Noninstantaneous Impulsive Fractional Quantum Hahn Integro-Difference Boundary Value Problems

Abstract

In this paper, we study the existence and uniqueness results for noninstantaneous impulsive fractional quantum Hahn integro-difference boundary value problems with integral boundary conditions, by using Banach contraction mapping principle and Leray–Schauder nonlinear alternative. Examples are included illustrating the obtained results. To the best of our knowledge, no work has reported on the existence of solutions to the Hahn-difference equation with noninstantaneous impulses.

1. Introduction and Preliminaries

In this paper, we investigate the existence and uniqueness of solutions for noninstantaneous impulsive fractional quantum Hahn integro-difference equation of the form:

$$\left\{\begin{array}{c}{}_{s_i}^C{D}_{q_i,{\omega }_i}^{{\alpha }_i}x\left(t\right)=f\left(t,x\left(t\right),\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\kappa }_i}x\right)\left(t\right)\right),t\in [s_i,t_{i+1}),i=0,1,2,\dots ,m,\hfill \\ x\left(t\right)=(t-t_i)g_i\left(t\right)+x\left(t_i\right),t\in [t_i,s_i),i=1,2,\dots ,m,\hfill \end{array}\right.$$

(1)

subject to integral boundary condition

$$\beta x\left(0\right)+\gamma x\left(T\right)=\sum _{z=0}^m{\mu }_z\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\delta }_z}x\right)\left(t_{z+1}\right)+\lambda ,$$

(2)

where ${}_{s_i}^C{D}_{q_i,{\omega }_i}^{{\alpha }_i}$, ${}_{s_i}{I}_{q_i,{\omega }_i}^{\varphi }$ are the fractional quantum Hahn difference and integral operators of orders ${\alpha }_i\in (0,1]$, $\varphi \in \{{\kappa }_i,{\delta }_i\}>0$, respectively, acting to the function on $[s_i,t_{i+1})$ with quantum numbers $q_i\in (0,1)$, ${\omega }_i\ge 0$, $i=0,1,2,\dots ,m,$ and $\beta$, $\gamma$, ${\mu }_z$$(z=0,1,\dots ,m)$, $\lambda$ are real constants. The functions $f:J\times {\mathbb{R}}^2\to \mathbb{R}$, $g_i:J\to \mathbb{R}$, are continuous, where $J:=[s_j,t_{j+1})\cup [t_i,s_i)\cup \left\{T\right\}$, $j=0,1,2,\dots ,m$, $i=1,2,3,\dots ,m$. The given impulsive points in J satisfy

$$0=s_0<t_1<s_1<t_2<s_2<\cdots <t_m<s_m<t_{m+1}=T.$$

Prompted by the application of fractional order derivatives and integrals to applied mathematics, analytical chemistry, neuron modeling and biological sciences, the theory of fractional calculus has attracted great interest from the mathematical science research community. For examples and recent development of the topic, see ([1,2,3,4,5,6,7]) and references cited therein.

Impulsive differential equations have become more important in some mathematical models of real phenomena, especially in control, biological, medical, and informational models. There are two types of impulses: instantaneous impulses in which the duration of these changes is relatively short, and non-instantaneous impulses in which an impulsive action, starting abruptly at a fixed point and continues on a finite time interval. Some examples of such processes can be found in physics, biology, population dynamics, ecology, pharmacokinetics, and others. For results with instantaneous impulses see, e.g., the monographs [8,9,10], the papers [11,12,13], and the references therein. Noninstantaneous impulsive differential equation was introduced by Hernández and O’Regan [14]. For some recent works, we refer the reader to [15,16,17,18] and references therein.

The notion of q-derivative was introduced in 1910 by Jackson [19] as

$$D_{q}f\left(t\right)=\left\{\begin{array}{cc}{\displaystyle \frac{f\left(t\right)-f\left(qt\right)}{t(1-q)},}\hfill & t\ne 0,\hfill \\ f^{\prime }\left(0\right),\hfill & t=0,\hfill \end{array}\right.$$

(3)

provided that $f^{\prime }\left(0\right)$ exists. The q-calculus appeared as a connection between mathematics and physics. The fundamental aspects of quantum calculus can be found in book [20]. For some recent results in quantum calculus we refer to the papers [21,22,23] and the references cited therein.

Hahn [24] established the difference operator $D_{q,\omega },$ in 1949,

$$D_{q,\omega }f\left(t\right)=\left\{\begin{array}{cc}{\displaystyle \frac{f(qt+\omega )-f\left(t\right)}{t(q-1)+\omega },}\hfill & t\ne {\omega }_0,\hfill \\ f^{\prime }\left({\omega }_0\right),\hfill & t={\omega }_0,\hfill \end{array}\right.$$

provided that f is differentiable at ${\omega }_0:=\omega /(1-q),$ where $q\in (0,1)$ and $\omega \ge 0$ are fixed constants. The Hahn difference operator unifies the Jackson q-difference derivative $D_q$, where $q\in (0,1),$ defined by (3), for $\omega =0,$ and the forward difference $D_{\omega }$ for $q\to 1,$ defined by

$$D_{\omega }f\left(t\right)=\frac{f(t+\omega )-f\left(t\right)}{\omega },$$

where $\omega >0$ is a fixed constant. The Hahn difference operator is used for constructing families of orthogonal polynomials and investigating some approximation problems, (cf. [25,26,27]).

In 2013, Tariboon and Ntouyas [28], presented the new concepts of quantum calculus on $[a,b]$ by defining

$${}_a{D}_{q}f\left(t\right)=\left\{\begin{array}{cc}{\displaystyle \frac{f\left(t\right)-f(qt+(1-q\left)a\right)}{(1-q)(t-a)},}\hfill & t\ne a,\hfill \\ \underset{t\to a}{lim}{}_a{D}_{q}f\left(t\right),\hfill & t=a.\hfill \end{array}\right.$$

(4)

With the help of definition (4), a series of quantum initial and boundary value problems contain impulses were studied. We refer the interested reader to the recent monograph [29] for details.

In 2016, Tariboon et al. [30], gave the generalization of Hahn difference operator on $[a,b]$ by

$${}_a{D}_{q,\omega }f\left(t\right)=\left\{\begin{array}{cc}{\displaystyle \frac{f\left(t\right)-f(qt+a(1-q)+\omega )}{(t-a)(1-q)-\omega },}\hfill & t\ne \theta ,\hfill \\ f^{\prime }\left(\theta \right),\hfill & t=\theta ,\hfill \end{array}\right.$$

provided that f is differentiable at $\theta :=a+(\omega /(1-q\left)\right)$.

Let $0<q_i<1$, ${\omega }_i\ge 0$ be quantum constants and ${\theta }_i$ be the point defined by

$${\theta }_i=\frac{{\omega }_i}{1-q_i}+s_i,$$

such that ${\theta }_i\in [s_i,t_{i+1})$, $i=0,1,2,\dots ,m$. The quantum shifting operator ${}_{{\theta }_i}{\mathrm{\Phi }}_{q_i}\left(t\right)$ is defined by

$${}_{{\theta }_i}{\mathrm{\Phi }}_{q_i}\left(t\right)=q_{i}t+(1-q_i){\theta }_i,t\in [s_i,t_{i+1}).$$

Please note that if ${\theta }_i=s_i$ and ${\theta }_i=0$ then (1) is reduced to

$${}_{s_i}{\mathrm{\Phi }}_{q_i}\left(t\right)=q_{i}t+(1-q_i)s_i,$$

(5)

and

$${}_0{\mathrm{\Phi }}_{q_i}\left(t\right)=q_{i}t,$$

(6)

respectively. The power function ${(t-r)}_{{\theta }_i}^{\left({\alpha }_i\right)}$, $t,r\in [s_i,t_{i+1})$, is defined by

$${(t-r)}_{{\theta }_i}^{\left({\alpha }_i\right)}=\prod _{k=0}^{\infty }\frac{(t-{}_{{\theta }_i}{\mathrm{\Phi }}_{q_i}^k\left(r\right))}{(t-{}_{{\theta }_i}{\mathrm{\Phi }}_{q_i}^{k+{\alpha }_i}\left(r\right))},$$

(7)

where ${}_{{\theta }_i}{\mathrm{\Phi }}_{q_i}^{\sigma }\left(r\right)=q_i^{\sigma }r+(1-q_i^{\sigma }){\theta }_i$, $\sigma \in \mathbb{R}$. If we put (5) and (6) in (7), then we obtain

$${(t-r)}_{s_i}^{\left({\alpha }_i\right)}=\prod _{k=0}^{\infty }\frac{(t-{}_{s_i}{\mathrm{\Phi }}_{q_i}^k\left(r\right))}{(t-{}_{s_i}{\mathrm{\Phi }}_{q_i}^{k+{\alpha }_i}\left(r\right))},$$

and

$${(t-r)}_0^{\left({\alpha }_i\right)}=\prod _{k=0}^{\infty }\frac{(t-q_i^{k}r)}{(t-q_i^{k+{\alpha }_i}r)},$$

(8)

respectively. The Gamma function in quantum calculus is defined by

$${\mathrm{\Gamma }}_q\left(\gamma \right)=\frac{{(1-q)}_0^{(\gamma -1)}}{{(1-q)}^{\gamma -1}},\gamma \in \mathbb{R}\setminus \{0,-1,-2,\dots \},$$

where ${(1-q)}_0^{(\gamma -1)}$ is defined by Formula (8). Indeed, ${\mathrm{\Gamma }}_q(\gamma +1)={\left[\gamma \right]}_q{\mathrm{\Gamma }}_q\left(\gamma \right)$, where ${\left[x\right]}_q=(1-q^x)/(1-q)$, $x\in \mathbb{R}$, is the quantum number or q-number. However, in our work, the number q will be replaced by $q_i$ on an interval $[s_i,t_{i+1})$, $i=0,1,2,\dots ,m$.

In the following definitions, we give the Riemann–Liouville fractional derivative and integral in Hahn calculus as well as the Caputo fractional derivative which can be found in [31,32,33].

Definition 1.

The fractional quantum Hahn difference operator of a Riemann–Liouville type of order ${\alpha }_i\ge 0$ on interval $[s_i,t_{i+1})$ is defined by $\left({}_{s_i}^{RL}{D}_{q_i,{\omega }_i}^{0}f\right)\left(t\right)=f\left(t\right)$ and

$$\left({}_{s_i}^{RL}{D}_{q_i,{\omega }_i}^{{\alpha }_i}f\right)\left(t\right)=\frac{1}{{\mathrm{\Gamma }}_{q_i}(n-{\alpha }_i)}{}_{s_i}{D}_{q_i,{\omega }_i}^n{\int }_{s_i}^t{(t-{}_{{\theta }_i}{\mathrm{\Phi }}_{q_i}\left(r\right))}_{{\theta }_i}^{(n-{\alpha }_i-1)}f\left(r\right){}_{s_i}{d}_{q_i,{\omega }_i}r,{\alpha }_i>0,$$

where n is the smallest integer greater than or equal to ${\alpha }_i$.

Definition 2.

Let ${\alpha }_i\ge 0$ and f be a function defined on $[s_i,t_{i+1})$. The fractional quantum Hahn integral operator of Riemann–Liouville type is given by $\left({}_{s_i}{I}_{q_i,{\omega }_i}^{0}f\right)\left(t\right)=f\left(t\right)$ and

$$\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\alpha }_i}f\right)\left(t\right)=\frac{1}{{\mathrm{\Gamma }}_{q_i}\left({\alpha }_i\right)}{\int }_{s_i}^t{(t-{}_{{\theta }_i}{\mathrm{\Phi }}_{q_i}\left(s\right))}_{{\theta }_i}^{({\alpha }_i-1)}f\left(s\right){}_{s_i}{d}_{q_i,{\omega }_i}s,{\alpha }_i>0,t\in [s_i,t_{i+1}).$$

Definition 3.

The fractional quantum Hahn difference operator of Caputo type ${\alpha }_i\ge 0$ on interval $[s_i,t_{i+1})$ is defined by $\left({}_{s_i}^C{D}_{q_i,{\omega }_i}^{0}f\right)\left(t\right)=f\left(t\right)$ and

$$\left({}_{s_i}^C{D}_{q_i,{\omega }_i}^{{\alpha }_i}f\right)\left(t\right)=\frac{1}{{\mathrm{\Gamma }}_{q_i}(n-{\alpha }_i)}{\int }_{s_i}^t{(t-{}_{{\theta }_i}{\mathrm{\Phi }}_{q_i}\left(r\right))}_{{\theta }_i}^{(n-{\alpha }_i-1)}{}_{s_i}{D}_{q_i,{\omega }_i}^{n}f\left(r\right){}_{s_i}{d}_{q_i,{\omega }_i}r,{\alpha }_i>0,$$

where n is the smallest integer greater than or equal to ${\alpha }_i$.

In our work, the problem (1) and (2) is based on fractional quantum Hahn calculus in Definitions 2 and 3. If ${\omega }_i=0$, that is ${\theta }_i=s_i$, then Definitions 1–3 are reduced to quantum calculus on the finite interval in the framework of Tariboon and Ntouyas [28] (see [29,30] for more details). Now we present some properties of fractional calculus on any interval $[s_i,t_{i+1})$, $i=0,1,2,\dots ,m$.

Theorem 1

([31]). Let $\alpha \in {\mathbb{R}}^{+}$, $\lambda \in (-1,\infty )$ and ${\theta }_i\in [s_i,t_{i+1})$ be given constants. The following formulas hold:

$\left(i\right)$

${\displaystyle \left({}_{s_i}{I}_{q_i,{\omega }_i}^{\alpha }{(x-s_i)}_{{\theta }_i}^{\left(\lambda \right)}\right)\left(t\right)=\frac{{\mathrm{\Gamma }}_{q_i}(\lambda +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +\lambda +1)}{(t-s_i)}_{{\theta }_i}^{(\alpha +\lambda )}}$;

$\left(ii\right)$

${\displaystyle \left({}_{s_i}^{RL}{D}_{q_i,{\omega }_i}^{\alpha }{(x-s_i)}_{{\theta }_i}^{\left(\lambda \right)}\right)\left(t\right)=\frac{{\mathrm{\Gamma }}_{q_i}(\lambda +1)}{{\mathrm{\Gamma }}_{q_i}(\lambda -\alpha +1)}{(t-s_i)}_{{\theta }_i}^{(\lambda -\alpha )}}$.

Theorem 2

([31]). Let $f\left(t\right)$ be a function defined on an interval $[s_i,t_{i+1})$ and constants $\beta ,\nu \in {\mathbb{R}}^{+}$, $\alpha \in (N-1,N)$ and ${\theta }_i\in [s_i,t_{i+1})$, $i=0,1,2\dots ,m$. Then, we have:

$\left(i\right)$

$\left({}_{s_i}{I}_{q_i,{\omega }_i}^{\beta }{}_{s_i}{I}_{q_i,{\omega }_i}^{\nu }f\right)\left(t\right)=\left({}_{s_i}{I}_{q_i,{\omega }_i}^{\nu }{}_{s_i}{I}_{q_i,{\omega }_i}^{\beta }f\right)\left(t\right)=\left({}_{s_i}{I}_{q_i,{\omega }_i}^{\beta +\nu }f\right)\left(t\right)$;

$\left(ii\right)$

$\left({}_{s_i}^{RL}{D}_{q_i,{\omega }_i}^{\beta }{}_{s_i}{I}_{q_i,{\omega }_i}^{\beta }f\right)\left(t\right)=\left({}_{s_i}^C{D}_{q_i,{\omega }_i}^{\beta }{}_{s_i}{I}_{q_i,{\omega }_i}^{\beta }f\right)=f\left(t\right)$;

$\left(iii\right)$

$\left({}_{s_i}{I}_{q_i,{\omega }_i}^{\alpha }{}_{s_i}^{RL}{D}_{q_i,{\omega }_i}^{\alpha }f\right)\left(t\right)=f\left(t\right)+c_1{(t-s_i)}_{{\theta }_i}^{(\alpha -1)}+c_2{(t-s_i)}_{{\theta }_i}^{(\alpha -2)}+\cdots +c_N{(t-s_i)}_{{\theta }_i}^{(\alpha -N)}$;

$\left(iv\right)$

$\left({}_{s_i}{I}_{q_i,{\omega }_i}^{\alpha }{}_{s_i}^C{D}_{q_i,{\omega }_i}^{\alpha }f\right)\left(t\right)=f\left(t\right)+d_0+d_1{(t-s_i)}_{{\theta }_i}^{\left(1\right)}+d_2{(t-s_i)}_{{\theta }_i}^{\left(2\right)}+\cdots +d_{N-1}{(t-s_i)}_{{\theta }_i}^{(N-1)},$for some$c_j,d_r\in \mathbb{R}$, $j=1,2,\dots ,N$, $r=0,1,\dots ,N-1$.

The rest of the paper is organized as follows. In Section 2, we first prove a basic lemma helping us to convert the boundary value problem (1) and (2) into an equivalent integral equation. Then we prove the main results, one existence and uniqueness result, via Banach contraction mapping principle and one existence result by using Leray–Schauder nonlinear alternative. Some special cases are discussed. Examples are also constructed to illustrate the main results in Section 3. The paper closes with a conclusion Section 4.

2. Main Results

In this section, we present our main results. The next lemma deals with a linear variant of the boundary value problem (1) and (2).

Lemma 1.

Let $q_i$, ${\omega }_i$, ${\alpha }_i$, ${\delta }_i$, ${\mu }_i$, $i=0,1,\dots ,m$, β, γ and λ be given constants which satisfy the problem (1) and (2). Assume that

$$\mathrm{\Omega }:=\beta +\gamma -\sum _{z=0}^m{\mu }_z\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\ne 0.$$

Then the following linear noninstantaneous impulsive fractional quantum Hahn difference equations with integral boundary conditions

$$\left\{\begin{array}{c}{}_{s_i}^C{D}_{q_i,{\omega }_i}^{{\alpha }_i}x\left(t\right)=h\left(t\right),t\in [s_i,t_{i+1}),i=0,1,2,\dots ,m,\hfill \\ x\left(t\right)=(t-t_i)g_i\left(t\right)+x\left(t_i\right),t\in [t_i,s_i),i=1,2,\dots ,m,\hfill \\ {\displaystyle \beta x\left(0\right)+\gamma x\left(T\right)=\sum _{z=0}^m{\mu }_z\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\delta }_z}x\right)\left(t_{z+1}\right)+\lambda ,}\hfill \end{array}\right.$$

(9)

has a unique solution $x\left(t\right)$, $t\in J$, of the form

$$x\left(t\right)=\left\{\begin{array}{c}{\displaystyle \frac{1}{\mathrm{\Omega }}\{\lambda +\sum _{z=0}^m{\mu }_z\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}h\right)\left(t_{z+1}\right)-\gamma \sum _{j=1}^m(s_j-t_j)g_j\left(s_j\right)-\gamma \sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right)}\hfill \\ {\displaystyle +\sum _{z=0}^m{\mu }_z\left[\sum _{j=1}^z(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{z-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right)\right]\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\}}\hfill \\ {\displaystyle +\sum _{j=1}^i(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{i-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right)+\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\alpha }_i}h\right)\left(t\right),}\hfill \\ t\in [s_i,t_{i+1}),i=0,1,2,\dots ,m,\hfill \\ {\displaystyle \frac{1}{\mathrm{\Omega }}\{\lambda +\sum _{z=0}^m{\mu }_z\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}h\right)\left(t_{z+1}\right)-\gamma \sum _{j=1}^m(s_j-t_j)g_j\left(s_j\right)-\gamma \sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right)}\hfill \\ {\displaystyle +\sum _{z=0}^m{\mu }_z\left[\sum _{j=1}^z(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{z-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right)\right]\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\}}\hfill \\ {\displaystyle +(t-t_i)g_i\left(t\right)+\sum _{j=1}^{i-1}(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{i-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right),}\hfill \\ {\displaystyle t\in [t_i,s_i),i=1,2,3,\dots ,m,}\hfill \end{array}\right.$$

(10)

with ${\sum }_a^b(·)=0$, if $b<a.$

Proof. 

Firstly, we take the quantum fractional Hahn integral of order ${\alpha }_0$ to the first equation of (9) on an interval $[s_0,t_1)$ and set $x\left(s_0^{+}\right)=C$, then

$${}_{s_0}{I}_{q_0,{\omega }_0}^{{\alpha }_0}\left({}_{s_0}{D}_{q_0,{\omega }_0}^{{\alpha }_0}x\right)\left(t\right)=x\left(t\right)=C+\left({}_{s_0}{I}_{q_0,{\omega }_0}^{{\alpha }_0}h\right)\left(t\right).$$

For $t=t_1$, we have

$$x\left(t_1\right)=C+\left({}_{s_0}{I}_{q_0,{\omega }_0}^{{\alpha }_0}h\right)\left(t_1\right).$$

(11)

On the first jump interval, $[t_1,s_1)$, calling the second equation of (9) with (11), we have

$$\begin{array}{ccc}\hfill x\left(t\right)& =& (t-t_1)g_1\left(t\right)+x\left(t_1\right)\hfill \\ & =& C+(t-t_1)g_1\left(t\right)+\left({}_{s_0}{I}_{q_0,{\omega }_0}^{{\alpha }_0}h\right)\left(t_1\right),t\in [t_1,s_1).\hfill \end{array}$$

(12)

Next consecutive subinterval $[s_1,t_2)$, again by applying the quantum fractional Hahn integral of order ${\alpha }_1$, we get

$$x\left(t\right)=x\left(s_1\right)+\left({}_{s_1}{I}_{q_1,{\omega }_1}^{{\alpha }_1}h\right)\left(t\right),$$

and then

$$x\left(t\right)=C+(s_1-t_1)g_1\left(s_1\right)+\left({}_{s_0}{I}_{q_0,{\omega }_0}^{{\alpha }_0}h\right)\left(t_1\right)+\left({}_{s_1}{I}_{q_1,{\omega }_1}^{{\alpha }_1}h\right)\left(t\right),$$

by (12).

Similar to the above method, we have for $t\in [t_2,s_2)$ as

$$x\left(t\right)=C+(t-t_2)g_2\left(t\right)+(s_1-t_1)g_1\left(s_1\right)+\left({}_{s_0}{I}_{q_0,{\omega }_0}^{{\alpha }_0}h\right)\left(t_1\right)+\left({}_{s_1}{I}_{q_1,{\omega }_1}^{{\alpha }_1}h\right)\left(t_2\right),$$

and for $t\in [s_2,t_3)$ as

$$\begin{array}{c}x\left(t\right)=C+(s_2-t_2)g_2\left(s_2\right)+(s_1-t_1)g_1\left(s_1\right)+\left({}_{s_0}{I}_{q_0,{\omega }_0}^{{\alpha }_0}h\right)\left(t_1\right)+\left({}_{s_1}{I}_{q_1,{\omega }_1}^{{\alpha }_1}h\right)\left(t_2\right)\hfill \\ +\left({}_{s_2}{I}_{q_2,{\omega }_2}^{{\alpha }_2}h\right)\left(t\right).\hfill \end{array}$$

Generally, we compute that

$$x\left(t\right)=\left\{\begin{array}{c}{\displaystyle C+\sum _{j=1}^i(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{i-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right)+\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\alpha }_i}h\right)\left(t\right),}\hfill \\ t\in [s_i,t_{i+1}),i=0,1,2,\dots ,m,\hfill \\ {\displaystyle C+(t-t_i)g_i\left(t\right)+\sum _{j=1}^{i-1}(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{i-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right),}\hfill \\ {\displaystyle t\in [t_i,s_i),i=1,2,3,\dots ,m,}\hfill \end{array}\right.$$

(13)

with ${\sum }_a^b(·)=0$, if $b<a$. From this, we have

$$x\left(T\right)=C+\sum _{j=1}^m(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right).$$

(14)

Now, taking the quantum Hahn fractional integral of order ${\delta }_z$ to (13) over $[s_z,t_{z+1})$, multiplying constant ${\mu }_z$ and summing for $z=0,1,2,\dots ,m$, we have

$$\begin{array}{ccc}\hfill \sum _{z=0}^m{\mu }_z\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\delta }_z}x\right)\left(t_{z+1}\right)& =& C\sum _{z=0}^m{\mu }_z\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}+\sum _{z=0}^m{\mu }_z\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}h\right)\left(t_{z+1}\right)\hfill \\ & & +\sum _{z=0}^m{\mu }_z\left[\sum _{j=1}^z(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{z-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right)\right]\hfill \\ & & \times \frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}.\hfill \end{array}$$

(15)

From (14) and (15) and boundary integral condition in (9), we get

$$\begin{array}{ccc}\hfill C& =& \frac{1}{\mathrm{\Omega }}\{\lambda +\sum _{z=0}^m{\mu }_z\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}h\right)\left(t_{z+1}\right)-\gamma \sum _{j=1}^m(s_j-t_j)g_j\left(s_j\right)-\gamma \sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right)\hfill \\ & & +\sum _{z=0}^m{\mu }_z\left[\sum _{j=1}^z(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{z-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}h\right)\left(t_{k+1}\right)\right]\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\}.\hfill \end{array}$$

(16)

Substituting (16) into (13) we get (10). The converse follows by direct computation. The proof is completed. □

Now, we establish the existence and uniqueness results for the boundary value problem (1) and (2) on $J=[0,T]$. We first define the Banach space $\mathcal{C}=C(J,\mathbb{R})$ equipped with the norm $\parallel x\parallel =sup\{|x\left(t\right)|:t\in J\}.$ Based on the Lemma 1, we define the operator $\mathcal{A}:\mathcal{C}\to \mathcal{C}$ by

$$\left(\mathcal{A}x\right)\left(t\right)=\left\{\begin{array}{c}{\displaystyle \frac{1}{\mathrm{\Omega }}\{\lambda +\sum _{z=0}^m{\mu }_z\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}{f}_{xy}\right)\left(t_{z+1}\right)-\gamma \sum _{j=1}^m(s_j-t_j)g_j\left(s_j\right)}\hfill \\ {\displaystyle -\gamma \sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}{f}_{xy}\right)\left(t_{k+1}\right)}\hfill \\ {\displaystyle +\sum _{z=0}^m{\mu }_z\left[\sum _{j=1}^z(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{z-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}{f}_{xy}\right)\left(t_{k+1}\right)\right]\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\}}\hfill \\ {\displaystyle +\sum _{j=1}^i(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{i-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}{f}_{xy}\right)\left(t_{k+1}\right)+\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\alpha }_i}{f}_{xy}\right)\left(t\right),}\hfill \\ t\in [s_i,t_{i+1}),i=0,1,2,\dots ,m,\hfill \\ {\displaystyle \frac{1}{\mathrm{\Omega }}\{\lambda +\sum _{z=0}^m{\mu }_z\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}{f}_{xy}\right)\left(t_{z+1}\right)-\gamma \sum _{j=1}^m(s_j-t_j)g_j\left(s_j\right)}\hfill \\ {\displaystyle -\gamma \sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}{f}_{xy}\right)\left(t_{k+1}\right)}\hfill \\ {\displaystyle +\sum _{z=0}^m{\mu }_z\left[\sum _{j=1}^z(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{z-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}{f}_{xy}\right)\left(t_{k+1}\right)\right]\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\}}\hfill \\ {\displaystyle +(t-t_i)g_i\left(t\right)+\sum _{j=1}^{i-1}(s_j-t_j)g_j\left(s_j\right)+\sum _{k=0}^{i-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}{f}_{xy}\right)\left(t_{k+1}\right),}\hfill \\ {\displaystyle t\in [t_i,s_i),i=1,2,3,\dots ,m,}\hfill \end{array}\right.$$

(17)

where the notation $f_{xy}=f(t,x,y)$, $y=\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\kappa }_i}x\right)\left(t\right)$ is used, that is,

$$f_{xy}=f\left(t,x\left(t\right),\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\kappa }_i}x\right)\left(t\right)\right),$$

which is convenient in our computations. The first result concerns the existence of a unique solution of the problem (1) and (2), and will be proved by using the Banach contraction mapping principle, involving the following constants:

$$\begin{array}{c}{\Lambda }_1=\sum _{k=0}^m\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)},{\Lambda }_2=\sum _{k=0}^m\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\kappa }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\kappa }_k+1)},\hfill \\ {\Lambda }_3=\sum _{k=0}^m\frac{|{\mu }_k|{(t_{k+1}-s_k)}_{{\theta }_k}^{({\alpha }_k+{\delta }_k)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+{\delta }_k+1)},{\Lambda }_4=\sum _{k=1}^m(s_k-t_k),\hfill \\ {\Lambda }_5=\sum _{z=0}^m|{\mu }_z|\left(\sum _{k=0}^{z-1}\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)}\right)\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)},\hfill \\ {\Lambda }_6=\sum _{z=0}^m|{\mu }_z|\left(\sum _{k=1}^z(s_k-t_k)\right)\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\hfill \\ {\Lambda }_7=\frac{1}{|\mathrm{\Omega }|}({\Lambda }_3+(|\gamma |+|\mathrm{\Omega }|){\Lambda }_1+{\Lambda }_5),{\Lambda }_8=\frac{1}{|\mathrm{\Omega }|}({\Lambda }_6+\left(|\gamma |+|\mathrm{\Omega }|\right){\Lambda }_4).\hfill \end{array}$$

Theorem 3.

Let $f:J\times {\mathbb{R}}^2\to \mathbb{R}$ and $g_i:J\to \mathbb{R}$, $i=1,2,\dots ,m$, be given continuous functions such that

$$|f(t,x_1,y_1)-f(t,x_2,y_2)|\le L_1|x_1-x_2|+L_2|y_1-y_2|,$$

(18)

where $L_1,L_2>0$, $\forall t\in J$ and $x_1,x_2,y_1,y_2\in \mathbb{R}$. If

$$(L_1+L_2{\Lambda }_2){\Lambda }_7<1,$$

(19)

then the boundary value problem of the noninstantaneous impulsive fractional quantum Hahn integro-difference in Equations (1) and (2) has a unique solution in J.

Proof. 

First of all, we will show that $\mathcal{A}{B}_r\subseteq B_r$,where $B_r=\{x\in \mathcal{C}:\parallel x\parallel \le r\}$, and the radius r is defined by

$$r\ge \frac{(|\lambda |/|\mathrm{\Omega }|)+M_1{\Lambda }_7+M_2{\Lambda }_8}{1-(L_1+L_2{\Lambda }_2){\Lambda }_7}.$$

(20)

Setting $M_1=sup\{|f(t,0,0)|:t\in J\}$ and $M_2=sup\{|g_i\left(t\right)|:t\in J,i=1,2,\dots ,m\}$, and using (18) we have

$$\begin{array}{ccc}\hfill |f_{xy}|:=|f(t,x,y)|& \le & |f(t,x,y)-f(t,0,0)|+|f(t,0,0)|\hfill \\ & \le & L_1|x|+L_2|y|+M_1\hfill \\ & \le & L_1|x|+L_2\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\kappa }_i}|x|\right)+M_1\hfill \\ & \le & L_{1}r+L_{2}r\sum _{l=0}^m\frac{{(t_{l+1}-s_l)}_{{\theta }_l}^{\left({\kappa }_l\right)}}{{\mathrm{\Gamma }}_{q_l}({\kappa }_l+1)}+M_1\hfill \\ & =& (L_1+L_2{\Lambda }_2)r+M_1.\hfill \end{array}$$

Then, for any $x\in B_r$, we obtain

$$\begin{array}{cc}& |\mathcal{A}x\left(t\right)|\le \underset{t\in J}{sup}|\mathcal{A}x\left(t\right)|\hfill \\ \le & {\displaystyle \frac{1}{|\mathrm{\Omega }|}\{|\lambda |+\sum _{z=0}^m|{\mu }_z|\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}|f_{xy}|\right)\left(t_{z+1}\right)+|\gamma |\sum _{j=1}^m(s_j-t_j)|g_j\left(s_j\right)|}\hfill \\ & +|\gamma |\sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}|f_{xy}|\right)\left(t_{k+1}\right)+\sum _{z=0}^m|{\mu }_z|[\sum _{j=1}^z(s_j-t_j)|g_j\left(s_j\right)|\hfill \\ & +\sum _{k=0}^{z-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}|f_{xy}|\right)\left(t_{k+1}\right)\left]\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\right\}\hfill \\ & {\displaystyle +\sum _{j=1}^m(s_j-t_j)|g_j\left(s_j\right)|+\sum _{k=0}^{m-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}|f_{xy}|\right)\left(t_{k+1}\right)+\left({}_{s_m}{I}_{q_m,{\omega }_m}^{{\alpha }_m}|f_{xy}|\right)\left(t_{m+1}\right)}\hfill \\ \le & {\displaystyle \frac{1}{|\mathrm{\Omega }|}\{|\lambda |+\left[\left(L_1+L_2\left(\sum _{l=0}^m\frac{{(t_{l+1}-s_l)}_{{\theta }_l}^{\left({\kappa }_l\right)}}{{\mathrm{\Gamma }}_{q_l}({\kappa }_l+1)}\right)\right)r+M_1\right]\sum _{z=0}^m|{\mu }_z|\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{({\alpha }_z+{\delta }_z)}}{{\mathrm{\Gamma }}_{q_z}({\alpha }_z+{\delta }_z+1)}}\hfill \\ & +|\gamma |M_2\sum _{j=1}^m(s_j-t_j)+|\gamma |\left[\left(L_1+L_2\left(\sum _{l=0}^m\frac{{(t_{l+1}-s_l)}_{{\theta }_l}^{\left({\kappa }_l\right)}}{{\mathrm{\Gamma }}_{q_l}({\kappa }_l+1)}\right)\right)r+M_1\right]\hfill \\ & \times \sum _{k=0}^m\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)}+M_2\sum _{z=0}^m|{\mu }_z|\left(\sum _{j=1}^z(s_j-t_j)\right)\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\hfill \\ & +\left[\left(L_1+L_2\left(\sum _{l=0}^m\frac{{(t_{l+1}-s_l)}_{{\theta }_l}^{\left({\kappa }_l\right)}}{{\mathrm{\Gamma }}_{q_l}({\kappa }_l+1)}\right)\right)r+M_1\right]\sum _{z=0}^m|{\mu }_z|\left(\sum _{k=0}^{z-1}\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)}\right)\hfill \\ & \times \frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\}+M_2\sum _{j=1}^m(s_j-t_j)\hfill \\ & {\displaystyle +\left[\left(L_1+L_2\left(\sum _{l=0}^m\frac{{(t_{l+1}-s_l)}_{{\theta }_l}^{\left({\kappa }_l\right)}}{{\mathrm{\Gamma }}_{q_l}({\kappa }_l+1)}\right)\right)r+M_1\right]\left(\sum _{k=0}^m\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)}\right)}\hfill \\ =& \frac{|\lambda |}{|\mathrm{\Omega }|}+r(L_1+L_2{\Lambda }_2){\Lambda }_7+M_1{\Lambda }_7+M_2{\Lambda }_8\le r,\hfill \end{array}$$

which holds from (19) and (20). This shows that $\mathcal{A}{B}_r\subseteq B_r$. To show that $\mathcal{A}$ is a contraction operator, we let $x_1,x_2\in B_r$ and $y_k=\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\beta }_i}{x}_k\right)\left(t\right)$, $k=1,2$, then

$$\begin{array}{cc}& |\mathcal{A}{x}_1\left(t\right)-\mathcal{A}{x}_2\left(t\right)|\le \underset{t\in J}{sup}|\mathcal{A}{x}_1\left(t\right)-\mathcal{A}{x}_2\left(t\right)|\hfill \\ \le & \frac{1}{|\mathrm{\Omega }|}\{\sum _{z=0}^m|{\mu }_z|\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}|f_{x_1{y}_1}-f_{x_2{y}_2}|\right)\left(t_{z+1}\right)+|\gamma |\sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}|f_{x_1{y}_1}-f_{x_2{y}_2}|\right)\left(t_{k+1}\right)\hfill \\ & +\sum _{z=0}^m|{\mu }_z|\left[\sum _{k=0}^{z-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}|f_{x_1{y}_1}-f_{x_2{y}_2}|\right)\left(t_{k+1}\right)\right]\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\}\hfill \\ & +\sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}|f_{x_1{y}_1}-f_{x_2{y}_2}|\right)\left(t_{k+1}\right)\hfill \\ \le & \frac{1}{|\mathrm{\Omega }|}\{\sum _{z=0}^m|{\mu }_z|(L_1+L_2{\Lambda }_2)\parallel x_1-x_2\parallel \left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}\left(1\right)\right)\left(t_{z+1}\right)\hfill \\ & +|\gamma |\sum _{k=0}^m(L_1+L_2{\Lambda }_2)\parallel x_1-x_2\parallel \left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}\left(1\right)\right)\left(t_{k+1}\right)\hfill \\ & +\sum _{z=0}^m|{\mu }_z|\left[\sum _{k=0}^{z-1}(L_1+L_2{\Lambda }_2)\parallel x_1-x_2\parallel \left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}\left(1\right)\right)\left(t_{k+1}\right)\right]\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\}\hfill \\ & +\sum _{k=0}^m(L_1+L_2{\Lambda }_2)\parallel x_1-x_2\parallel \left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}\left(1\right)\right)\left(t_{k+1}\right)\hfill \\ \le & \frac{1}{|\mathrm{\Omega }|}\{\sum _{z=0}^m\frac{|{\mu }_k|{(t_{k+1}-s_k)}_{{\theta }_k}^{({\alpha }_k+{\delta }_k)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+{\delta }_k+1)}+|\gamma |\sum _{k=0}^m\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)}\hfill \\ & +\sum _{z=0}^m|{\mu }_z|\left(\sum _{k=0}^{z-1}\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)}\right)\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\hfill \\ & +|\mathrm{\Omega }|\sum _{k=0}^m\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)}\}(L_1+L_2{\Lambda }_2)\parallel x_1-x_2\parallel \hfill \\ =& (L_1+L_2{\Lambda }_2){\Lambda }_7\parallel x_1-x_2\parallel ,\hfill \end{array}$$

since

$$\begin{array}{ccc}\hfill |f_{x_1{y}_1}-f_{x_2{y}_2}|& =& \left|f(t,x_1,\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\kappa }_i}{x}_1\right))-f(t,x_2,\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\kappa }_i}{x}_2\right))\right|\hfill \\ & \le & L_1|x_1-x_2|+L_2|x_1-x_2|\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\kappa }_i}\left(1\right)\right)\hfill \\ & \le & (L_1+L_2{\Lambda }_2)|x_1-x_2|.\hfill \end{array}$$

Therefore, the operator $\mathcal{A}$ satisfies $\parallel \mathcal{A}{x}_1-\mathcal{A}{x}_2\parallel \le (L_1+L_2{\Lambda }_2)\parallel x_1-x_2\parallel$. As, $(L_1+L_2{\Lambda }_2)<1$, we can conclude that the operator $\mathcal{A}$ is a contraction mapping which has a unique fixed point in $B_r$. Hence the problem (1) and (2) has a unique solution on J. This completes the proof. □

If we set $\gamma =0$, ${\mu }_z=0$, $z=0,1,2,\dots ,m$, and $\beta \ne 0$ in (2), then we have

$$x\left(0\right)=\frac{\lambda }{\beta },$$

(21)

which leads to the initial value problem (1)–(21). In addition, some constants are reduced to $\mathrm{\Omega }=\beta$, ${\Lambda }_3={\Lambda }_5={\Lambda }_6=0$ and ${\Lambda }_7={\Lambda }_1$. Then we get the following corollary.

Corollary 1.

Suppose that f and g satisfy the conditions of Theorem 3. If

$$(L_1+L_2{\Lambda }_2){\Lambda }_1<1,$$

then the initial value problem of the noninstantaneous impulsive fractional quantum Hahn integro-difference Equations (1)–(21) has a unique solution on J.

The next lemma is called the nonlinear alternative for single valued maps [34] which will be used to prove the next result.

Lemma 2.

Let X be a Banach space, U a closed, convex subset of $X,$V an open subset of U and $0\in V.$ Suppose that $B:\overline{V}\to C$ is a continuous, compact (that is, $B\left(\overline{V}\right)$ is a relatively compact subset of U) map. Then either

(i) 

B has a fixed point in $\overline{V},$ or

(ii) 

there is a $x\in \partial V$ (the boundary of V in U) and $\lambda \in (0,1)$ with $x=\lambda B\left(x\right).$

Theorem 4.

Assume that $f:J\times {\mathbb{R}}^2\to \mathbb{R}$ and $g_i:J\to \mathbb{R}$, $i=1,2,\dots ,m$, are continuous functions. In addition, we suppose that:

$\left(A_1\right)$

there exist a continuous nondecreasing function $\psi :[0,\infty )\to (0,\infty )$ and three continuous functions $p_1,p_2,p_3:J\to {\mathbb{R}}^{+}$ such that

$$|f(t,x,y)|\le p_1\left(t\right)\psi (|x|)+p_2\left(t\right)|y|,\mathrm{and}|g_i\left(t\right)|\le p_3\left(t\right),$$

(22)

for all $t\in J$ and $x,y\in \mathbb{R}$, $i=1,2,\dots ,m;$

$\left(A_2\right)$

there exists a constant $K>0$ such that

$${\displaystyle \frac{(1-\parallel p_2\parallel {\Lambda }_2{\Lambda }_7)K}{|\lambda |/|\mathrm{\Omega }|+\parallel p_1\parallel {\Lambda }_7\psi \left(K\right)+\parallel p_3\parallel {\Lambda }_8}>1,}$$

(23)

with $\parallel p_2\parallel {\Lambda }_2{\Lambda }_7<1.$

Then the boundary value problem of the noninstantaneous impulsive fractional quantum Hahn integro-difference Equations (1) and (2) has at least one solution on J.

Proof. 

Let $\mathcal{A}$ be the operator defined in (17) and now we are going to prove that the operator $\mathcal{A}$ is compact on a bounded ball $B_{\rho }$, where $B_{\rho }=\{x\in \mathcal{C}$ : $\parallel x\parallel \le \rho \}$. For any $x\in B_{\rho }$, we have

$$\begin{array}{cc}& |\mathcal{A}x\left(t\right)|\le \underset{t\in J}{sup}|\mathcal{A}x\left(t\right)|\hfill \\ \le & {\displaystyle \frac{1}{|\mathrm{\Omega }|}\{|\lambda |+\sum _{z=0}^m|{\mu }_z|\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}|f_{xy}|\right)\left(t_{z+1}\right)+|\gamma |\sum _{j=1}^m(s_j-t_j)|g_j\left(s_j\right)|}\hfill \\ & +|\gamma |\sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}|f_{xy}|\right)\left(t_{k+1}\right)+\sum _{z=0}^m|{\mu }_z|[\sum _{j=1}^z(s_j-t_j)|g_j\left(s_j\right)|\hfill \\ & +\sum _{k=0}^{z-1}\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}|f_{xy}|\right)\left(t_{k+1}\right)\left]\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\right\}\hfill \\ & {\displaystyle +\sum _{j=1}^m(s_j-t_j)|g_j\left(s_j\right)|+\sum _{k=0}^m\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}|f_{xy}|\right)\left(t_{k+1}\right)}\hfill \\ \le & {\displaystyle \frac{1}{|\mathrm{\Omega }|}\{|\lambda |+\sum _{z=0}^m|{\mu }_z|(\parallel p_1\parallel \psi \left(\rho \right)+\parallel p_2\parallel \rho {\Lambda }_2)\left({}_{s_z}{I}_{q_z,{\omega }_z}^{{\alpha }_z+{\delta }_z}\left(1\right)\right)\left(t_{z+1}\right)+\parallel p_3\parallel |\gamma |\sum _{j=1}^m(s_j-t_j)}\hfill \\ & +|\gamma |\sum _{k=0}^m(\parallel p_1\parallel \psi \left(\rho \right)+\parallel p_2\parallel \rho {\Lambda }_2)\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}\left(1\right)\right)\left(t_{k+1}\right)+\sum _{z=0}^m|{\mu }_z|[\parallel p_3\parallel \sum _{j=1}^z(s_j-t_j)\hfill \\ & +\sum _{k=0}^{z-1}(\parallel p_1\parallel \psi \left(\rho \right)+\parallel p_2\parallel \rho {\Lambda }_2)\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}\left(1\right)\right)\left(t_{k+1}\right)\left]\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\right\}\hfill \\ & +\parallel p_3\parallel \sum _{j=1}^m(s_j-t_j)+\sum _{k=0}^m(\parallel p_1\parallel \psi \left(\rho \right)+\parallel p_2\parallel \rho {\Lambda }_2)\left({}_{s_k}{I}_{q_k,{\omega }_k}^{{\alpha }_k}\left(1\right)\right)\left(t_{k+1}\right)\hfill \\ =& \frac{|\lambda |}{|\mathrm{\Omega }|}+(\parallel p_1\parallel \psi \left(\rho \right)+\parallel p_2\parallel \rho {\Lambda }_2)\frac{1}{|\mathrm{\Omega }|}[\sum _{z=0}^m\frac{|{\mu }_z|{(t_{z+1}-s_z)}_{{\theta }_z}^{({\alpha }_z+{\delta }_z)}}{{\mathrm{\Gamma }}_{q_z}({\alpha }_z+{\delta }_z+1)}\hfill \\ & +|\gamma |\sum _{k=0}^m\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)}+\sum _{z=0}^m|{\mu }_z|\left(\sum _{k=0}^{z-1}\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)}\right)\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}\hfill \\ & +|\mathrm{\Omega }|\sum _{k=0}^m\frac{{(t_{k+1}-s_k)}_{{\theta }_k}^{\left({\alpha }_k\right)}}{{\mathrm{\Gamma }}_{q_k}({\alpha }_k+1)}]+\parallel p_3\parallel \frac{1}{|\mathrm{\Omega }|}\left[\right(|\gamma |+|\mathrm{\Omega }|)\sum _{j=1}^m(s_j-t_j)\hfill \\ & +\sum _{z=0}^m|{\mu }_z|\left(\sum _{j=1}^z(s_j-t_j)\right)\frac{{(t_{z+1}-s_z)}_{{\theta }_z}^{\left({\delta }_z\right)}}{{\mathrm{\Gamma }}_{q_z}({\delta }_z+1)}]\hfill \\ =& \frac{|\lambda |}{|\mathrm{\Omega }|}+(\parallel p_1\parallel \psi \left(\rho \right)+\parallel p_2\parallel \rho {\Lambda }_2){\Lambda }_7+\parallel p_3\parallel {\Lambda }_8,\hfill \end{array}$$

which yields $\parallel \mathcal{A}x\parallel \le (|\lambda |/|\mathrm{\Omega }|)+(\parallel p_1\parallel \psi \left(\rho \right)+\parallel p_2\parallel \rho {\Lambda }_2){\Lambda }_7+\parallel p_3\parallel {\Lambda }_8$. Then the set $\mathcal{A}{B}_{\rho }$ is uniformly bounded. In compactness, we need to prove that $\mathcal{A}{B}_{\rho }$ is an equicontinuous set. Let ${\tau }_1$ and ${\tau }_2$ be two points such that ${\tau }_1<{\tau }_2$. Then, for any $x\in B_{\rho }$, we can compute that

$$\begin{array}{ccc}\hfill |\mathcal{A}x\left({\tau }_2\right)-\mathcal{A}x\left({\tau }_1\right)|& =& \left|\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\alpha }_i}{f}_{xy}\right)\left({\tau }_2\right)-\left({}_{s_i}{I}_{q_i,{\omega }_i}^{{\alpha }_i}{f}_{xy}\right)\left({\tau }_1\right)\right|\hfill \\ & \le & \frac{(\parallel p_1\parallel \psi \left(\rho \right)+\parallel p_2\parallel \rho {\Lambda }_2)}{{\mathrm{\Gamma }}_{q_i}\left({\alpha }_i\right)}|{\int }_{s_i}^{{\tau }_2}\left({\tau }_2-{}_{{\theta }_i}{\mathrm{\Phi }}_{q_i}\left(s\right)\right){}_{{\theta }_i}^{({\alpha }_i-1)}{}_{s_i}{d}_{q_i,{\omega }_i}s\hfill \\ & & -{\int }_{s_i}^{{\tau }_1}\left({\tau }_1-{}_{{\theta }_i}{\mathrm{\Phi }}_{q_i}\left(s\right)\right){}_{{\theta }_i}^{({\alpha }_i-1)}{}_{s_i}{d}_{q_i,{\omega }_i}s|\hfill \\ & =& \frac{(\parallel p_1\parallel \psi \left(\rho \right)+\parallel p_2\parallel \rho {\Lambda }_2)}{{\mathrm{\Gamma }}_{q_i}({\alpha }_i+1)}\left|\left({\tau }_2-s_i\right){}_{{\theta }_i}^{\left({\alpha }_i\right)}-\left({\tau }_1-s_i\right){}_{{\theta }_i}^{\left({\alpha }_i\right)}\right|,\hfill \end{array}$$

for ${\tau }_1,{\tau }_2\in [s_i,t_{i+1})$, $i=0,1,2,\dots ,m$. Therefore, we have $|\mathcal{A}x\left({\tau }_2\right)-\mathcal{A}x\left({\tau }_1\right)|\to 0$ as ${\tau }_1\to {\tau }_2$. For the case ${\tau }_1,{\tau }_2\in [t_i,s_i)$, $i=1,2,3,\dots ,m$, we have

$$|\mathcal{A}x\left({\tau }_2\right)-\mathcal{A}x\left({\tau }_1\right)|=\left|({\tau }_2-t_i)g_i\left({\tau }_2\right)-({\tau }_1-t_i)g_i\left({\tau }_1\right)\right|\to 0\mathrm{as}{\tau }_1\to {\tau }_2.$$

Then $\mathcal{A}{B}_{\rho }$ is an equicontinuous set. Therefore, we can conclude that the set $\mathcal{A}{B}_{\rho }$ is relative compact. Thus, applying Arzelá-Ascoli theorem, the operator $\mathcal{A}:\mathcal{C}\to \mathcal{C}$ is completely continuous.

Finally, we will show that there exists an open set $U\subseteq B_{\rho }$ and $0\in U$ such that $x\ne \lambda \mathcal{A}x$, where $\lambda \in (0,1)$ and $x\in \partial U$. Let $x\in \mathcal{C}$ and $x=\lambda \mathcal{A}x$ for some $\lambda \in (0,1)$. Then for any $t\in J$, using the computation in the first step, we obtain

$$\begin{array}{ccc}\hfill |x(t)|& =& \lambda |\mathcal{A}x\left(t\right)|\le \underset{t\in J}{sup}|\mathcal{A}x\left(t\right)|\hfill \\ & \le & \frac{|\lambda |}{|\mathrm{\Omega }|}+(\parallel p_1\parallel \psi (\parallel x\parallel )+\parallel p_2\parallel {\Lambda }_2\parallel x\parallel ){\Lambda }_7+\parallel p_3\parallel {\Lambda }_8,\hfill \end{array}$$

which can be written as

$${\displaystyle \frac{(1-\parallel p_2\parallel {\Lambda }_2{\Lambda }_7)\parallel x\parallel }{|\lambda |/|\mathrm{\Omega }|+\parallel p_1\parallel {\Lambda }_7\psi (\parallel x\parallel )+\parallel p_3\parallel {\Lambda }_8}\le 1.}$$

From $\left(A_2\right)$, there exists a positive constant K such that $\parallel x\parallel \ne K$. Then we define $U=\{x\in \mathcal{C}$ : $\parallel x\parallel <K\}$. From above, the operator $\mathcal{A}:\overline{U}\to \mathcal{C}$ is continuous and completely continuous. Therefore, there is no $x\in \partial U$ such that $x=\lambda \mathcal{A}x$ with $0<\lambda <1$. Applying Lemma 2 we get that the operator $\mathcal{A}$ has a fixed point $x\in \overline{U}$, which obviously is a solution of problem (1)–(2) on J. The proof is completed. □

Please note that the nonlinear condition (22) of functions f and g is a very general condition. However, we can reduce it to be a linear one by

$$|f(t,x,y)|\le K_1|x|+K_2,\mathrm{and}|g_i\left(t\right)|\le K_3,$$

(24)

by choosing $p_1\left(t\right)\equiv 1$, $p_2\left(t\right)\equiv 0$, $p_3\left(t\right)\equiv K_3$ and $\psi (|x|)=K_1|x|+K_2$, where constants $K_1\ge 0$ and $K_2,K_3>0$.

Corollary 2.

If f and $g_i$, $i=1,2,3,\dots ,m$, satisfy (24) and if $K_1{\Lambda }_7<1$, then the problem (1) and (2) has at least one solution on J.

If the function f is bounded by linear integro-term, i.e.,

$$|f(t,x,y)|\le K_2|y|+K_1,$$

(25)

by setting $p_1\left(t\right)\equiv K_1$, $p_2\left(t\right)=K_2$, $\psi (·)\equiv 1,$ then we have the corollary.

Corollary 3.

Let $g_i$ be bounded, f satisfies condition (25) and $K_2{\Lambda }_2{\Lambda }_7<1$. Then the problem (1) and (2) has at least one solution on J.

Finally, we state the corresponding existence result for the initial value problem discussed in Corollary 1.

Corollary 4.

Assume that f and g satisfy condition (22). If there exists a positive constant K satisfying

$${\displaystyle \frac{(1-\parallel p_2\parallel {\Lambda }_2{\Lambda }_1)K}{|\lambda |/|\beta |+\parallel p_1\parallel \psi \left(K\right){\Lambda }_1+\parallel p_3\parallel {\Lambda }_4}>1,}$$

with $\parallel p_2\parallel {\Lambda }_2{\Lambda }_1<1,$ then the initial value problem (1)–(21) has at least one solution on J.

3. Examples

In this section, we give some examples to illustrate the usefulness of our main results.

Example 1.

Consider the following noninstantaneous impulsive fractional quantum Hahn integro-difference equation with integral boundary conditions of the form:

$$\left\{\begin{array}{c}{\displaystyle {}_{\left(2i\right)}^C{D}_{\left(\frac{i+2}{i+3}\right),\left(\frac{1}{i+5}\right)}^{\left(\frac{i+1}{i+2}\right)}x\left(t\right)=f\left(t,x\left(t\right),\left({}_{\left(2i\right)}{I}_{\left(\frac{i+2}{i+3}\right),\left(\frac{1}{i+5}\right)}^{\left(\frac{2}{2i+3}\right)}x\right)\left(t\right)\right),t\in [2i,2i+1),i=0,1,2,3,}\hfill \\ {\displaystyle x\left(t\right)=(t-(2i-1))\frac{1}{30}{e}^{-t\sqrt{i}}+x(2i-1),t\in [2i-1,2i),i=1,2,3,}\hfill \\ {\displaystyle \frac{1}{4}x\left(0\right)+\frac{2}{3}x\left(7\right)=\sum _{z=0}^3\left(\frac{2z+1}{2z+2}\right)\left({}_{\left(2z\right)}{I}_{\left(\frac{z+2}{z+3}\right),\left(\frac{1}{z+5}\right)}^{\left(\frac{2z+1}{2}\right)}x\right)(2z+1)+\frac{1}{35}.}\hfill \end{array}\right.$$

(26)

Setting ${\alpha }_i=(i+1)/(i+2)$, $q_i=(i+2)/(i+3)$, ${\omega }_i=1/(i+5)$, ${\kappa }_i=2/(2i+3)$, ${\delta }_i=(2i+1)/2$, ${\mu }_i=(2i+1)/(2i+2)$, $i=0,1,2,3$, $\beta =1/4$, $\gamma =2/3$, $\lambda =1/35$, $t_j=2j-1$, $j=1,2,3,4$, $s_r=2r$, $r=0,1,2,3$ and $g_i\left(t\right)=(1/30)e^{-t\sqrt{i}}$, $i=1,2,3$. From above information, we can compute that ${\theta }_i=(2i^2+11i+3)/(i+5)$, $i=0,1,2,3$, $\mathrm{\Omega }\approx 0.42679$, ${\Lambda }_1\approx 1.43076$, ${\Lambda }_2\approx 2.20598$, ${\Lambda }_3\approx 0.33673$, ${\Lambda }_4=3$, ${\Lambda }_5\approx 0.05867$, ${\Lambda }_6\approx 0.07359$, ${\Lambda }_7\approx 4.59212$ and ${\Lambda }_8\approx 7.85859$. Please note that ${\theta }_i\in [2i,2i+1)$, $i=0,1,2,3$.

$\left(i\right)$ Let the nonlinear function f be defined by

$${\displaystyle f\left(t,x,\left({}_{\left(2i\right)}{I}_{\left(\frac{i+2}{i+3}\right),\left(\frac{1}{i+5}\right)}^{\left(\frac{2}{2i+3}\right)}x\right)\right)=\frac{e^{-t}}{32}\left(\frac{x^2+2|x|}{1+|x|}\right)+\frac{1}{(t^2+15)}sin\left({}_{\left(2i\right)}{I}_{\left(\frac{i+2}{i+3}\right),\left(\frac{1}{i+5}\right)}^{\left(\frac{2}{2i+3}\right)}x\right).}$$

(27)

Now we see that

$${\displaystyle |f\left(t,x_1,y_1\right)-f(t,x_2,y_2)|\le \frac{1}{16}|x_1-x_2|+\frac{1}{15}|y_1-y_2|,}$$

which is Lipschitz with Lipschitz constants $L_1=1/16$ and $L_2=1/15$. Then we obtain

$$(L_1+L_2{\Lambda }_2){\Lambda }_7\approx 0.96235<1,$$

which implies by the conclusion of Theorem 3 that the problem (26) and (27) has a unique solution on $[0,7]$.

$\left(ii\right)$ Consider now the function f defines as

$${\displaystyle f\left(t,x,\left({}_{\left(2i\right)}{I}_{\left(\frac{i+2}{i+3}\right),\left(\frac{1}{i+5}\right)}^{\left(\frac{2}{2i+3}\right)}x\right)\right)=\frac{{cos}^{2}t}{10}\left(\frac{x^{16}}{x^{14}+1}\right)+\frac{1}{2{(t+5)}^2}\left({}_{\left(2i\right)}{I}_{\left(\frac{i+2}{i+3}\right),\left(\frac{1}{i+5}\right)}^{\left(\frac{2}{2i+3}\right)}x\right).}$$

(28)

It is easy to observe that

$${\displaystyle |f\left(t,x,y\right)|\le \frac{{cos}^{2}t}{10}{x}^2+\frac{1}{2{(t+5)}^2}|y|.}$$

Thus, we set $p_1\left(t\right)=(1/10){cos}^{2}t$, $\psi \left(x\right)=x^2$, $p_2\left(t\right)=1/\left(2{(t+5)}^2\right)$. From (26), we also set $p_3\left(t\right)=(1/30)e^{-t}$. Then we get $\parallel p_1\parallel =1/10$, $\parallel p_2\parallel =1/50$ and $\parallel p_3\parallel =1/30$ which can be computed that there exists a positive constant $K\in (0.67434,1.06211)$ satisfying inequality in (23). Therefore the problem (26)–(28) satisfies all conditions of Theorem 4 which leads to conclude that there exists at least one solution on $[0,7]$.

$\left(iii\right)$ If f is defined by

$${\displaystyle f\left(t,x,\left({}_{\left(2i\right)}{I}_{\left(\frac{i+2}{i+3}\right),\left(\frac{1}{i+5}\right)}^{\left(\frac{2}{2i+3}\right)}x\right)\right)=\frac{x^2{sin}^{4}x}{(\sqrt{t}+5)(1+|x|)}+\frac{1}{t+4}{e}^{-\left|\left({}_{\left(2i\right)}{I}_{\left(\frac{i+2}{i+3}\right),\left(\frac{1}{i+5}\right)}^{\left(\frac{2}{2i+3}\right)}x\right)\right|},}$$

(29)

then we get

$${\displaystyle |f\left(t,x,y\right)|\le \frac{1}{5}|x|+\frac{1}{4},}$$

and $|g_i\left(t\right)|\le 1/3$. By setting $K_1=1/5$, $K_2=1/4$ and $K_3=1/3$, we obtain $K_1{\Lambda }_7\approx 0.91842<1$ which implies by Corollary 2 that the problem (26)–(29) has least one solution on $[0,7]$.

$\left(iv\right)$ Put the function f by

$${\displaystyle f\left(t,x,\left({}_{\left(2i\right)}{I}_{\left(\frac{i+2}{i+3}\right),\left(\frac{1}{i+5}\right)}^{\left(\frac{2}{2i+3}\right)}x\right)\right)=\frac{{tan}^{-1}|x|}{2+ln(t+1)}+\frac{e^{-t}}{11}\left({}_{\left(2i\right)}{I}_{\left(\frac{i+2}{i+3}\right),\left(\frac{1}{i+5}\right)}^{\left(\frac{2}{2i+3}\right)}x\right).}$$

(30)

It is obvious that

$${\displaystyle |f\left(t,x,y\right)|\le \frac{\pi }{4}+\frac{1}{11}|y|.}$$

From (25), we set $K_1=\pi /4$ and $K_2=1/11$. Since $K_2{\Lambda }_2{\Lambda }_7\approx 0.92092<1$, we deduce by Corollary 3 that the problem (26)–(30) has least one solution on $[0,7]$.

4. Conclusions

In this paper, we establish existence and uniqueness of solutions for a boundary value problem for fractional quantum Hahn integro-difference equations with noninstantaneous impulses, supplemented with integral boundary conditions. The classical Banach fixed point theorem is used to prove the existence and uniqueness result, while the existence result is proved via Leray–Schauder nonlinear alternative. Examples are included illustrating the obtained results.