# Impulsive quantum $$(p,q)$$-difference equations

## Abstract

In this paper we study quantum $$(p,q)$$-difference equations with impulse and initial or boundary conditions. We consider first order impulsive $$(p,q)$$-difference boundary value problems and second order impulsive $$(p,q)$$-difference initial value problems. Existence and uniqueness results are proved via Banach’s fixed point theorem.

## Introduction and preliminaries

Let p, q be quantum constants satisfying $$0< q< p\leq1$$. The $$(p,q)$$-number, $$[n]_{p,q}$$, is defined by

$$[n]_{p,q}=\frac{p^{n}-q^{n}}{p-q}.$$

If n is a positive integer, then

$$[n]_{p,q}=p^{n-1}+p^{n-2}q+\cdots +pq^{n-2}+q^{n-1}\quad \text{and} \quad\lim_{(p,q)\to(1,1)}[n]_{p,q}=n.$$

The $$(p,q)$$-difference of a function f on $$[0,\infty)$$ is defined by

$$D_{p,q}f(t)=\frac{f(pt)-f(qt)}{(p-q)t}, \quad t\neq0,$$
(1.1)

and $$D_{p,q}f(0)=f'(0)$$. If $$f(t)=t^{\alpha}$$, $$\alpha\geq0$$, then we have

$$D_{p,q}t^{\alpha}=[\alpha]_{p,q}t^{\alpha-1}.$$
(1.2)

Note that if the function f is defined on $$[0,T]$$, then the function $$D_{p,q}f(t)$$ is defined on $$[0, T/p]$$. For some details of the shifting property and nonlocal boundary value problems for first-order $$(p,q)$$-difference equations, we refer the reader to . In addition, in , the authors defined the second-order $$(p,q)$$-difference by

$$D_{p,q}^{2}f(t)=\frac{qf(p^{2}t)-(p+q)f(pqt)+pf(q^{2}t)}{pq(p-q)^{2}t^{2}}.$$

Then we see that if $$f(t)$$ is defined on $$[0,T]$$ then the function $$D_{p,q}^{2}f(t)$$ is defined on $$[0, T/p^{2}]$$.

The $$(p,q)$$-integral of a function f on $$[0,\infty)$$ is defined by

$$\int_{0}^{t}f(s)\,d_{p,q}s =(p-q)t\sum _{n=0}^{\infty}\frac {q^{n}}{p^{n+1}}f \biggl( \frac{q^{n}}{p^{n+1}}t \biggr).$$
(1.3)

If $$f(t)=t^{\alpha}$$, $$\alpha>0$$, then we have the formula

$$\int_{0}^{t}s^{\alpha}\,d_{p,q}s = \frac{p-q}{p^{\alpha+1}-q^{\alpha +1}}t^{\alpha+1}.$$
(1.4)

Now we observe that if the function f is defined on a finite interval $$[0,T]$$ then the function $$\int_{0}^{t}f(s)\,d_{p,q}s$$ is defined on $$[0, pT]$$. In , the authors gave the formula of the double $$(p,q)$$-integral

\begin{aligned} \int_{0}^{t} \int_{0}^{s}f(r)\,d_{p,q}r\,d_{p,q}s =& \frac{1}{p} \int _{0}^{t}(t-qs)f \biggl(\frac{1}{p}s \biggr)\,d_{p,q}s \\ =& \frac{1}{p}(p-q)t^{2}\sum_{n=0}^{\infty} \frac{q^{n}}{p^{2n+2}} \bigl(p^{n+1}-q^{n+1} \bigr)f \biggl( \frac{q^{n}}{p^{n+2}}t \biggr), \end{aligned}

which implies that if f is defined on $$[0,T]$$, then the function $$\int_{0}^{t}\int_{0}^{s}f(r)\,d_{p,q}r\,d_{p,q}s$$ is defined on $$[0, p^{2}T]$$.

The $$(p,q)$$-calculus was introduced in . For some recent results, see  and references cited therein. For $$p=1$$, the $$(p,q)$$-calculus is reduced to the classical q-calculus initiated by Jackson [11, 12]. See also [13, 14].

In [15, 16], M. Tunç and E. Göv defined the quantum $$(p,q)$$-difference of a function f on the finite interval $$[a,b]$$ by

$${_{a}}D_{p,q}f(t)=\frac{f(pt+(1-p)a)-f(qt+(1-q)a)}{(p-q)(t-a)}, \quad t \neq a,$$
(1.5)

and $${_{a}}D_{p,q}f(a)=f'(a)$$. The $$(p,q)$$-difference of a power function $$f(t)=(t-a)^{\alpha}$$, $$\alpha\geq0$$, is given by

$$D_{p,q}(t-a)^{\alpha}=[\alpha]_{p,q}(t-a)^{\alpha-1}.$$
(1.6)

Furthermore, they defined the $$(p,q)$$-integral of a function f on $$[a,b]$$ as

$$\int_{a}^{t}f(s)\,{_{a}}d_{p,q}s=(p-q) (t-a)\sum_{n=0}^{\infty}\frac {q^{n}}{p^{n+1}}f \biggl(\frac{q^{n}}{p^{n+1}}t+ \biggl(1-\frac {q^{n}}{p^{n+1}} \biggr)a \biggr).$$
(1.7)

As is customary, we put the following relation:

$$\int_{a}^{t}(s-a)^{\alpha}\,{_{a}}d_{p,q}s =\frac{p-q}{ (p^{\alpha +1}-q^{\alpha+1} )}(t-a)^{\alpha+1},\quad\alpha\geq0.$$
(1.8)

It is obvious that if $$a=0$$, then equations (1.5)–(1.8) are reduced to (1.1)–(1.4), respectively.

The domain-shift properties of the $$(p,q)$$-difference and $$(p,q)$$-integral operators for a function $$f(t)$$, $$t\in[a,b]$$ are respectively given by

$${_{a}}D_{p,q}f(t),\quad t\in \biggl[a,\frac{1}{p}(b-a)+a \biggr]\quad\text{and}\quad \int_{a}^{t}f(s)\,{_{a}}d_{p,q}s,\quad t\in \bigl[a,p(b-a)+a \bigr].$$

Also we remark that if $$p=1$$, then both domains are reduced to $$[a,b]$$. For the shifting of the second order $$(p,q)$$-difference and integral domains, we consider the following result.

### Lemma 1.1

Letfbe a function defined on an interval$$[a,b]$$with$$a\geq0$$. The domains of$${_{a}}D_{p,q}^{2}f$$and$$\int_{a}^{t}\int_{a}^{r}f(s) \,{_{a}}d_{p,q}s \,{_{a}}d_{p,q}r$$are

$$\biggl[a,\frac{1}{p^{2}}(b-a)+a \biggr]\quad\textit{and}\quad \bigl[a,p^{2}(b-a)+a \bigr],$$

respectively.

### Proof

We have

\begin{aligned} {_{a}}D_{p,q}^{2}f(t) =&{_{a}}D_{p,q} ({_{a}}D_{p,q}f ) (t) ={_{a}}D_{p,q} \biggl(\frac {f(pt+(1-p)a)-f(qt+(1-q)a)}{(p-q)(t-a)} \biggr) \\ =& \biggl\{ \frac {f(p(pt+(1-p)a)+(1-p)a)-f(q(pt+(1-p)a)+(1-q)a)}{(p-q)((pt+(1-p)a)-a)} \\ &{}- \frac {f(p(qt+(1-q)a)+(1-p)a)-f(q(qt+(1-q)a)+(1-q)a)}{(p-q)((qt+(1-q)a)-a)} \biggr\} \\ &{} /(p-q) (t-a) \\ =&\frac {qf(p^{2}t+(1-p^{2})a)-(p+q)f(pqt+(1-pq)a)+pf(q^{2}t+(1-q^{2})a)}{pq(p-q)^{2}(t-a)^{2}}. \end{aligned}

Setting $$p^{2}t+(1-p^{2})a=b$$, we have

$$t=\frac{1}{p^{2}}(b-a)+a.$$

Then $${_{a}}D_{p,q}^{2}f$$ is defined on $$[a,(b-a)/p^{2}+a]$$.

Next we write the double $$(p,q)$$-integral in the form of an infinite sum of a function f defined on $$[a,b]$$. We have

\begin{aligned} \int_{a}^{t} \int_{a}^{s}f(r) \,{_{a}}d_{p,q}r \,{_{a}}d_{p,q}s &= \int_{a}^{t} \Biggl[(p-q) (s-a)\sum _{n=0}^{\infty}\frac {q^{n}}{p^{n+1}}f \biggl( \frac{q^{n}}{p^{n+1}}s+ \biggl\{ 1-\frac {q^{n}}{p^{n+1}} \biggr\} a \biggr)\,{_{a}}d_{p,q}s \Biggr] \\ &=(p-q)\sum_{n=0}^{\infty}\frac{q^{n}}{p^{n+1}} \biggl[ \int _{a}^{t}(s-a)f \biggl(\frac{q^{n}}{p^{n+1}}s+ \biggl\{ 1-\frac {q^{n}}{p^{n+1}} \biggr\} a \biggr)\,{_{a}}d_{p,q}s \biggr].\end{aligned}

Now we consider

\begin{aligned} & \int_{a}^{t}(s-a)f \biggl(\frac{q^{n}}{p^{n+1}}s+ \biggl\{ 1-\frac {q^{n}}{p^{n+1}} \biggr\} a \biggr)\,{_{a}}d_{p,q}s \\ &\quad=(p-q) (t-a)\sum_{m=0}^{\infty} \frac{q^{m}}{p^{m+1}} \biggl(\frac {q^{m}}{p^{m+1}}t+ \biggl\{ 1-\frac{q^{m}}{p^{m+1}} \biggr\} a-a \biggr) \\ &\qquad{}\times f \biggl(\frac{q^{n}}{p^{n+1}} \biggl[\frac {q^{m}}{p^{m+1}}t+ \biggl\{ 1- \frac{q^{m}}{p^{m+1}} \biggr\} a \biggr]+ \biggl\{ 1-\frac{q^{n}}{p^{n+1}} \biggr\} a \biggr) \\ &\quad=(p-q) (t-a)^{2}\sum_{m=0}^{\infty} \frac{q^{2m}}{p^{2m+2}}f \biggl(\frac{q^{m+n}}{p^{m+n+2}}t+ \biggl\{ 1-\frac{q^{m+n}}{p^{m+n+2}} \biggr\} a \biggr), \end{aligned}

\begin{aligned} & \int_{a}^{t} \int_{a}^{s}f(r) \,{_{a}}d_{p,q}r \,{_{a}}d_{p,q}s \\ &\quad=(p-q)^{2}(t-a)^{2}\sum_{n=0}^{\infty} \sum_{m=0}^{\infty}\frac {q^{2m+n}}{p^{2m+n+3}} f \biggl( \frac{q^{m+n}}{p^{m+n+2}}t+ \biggl\{ 1-\frac {q^{m+n}}{p^{m+n+2}} \biggr\} a \biggr). \end{aligned}
(1.9)

For $$m=n=0$$ and setting

$$\frac{1}{p^{2}}t+ \biggl\{ 1-\frac{1}{p^{2}} \biggr\} a=b,$$

we obtain $$t=p^{2}(b-a)+a$$, which implies that $$\int_{a}^{t}\int_{a}^{r}f(s) \,{_{a}}d_{p,q}s \,{_{a}}d_{p,q}r$$ is valid on $$[a, p^{2}(b-a)+a]$$. The proof is completed. □

Before going to the next result, we would like to recall the operator $${_{a}}\varPhi_{r}$$ defined by

$${_{a}}\varPhi_{r}(m)=rm+(1-r)a,$$

where $$m,a\in{\mathbb{R}}$$ and $$r\in[0,1]$$. Some properties of this operator can be found in .

### Lemma 1.2

Letfbe a function defined on$$[a,b]$$. Then the double$$(p,q)$$-integral offcan be written as a single one by

$$\int_{a}^{t} \int_{a}^{s}f(r) \,{_{a}}d_{p,q}r \,{_{a}}d_{p,q}s=\frac{1}{p} \int _{a}^{t}\bigl(t-{_{a}} \varPhi_{q}(s)\bigr)f \bigl({_{a}}\varPhi_{\frac{1}{p}}(s) \bigr)\,{_{a}}d_{p,q}s,\quad t\in\bigl[a,p^{2}(b-a)+a \bigr].$$
(1.10)

### Proof

The double summation in (1.9) can be formulated by a single summation as

\begin{aligned} &\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{q^{2m+n}}{p^{2m+n+3}} f \biggl( \frac{q^{m+n}}{p^{m+n+2}}t+ \biggl\{ 1-\frac {q^{m+n}}{p^{m+n+2}} \biggr\} a \biggr) \\ &\quad=\sum_{n=0}^{\infty} \biggl[ \frac{q^{n}}{p^{n+3}} f \biggl(\frac{q^{n}}{p^{n+2}}t+ \biggl\{ 1-\frac{q^{n}}{p^{n+2}} \biggr\} a \biggr)+\frac{q^{n+2}}{p^{n+5}} f \biggl(\frac{q^{n+1}}{p^{n+3}}t+ \biggl\{ 1- \frac {q^{n+1}}{p^{n+3}} \biggr\} a \biggr) \\ &\quad\quad{}+ \frac{q^{n+4}}{p^{n+7}} f \biggl(\frac{q^{n+2}}{p^{n+4}}t+ \biggl\{ 1- \frac {q^{n+2}}{p^{n+4}} \biggr\} a \biggr)+\frac{q^{n+6}}{p^{n+9}} f \biggl( \frac{q^{n+3}}{p^{n+5}}t+ \biggl\{ 1-\frac {q^{n+3}}{p^{n+5}} \biggr\} a \biggr)+\cdots \biggr] \\ &\quad=\frac{1}{p^{3}}f \biggl(\frac{1}{p^{2}}t+ \biggl\{ 1-\frac {1}{p^{2}} \biggr\} a \biggr)+\frac{q}{p^{4}} \biggl(1+\frac{q}{p} \biggr) f \biggl( \frac{q}{p^{3}}t+ \biggl\{ 1-\frac{q}{p^{3}} \biggr\} a \biggr) \\ &\qquad{}+ \frac{q^{2}}{p^{5}} \biggl(1+\frac{q}{p}+\frac{q^{2}}{p^{2}} \biggr) f \biggl(\frac{q^{2}}{p^{4}}t+ \biggl\{ 1-\frac{q^{2}}{p^{4}} \biggr\} a \biggr)+\cdots \\ &\quad=\sum_{n=0}^{\infty}\frac{q^{n}}{p^{n+3}} \biggl(\frac {p^{n+1}-q^{n+1}}{p^{n}(p-q)} \biggr)f \biggl(\frac{q^{n}}{p^{n+2}}t+ \biggl\{ 1- \frac{q^{n}}{p^{n+2}} \biggr\} a \biggr) \\ &\quad=\frac{1}{p-q}\sum_{n=0}^{\infty} \frac{q^{n}}{p^{n+1}} \biggl(\frac {1}{p}-\frac{q^{n+1}}{p^{n+2}} \biggr)f \biggl( \frac {q^{n}}{p^{n+2}}t+ \biggl\{ 1-\frac{q^{n}}{p^{n+2}} \biggr\} a \biggr). \end{aligned}

Substituting into (1.9) yields

\begin{aligned} & \int_{a}^{t} \int_{a}^{s}f(r) \,{_{a}}d_{p,q}r \,{_{a}}d_{p,q}s \\ &\quad=\frac{1}{p}(p-q) (t-a)\sum_{n=0}^{\infty} \frac{q^{n}}{p^{n+1}} \biggl(t- \biggl[\frac{q^{n+1}}{p^{n+1}}t+ \biggl\{ 1- \frac {q^{n+1}}{p^{n+1}} \biggr\} a \biggr] \biggr) \\ &\qquad{}\times f \biggl(\frac{q^{n}}{p^{n+2}}t+ \biggl\{ 1-\frac {q^{n}}{p^{n+2}} \biggr\} a \biggr) \\ &\quad=\frac{1}{p}(p-q) (t-a)\sum_{n=0}^{\infty} \frac{q^{n}}{p^{n+1}} \biggl(t-{_{a}}\varPhi_{q} \biggl( \biggl[\frac{q^{n}}{p^{n+1}}t+ \biggl\{ 1-\frac {q^{n}}{p^{n+1}} \biggr\} a \biggr] \biggr) \biggr) \\ &\qquad{}\times f \biggl({_{a}}\varPhi_{\frac{1}{p}} \biggl( \frac {q^{n}}{p^{n+1}}t+ \biggl\{ 1-\frac{q^{n}}{p^{n+1}} \biggr\} a \biggr) \biggr) \\ &\quad=\frac{1}{p} \int_{a}^{t}\bigl(t-{_{a}} \varPhi_{q}(s)\bigr)f \bigl({_{a}}\varPhi _{\frac{1}{p}}(s) \bigr)\,{_{a}}d_{p,q}s, \end{aligned}

which is completed the proof. □

### Remark 1.3

If $$a=0$$, then (1.10) is reduced to a result of Theorem 3 in .

The following theorem has been proved in .

### Theorem 1.4

The fundamental relations of$$(p,q)$$-calculus can be stated as

1. (i)

$${_{a}}D_{p,q}\int_{a}^{t} f(s) \,{_{a}}d_{p,q}s=f(t)$$;

2. (ii)

$$\int_{a}^{t}{_{a}}D_{p,q}f(s) \,{_{a}}d_{p,q}s=f(t)-f(a)$$.

In this paper we study the impulsive $$(p,q)$$-difference equations with initial and boundary conditions. We consider four types of problems, two impulsive $$(p,q)$$-difference equations of type I and two impulsive $$(p,q)$$-difference equations of type II (explained in the next section). Existence and uniqueness results are proved via Banach’s contraction mapping principle. Examples illustrating the obtained results are also constructed.

## Impulsive $$(p,q)$$-difference equations

In this section, we consider the first and second order $$(p,q)$$-difference equations with initial or boundary conditions and also prove the existence and uniqueness of solutions for impulsive problems. Firstly, let $$t_{k}$$, $$k=1,\ldots,m$$, be the impulsive points such that $$0=t_{0}< t_{1}<\cdots<t_{k}<\cdots<t_{m}<t_{m+1}=T$$ and $$J_{k}=(t_{k},t_{k+1}]$$, $$k=1,\ldots,m$$, $$J_{0}=[0,t_{1}]$$ be the intervals such that $$\bigcup_{k=0}^{m}J_{k}=[0,T]:=J$$. The investigations are based on $$(p,q)$$-calculus introduced in the previous section by replacing a point a by $$t_{k}$$, quantum numbers p by $$p_{k}$$ and q by $$q_{k}$$, $$k=0,1,\ldots,m$$, and also applying the $$(p_{k},q_{k})$$-difference and $$(p_{k},q_{k})$$-integral operators only on a finite subinterval of J. In addition, the consecutive subintervals can be related with jump conditions which provide a meaning of quantum difference equations with impulse effects. There are two types of impulsive problems which will be established in the next two subsections. The consecutive domains of impulsive $$(p,q)$$-difference equations of type I are overlapped, while the unknown functions of impulsive equations of type II are defined on disconnected consecutive domains.

### Impulsive $$(p,q)$$-difference equations of type I

Consider the first-order impulsive $$(p,q)$$-difference impulsive boundary value problem of the form

$$\textstyle\begin{cases} {_{t_{k}}}D_{p_{k},q_{k}}x(t)=f(t,x(t)),\quad t\in (t_{k}, \frac {1}{p_{k}}(t_{k+1}-t_{k})+t_{k} ], k=0,1,\ldots,m,\\ \Delta x(t_{k})=\varphi_{k}(x(t_{k})),\quad k=1,2,\ldots, m,\\ \alpha x(0)+\beta x(T)=\gamma, \end{cases}$$
(2.1)

where α, β, and γ are real constants with $$\alpha\neq-\beta$$, the quantum numbers $$p_{k}$$, $$q_{k}$$ satisfy $$0< q_{k}< p_{k}\leq1$$, $$k=0,1,\ldots,m$$, $$f:[0, ((T-t_{m})/p_{m})+t_{m}]\times {\mathbb{R}}\to{\mathbb{R}}$$ and $$\varphi_{k}:{\mathbb{R}}\to {\mathbb{R}}$$, $$k=1,2,\ldots,m$$, are given functions, and $${_{t_{k}}}D_{p_{k},q_{k}}$$ is the quantum $$(p_{k},q_{k})$$-difference operator starting at a point $$t_{k}$$, $$k=0,1,\ldots,m$$.

We remark that there are some overlapped intervals of domains of the first equation in (2.1). For example, if the unknown function $$x(t)$$ is defined on $$J=[0,2]$$ and if there is an impulse point $$t_{1}=1$$, that is, $$x(1^{+})\neq x(1^{-})$$, with $$p_{0}=1/2$$, $$q_{0}=1/3$$, $$p_{1}=1/4$$, and $$q_{1}=1/5$$. Then we have the $$(p,q)$$-difference equations

$${_{0}}D_{\frac{1}{2},\frac{1}{3}}x(t)=f\bigl(t,x(t)\bigr),\quad t\in (0, 2 ] \quad\text{and}\quad{_{1}}D_{\frac{1}{4},\frac {1}{5}}x(t)=f\bigl(t,x(t)\bigr), \quad t\in (1, 5 ].$$

However, by the shifting property of $$(p,q)$$-integration applied to the two above equations, we have

$$x(t)=x(0)+ \int_{0}^{t}f\bigl(s,x(s)\bigr) \,{_{0}}d_{\frac{1}{2},\frac{1}{3}}s \quad t\in (0, 1 ],$$

and

$$x(t)=x\bigl(1^{+}\bigr)+ \int_{1}^{t}f\bigl(s,x(s)\bigr)\, {_{1}}d_{\frac{1}{4},\frac{1}{5}}s,\quad t\in (1, 2 ],$$

respectively.

### Theorem 2.1

The nonlinear first-order$$(p,q)$$-difference boundary value problem (2.1) can be transformed into an integral equation

\begin{aligned}[b] x(t)&= \frac{\gamma}{(\alpha+\beta)}-\frac{\beta}{(\alpha+\beta )} \Biggl(\sum _{i=0}^{m} \int _{t_{i}}^{t_{i+1}}f\bigl(s,x(s)\bigr)\,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{m}\varphi _{j} \bigl(x(t_{j})\bigr) \Biggr) \\ &\quad+ \sum_{i=0}^{k-1} \int_{t_{i}}^{t_{i+1}}f\bigl(s,x(s)\bigr) \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{k} \varphi_{j}\bigl(x(t_{j})\bigr)+ \int _{t_{k}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{k}}}d_{p_{k},q_{k}}s,\quad t\in J,\end{aligned}
(2.2)

with$$\sum_{a}^{b}(\cdot)=0$$, if$$b< a$$.

### Proof

From $${_{t_{0}}}D_{p_{0},q_{0}}x(t)=f(t,x(t))$$, $$t\in(t_{0}, (1/p_{0})(t_{1}-t_{0})+t_{0}]$$, by taking the $$(p_{0},q_{0})$$-integral, we obtain

$$x(t)=x(0)+ \int_{t_{0}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s, \quad t\in(t_{0},t_{1}],$$

by using Theorem 1.4 and the shifting property. Next, for $${_{t_{1}}}D_{p_{1},q_{1}}x(t)=f(t,x(t))$$, $$t\in(t_{1}, (1/p_{1})(t_{2}-t_{1})+t_{1}]$$, where $$t_{1}$$ is the first impulsive point in J, we also obtain by applying the $$(p_{1},q_{1})$$-integration,

$$x(t)=x\bigl(t_{1}^{+}\bigr)+ \int_{t_{1}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{1}}}d_{p_{1},q_{1}}s, \quad t\in(t_{1},t_{2}].$$

By the impulsive condition $$x(t_{1}^{+})=x(t_{1})+\varphi_{1}(x(t_{1}))$$, it follows, for $$t\in(t_{1},t_{2}]$$, that

$$x(t)=x(0)+ \int_{t_{0}}^{t_{1}}f\bigl(s,x(s)\bigr)\,{_{t_{0}}}d_{p_{0},q_{0}}s+ \varphi _{1}\bigl(x(t_{1})\bigr)+ \int_{t_{1}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{1}}}d_{p_{1},q_{1}}s.$$

For $${_{t_{2}}}D_{p_{2},q_{2}}x(t)=f(t,x(t))$$, $$t\in(t_{2}, (1/p_{2})(t_{3}-t_{2})+t_{2}]$$, we get

$$x(t)=x\bigl(t_{2}^{+}\bigr)+ \int_{t_{2}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{2}}}d_{p_{2},q_{2}}s, \quad t\in(t_{2},t_{3}],$$

by $$(p_{2},q_{2})$$-integration and

\begin{aligned} x(t) =& x(0)+ \int_{t_{0}}^{t_{1}}f\bigl(s,x(s)\bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s+ \int _{t_{1}}^{t_{2}}f\bigl(s,x(s)\bigr) \,{_{t_{1}}}d_{p_{1},q_{1}}s \\ &{}+ \varphi_{1}\bigl(x(t_{1})\bigr)+\varphi_{2} \bigl(x(t_{2})\bigr)+ \int_{t_{2}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{2}}}d_{p_{2},q_{2}}s, \quad t\in(t_{2},t_{3}], \end{aligned}

due to the impulsive condition $$x(t_{2}^{+})=x(t_{2}^{-})+\varphi_{2}(x(t_{2}))$$.

Repeating this process, we obtain, for $$t\in J_{k}$$, $$k=0,1,\ldots,m$$, that

$$x(t)=x(0)+\sum_{i=0}^{k-1} \int_{t_{i}}^{t_{i+1}}f\bigl(s,x(s)\bigr) \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{k} \varphi_{j}\bigl(x(t_{j})\bigr)+ \int _{t_{k}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{k}}}d_{p_{k},q_{k}}s.$$
(2.3)

After that from the boundary condition $$\alpha x(0)+\beta x(T)=\gamma$$, we have

$$x(0)=\frac{\gamma}{(\alpha+\beta)}-\frac{\beta}{(\alpha+\beta )} \Biggl(\sum _{i=0}^{m} \int_{t_{i}}^{t_{i+1}}f\bigl(s,x(s)\bigr) \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{m} \varphi_{j}\bigl(x(t_{j})\bigr) \Biggr).$$

Putting the value of $$x(0)$$ into (2.3), shows that (2.2) is true and the proof is completed. □

### Remark 2.2

If $$\alpha\neq0$$ and $$\beta=0$$, then the boundary value problem (2.1) can be reduced to the initial value problem with initial condition $$x(0)=\gamma/\alpha$$.

Before going to the second-order impulsive problem, we define

$$\tau_{k}=\frac{1}{p_{k-1}}(t_{k}-t_{k-1})+t_{k-1}, \quad k=1,2,\ldots,m,$$

which are impulsive shifting points of the $$(p_{k},q_{k})$$-derivative of the unknown function in our system. In addition, we introduce a notation

$$\langle t_{i+1}\rangle_{k}= \textstyle\begin{cases} t_{i+1},& t_{i+1}\leq t_{k},\\ t,& t_{i+1}> t_{k}. \end{cases}$$

For example,

\begin{aligned} \sum_{i=0}^{2} \bigl(\langle t_{i+1}\rangle_{2}-t_{i} \bigr)K_{i} =& \bigl(\langle t_{1}\rangle_{2}-t_{0} \bigr)K_{0}+ \bigl(\langle t_{2}\rangle _{2}-t_{2} \bigr)K_{1}+ \bigl(\langle t_{3}\rangle_{2}-t_{2} \bigr)K_{2} \\ =&(t_{1}-t_{0})K_{0}+(t_{2}-t_{1})K_{1}+(t-t_{2})K_{2}, \end{aligned}

where $$K_{i}\in{\mathbb{R}}$$, $$i=0,1,2$$.

Now, we consider the second-order impulsive $$(p,q)$$-difference initial value problem of the form

$$\textstyle\begin{cases} {_{t_{k}}}D_{p_{k},q_{k}}^{2}x(t)=f(t,x(t)),\quad t\in (t_{k}, \frac {1}{p_{k}^{2}}(t_{k+1}-t_{k})+t_{k} ], k=0,1,\ldots,m,\\ \Delta x(t_{k})=\varphi_{k}(x(t_{k})),\quad k=1,2,\ldots, m,\\ {_{t_{k}}}D_{p_{k},q_{k}}x(t_{k}^{+})-{_{t_{k-1}}}D_{p_{k-1},q_{k-1}}x(\tau _{k})=\varphi_{k}^{\ast}(x(t_{k})),\quad k=1,2,\ldots, m,\\ x(0)=\lambda_{1},\qquad{_{t_{0}}}D_{p_{0},q_{0}}x(0)=\lambda_{2}, \end{cases}$$
(2.4)

where $$f:[0,((T-t_{m})/p_{m}^{2})+t_{m}]\times{\mathbb{R}}\to{\mathbb{R}}$$, $$\varphi_{k}:{\mathbb{R}}\to{\mathbb{R}}$$ and $$\varphi_{k}^{\ast }:{\mathbb{R}}\to{\mathbb{R}}$$, are given functions, $$\lambda_{1}$$, $$\lambda_{2}$$ are given constants. Observe that the distance between the impulsive points $$t_{k}$$ and $$\tau_{k}$$ in the third equation of (2.4) depends on the value of $$p_{k-1}$$ for $$k=1,2,\ldots,m$$. Indeed,

$$\tau_{k}-t_{k}=\frac{1}{p_{k-1}}(t_{k}-t_{k-1})+t_{k-1}-t_{k}= \frac {(1-p_{k-1})}{p_{k-1}}(t_{k}-t_{k-1}),$$

which has appeared by the shifting property of $$(p,q)$$-calculus as discussed in the previous section.

### Theorem 2.3

The impulsive initial value problem of type I given by the$$(p,q)$$-difference equation (2.4) can be expressed as an integral equation of the form

\begin{aligned} x(t) =& \lambda_{1}+\sum_{i=0}^{k} \bigl(\langle t_{i+1}\rangle _{k}-t_{i} \bigr) \Biggl[\lambda_{2}+\sum_{j=0}^{i-1} \biggl\{ \int _{t_{j}}^{\tau_{j+1}}f\bigl(s,x(s)\bigr) \,{_{t_{j}}}d_{p_{j},q_{j}}s+\varphi^{\ast }_{j+1} \bigl(x(t_{j+1})\bigr) \biggr\} \Biggr] \\ &{}+ \sum_{r=0}^{k-1} \biggl\{ \frac{1}{p_{r}} \int_{t_{r}}^{t_{r+1}} \bigl(t_{r+1}-{_{t_{r}}} \varPhi_{q_{r}}(s) \bigr)f_{x} \bigl({_{t_{r}}}\varPhi _{\frac{1}{p_{r}}}(s) \bigr) \,{_{t_{r}}}d_{p_{r},q_{r}}s + \varphi_{r+1}\bigl(x(t_{r+1})\bigr) \biggr\} \\ &{}+ \frac{1}{p_{k}} \int_{t_{k}}^{t} \bigl(t-{_{t_{k}}}\varPhi _{q_{k}}(s) \bigr)f_{x} \bigl({_{t_{k}}} \varPhi_{\frac{1}{p_{k}}}(s) \bigr) \,{_{t_{k}}}d_{p_{k},q_{k}}s,\quad t\in J_{k}, k=0,1,\ldots,m, \end{aligned}
(2.5)

where$$f_{x} ({_{t_{r}}}\varPhi_{\frac{1}{p_{r}}}(s) )=f ({_{t_{r}}}\varPhi_{\frac{1}{p_{r}}}(s),x ({_{t_{r}}}\varPhi_{\frac {1}{p_{r}}}(s) ) )$$, $$r=0,1,\ldots,k$$, and$$\sum_{a}^{b}(\cdot )=0$$, when$$b< a$$.

### Proof

By computing the $$(p_{0},q_{0})$$-integral of both sides of the first equation of (2.4), we get

$${_{t_{0}}}D_{p_{0},q_{0}}x(t)={_{t_{0}}}D_{p_{0},q_{0}}x(0)+ \int_{t_{0}}^{t}f\bigl(s, x(s)\bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s,\quad t\in \biggl(0,\frac{1}{p_{0}}t_{1} \biggr].$$

Applying another $$(p_{0},q_{0})$$-integration, we obtain, for $$t\in(0,t_{1}]$$,

\begin{aligned} x(t) =& x(0)+t{_{t_{0}}}D_{p_{0},q_{0}}x(0)+ \int_{t_{0}}^{t} \int_{t_{0}}^{r}f\bigl(s, x(s)\bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s \,{_{t_{0}}}d_{p_{0},q_{0}}r \\ =&\lambda_{1}+\lambda_{2}t+\frac{1}{p_{0}} \int_{t_{0}}^{t} \bigl(t-{_{t_{0}}} \varPhi_{q_{0}}(s) \bigr)f_{x} \bigl({_{t_{0}}} \varPhi_{\frac {1}{p_{0}}}(s) \bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s. \end{aligned}

For $$t\in(t_{1},((t_{2}-t_{1})/p_{1}^{2})+t_{1}]$$, applying the double $$(p_{1},q_{1})$$-integration to both sides of the first equation of (2.4), we have

$$x(t)=x\bigl(t_{1}^{+}\bigr)+(t-t_{1}){_{t_{1}}}D_{p_{1},q_{1}}x \bigl(t_{1}^{+}\bigr)+\frac{1}{p_{1}} \int _{t_{1}}^{t} \bigl(t-{_{t_{1}}} \varPhi_{q_{1}}(s) \bigr)f_{x} \bigl({_{t_{1}}} \varPhi_{\frac{1}{p_{1}}}(s) \bigr) \,{_{t_{1}}}d_{p_{1},q_{1}}s,$$

where $$t\in(t_{1},t_{2}]$$. Due to the impulsive conditions

\begin{aligned} x\bigl(t_{1}^{+}\bigr) =& x(t_{1})+\varphi_{1} \bigl(x(t_{1})\bigr) \\ =&\lambda_{1}+\lambda_{2}t_{1}+ \frac{1}{p_{0}} \int_{t_{0}}^{t_{1}} \bigl(t_{1}-{_{t_{0}}} \varPhi_{q_{0}}(s) \bigr)f_{x} \bigl({_{t_{0}}} \varPhi_{\frac {1}{p_{0}}}(s) \bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s+ \varphi_{1}\bigl(x(t_{1})\bigr) \end{aligned}

and

\begin{aligned} {_{t_{1}}}D_{p_{1},q_{1}}x\bigl(t_{1}^{+}\bigr) =& {_{t_{0}}}D_{p_{0},q_{0}}x(\tau_{1})+\varphi _{1}^{\ast}\bigl(x(t_{1})\bigr) \\ =&\lambda_{2}+ \int_{t_{0}}^{\tau_{1}}f\bigl(s, x(s)\bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s+\varphi_{1}^{\ast} \bigl(x(t_{1})\bigr), \end{aligned}

we have

\begin{aligned} x(t)&=\lambda_{1}+\lambda_{2}t_{1}+ \frac{1}{p_{0}} \int_{t_{0}}^{t_{1}} \bigl(t_{1}-{_{t_{0}}} \varPhi_{q_{0}}(s) \bigr)f_{x} \bigl({_{t_{0}}} \varPhi_{\frac {1}{p_{0}}}(s) \bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s+ \varphi_{1}\bigl(x(t_{1})\bigr) \\ &\quad+ (t-t_{1}) \biggl[ \lambda_{2}+ \int_{t_{0}}^{\tau_{1}}f\bigl(s, x(s)\bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s+\varphi_{1}^{\ast} \bigl(x(t_{1})\bigr) \biggr] \\ &\quad+ \frac{1}{p_{1}} \int_{t_{1}}^{t} \bigl(t-{_{t_{1}}} \varPhi_{q_{1}}(s) \bigr)f_{x} \bigl({_{t_{1}}} \varPhi_{\frac{1}{p_{1}}}(s) \bigr) \,{_{t_{1}}}d_{p_{1},q_{1}}s, \quad t \in(t_{1},t_{2}].\end{aligned}

Similarly, we deduce the integral equation (2.5), as desired. □

Now, the existence and uniqueness results for problems (2.1) and (2.4) will be proved by using the Banach’s contraction mapping principle. Let us define the space $$\mathit{PC}(J,{\mathbb{R}})=\{x:J\to{\mathbb{R}}$$: $$x(t)$$ is continuous everywhere except for some $$t_{k}$$ in which $$x(t_{k}^{+})$$ and $$x(t_{k}^{-})$$ exist and $$x(t_{k}^{-})=x(t_{k})$$, $$k=1,2,\ldots,m\}$$. The set $$\mathit{PC}(J,{\mathbb{R}})$$ is a Banach space equipped with the norm $$\|x\| =\sup\{|x(t)|: t\in J\}$$. For convenience, we put

\begin{aligned}& \varOmega_{1}=\frac{ \vert \beta \vert + \vert \alpha+\beta \vert }{ \vert \alpha+\beta \vert }\sum_{i=0}^{m}(t_{i+1}-t_{i}), \\& \varOmega_{2}=m \biggl(\frac{ \vert \beta \vert + \vert \alpha+\beta \vert }{ \vert \alpha+\beta \vert } \biggr), \\& \varOmega_{3}=\sum_{i=0}^{m} \Biggl\{ (t_{i+1}-t_{i})\sum_{j=0}^{i-1}( \tau _{j+1}-t_{j}) \Biggr\} +\sum _{r=0}^{m}\frac{(t_{r+1}-t_{r})^{2}}{p_{r}+q_{r}}, \\& \varOmega_{4}=\sum_{i=0}^{m}(t_{i+1}-t_{i})i. \end{aligned}

### Theorem 2.4

Let$$f:[0,((T-t_{m})/p_{m})+t_{m}]\times {\mathbb{R}}\to{\mathbb{R}}$$and$$\varphi_{k}:{\mathbb{R}}\to {\mathbb{R}}$$, $$k=1,2,\ldots,m$$, be given functions satisfying

$$(H_{1})$$:

There exist positive constants$$L_{1}$$and$$L_{2}$$such that

\begin{aligned} \bigl\vert f(t, x) - f(t, y) \bigr\vert \leq L_{1} \vert x - y \vert \quad\textit{and}\quad \bigl\vert \varphi _{k}(x) - \varphi_{k}(y) \bigr\vert \leq L_{2} \vert x - y \vert , \end{aligned}

for all$$t\in[0,((T-t_{m})/p_{m})+t_{m}]$$, $$x,y\in{\mathbb{R}}$$and$$k=1,2,\ldots,m$$.

If

$$L_{1}\varOmega_{1}+L_{2} \varOmega_{2}< 1,$$
(2.6)

then the boundary value problem (2.1) has a unique solution onJ.

### Proof

In view of Theorem 2.1, we define the operator $${\mathcal{A}}:\mathit{PC}(J,{\mathbb{R}})\to \mathit{PC}(J,{\mathbb{R}})$$ by

\begin{aligned} {\mathcal{A}}x(t) =& \frac{\gamma}{(\alpha+\beta)}-\frac{\beta }{(\alpha+\beta)} \Biggl(\sum _{i=0}^{m} \int_{t_{i}}^{t_{i+1}}f\bigl(s,x(s)\bigr) \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{m} \varphi_{j}\bigl(x(t_{j})\bigr) \Biggr) \\ &{}+ \sum_{i=0}^{k-1} \int_{t_{i}}^{t_{i+1}}f\bigl(s,x(s)\bigr) \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{k} \varphi_{j}\bigl(x(t_{j})\bigr)+ \int _{t_{k}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{k}}}d_{p_{k},q_{k}}s,\quad t\in J. \end{aligned}

Define the ball $$B_{r_{1}}=\{x\in \mathit{PC}(J,{\mathbb{R}}):\|x\|\leq r_{1}\}$$ where the positive constant $$r_{1}$$ is defined by

$$r_{1}>\frac{ ( \vert \gamma \vert / \vert \alpha+\beta \vert )+M_{1}\varOmega _{1}+M_{2}\varOmega_{2}}{1-(L_{1}\varOmega_{1}+L_{2}\varOmega_{2})}.$$

The Banach contraction mapping principle is used to claim that there exists a unique fixed point of an operator equation $$x={\mathcal{A}}x$$ in $$B_{r_{1}}$$. By setting $$\sup_{t\in J}|f(t,0)|=M_{1}$$, and $$\sup\{ |\varphi_{i}(0)|, i=1,2,\ldots, m\}=M_{2}$$ and using the inequalities $$|f(t,x)|\leq|f(t,x)-f(t,0)|+|f(t,0)|\leq L_{1}r_{1}+M_{1}$$ and $$|\varphi _{i}(x)|\leq|\varphi_{i}(x)-\varphi_{i}(0)|+|\varphi_{i}(0)|\leq L_{1}r_{1}+M_{2}$$, $$i=1,2,\ldots, m$$, we have

\begin{aligned} \bigl\vert {\mathcal{A}}x(t) \bigr\vert \leq& \frac{ \vert \gamma \vert }{ \vert \alpha+\beta \vert }+ \frac { \vert \beta \vert }{ \vert \alpha+\beta \vert } \Biggl(\sum_{i=0}^{m} \int _{t_{i}}^{t_{i+1}} \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum _{j=1}^{m} \bigl\vert \varphi _{j} \bigl(x(t_{j})\bigr) \bigr\vert \Biggr) \\ &{}+ \sum_{i=0}^{k-1} \int_{t_{i}}^{t_{i+1}} \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{k} \bigl\vert \varphi_{j}\bigl(x(t_{j})\bigr) \bigr\vert + \int _{t_{k}}^{t} \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert \,{_{t_{k}}}d_{p_{k},q_{k}}s \\ \leq&\frac{ \vert \gamma \vert }{ \vert \alpha+\beta \vert }+\frac{ \vert \beta \vert }{ \vert \alpha +\beta \vert } \Biggl(\sum _{i=0}^{m}(L_{1}r_{1}+M_{1}) \int _{t_{i}}^{t_{i+1}}(1)\,{_{t_{i}}}d_{p_{i},q_{i}}s +(L_{2}r_{1}+M_{2})\sum _{j=1}^{m}(1) \Biggr) \\ & {}+(L_{1}r_{1}+M_{1})\sum _{i=0}^{m-1} \int _{t_{i}}^{t_{i+1}}(1)\,{_{t_{i}}}d_{p_{i},q_{i}}s+(L_{2}r_{1}+M_{2}) \sum_{j=1}^{m}(1) \\ &{}+(L_{1}r_{1}+M_{1}) \int_{t_{m}}^{t_{m+1}}(1)\,{_{t_{k}}}d_{p_{k},q_{k}}s \\ =&\frac{ \vert \gamma \vert }{ \vert \alpha+\beta \vert }+\frac{ \vert \beta \vert }{ \vert \alpha+\beta \vert } \Biggl((L_{1}r_{1}+M_{1}) \sum_{i=0}^{m}(t_{i+1}-t_{i}) +m(L_{2}r_{1}+M_{2}) \Biggr) \\ &{}+ (L_{1}r_{1}+M_{1})\sum _{i=0}^{m-1}(t_{i+1}-t_{i}) +m(L_{2}r_{1}+M_{2})+(L_{1}r_{1}+M_{1}) (t_{m+1}-t_{m}) \\ =&\frac{ \vert \gamma \vert }{ \vert \alpha+\beta \vert }+L_{1}\varOmega_{1}r_{1}+L_{2} \varOmega _{2}r_{1}+M_{1}\varOmega_{1}+M_{2} \varOmega_{2}< r_{1}, \end{aligned}

which leads to $${\mathcal{A}}B_{r_{1}}\subset B_{r_{1}}$$. To prove that $${\mathcal{A}}$$ is a contraction, we let $$x,y\in B_{r_{1}}$$. Then we have

\begin{aligned}& \bigl\vert {\mathcal{A}}x(t)-{\mathcal{A}}y(t) \bigr\vert \\& \quad\leq \frac{ \vert \beta \vert }{ \vert \alpha+\beta \vert } \Biggl(\sum_{i=0}^{m} \int _{t_{i}}^{t_{i+1}} \bigl\vert f\bigl(s,x(s)\bigr)-f \bigl(s,y(s)\bigr) \bigr\vert \,{_{t_{i}}}d_{p_{i},q_{i}}s+ \sum_{j=1}^{m} \bigl\vert \varphi_{j}\bigl(x(t_{j})\bigr)-\varphi_{j} \bigl(y(t_{j})\bigr) \bigr\vert \Biggr) \\& \qquad{}+\sum _{i=0}^{k-1} \int_{t_{i}}^{t_{i+1}} \bigl\vert f\bigl(s,x(s)\bigr)-f \bigl(s,y(s)\bigr) \bigr\vert \,{_{t_{i}}}d_{p_{i},q_{i}}s \\& \qquad{}+ \sum_{j=1}^{k} \bigl\vert \varphi_{j}\bigl(x(t_{j})\bigr)-\varphi _{j} \bigl(y(t_{j})\bigr) \bigr\vert + \int_{t_{k}}^{t} \bigl\vert f\bigl(s,x(s)\bigr)-f \bigl(s,y(s)\bigr) \bigr\vert \,{_{t_{k}}}d_{p_{k},q_{k}}s \\& \quad\leq\frac{ \vert \beta \vert }{ \vert \alpha+\beta \vert } \Biggl(L_{1} \Vert x-y \Vert \sum _{i=0}^{m}(t_{i+1}-t_{i}) +mL_{2} \Vert x-y \Vert \Biggr) \\& \qquad{}+ L_{1} \Vert x-y \Vert \sum_{i=0}^{m}(t_{i+1}-t_{i}) +mL_{2} \Vert x-y \Vert \\& \quad=(L_{1}\varOmega_{1}+L_{2} \varOmega_{2}) \Vert x-y \Vert . \end{aligned}

Therefore, $$\|{\mathcal{A}}x-{\mathcal{A}}y\|\leq(L_{1}\varOmega _{1}+L_{2}\varOmega_{2})\|x-y\|$$. By means of the Banach contraction mapping principle, the operator $${\mathcal{A}}$$ has a unique fixed point in $$B_{r_{1}}$$ which is a unique solution of boundary value problem (2.1). The proof is completed. □

### Theorem 2.5

Assume that the functions$$f:[0,((T-t_{m})/p_{m}^{2})+t_{m}]\times{\mathbb {R}}\to{\mathbb{R}}$$and$$\varphi_{k}:{\mathbb{R}}\to{\mathbb{R}}$$, $$k=1,2,\ldots,m$$, satisfy$$(H_{1})$$. In addition, we suppose that the functions$$\varphi_{k}^{\ast}:{\mathbb{R}}\to{\mathbb{R}}$$, $$k=1,2,\ldots,m$$, satisfy

$$(H_{2})$$:

There exists a positive constant$$L_{3}$$such that

\begin{aligned} \bigl\vert \varphi_{k}^{\ast}(x) - \varphi_{k}^{\ast}(y) \bigr\vert \leq L_{3} \vert x - y \vert , \end{aligned}
for all$$x,y\in{\mathbb{R}}$$.

If

$$L_{1}\varOmega_{3}+L_{2}m+L_{3} \varOmega_{4}< 1,$$
(2.7)

then the boundary value problem (2.4) has a unique solution on$$[0,T]$$.

### Proof

The proof is similar to that of Theorem 2.4 and is omitted. □

### Example 2.6

Consider the following first-order impulsive quantum $$(p,q)$$-difference equation of type I subject to the boundary condition of the form:

$$\textstyle\begin{cases} {_{k}}D_{\frac{1}{k+2},\frac{1}{k+3}} x(t)=\frac{1}{18+t^{2}} (\frac{x^{2}(t)+2 \vert x(t) \vert }{1+ \vert x(t) \vert } )+\frac{3}{2},\quad t\in (k, 2k+2 ], k=0,1,2,\\ \Delta x(k)=\frac{1}{6k}\sin x(t_{k}),\quad k=1,2,\\\frac{1}{2} x(0)+\frac{1}{3} x(3)=\frac{1}{4}. \end{cases}$$
(2.8)

Here $$p_{k}=1/(k+2)$$, $$q_{k}=1/(k+3)$$, $$k=0,1,2$$, $$\alpha=1/2$$, $$\beta =1/3$$, $$\gamma=1/4$$, $$t_{k}=k$$, $$k=1,2$$, $$T=3$$, and $$m=2$$. The given data leads to constants $$\varOmega_{1}=21/5$$, $$\varOmega_{2}=14/5$$. Setting

$$f(t,x)=\frac{1}{18+t^{2}} \biggl(\frac{x^{2}+2 \vert x \vert }{1+ \vert x \vert } \biggr)+\frac {3}{2} \quad\text{and}\quad\varphi_{k}(x)=\frac{1}{6k}\sin x,$$

we have $$|f(t,x)-f(t,y)|\leq(1/9)|x-y|$$ and $$|\varphi_{k}(x)-\varphi _{k}(y)|\leq(1/6)|x-y|$$ which satisfy Condition $$(H_{1})$$ in Theorem 2.4 with $$L_{1}=1/9$$ and $$L_{2}=1/6$$. Since $$L_{1}\varOmega _{1}+L_{2}\varOmega_{2}=14/15<1$$, by Theorem 2.4, the boundary value problem (2.8) has a unique solution x on $$[0, 3]$$.

### Example 2.7

Consider the following second-order impulsive quantum $$(p,q)$$-difference equation of type I with the initial conditions of the form:

$$\textstyle\begin{cases} {_{k}}D_{\frac{1}{k+2},\frac{1}{k+3}}^{2} x(t)=\frac{1}{5(t+5)}\tan ^{-1} \vert x(t) \vert +\frac{1}{2},\quad t\in (k, k^{2}+5k+4 ], k=0,1,2,\\ \Delta x(k)=\frac{ \vert x(t_{k}) \vert }{10k(1+ \vert x(t_{k}) \vert )},\quad k=1,2,\\ {_{k}}D_{\frac{1}{k+2},\frac{1}{k+3}} x(k)-{_{(k-1)}}D_{\frac {1}{k+1}, \frac{1}{k+2}}x(2k)=\frac{1}{15k^{2}}\sin \vert x(t_{k}) \vert ,\quad k=1,2,\\x(0)=\frac{3}{5},\qquad{_{0}}D_{\frac{1}{2},\frac{1}{3}}x(0)=\frac{5}{7}. \end{cases}$$
(2.9)

Here the quantum constants $$p_{k}$$, $$q_{k}$$ and impulsive points $$t_{k}$$, are as in Example 2.6. In addition, $$\tau_{k}=2k$$, $$k=1,2$$, and initial constants $$\lambda_{1}=3/5$$, $$\lambda_{2}=5/7$$. Next, we can compute that $$\varOmega_{3}=12.1365$$ and $$\varOmega_{4}=3$$. Set

$$f(t,x)=\frac{1}{5(t+5)}\tan^{-1} \vert x \vert + \frac{1}{2},\qquad\varphi _{k}(x)=\frac{ \vert x \vert }{10k(1+ \vert x \vert )},\quad \text{and}\quad\varphi^{\ast }_{k}(x)=\frac{1}{15k^{2}}\sin \vert x \vert .$$

It is easy to see that f, $$\varphi_{k}$$, and $$\varphi_{k}^{\ast}$$ satisfy $$(H_{1})$$ and $$(H_{2})$$ with $$L_{1}=1/25$$, $$L_{2}=1/10$$, and $$L_{3}=1/15$$. Therefore, we have $$L_{1}\varOmega_{3}+L_{2}m+L_{3}\varOmega _{4}=0.8855<1$$. Hence the boundary value problem (2.9) has a unique solution x on $$[0, 3]$$ by Theorem 2.5.

### Impulsive $$(p,q)$$-difference equations of type II

Now we study the first-order impulsive $$(p,q)$$-difference boundary value problem of the form

$$\textstyle\begin{cases} {_{t_{k}}}D_{p_{k},q_{k}}x(t)=f(t,x(t)),\quad t\in (t_{k}, t_{k+1} ], k=0,1,\ldots,m,\\ x(t_{k}^{+})-x(\rho_{k})=\varphi_{k}(x(\rho_{k})),\quad k=1,2,\ldots, m,\\ \alpha x(0)+\beta x(\rho_{m+1})=\gamma, \end{cases}$$
(2.10)

where $$f:J\times{\mathbb{R}}\to{\mathbb{R}}$$ and the functions $$\varphi_{k}$$, $$k=1,2,\ldots,m$$, and constants α, β, γ are defined as in Sect. 2.1. The constant $$\rho_{k}$$ is defined by

$$\rho_{k}=p_{k-1}(t_{k}-t_{k-1})+t_{k-1}, \quad k=1,2,\ldots,m, m+1.$$

Then the lagging distance is $$t_{k}-\rho_{k}=(1-p_{k-1})(t_{k}-t_{k-1})$$ which depends on the value of $$p_{k-1}\in(0,1]$$.

To observe the special characteristic of this type, by the shifting property of the $$(p,q)$$-derivative, we see that the unknown function $$x(t)$$ is defined on $$[t_{0},\rho_{1}]\cup(t_{k},\rho_{k+1}]$$, $$k=1,2,\ldots,m$$.

### Example 2.8

Let $$J=[0,2]$$ and $$t_{1}=1$$ be an impulsive point. Then

$${_{0}}D_{\frac{1}{2}, \frac{1}{3}}x(t)=f\bigl(t,x(t)\bigr),\quad t\in[0,1],$$

and

$${_{1}}D_{\frac{1}{4}, \frac {1}{5}}x(t)=f\bigl(t,x(t)\bigr), \quad t\in(1,2],$$

can be presented as

$$x(t)=x(0)+ \int_{0}^{t}f\bigl(s,x(s)\bigr) \,{_{0}}d_{\frac{1}{2}, \frac{1}{3}}, \quad t\in \biggl[0, \frac{1}{2} \biggr],$$

and

$$x(t)=x\bigl(1^{+}\bigr)+ \int_{1}^{t}f\bigl(s,x(s)\bigr) \,{_{1}}d_{\frac{1}{4}, \frac {1}{5}}, \quad t\in \biggl(1, \frac{5}{4} \biggr].$$

### Theorem 2.9

The first-order type II$$(p,q)$$-difference boundary value problem (2.10) can be expressed as an integral equation

\begin{aligned} x(t) =& \frac{\gamma}{(\alpha+\beta)}-\frac{\beta}{(\alpha+\beta )} \Biggl(\sum _{i=0}^{m} \int_{t_{i}}^{\rho_{i+1}}f\bigl(s,x(s)\bigr) \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{m} \varphi_{j}\bigl(x(\rho_{j})\bigr) \Biggr) \\ &{}+ \sum_{i=0}^{k-1} \int_{t_{i}}^{\rho_{i+1}}f\bigl(s,x(s)\bigr) \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{k} \varphi_{j}\bigl(x(\rho_{j})\bigr)+ \int _{t_{k}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{k}}}d_{p_{k},q_{k}}s, \end{aligned}
(2.11)

with$$\sum_{a}^{b}(\cdot)=0$$, if$$b< a$$.

### Proof

Firstly, the $$(p_{0},q_{0})$$-integration of the first equation in (2.10) yields

$$x(t)=x(0)+ \int_{t_{0}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s, \quad t\in (t_{0},\rho_{1}].$$

In particular, for $$t=\rho_{1}$$, we have

$$x(\rho_{1})=x(0)+ \int_{t_{0}}^{\rho_{1}}f\bigl(s,x(s)\bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s.$$

For $$k=1$$, by $$(p_{1},q_{1})$$-integration, we obtain

$$x(t)=x\bigl(t_{1}^{+}\bigr)+ \int_{t_{1}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{1}}}d_{p_{1},q_{1}}s, \quad t\in (t_{1},\rho_{2}],$$

$$x(t)=x(0)+ \int_{t_{0}}^{\rho_{1}}f\bigl(s,x(s)\bigr)\,{_{t_{0}}}d_{p_{0},q_{0}}s+ \varphi _{1}\bigl(x(\rho_{1})\bigr)+ \int_{t_{1}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{1}}}d_{p_{1},q_{1}}s,$$

by using the impulse condition $$x(t_{1}^{+})=x(\rho_{1})+\varphi_{1}(x(\rho_{1}))$$.

Repeating the process for any $$t\in(t_{k},\rho_{k+1}]$$, we get

$$x(t)=x(0)+\sum_{i=0}^{k-1} \int_{t_{i}}^{\rho_{i+1}}f\bigl(s,x(s)\bigr) \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{k} \varphi_{j}\bigl(x(\rho_{j})\bigr)+ \int _{t_{k}}^{t}f\bigl(s,x(s)\bigr) \,{_{t_{k}}}d_{p_{k},q_{k}}s.$$

Since

$$x(\rho_{m+1})=x(0)+\sum_{i=0}^{m} \int_{t_{i}}^{\rho_{i+1}}f\bigl(s,x(s)\bigr) \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{m} \varphi_{j}\bigl(x(\rho_{j})\bigr),$$

by the boundary condition, we have

$$x(0)=\frac{\gamma}{(\alpha+\beta)}-\frac{\beta}{(\alpha+\beta )} \Biggl(\sum _{i=0}^{m} \int_{t_{i}}^{\rho_{i+1}}f\bigl(s,x(s)\bigr) \,{_{t_{i}}}d_{p_{i},q_{i}}s +\sum_{j=1}^{m} \varphi_{j}\bigl(x(\rho_{j})\bigr) \Biggr),$$

which implies that (2.11) holds. This completes the proof. □

Next we define the points $$\rho^{\ast }_{k}=p_{k-1}^{2}(t_{k}-t_{k-1})+t_{k-1}$$, $$k=1,2,\ldots,m,m+1$$. Now we consider the second-order type II impulsive $$(p,q)$$-difference initial value problem of the form

$$\textstyle\begin{cases} {_{t_{k}}}D_{p_{k},q_{k}}^{2}x(t)=f(t,x(t)),\quad t\in (t_{k}, t_{k+1} ], k=0,1,\ldots,m,\\ x(t_{k}^{+})-x(\rho_{k}^{\ast})=\varphi_{k}(x(\rho^{\ast}_{k})),\quad k=1,2,\ldots, m,\\ {_{t_{k}}}D_{p_{k},q_{k}}x(t_{k}^{+})-{_{t_{k-1}}}D_{p_{k-1},q_{k-1}}x(\rho _{k})=\varphi_{k}^{\ast}(x(\rho^{\ast}_{k})),\quad k=1,2,\ldots, m,\\ x(0)=\lambda_{1},\qquad{_{t_{0}}}D_{p_{0},q_{0}}x(0)=\lambda_{2}, \end{cases}$$
(2.12)

where $$f:J\times{\mathbb{R}}\to{\mathbb{R}}$$, while other functions and constants are defined as in Sect. 2.1. Since $$0< p_{k}\leq1$$, we have $$\rho^{\ast}_{k}\leq t_{k}$$, and consequently $$(t_{k},\rho^{\ast }_{k}]\subseteq(t_{k},t_{k+1}]$$ for all $$k=0,1,\ldots,m$$. By Lemma 1.1, the unknown function $$x(t)$$ of problem (2.12) is defined on $$[t_{0},\rho^{\ast}_{1}]\bigcup_{k=1}^{m}(t_{k}, \rho^{\ast}_{k+1}]$$.

### Theorem 2.10

The initial value problem (2.12) of the impulsive$$(p,q)$$-difference equation of type II can be stated as an integral equation of the form

\begin{aligned} x(t) =& \lambda_{1}+\sum_{i=0}^{k} \bigl(\bigl\langle \rho^{\ast }_{i+1}\bigr\rangle _{k}-t_{i} \bigr) \Biggl[\lambda_{2}+\sum_{j=0}^{i-1} \biggl\{ \int_{t_{j}}^{\rho_{j+1}}f\bigl(s,x(s)\bigr) \,{_{t_{j}}}d_{p_{j},q_{j}}s+\varphi^{\ast }_{j+1} \bigl(x\bigl(\rho^{\ast}_{j+1}\bigr)\bigr) \biggr\} \Biggr] \\ &{}+ \sum_{r=0}^{k-1} \biggl\{ \frac{1}{p_{r}} \int_{t_{r}}^{\rho^{\ast }_{r+1}} \bigl(\rho^{\ast}_{r+1}-{_{t_{r}}} \varPhi_{q_{r}}(s) \bigr)f_{x} \bigl({_{t_{r}}} \varPhi_{\frac{1}{p_{r}}}(s) \bigr) \,{_{t_{r}}}d_{p_{r},q_{r}}s + \varphi_{r+1}\bigl(x\bigl(\rho^{\ast}_{r+1}\bigr) \bigr) \biggr\} \\ &{}+ \frac{1}{p_{k}} \int_{t_{k}}^{t} \bigl(t-{_{t_{k}}}\varPhi _{q_{k}}(s) \bigr)f_{x} \bigl({_{t_{k}}} \varPhi_{\frac{1}{p_{k}}}(s) \bigr) \,{_{t_{k}}}d_{p_{k},q_{k}}s,\quad t \in(t_{k},\rho^{\ast}_{k+1}], k=0,1,\ldots ,m. \end{aligned}
(2.13)

### Proof

The mathematical induction will be used to prove that (2.13) holds. To do this, by applying the double $$(p_{0},q_{0})$$-integration to the first equation of (2.12), we obtain

$$x(t)=\lambda_{1}+\lambda_{2}t+\frac{1}{p_{0}} \int_{t_{0}}^{t} \bigl(t-{_{t_{0}}} \varPhi_{q_{0}}(s) \bigr)f_{x} \bigl({_{t_{0}}} \varPhi_{\frac {1}{p_{0}}}(s) \bigr) \,{_{t_{0}}}d_{p_{0},q_{0}}s,\quad t \in(t_{0},\rho^{\ast}_{1}],$$

which implies that (2.13) is true for $$k=0$$. In the next step, we suppose that (2.13) holds for $$t\in (t_{k},\rho^{\ast}_{k+1}]$$. By mathematical induction, we shall show that (2.13) holds on $$(t_{k+1}, \rho^{\ast}_{k+2}]$$. Now, the double $$(p_{0},q_{0})$$-integration of the first equation of (2.12) yields on $$t\in(t_{k+1},\rho^{\ast}_{k+2}]$$ that

\begin{aligned} x(t) =&x\bigl(t_{k+1}^{+}\bigr)+(t-t_{k+1}) {_{t_{k+1}}}D_{p_{k+1},q_{k+1}}x\bigl(t_{k+1}^{+}\bigr) \\ &{} +\frac{1}{p_{k+1}} \int_{t_{k+1}}^{t} \bigl(t-{_{t_{k+1}}}\varPhi _{q_{k+1}}(s) \bigr)f_{x} \bigl({_{t_{k+1}}} \varPhi_{\frac {1}{p_{k+1}}}(s) \bigr) \,{_{t_{k+1}}}d_{p_{k+1},q_{k+1}}s. \end{aligned}
(2.14)

We have

\begin{aligned} x\bigl(t_{k+1}^{+}\bigr) =&x\bigl(\rho_{k+1}^{\ast} \bigr)+\varphi_{k+1}\bigl(x\bigl(\rho_{k}^{\ast }\bigr) \bigr) \\ =& \lambda_{1}+\sum_{i=0}^{k} \bigl( \rho^{\ast}_{i+1}-t_{i} \bigr) \Biggl[ \lambda_{2}+\sum_{j=0}^{i-1} \biggl\{ \int_{t_{j}}^{\rho_{j+1}}f\bigl(s,x(s)\bigr) \,{_{t_{j}}}d_{p_{j},q_{j}}s+\varphi^{\ast}_{j+1} \bigl(x\bigl(\rho^{\ast }_{j+1}\bigr)\bigr) \biggr\} \Biggr] \\ &{}+ \sum_{r=0}^{k} \biggl\{ \frac{1}{p_{r}} \int_{t_{r}}^{\rho^{\ast }_{r+1}} \bigl(\rho^{\ast}_{r+1}-{_{t_{r}}} \varPhi_{q_{r}}(s) \bigr)f_{x} \bigl({_{t_{r}}} \varPhi_{\frac{1}{p_{r}}}(s) \bigr) \,{_{t_{r}}}d_{p_{r},q_{r}}s + \varphi_{r+1}\bigl(x\bigl(\rho^{\ast}_{r+1}\bigr) \bigr) \biggr\} \end{aligned}

and

\begin{aligned} {_{t_{k+1}}}D_{p_{k+1},q_{k+1}}x\bigl(t_{k+1}^{+}\bigr) =&{_{t_{k}}}D_{p_{k},q_{k}}x(\rho_{k+1})+ \varphi_{k+1}^{\ast }\bigl(x\bigl(\rho_{k}^{\ast} \bigr)\bigr) \\ =& \lambda_{2}+\sum_{j=0}^{k-1} \biggl\{ \int_{t_{j}}^{\rho _{j+1}}f\bigl(s,x(s)\bigr) \,{_{t_{j}}}d_{p_{j},q_{j}}s+\varphi^{\ast}_{j+1} \bigl(x\bigl(\rho ^{\ast}_{j+1}\bigr)\bigr) \biggr\} \\ &{}+ \int_{t_{k}}^{\rho_{k+1}}f\bigl(s,x(s)\bigr) \,{_{t_{k}}}d_{p_{k},q_{k}}s+\varphi ^{\ast}_{k+1}\bigl(x \bigl(\rho^{\ast}_{k+1}\bigr)\bigr) \\ =&\lambda_{2}+\sum_{j=0}^{k} \biggl\{ \int_{t_{j}}^{\rho_{j+1}}f\bigl(s,x(s)\bigr) \,{_{t_{j}}}d_{p_{j},q_{j}}s+\varphi^{\ast}_{j+1} \bigl(x\bigl(\rho^{\ast }_{j+1}\bigr)\bigr) \biggr\} . \end{aligned}

Substituting above two values into (2.14), we obtain

\begin{aligned} x(t) =&\lambda_{1}+\sum_{i=0}^{k} \bigl( \rho^{\ast}_{i+1}-t_{i} \bigr) \Biggl[ \lambda_{2}+\sum_{j=0}^{i-1} \biggl\{ \int_{t_{j}}^{\rho _{j+1}}f\bigl(s,x(s)\bigr) \,{_{t_{j}}}d_{p_{j},q_{j}}s+\varphi^{\ast}_{j+1} \bigl(x\bigl(\rho ^{\ast}_{j+1}\bigr)\bigr) \biggr\} \Biggr] \\ &{}+ \sum_{r=0}^{k} \biggl\{ \frac{1}{p_{r}} \int_{t_{r}}^{\rho^{\ast }_{r+1}} \bigl(\rho^{\ast}_{r+1}-{_{t_{r}}} \varPhi_{q_{r}}(s) \bigr)f_{x} \bigl({_{t_{r}}} \varPhi_{\frac{1}{p_{r}}}(s) \bigr) \,{_{t_{r}}}d_{p_{r},q_{r}}s + \varphi_{r+1}\bigl(x\bigl(\rho^{\ast}_{r+1}\bigr) \bigr) \biggr\} \\ &{}+ (t-t_{k+1}) \Biggl(\lambda_{2}+\sum _{j=0}^{k} \biggl\{ \int _{t_{j}}^{\rho_{j+1}}f\bigl(s,x(s)\bigr) \,{_{t_{j}}}d_{p_{j},q_{j}}s+\varphi^{\ast }_{j+1} \bigl(x\bigl(\rho^{\ast}_{j+1}\bigr)\bigr) \biggr\} \Biggr) \\ &{}+ \frac{1}{p_{k+1}} \int_{t_{k+1}}^{t} \bigl(t-{_{t_{k+1}}}\varPhi _{q_{k+1}}(s) \bigr)f_{x} \bigl({_{t_{k+1}}} \varPhi_{\frac {1}{p_{k+1}}}(s) \bigr) \,{_{t_{k+1}}}d_{p_{k+1},q_{k+1}}s \\ =&\lambda_{1}+\sum_{i=0}^{k+1} \bigl(\bigl\langle \rho^{\ast }_{i+1}\bigr\rangle _{k+1}-t_{i} \bigr) \Biggl[\lambda_{2}+\sum_{j=0}^{i-1} \biggl\{ \int_{t_{j}}^{\rho_{j+1}}f\bigl(s,x(s)\bigr) \,{_{t_{j}}}d_{p_{j},q_{j}}s+\varphi ^{\ast}_{j+1} \bigl(x\bigl(\rho^{\ast}_{j+1}\bigr)\bigr) \biggr\} \Biggr] \\ &{}+ \sum_{r=0}^{k} \biggl\{ \frac{1}{p_{r}} \int_{t_{r}}^{\rho^{\ast }_{r+1}} \bigl(\rho^{\ast}_{r+1}-{_{t_{r}}} \varPhi_{q_{r}}(s) \bigr)f_{x} \bigl({_{t_{r}}} \varPhi_{\frac{1}{p_{r}}}(s) \bigr) \,{_{t_{r}}}d_{p_{r},q_{r}}s + \varphi_{r+1}\bigl(x\bigl(\rho^{\ast}_{r+1}\bigr) \bigr) \biggr\} \\ &{}+ \frac{1}{p_{k+1}} \int_{t_{k+1}}^{t} \bigl(t-{_{t_{k+1}}}\varPhi _{q_{k+1}}(s) \bigr)f_{x} \bigl({_{t_{k+1}}} \varPhi_{\frac {1}{p_{k+1}}}(s) \bigr) \,{_{t_{k+1}}}d_{p_{k+1},q_{k+1}}s, \end{aligned}

which holds for $$(t_{k+1}, \rho^{\ast}_{k+2}]$$. This completes the proof. □

To investigate the impulsive $$(p,q)$$-difference equations of type II, we define intervals of solutions as $$\varLambda_{1}= (\bigcup_{k=0}^{m}(t_{k},\rho_{k+1}] )\cup\{0\}$$ and $$\varLambda_{2}= (\bigcup_{k=0}^{m}(t_{k},\rho_{k+1}^{\ast}] )\cup\{0\}$$, and also the spaces $$\mathit{PC}_{1}(\varLambda_{1}, {\mathbb{R}})=\{x:\varLambda_{1}\to {\mathbb{R}}$$: $$x(t)$$ is continuous everywhere on $$\varLambda_{1}$$ such that $$x(t_{k}^{+})$$ and $$x(\rho_{k+1})$$ exist for all $$k=0,1,\ldots,m\}$$ and $$\mathit{PC}_{2}(\varLambda_{2}, {\mathbb{R}})=\{x:\varLambda_{2}\to{\mathbb {R}}$$: $$x(t)$$ is continuous everywhere on $$\varLambda_{2}$$ such that $$x(t_{k}^{+})$$ and $$x(\rho_{k+1}^{\ast})$$ exist for all $$k=0,1,\ldots,m\}$$. Both of them are Banach spaces equipped with the norms $$\|x\|_{1}=\sup \{|x(t)|, t\in\varLambda_{1}\}$$ and $$\|x\|_{2}=\sup\{|x(t)|, t\in \varLambda_{2}\}$$.

In proving our next results, we use the constants:

\begin{aligned}& \varOmega_{5}=\frac{ \vert \beta \vert + \vert \alpha+\beta \vert }{ \vert \alpha+\beta \vert }\sum_{i=0}^{m}( \rho_{i+1}-t_{i}), \\& \varOmega_{6}:=\sum_{i=0}^{m} \Biggl\{ \bigl(\rho_{i+1}^{\ast}-t_{i}\bigr)\sum _{j=0}^{i-1}(\rho_{j+1}-t_{j}) \Biggr\} +\sum_{r=0}^{m}\frac{(\rho _{r+1}^{\ast}-t_{r})^{2}}{p_{r}+q_{r}}, \\& \varOmega_{7}:=\sum_{i=0}^{m} \bigl(\rho_{i+1}^{\ast}-t_{i}\bigr)i. \end{aligned}

Applying Theorem 2.9 to define the operator on $$\mathit{PC}_{1}(\varLambda_{1}, {\mathbb{R}})$$ and following the method of Theorem 2.4, we can easily prove the existence of a unique solution of problem (2.10).

### Theorem 2.11

Assume that the functions$$f:[0,T]\times{\mathbb{R}}\to{\mathbb {R}}$$and$$\varphi_{k}:{\mathbb{R}}\to{\mathbb{R}}$$, $$k=1,2,\ldots ,m$$, satisfy condition$$(H_{1})$$. If

$$L_{1}\varOmega_{5}+L_{2} \varOmega_{2}< 1,$$
(2.15)

then the boundary value problem of type II (2.10) has a unique solution on$$\varLambda_{1}$$.

### Theorem 2.12

Assume that the functions$$f:[0,T]\times{\mathbb{R}}\to{\mathbb {R}}$$, $$\varphi_{k}:{\mathbb{R}}\to{\mathbb{R}}$$and$$\varphi _{k}^{\ast}:{\mathbb{R}}\to{\mathbb{R}}$$, $$k=1,2,\ldots,m$$, satisfy$$(H_{1})$$$$(H_{2})$$. If

$$L_{1}\varOmega_{6}+L_{2}m+L_{3} \varOmega_{7}< 1,$$
(2.16)

then the problem of type II (2.12) has a unique solution on$$\varLambda_{2}$$.

### Proof

To show the technique of computation of constants $$\varOmega_{6}$$ and $$\varOmega_{7}$$, we give a short proof. Now we prove that the operator equation $$x={\mathcal{B}}x$$ has a unique fixed point, where the operator $${\mathcal{B}}:\mathit{PC}_{2}(\varLambda _{2}, {\mathcal{R}})\to \mathit{PC}_{2}(\varLambda_{2}, {\mathcal{R}})$$ is defined, in view of Theorem 2.10, by

\begin{aligned} {\mathcal{B}}x(t) =& \lambda_{1}+\sum_{i=0}^{k} \bigl(\bigl\langle \rho ^{\ast}_{i+1}\bigr\rangle _{k}-t_{i} \bigr) \Biggl[\lambda_{2}+\sum_{j=0}^{i-1} \biggl\{ \int_{t_{j}}^{\rho_{j+1}}f\bigl(s,x(s)\bigr) \,{_{t_{j}}}d_{p_{j},q_{j}}s+\varphi^{\ast}_{j+1} \bigl(x\bigl(\rho^{\ast }_{j+1}\bigr)\bigr) \biggr\} \Biggr] \\ &{}+ \sum_{r=0}^{k-1} \biggl\{ \frac{1}{p_{r}} \int_{t_{r}}^{\rho^{\ast }_{r+1}} \bigl(\rho^{\ast}_{r+1}-{_{t_{r}}} \varPhi_{q_{r}}(s) \bigr)f_{x} \bigl({_{t_{r}}} \varPhi_{\frac{1}{p_{r}}}(s) \bigr) \,{_{t_{r}}}d_{p_{r},q_{r}}s + \varphi_{r+1}\bigl(x\bigl(\rho^{\ast}_{r+1}\bigr) \bigr) \biggr\} \\ &{}+ \frac{1}{p_{k}} \int_{t_{k}}^{t} \bigl(t-{_{t_{k}}}\varPhi _{q_{k}}(s) \bigr)f_{x} \bigl({_{t_{k}}} \varPhi_{\frac{1}{p_{k}}}(s) \bigr) \,{_{t_{k}}}d_{p_{k},q_{k}}s,\quad t \in(t_{k},\rho^{\ast}_{k+1}], k=0,1,\ldots,m. \end{aligned}

By a similar method as in Theorem 2.4, we can show that the operator $${\mathcal{B}}$$ maps a subset of $$\mathit{PC}_{2}(\varLambda_{2}, {\mathcal{R}})$$ into subset of $$\mathit{PC}_{2}(\varLambda_{2}, {\mathcal{R}})$$. Next, we will prove that $${\mathcal{B}}$$ is a contraction. Let $$x,y\in \mathit{PC}_{2}(\varLambda_{2}, {\mathcal{R}})$$. Then we have

\begin{aligned} & \bigl\vert {\mathcal{B}}x(t)-{\mathcal{B}}y(t) \bigr\vert \\ &\quad\leq \sum_{i=0}^{k} \bigl(\bigl\langle \rho^{\ast}_{i+1}\bigr\rangle _{k}-t_{i} \bigr) \Biggl[\sum_{j=0}^{i-1} \biggl\{ \int_{t_{j}}^{\rho _{j+1}} \bigl\vert f\bigl(s,x(s)\bigr)-f \bigl(s,y(s)\bigr) \bigr\vert \,{_{t_{j}}}d_{p_{j},q_{j}}s \\ &\qquad{}+ \bigl\vert \varphi^{\ast}_{j+1}\bigl(x\bigl( \rho^{\ast}_{j+1}\bigr)\bigr)-\varphi^{\ast }_{j+1} \bigl(y\bigl(\rho^{\ast}_{j+1}\bigr)\bigr) \bigr\vert \biggr\} \Biggr] \\ &\qquad{}+ \sum_{r=0}^{k-1} \biggl\{ \frac{1}{p_{r}} \int_{t_{r}}^{\rho^{\ast }_{r+1}} \bigl(\rho^{\ast}_{r+1}-{_{t_{r}}} \varPhi_{q_{r}}(s) \bigr) \bigl\vert f_{x} \bigl({_{t_{r}}}\varPhi_{\frac{1}{p_{r}}}(s) \bigr) -f_{y} \bigl({_{t_{r}}}\varPhi_{\frac{1}{p_{r}}}(s) \bigr) \bigr\vert \,{_{t_{r}}}d_{p_{r},q_{r}}s \\ &\qquad{}+ \bigl\vert \varphi_{r+1}\bigl(x\bigl(\rho^{\ast}_{r+1} \bigr)\bigr)-\varphi_{r+1}\bigl(y\bigl(\rho ^{\ast}_{r+1} \bigr)\bigr) \bigr\vert \biggr\} \\ &\qquad{}+ \frac{1}{p_{k}} \int_{t_{k}}^{t} \bigl(t-{_{t_{k}}}\varPhi _{q_{k}}(s) \bigr) \bigl\vert f_{x} \bigl({_{t_{k}}} \varPhi_{\frac {1}{p_{k}}}(s) \bigr)-f_{y} \bigl({_{t_{k}}} \varPhi_{\frac {1}{p_{k}}}(s) \bigr) \bigr\vert \,{_{t_{k}}}d_{p_{k},q_{k}}s \\ &\leq \sum_{i=0}^{m} \bigl( \rho^{\ast}_{i+1}-t_{i} \bigr) \Biggl[\sum _{j=0}^{i-1} \biggl\{ L_{1} \Vert x-y \Vert _{2} \int_{t_{j}}^{\rho_{j+1}}(1) \,{_{t_{j}}}d_{p_{j},q_{j}}s+L_{3} \Vert x-y \Vert _{2} \biggr\} \Biggr] \\ &\qquad{}+ \sum_{r=0}^{m-1} \biggl\{ \frac{1}{p_{r}}L_{1} \Vert x-y \Vert _{2} \int _{t_{r}}^{\rho^{\ast}_{r+1}} \bigl(\rho^{\ast}_{r+1}-{_{t_{r}}} \varPhi _{q_{r}}(s) \bigr) (1) \,{_{t_{r}}}d_{p_{r},q_{r}}s+L_{2} \Vert x-y \Vert _{2} \biggr\} \\ &\qquad{}+ \frac{1}{p_{m}}L_{1} \Vert x-y \Vert _{2} \int_{t_{m}}^{\rho_{m+1}^{\ast}} \bigl(\rho_{m+1}^{\ast}-{_{t_{m}}} \varPhi_{q_{m}}(s) \bigr) (1) \,{_{t_{m}}}d_{p_{m},q_{m}}s \\ &\quad= \sum_{i=0}^{m} \bigl( \rho^{\ast}_{i+1}-t_{i} \bigr) \Biggl[\sum _{j=0}^{i-1} \bigl\{ L_{1} \Vert x-y \Vert _{2}(\rho_{j+1}-t_{j})+L_{3} \Vert x-y \Vert _{2} \bigr\} \Biggr] \\ &\qquad{}+ \sum_{r=0}^{m-1} \biggl\{ L_{1} \Vert x-y \Vert _{2}\frac{(\rho_{r+1}^{\ast }-t_{r})^{2}}{p_{r}+q_{r}}+L_{2} \Vert x-y \Vert _{2} \biggr\} +L_{1} \Vert x-y \Vert _{2}\frac{(\rho _{m+1}^{\ast}-t_{m})^{2}}{p_{m}+q_{m}} \\ &\quad=(L_{1}\varOmega_{6}+L_{2}m+L_{3} \varOmega_{7}) \Vert x-y \Vert _{2}, \end{aligned}

which implies that $$\|{\mathcal{B}}x-{\mathcal{B}}y\|_{2}\leq (L_{1}\varOmega_{6}+L_{2}m+L_{3}\varOmega_{7})\|x-y\|_{2}$$. Condition (2.16) and the Banach contraction mapping principle guarantee that the impulsive $$(p,q)$$-difference initial value problem of type II (2.12) has a unique solution on $$\varLambda_{2}$$. The proof is completed. □

### Example 2.13

Consider the following first-order impulsive $$(p,q)$$-difference equation of type II subject to the boundary condition of the form:

$$\textstyle\begin{cases} {_{k}}D_{\frac{k+1}{k+2},\frac{k+1}{k+3}} x(t)=\frac {5}{6(3+t)^{2}} (\frac{x^{2}(t)+2 \vert x(t) \vert }{1+ \vert x(t) \vert } )+\frac {3}{4},\quad t\in (k, k+1 ], k=0,1,2,\\ x(k)-x (\frac{k^{2}+k-1}{k+1} )=\frac{1}{6k}\tan^{-1} (x (\frac{k^{2}+k-1}{k+1} ) ),\quad k=1,2,\\\frac{1}{2} x(0)+\frac{1}{3} x (\frac{11}{4} )=\frac{1}{4}. \end{cases}$$
(2.17)

Here the quantum numbers are $$p_{k}=(k+1)/(k+2)$$, $$q_{k}=(k+1)/(k+3)$$, $$k=0,1,2$$, $$J=[0,3]$$, $$t_{k}=k$$, $$k=1,2$$, $$\alpha=1/2$$, $$\beta=1/3$$, $$\gamma=1/4$$, and $$\rho_{k}=(k^{2}+k-1)/(k+1)$$. We can find that $$\varOmega_{2}=2.8000$$, $$\varOmega_{5}=2.6833$$, and

$$\varLambda_{1}= \biggl[0,\frac{1}{2} \biggr]\cup \biggl(1, \frac {5}{3} \biggr]\cup \biggl(2,\frac{11}{4} \biggr].$$

By setting

$$f(t,x)=\frac{5}{6(3+t)^{2}} \biggl(\frac{x^{2}+2 \vert x \vert }{1+ \vert x \vert } \biggr)+\frac {3}{4} \quad\text{and}\quad\varphi_{k}(x)=\frac{1}{6k} \tan^{-1} (x ),$$

we see that the functions f and $$\varphi_{k}$$ satisfy $$(H_{1})$$ with $$L_{1}=5/27$$ and $$L_{2}=1/6$$, respectively. Then we get $$L_{1}\varOmega _{5}+L_{2}\varOmega_{2}=0.9543<1$$. Therefore, by Theorem 2.11, the boundary value problem (2.17) has a unique solution x on $$\varLambda_{1}$$.

### Example 2.14

Consider the following second-order impulsive $$(p,q)$$-difference equation of type II with the initial conditions of the form:

$$\textstyle\begin{cases} {_{k}}D_{\frac{k+1}{k+2},\frac{k+1}{k+3}}^{2} x(t)=\frac {1}{10(t+6)}\sin \vert x(t) \vert +\frac{5}{6},\quad t\in (k, k+1 ], k=0,1,2,\\ x(k^{+})-x (\frac{k^{3}+2k^{2}-k-1}{(k+1)^{2}} )=\frac {3}{5(k+1)^{2}}\tan^{-1} (x (\frac {k^{3}+2k^{2}-k-1}{(k+1)^{2}} ) ),\quad k=1,2,\\ {_{k}}D_{\frac{k+1}{k+2},\frac{k+1}{k+3}} x(k^{+})-{_{(k-1)}}D_{\frac {k}{k+1}, \frac{k}{k+2}}x (\frac{k^{2}+k-1}{k+1} )=\frac {1}{5k^{3}} \vert x (\frac{k^{3}+2k^{2}-k-1}{(k+1)^{2}} ) \vert ,\quad k=1,2,\\ x(0)=\frac{3}{5},\qquad{_{0}}D_{\frac{1}{2},\frac{1}{3}}x(0)=\frac{5}{7}. \end{cases}$$
(2.18)

The quantum numbers $$p_{k}$$, $$q_{k}$$, impulsive points $$t_{k}$$, $$\rho_{k}$$, and interval J are defined the same as in Example 2.13. We have the constants $$\lambda_{1}=3/5$$, $$\lambda_{2}=5/7$$, and points $$\rho _{k}^{\ast}=(k^{3}+2k^{2}-k-1)/(k+1)^{2}$$. Next we can find that $$\varOmega _{6}=18.4273$$, $$\varOmega_{7}=1.5694$$, and

$$\varLambda_{2}= \biggl[0,\frac{1}{4} \biggr]\cup \biggl(1, \frac {13}{9} \biggr]\cup \biggl(2,\frac{41}{16} \biggr].$$

By setting

$$f(t,x)=\frac{1}{10(t+6)}\sin \vert x \vert +\frac{5}{6},\qquad \varphi _{k}(x)=\frac{3}{5(k+1)^{2}}\tan^{-1} (x ),\quad\text{and}\quad\varphi^{\ast}_{k}(x)=\frac{1}{5k^{3}} \vert x \vert ,$$

we deduce that $$(H_{1})$$$$(H_{2})$$ are fulfilled with $$L_{1}=1/60$$, $$L_{2}=3/20$$, and $$L_{3}=1/5$$. Hence, it follows that $$L_{1}\varOmega _{6}+L_{2}m+L_{3}\varOmega_{7}=0.9210<1$$. Therefore, by applying Theorem 2.12, the boundary value problem (2.18) has a unique solution x on $$\varLambda_{2}$$.

## Conclusion

In this research, we initiated the study of the first and second order $$(p,q)$$-difference equations with initial or boundary conditions. Firstly, we let $$t_{k}$$, $$k=1,\ldots,m$$, be the impulsive points such that $$0=t_{0}< t_{1}<\cdots<t_{k}<\cdots<t_{m}<t_{m+1}=T$$ and $$J_{k}=(t_{k},t_{k+1}]$$, $$k=1,\ldots,m$$, $$J_{0}=[0,t_{1}]$$ be the intervals such that $$\bigcup_{k=0}^{m}J_{k}=[0,T]:=J$$. The investigations were based on $$(p,q)$$-calculus introduced in the first section of this paper, by replacing a point a by $$t_{k}$$, quantum numbers p by $$p_{k}$$ and q by $$q_{k}$$, $$k=0,1,\ldots,m$$, and also applying the $$(p_{k},q_{k})$$-difference and $$(p_{k},q_{k})$$-integral operators only on a finite subinterval of J. In addition, the consecutive subintervals could be related with jump conditions which led to a meaning of quantum difference equations with impulse effects. There are two types of impulsive problems. The consecutive domains of impulsive $$(p,q)$$-difference equations of type I are overlapped, while the unknown functions of impulsive equations of type II are defined on disjoint consecutive domains. Four types of problems were considered, two impulsive $$(p,q)$$-difference equations of type I and two impulsive $$(p,q)$$-difference equations of type II. Existence and uniqueness results were proved via Banach’s contraction mapping principle. Examples illustrating the obtained results were also presented.

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### Availability of data and materials

Data sharing not applicable to this article as no data sets were generated or analyzed during the current study.

## Funding

This work was financially supported by Rajamangala University of Technology Rattanakosin, Wang Klai Kangwon Campus, Thailand (Grant No. A-51/2561).

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All authors contributed equally to this work. All authors read and approved the final manuscript.

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The authors declare that they have no competing interests. 