Theory and Modern Applications

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.

References

1. 1.

Kamsrisuk, N., Promsakon, C., Ntouyas, S.K., Tariboon, J.: Nonlocal boundary value problems for $$(p,q)$$-difference equations. Differ. Equ. Appl. 10(2), 183–195 (2018)

2. 2.

Promsakon, C., Kamsrisuk, N., Ntouyas, S.K., Tariboon, J.: On the second-order quantum $$(p,q)$$-difference equations with separated boundary conditions. Adv. Math. Phys. 2018, Article ID 9089865 (2018)

3. 3.

Chakrabarti, R., Jagannathan, R.: A $$(p,q)$$-oscillator realization of two-parameter quantum algebras. J. Phys. A, Math. Gen. 24, L711–L718 (1991)

4. 4.

Jagannathan, R., Srinivasa Rao, K.: Two-parameter quantum algebras, twin-basic number, and associated generalized hypergeometric series. arXiv:math/0602613v1

5. 5.

Sahai, V., Yadav, S.: Representations of two parameter quantum algebras and $$(p,q)$$-special functions. J. Math. Anal. Appl. 335, 268–279 (2007)

6. 6.

Burban, I.: Two-parameter deformation of the oscillator algebra and $$(p,q)$$-analog of two-dimensional conformal field theory. J. Nonlinear Math. Phys. 2(3–4), 384–391 (2013)

7. 7.

Njionou Sadjang, P.: On the fundamental theorem of $$(p,q)$$-calculus and some $$(p,q)$$-Taylor formulas. arXiv:1309.3934v1

8. 8.

Hounkonnou, M.N., Bukweli Kyemba, J.D.: $$R(p,q)$$-Calculus differentiation and integration. J. Math. 49(2), 145–167 (2013)

9. 9.

Mursaleen, M., Ansari, K.J., Khan, A.: On $$(p,q)$$-analogue of Bernstein operators. Appl. Math. Comput. 266, 874–882 (2015)

10. 10.

Araci, S., Duran, U., Acikgoz, M., Srivastava, H.M.: A certain $$(p,q)$$-derivative operator and associated divided differences. J. Inequal. Appl. 2016, Article ID 301 (2016)

11. 11.

Jackson, H.F.: On q-definite integral. Q. J. Pure Appl. Math. 41, 193–203 (1910)

12. 12.

Jackson, H.F.: q-Difference equations. Am. J. Math. 32, 305–314 (1910)

13. 13.

Kac, V., Cheung, P.: Quantum Calculus. Springer, New York (2002)

14. 14.

Ahmad, B., Ntouyas, S.K., Tariboon, J.: Quantum Calculus: New Concepts, Impulsive IVPs and BVPs, Inequalities. Trends in Abstract and Applied Analysis, vol. 4. World Scientific, Singapore (2016)

15. 15.

Tunç, M., Göv, E.: $$(p,q)$$-Integral inequalities. RGMIA Res. Rep. Collect. 19, Article ID 97 (2016)

16. 16.

Tunç, M., Göv, E.: Some integral inequalities via $$(p,q)$$-calculus on finite intervals. RGMIA Res. Rep. Collect. 19, Article ID 95 (2016)

17. 17.

Tariboon, J., Ntouyas, S.K., Agarwal, P.: New concepts of fractional quantum calculus and applications to impulsive fractional q-difference equations. Adv. Differ. Equ. 2015, Article ID 18 (2015)

Not applicable.

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).

Author information

Authors

Contributions

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

Ethics declarations

Competing interests

The authors declare that they have no competing interests. 