ,x(\sigma ))d_{q_j}\sigma d_{q_j}s-\frac{1}{T-\eta }{\int }_{t_m}^T{\int }_{t_m}^{s}f(\sigma ,x(\sigma ))d_{q_m}\sigma d_{q_m}s.\end{array}$$

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

In view of Lemma 3.1, we define an operator $\mathcal{A}:PC(J,\mathbb{R})\to PC(J,\mathbb{R})$ by

$$\begin{array}{rl}(\mathcal{A}x)(t)=& -t\sum _{k=1}^j({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))\\ -\frac{t}{T-\eta }\sum _{k=j+1}^m({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^{s}f(\sigma ,x(\sigma ))d_{q_{k-1}}\sigma d_{q_{k-1}}s+I_k(x(t_k)))\\ -\frac{t}{T-\eta }\sum _{k=j+1}^m({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))(T-t_k)\\ +\frac{t}{T-\eta }{\int }_{t_j}^{\eta }{\int }_{t_j}^{s}f(\sigma ,x(\sigma ))d_{q_j}\sigma d_{q_j}s\\ -\frac{t}{T-\eta }{\int }_{t_m}^T{\int }_{t_m}^{s}f(\sigma ,x(\sigma ))d_{q_m}\sigma d_{q_m}s\\ +\sum _{0<t_k<t}({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^{s}f(\sigma ,x(\sigma ))d_{q_{k-1}}\sigma d_{q_{k-1}}s+I_k(x(t_k)))\\ +\sum _{0<t_k<t}({\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+I_k^{\ast }(x(t_k)))(t-t_k)\\ +{\int }_{t_k}^t{\int }_{t_k}^{s}f(\sigma ,x(\sigma ))d_{q_k}\sigma d_{q_k}s.\end{array}$$

(3.8)

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

For convenience, we set

$${\mathrm{\Phi }}_k=[(t_k-t_{k-1})(T-t_k)+\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}]M_1+M_2+(T-t_k)M_3,$$

(3.9)

$${\mathrm{\Psi }}_k=[(t_k-t_{k-1})(T-t_k)+\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}]L_1+L_2+(T-t_k)L_3,$$

(3.10)

for $k=1,\dots ,m$.

Theorem 3.2 Assume that:

(H₁) The function $f:[0,T]\times \mathbb{R}\to \mathbb{R}$ is a continuous and there exists a constant $L_1>0$ such that $|f(t,x)-f(t,y)|\le L_1|x-y|$, for each $t\in J$ and $x,y\in \mathbb{R}$.

(H₂) The functions $I_k,I_k^{\ast }:\mathbb{R}\to \mathbb{R}$ are continuous and there exist constants $L_2,L_3>0$ such that $|I_k(x)-I_k(y)|\le L_2|x-y|$ and $|I_k^{\ast }(x)-I_k^{\ast }(y)|\le L_3|x-y|$ for each $x,y\in \mathbb{R}$, $k=1,2,\dots ,m$.

If

$$\begin{array}{rcl}\mathrm{\Lambda }& :=& T\sum _{k=1}^j[(t_k-t_{k-1})L_1+L_3]+\frac{T}{T-\eta }\sum _{k=j+1}^m{\mathrm{\Psi }}_k+\frac{T{L}_1}{T-\eta }(\frac{{(\eta -t_j)}^2}{1+q_j}+\frac{{(T-t_m)}^2}{1+q_m})\\ +\sum _{k=1}^m{\mathrm{\Psi }}_k+\frac{{(T-t_m)}^2}{1+q_m}{L}_1\le \delta <1,\end{array}$$

(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, $x=\mathcal{A}x$, where the operator $\mathcal{A}$ is defined by (3.8). By using Banach’s contraction principle, we shall show that $\mathcal{A}$ has a fixed point which is the unique solution of problem (1.1).

Set ${sup}_{t\in J}|f(t,0)|=M_1<\mathrm{\infty }$, $sup\{|I_k(0)|:k=1,2,\dots ,m\}=M_2<\mathrm{\infty }$, $sup\{|I_k^{\ast }(0)|:k=1,2,\dots ,m\}=M_3<\mathrm{\infty }$ and a constant

$$\begin{array}{rl}\rho =& T\sum _{k=1}^j[(t_k-t_{k-1})M_1+M_3]+\frac{T}{T-\eta }\sum _{k=j+1}^m{\mathrm{\Phi }}_k\\ +\frac{T{M}_1}{T-\eta }(\frac{{(\eta -t_j)}^2}{1+q_j}+\frac{{(T-t_m)}^2}{1+q_m})\\ +\sum _{k=1}^m{\mathrm{\Phi }}_k+\frac{{(T-t_m)}^2}{1+q_m}{M}_1.\end{array}$$

(3.12)

Choosing $r\ge \frac{\rho }{1-\epsilon }$, where $\delta \le \epsilon <1$, we show that $\mathcal{A}{B}_r\subset B_r$, where $B_r=\{x\in PC(J,\mathbb{R}):\parallel x\parallel \le r\}$. For $x\in B_r$, we have

$$\begin{array}{r}\parallel \mathcal{A}x\parallel \\ \le \underset{t\in J}{sup}\{t\sum _{k=1}^j\left({\int }_{t_{k-1}}^{t_k}\right|f(s,x(s))|d_{q_{k-1}}s+|I_k^{\ast }(x(t_k))\left|\right)\\ +\frac{t}{T-\eta }\sum _{k=j+1}^m\left({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^s\right|f(\sigma ,x(\sigma ))|d_{q_{k-1}}\sigma d_{q_{k-1}}s+|I_k(x(t_k))\left|\right)\\ +\frac{t}{T-\eta }\sum _{k=j+1}^m\left({\int }_{t_{k-1}}^{t_k}\right|f(s,x(s))|d_{q_{k-1}}s+|I_k^{\ast }(x(t_k))\left|\right)(T-t_k)\\ +\frac{t}{T-\eta }{\int }_{t_j}^{\eta }{\int }_{t_j}^s\left|f(\sigma ,x(\sigma ))\right|d_{q_j}\sigma d_{q_j}s+\frac{t}{T-\eta }{\int }_{t_m}^T{\int }_{t_m}^s\left|f(\sigma ,x(\sigma ))\right|d_{q_m}\sigma d_{q_m}s\\ +\sum _{0<t_k<t}\left({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^s\right|f(\sigma ,x(\sigma ))|d_{q_{k-1}}\sigma d_{q_{k-1}}s+|I_k(x(t_k))\left|\right)\\ +\sum _{0<t_k<t}\left({\int }_{t_{k-1}}^{t_k}\right|f(s,x(s))|d_{q_{k-1}}s+|I_k^{\ast }(x(t_k))\left|\right)(t-t_k)\\ +{\int }_{t_k}^t{\int }_{t_k}^s\left|f(\sigma ,x(\sigma ))\right|d_{q_k}\sigma d_{q_k}s\}\\ \le T\sum _{k=1}^j({\int }_{t_{k-1}}^{t_k}\left(\right|f(s,x(s))-f(s,0)|+|f(s,0)\left|\right)d_{q_{k-1}}s+|I_k^{\ast }(x(t_k))-I_k^{\ast }(0)|+|I_k^{\ast }(0)\left|\right)\\ +\frac{T}{T-\eta }\sum _{k=j+1}^m({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^s\left(\right|f(\sigma ,x(\sigma ))-f(\sigma ,0)|+|f(\sigma ,0)\left|\right)d_{q_{k-1}}\sigma d_{q_{k-1}}s\\ +|I_k(x(t_k))-I_k(0)|+|I_k(0)|)\\ +\frac{T}{T-\eta }\sum _{k=j+1}^m({\int }_{t_{k-1}}^{t_k}\left(\right|f(s,x(s))-f(s,0)|+|f(s,0)\left|\right)d_{q_{k-1}}s\\ +|I_k^{\ast }(x(t_k))-I_k^{\ast }(0)|+|I_k^{\ast }(0)|)(T-t_k)\\ +\frac{T}{T-\eta }{\int }_{t_j}^{\eta }{\int }_{t_j}^s\left(\right|f(\sigma ,x(\sigma ))-f(\sigma ,0)|+|f(\sigma ,0)\left|\right)d_{q_j}\sigma d_{q_j}s\\ +\frac{T}{T-\eta }{\int }_{t_m}^T{\int }_{t_m}^s\left(\right|f(\sigma ,x(\sigma ))-f(\sigma ,0)|+|f(\sigma ,0)\left|\right)d_{q_m}\sigma d_{q_m}s\\ +\sum _{k=1}^m({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^s\left(\right|f(\sigma ,x(\sigma ))-f(\sigma ,0)|+|f(\sigma ,0)\left|\right)d_{q_{k-1}}\sigma d_{q_{k-1}}s\\ +|I_k(x(t_k))-I_k(0)|+|I_k(0)|)\\ +\sum _{k=1}^m({\int }_{t_{k-1}}^{t_k}\left(\right|f(s,x(s))-f(s,0)|+|f(s,0)\left|\right)d_{q_{k-1}}s\\ +|I_k^{\ast }(x(t_k))-I_k^{\ast }(0)|+|I_k^{\ast }(0)|)(T-t_k)\\ +{\int }_{t_m}^T{\int }_{t_m}^s\left(\right|f(\sigma ,x(\sigma ))-f(\sigma ,0)|+|f(\sigma ,0)\left|\right)d_{q_m}\sigma d_{q_m}s\\ \le T\sum _{k=1}^j((t_k-t_{k-1})(L_{1}r+M_1)+L_{3}r+M_3)\\ +\frac{T}{T-\eta }\sum _{k=j+1}^m(\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}(L_{1}r+M_1)+L_{2}r+M_2)\\ +\frac{T}{T-\eta }\sum _{k=j+1}^m((t_k-t_{k-1})(L_{1}r+M_1)+L_{3}r+M_3)(T-t_k)\\ +\frac{T}{T-\eta }(\frac{{(\eta -t_j)}^2}{1+q_j}+\frac{{(T-t_m)}^2}{1+q_m})(L_{1}r+M_1)\\ +\sum _{k=1}^m(\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}(L_{1}r+M_1)+L_{2}r+M_2)\\ +\sum _{k=1}^m((t_k-t_{k-1})(L_{1}r+M_1)+L_{3}r+M_3)(T-t_k)+\frac{{(T-t_m)}^2}{1+q_m}(L_{1}r+M_1)\\ =r\mathrm{\Lambda }+\rho \le (\delta +1-\epsilon )r\le r.\end{array}$$

It follows that $\mathcal{A}{B}_r\subset B_r$.

For $x,y\in PC(J,\mathbb{R})$ and for each $t\in J$, we have

$$\begin{array}{r}\parallel \mathcal{A}x-\mathcal{A}y\parallel \\ \le \underset{t\in J}{sup}\{t\sum _{k=1}^j\left({\int }_{t_{k-1}}^{t_k}\right|f(s,x(s))-f(s,y(s))|d_{q_{k-1}}s+|I_k^{\ast }(x(t_k))-I_k^{\ast }(y(t_k))\left|\right)\\ +\frac{t}{T-\eta }\sum _{k=j+1}^m\left({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^s\right|f(\sigma ,x(\sigma ))-f(\sigma ,y(\sigma ))|d_{q_{k-1}}\sigma d_{q_{k-1}}s\\ +|I_k(x(t_k))-I_k(y(t_k))|)\\ +\frac{t}{T-\eta }\sum _{k=j+1}^m\left({\int }_{t_{k-1}}^{t_k}\right|f(s,x(s))-f(s,y(s))|d_{q_{k-1}}s+|I_k^{\ast }(x(t_k))-I_k^{\ast }(y(t_k))\left|\right)(T-t_k)\\ +\frac{t}{T-\eta }{\int }_{t_j}^{\eta }{\int }_{t_j}^s|f(\sigma ,x(\sigma ))-f(\sigma ,y(\sigma ))|d_{q_j}\sigma d_{q_j}s\\ +\frac{t}{T-\eta }{\int }_{t_m}^T{\int }_{t_m}^s|f(\sigma ,x(\sigma ))-f(\sigma ,y(\sigma ))|d_{q_m}\sigma d_{q_m}s\\ +\sum _{0<t_k<t}\left({\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^s\right|f(\sigma ,x(\sigma ))-f(\sigma ,y(\sigma ))|d_{q_{k-1}}\sigma d_{q_{k-1}}s+|I_k(x(t_k))-I_k(y(t_k))\left|\right)\\ +\sum _{0<t_k<t}\left({\int }_{t_{k-1}}^{t_k}\right|f(s,x(s))-f(s,y(s))|d_{q_{k-1}}s+|I_k^{\ast }(x(t_k))-I_k^{\ast }(y(t_k))\left|\right)(t-t_k)\\ +{\int }_{t_k}^t{\int }_{t_k}^s|f(\sigma ,x(\sigma ))-f(\sigma ,y(\sigma ))|d_{q_k}\sigma d_{q_k}s\}\\ \le T\parallel x-y\parallel \sum _{k=1}^j[(t_k-t_{k-1})L_1+L_3]+\frac{T\parallel x-y\parallel }{T-\eta }\sum _{k=j+1}^m(\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}{L}_1+L_2)\\ +\frac{T\parallel x-y\parallel }{T-\eta }\sum _{k=j+1}^m((t_k-t_{k-1})L_1+L_3)(T-t_k)\\ +\frac{T\parallel x-y\parallel }{T-\eta }(\frac{{(\eta -t_j)}^2}{1+q_j}+\frac{{(T-t_m)}^2}{1+q_m})L_1\\ +\sum _{k=1}^m(\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}{L}_1+L_2)\parallel x-y\parallel +\sum _{k=1}^m((t_k-t_{k-1})L_1+L_3)(T-t_k)\parallel x-y\parallel \\ +\frac{{(T-t_m)}^2}{1+q_m}{L}_1\parallel x-y\parallel \\ =\mathrm{\Lambda }\parallel x-y\parallel .\end{array}$$

As $\mathrm{\Lambda }<1$, $\mathcal{A}$ is a contraction. Hence, by Banach’s contraction mapping principle, we find that $\mathcal{A}$ 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) $Ax+By\in M$ whenever $x,y\in M$; (b) A is compact and continuous; (c) B is a contraction mapping. Then there exists $z\in M$ such that $z=Az+Bz$.

Further, we use the notation

$$\begin{array}{rcl}{\theta }_1& =& T\sum _{k=1}^j(t_k-t_{k-1})+\frac{T}{T-\eta }\sum _{k=j+1}^m\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}\\ +\frac{T}{T-\eta }\sum _{k=j+1}^m(T-t_k)(t_k-t_{k-1})+\frac{T{(\eta -t_j)}^2}{(T-\eta )(1+q_j)}\\ +\frac{T{(T-t_m)}^2}{(T-\eta )(1+q_m)}+\sum _{k=1}^{m+1}\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}+\sum _{k=1}^m(T-t_k)(t_k-t_{k-1}),\end{array}$$

(3.13)

and

$${\theta }_2=jT{N}_2+\frac{(m-j)T{N}_1}{T-\eta }+m{N}_1+N_2\sum _{k=1}^m(T-t_k)+\frac{T{N}_2}{T-\eta }\sum _{j+1}^m(T-t_k).$$

(3.14)

Theorem 3.4 Let $f:J\times \mathbb{R}\to \mathbb{R}$ be a continuous function. Assume that (H₂) holds and in addition suppose that:

(H₃) $|f(t,x)|\le \mu (t)$, $\mathrm{\forall }(t,x)\in J\times \mathbb{R}$, and $\mu \in C(J,{\mathbb{R}}^{+})$.

(H₄) There exist constants $N_1,N_2>0$ such that $|I_k(x)|\le N_1$ and $|I_k^{\ast }(x)|\le N_2$ for all $x\in \mathbb{R}$, for $k=1,2,\dots ,m$.

Then the impulsive $q_k$-difference boundary value problem (1.1) has at least one solution on J provided that

$$jT{L}_3+m{L}_2+\frac{T(m-j)L_2}{T-\eta }+L_3\sum _{k=1}^m(T-t_k)<1.$$

(3.15)

Proof Firstly, we define ${sup}_{t\in J}|\mu (t)|=\parallel \mu \parallel$. Choosing a suitable ball $B_R=\{x\in PC(J,\mathbb{R}):\parallel x\parallel \le R\}$, where

$$R\ge \parallel \mu \parallel {\theta }_1+{\theta }_2,$$

(3.16)

and ${\theta }_1$, ${\theta }_2$ are defined by (3.13), (3.14), respectively, we define the operators ${\mathcal{S}}_1$ and ${\mathcal{S}}_2$ on $B_R$ by

$$\begin{array}{r}({\mathcal{S}}_{1}x)(t)\\ =-t\sum _{k=1}^j{\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s-\frac{t}{T-\eta }\sum _{k=j+1}^m{\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^{s}f(\sigma ,x(\sigma ))d_{q_{k-1}}\sigma d_{q_{k-1}}s\\ -\frac{t}{T-\eta }\sum _{k=j+1}^m(T-t_k){\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+\frac{t}{T-\eta }{\int }_{t_j}^{\eta }{\int }_{t_j}^{s}f(\sigma ,x(\sigma ))d_{q_j}\sigma d_{q_j}s\\ -\frac{t}{T-\eta }{\int }_{t_m}^T{\int }_{t_m}^{s}f(\sigma ,x(\sigma ))d_{q_m}\sigma d_{q_m}s+\sum _{0<t_k<t}{\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^{s}f(\sigma ,x(\sigma ))d_{q_{k-1}}\sigma d_{q_{k-1}}s\\ +\sum _{0<t_k<t}(t-t_k){\int }_{t_{k-1}}^{t_k}f(s,x(s))d_{q_{k-1}}s+{\int }_{t_k}^t{\int }_{t_k}^{s}f(\sigma ,x(\sigma ))d_{q_k}\sigma d_{q_k}s,t\in [0,T],\end{array}$$

and

$$\begin{array}{rl}({\mathcal{S}}_{2}x)(t)=& -t\sum _{k=1}^j{I}_k^{\ast }(x(t_k))-\frac{t}{T-\eta }\sum _{k=j+1}^m{I}_k(x(t_k))-\frac{t}{T-\eta }\sum _{k=j+1}^m(T-t_k)I_k^{\ast }(x(t_k))\\ +\sum _{0<t_k<t}{I}_k(x(t_k))+\sum _{0<t_k<t}(t-t_k)I_k^{\ast }(x(t_k)),t\in [0,T].\end{array}$$

For any $x,y\in B_R$, we have

$$\begin{array}{rcl}\parallel {\mathcal{S}}_{1}x+{\mathcal{S}}_{2}y\parallel & \le & \parallel \mu \parallel [T\sum _{k=1}^j(t_k-t_{k-1})+\frac{T}{T-\eta }\sum _{k=j+1}^m\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}\\ +\frac{T}{T-\eta }\sum _{k=j+1}^m(T-t_k)(t_k-t_{k-1})+\frac{T{(\eta -t_j)}^2}{(T-\eta )(1+q_j)}\\ +\frac{T{(T-t_m)}^2}{(T-\eta )(1+q_m)}+\sum _{k=1}^{m+1}\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}+\sum _{k=1}^m(T-t_k)(t_k-t_{k-1})]\\ +jT{N}_2+\frac{(m-j)T{N}_1}{T-\eta }+m{N}_1+N_2\sum _{k=1}^m(T-t_k)+\frac{T{N}_2}{T-\eta }\sum _{j+1}^m(T-t_k)\\ =& \parallel \mu \parallel {\theta }_1+{\theta }_2\\ \le & R.\end{array}$$

Hence, ${\mathcal{S}}_{1}x+{\mathcal{S}}_{2}y\in B_R$.

To show that ${\mathcal{S}}_2$ is a contraction, for $x,y\in PC(J,\mathbb{R})$, we have

$$\begin{array}{rcl}\parallel {\mathcal{S}}_{2}x-{\mathcal{S}}_{2}y\parallel & \le & T\sum _{k=1}^j|I_k^{\ast }(x(t_k))-I_k^{\ast }(y(t_k))|+\frac{T}{T-\eta }\sum _{k=j+1}^m|I_k(x(t_k))-I_k(y(t_k))|\\ +\sum _{k=1}^m|I(x(t_k))-I_k(y(t_k))|+\sum _{k=1}^m(t-t_k)|I_k^{\ast }(x(t_k))-I_k^{\ast }(y(t_k))|\\ \le & [jT{L}_3+m{L}_2+\frac{T(m-j)L_2}{T-\eta }+L_3\sum _{k=1}^m(T-t_k)]\parallel x-y\parallel .\end{array}$$

From (3.15), it follows that ${\mathcal{S}}_2$ is a contraction.

Next, the continuity of f implies that the operator ${\mathcal{S}}_1$ is continuous. Further, ${\mathcal{S}}_1$ is uniformly bounded on $B_R$ by

$$\parallel {\mathcal{S}}_{1}x\parallel \le \parallel \mu \parallel {\theta }_1.$$

Now we shall prove the compactness of ${\mathcal{S}}_1$. Setting ${sup}_{(t,x)\in J\times B_R}|f(t,x)|=f^{\ast }<\mathrm{\infty }$, then for each ${\tau }_1,{\tau }_2\in (t_l,t_{l+1})$ for some $l\in \{0,1,\dots ,m\}$ with ${\tau }_2>{\tau }_1$, we have

$$\begin{array}{r}|({\mathcal{S}}_{1}x)({\tau }_2)-({\mathcal{S}}_{1}x)({\tau }_1)|\\ \le |{\tau }_2-{\tau }_1|\sum _{k=1}^j{\int }_{t_{k-1}}^{t_k}\left|f(s,x(s))\right|d_{q_{k-1}}s\\ +\frac{|{\tau }_2-{\tau }_1|}{T-\eta }\sum _{k=j+1}^m{\int }_{t_{k-1}}^{t_k}{\int }_{t_{k-1}}^s\left|f(\sigma ,x(\sigma ))\right|d_{q_{k-1}}\sigma d_{q_{k-1}}s\\ +\frac{|{\tau }_2-{\tau }_1|}{T-\eta }\sum _{k=j+1}^m(T-t_k){\int }_{t_{k-1}}^{t_k}\left|f(s,x(s))\right|d_{q_{k-1}}s\\ +\frac{|{\tau }_2-{\tau }_1|}{T-\eta }{\int }_{t_j}^{\eta }{\int }_{t_j}^s\left|f(\sigma ,x(\sigma ))\right|d_{q_j}\sigma d_{q_j}s\\ +\frac{|{\tau }_2-{\tau }_1|}{T-\eta }{\int }_{t_m}^T{\int }_{t_m}^s\left|f(\sigma ,x(\sigma ))\right|d_{q_m}\sigma d_{q_m}s+|{\tau }_2-{\tau }_1|\sum _{k=1}^l{\int }_{t_{k-1}}^{t_k}\left|f(s,x(s))\right|d_{q_{k-1}}s\\ +\left|{\int }_{t_l}^{{\tau }_2}{\int }_{t_l}^s\right|f(\sigma ,x(\sigma ))|d_{q_l}\sigma d_{q_l}s-{\int }_{t_l}^{{\tau }_1}{\int }_{t_l}^s|f(\sigma ,x(\sigma ))\left|d_{q_l}\sigma d_{q_l}s\right|\\ \le |{\tau }_2-{\tau }_1|f^{\ast }[\sum _{k=1}^j(t_k-t_{k-1})+\frac{1}{T-\eta }\sum _{k=j+1}^m\frac{{(t_k-t_{k-1})}^2}{1+q_{k-1}}+\frac{{(\eta -t_j)}^2}{(T-\eta )(1+q_j)}\\ +\frac{{(T-t_m)}^2}{(T-\eta )(1+q_m)}+\frac{1}{T-\eta }\sum _{k=j+1}^m(T-t_k)(t_k-t_{k-1})\\ +\sum _{k=1}^l(t_k-t_{k-1})+\frac{({\tau }_1+{\tau }_2+2t_l)}{1+q_l}].\end{array}$$

As ${\tau }_1\to {\tau }_2$, the right hand side above (which is independent of x) tends to zero. Therefore, the operator ${\mathcal{S}}_1$ is equicontinuous. Since ${\mathcal{S}}_1$ maps bounded subsets into relatively compact subsets, it follows that ${\mathcal{S}}_1$ is relative compact on $B_R$. Hence, by the Arzelá-Ascoli theorem, ${\mathcal{S}}_1$ 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:

$$\{\begin{array}{c}{D}_{\frac{4}{5+k}}^{2}x(t)=\frac{e^{-{cos}^{2}t}|x(t)|}{{(6+t)}^2(1+|x(t)|)},t\in J=[0,1],t\ne t_k=\frac{k}{10},\hfill \\ \mathrm{\Delta }x(t_k)=\frac{|x(t_k)|}{8(7+|x(t_k)|)},k=1,2,\dots ,9,\hfill \\ D_{\frac{4}{5+k}}x(t_k^{+})-D_{\frac{4}{4+k}}x(t_k)=\frac{1}{6}{tan}^{-1}(\frac{1}{8}x(t_k)),k=1,2,\dots ,9,\hfill \\ x(0)=0,x(1)=x(\frac{1}{4}).\hfill \end{array}$$

(4.1)

Here $q_k=4/(5+k)$ for $k=0,1,2,\dots ,9$, $m=9$, $T=1$, $\eta =1/4$, $j=2$, $f(t,x)=(e^{-{cos}^{2}t}|x|)/({(6+t)}^2(1+|x|))$, $I_k(x)=|x|/(8(7+|x|))$ and $I_k^{\ast }(x)=(1/6){tan}^{-1}(x/8)$. Since

$$\begin{array}{r}|f(t,x)-f(t,y)|\le (1/36)|x-y|,\\ |I_k(x)-I_k(y)|\le (1/56)|x-y|\text{and}|I_k^{\ast }(x)-I_k^{\ast }(y)|\le (1/48)|x-y|,\end{array}$$

then (H₁) and (H₂) are satisfied with $L_1=(1/36)$, $L_2=(1/56)$, $L_3=(1/48)$. We can show that

$$\mathrm{\Lambda }\approx 0.5730986482<1.$$

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:

$$\{\begin{array}{c}{D}_{\frac{3}{6+k}}^{2}x(t)=\frac{{sin}^2(\pi t)}{{(t+4)}^2}\frac{|x(t)|}{(1+|x(t)|)},t\in J=[0,1],t\ne t_k=\frac{k}{10},\hfill \\ \mathrm{\Delta }x(t_k)=\frac{|x(t_k)|}{9(7+|x(t_k)|)},k=1,2,\dots ,9,\hfill \\ D_{\frac{3}{6+k}}x(t_k^{+})-D_{\frac{3}{5+k}}x(t_k)=\frac{|x(t_k)|}{4(5+|x(t_k)|)},k=1,2,\dots ,9,\hfill \\ x(0)=0,x(1)=x(\frac{9}{20}).\hfill \end{array}$$

(4.2)

Set $q_k=3/(6+k)$ for $k=0,1,2,\dots ,9$, $m=9$, $T=1$, $\eta =9/20$, $j=4$, $f(t,x)=({sin}^2(\pi t)|x|)/({(t+4)}^2(1+|x|))$, $I_k(x)=|x|/(9(7+|x|))$ and $I_k^{\ast }(x)=|x|/(4(5+|x|))$. Since

$$|I_k(x)-I_k(y)|\le (1/63)|x-y|\text{and}|I_k^{\ast }(x)-I_k^{\ast }(y)|\le (1/20)|x-y|,$$

then (H₂) is satisfied with $L_2=(1/63)$, $L_3=(1/20)$. It is easy to verify that $|f(t,x)|\le \mu (t)\equiv 1$, $I_k(x)\le N_1=1/9$ and $I_k^{\ast }(x)\le N_2=1/4$ for all $t\in [0,1]$, $x\in \mathbb{R}$, $k=1,\dots ,m$. Thus (H₃) and (H₄) are satisfied. We can show that

$$jT{L}_3+m{L}_2+\frac{T(m-j)L_2}{T-\eta }+L_3\sum _{k=1}^m(T-t_k)=\frac{19741}{27720}<1.$$

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.