# On the solutions of a nonlinear system of q-difference equations

## Abstract

In this paper, we examine the existence and uniqueness of solutions for a system of the first-order q-difference equations with multi-point and q-integral boundary conditions using various fixed point (fp) theorems. Also, we give two examples to support our results.

## 1 Introduction

Multi-point boundary value problems and initial value problems occur in different subfields of applied mathematics and physics. They can be used to calculate the vibration of a guy wire consisting of a finite number of segments with uniform cross-section and different densities [1]. Additionally, many problems in elastic stability theory can be treated as multi-point problems [2]. Integral boundary conditions and boundary value problems include double, triple, multiple, and nonlocal problems, and many studies have been conducted on this subject (see [38]).

Fixed point (fp) theory is used to show the existence and uniqueness of the solutions of functional equations, integral equations, and problems such as fractional-order difference equations and initial value problems. For example, Şahin [9] obtained the solution of a functional equation in a Banach space using an iterative process. Şahin [10] investigated the sufficient conditions for the existence of a solution to nonlinear Fredholm integral equations. Khatoon et al. [11] studied the approximation of fixed points and applied it to fractional-order differential equations. Şahin [12] gave existence and uniqueness results for nonlinear fractional-order differential equations with the help of a newly defined Q-function. Almarri [13] introduced an fp theorem for M-metric spaces and gave an important application to solve the initial value problem.

On the other hand, q-boundary value problems have been obtained using the q-derivative defined by

$$D_{q}x(t)=\frac{x(t)-x(qt)}{(1-q)t} \quad \text{for} \quad t\neq 0,q\neq 1$$

and $$D_{q}x(0)=\displaystyle \lim _{t \to 0}D_{q}x(t)$$ (see [14]) instead of the classical derivative, which attracted much attention. For instance, Ahmad and Ntouyas [15] investigated a q-boundary value problem with nonlocal and integral boundary conditions. Thiramanus and Tariboon [16] gave the existence and uniqueness of solutions for the nonlinear second-order q-difference equation with three-point boundary conditions. Ma and Yang [17] provided solutions for nonlinear fractional-order q-difference equations with multi-point boundary conditions using standard fp theorems. Also, the nonlinear fractional-order q-difference equations seem to have an important place in the mathematical modeling of many phenomena in engineering and science. For further studies on this subject, we refer readers to [1823]. However, there is still not enough work on multi-point boundary value problems and initial value problems in q-calculus, and there are still problems worth studying on the subject.

Recently, Ngoc and Long [24] studied the nonlinear system associated with multi-point and integral initial conditions as follows:

\begin{aligned}& \left \{ \textstyle\begin{array}{rrc} x'(t)=g(t,x(t),y(t)),& & t\in (0,T), \\ y'(t)=h(t,x(t),y(t)),& & t\in (0,T), \end{array}\displaystyle \right . \end{aligned}
(1)
\begin{aligned}& \left \{ \textstyle\begin{array}{llc} x(0)=&x_{0}, \\ y(0)=&\sum _{j=1}^{N}C_{j}.y(T_{j})+\int _{0}^{T}K(t).y(t)\,dt. \end{array}\displaystyle \right . \end{aligned}
(2)

Let $$0< q<1$$, $$S=[0,T]\cap q^{\bar{\mathbb{N}}}$$ and $$S^{*}=(0,T)\cap q^{\bar{\mathbb{N}}}$$, where $$q^{\bar{\mathbb{N}}}=\{q^{n}:n\in \mathbb{N}\}\cup \{0\}$$ and $$T\in q^{\bar{\mathbb{N}}}$$ is a constant. In this paper, by acting on all the mentioned studies, we discuss the existence and uniqueness results for the following nonlinear system:

$$\left \{ \textstyle\begin{array}{llc} D_{q}x(t)=g(t,x(t),y(t)),& & t\in S^{*}, \\ D_{q}y(t)=h(t,x(t),y(t)),& & t\in S^{*}, \end{array}\displaystyle \right .$$
(3)

associated with multi-point and q-integral initial conditions as follows:

$$\left \{ \textstyle\begin{array}{llc} x(0)=&x_{0}, \\ y(0)=&\sum _{j=1}^{N}C_{j}.y(T_{j})+\int _{0}^{T}K(t).y(t) \,d_{q}t. \end{array}\displaystyle \right .$$
(4)

System (3)–(4) is reduced to system (1)–(2) when $$q\rightarrow 1$$.

Our results are based on the Schaefer fp theorem, the Krasnoselskii fp theorem, and the Banach fp theorem. This paper consists of four sections. Section 2 outlines some essential tools and concepts needed. In Sect. 3, we prove that the solution of system (3)–(4) is equivalent to the solution of a system of integral equations and rewrite the obtained solutions by using an appropriate Green function. We also prove an essential property of the Green function, which will be used later in the paper. In Sect. 4, we investigate the existence and uniqueness of solutions and present three main conclusions. Moreover, we provide two numerical examples to help the reader better understand the primary outcomes. Our results extend and improve many known results in the literature.

## 2 Preliminaries

Let us recall the concept of the cone.

### Definition 2.1

[25] Let E be a real Banach space. A subset P of E is called a cone if it satisfies the following conditions:

1. (i)

$$P\neq \varnothing$$, $$P\neq \{0\}$$, and P is closed.

2. (ii)

$$a.u+b.v \in P$$ for all $$u,v \in P$$ and $$a,b \in \mathbb{R}$$ with $$a,b\geq 0$$.

3. (iii)

If $$u \in P$$ and $$-u \in P$$, then $$u=0$$.

According to this definition, we recall the following cones in $$\mathbb{R}^{n}$$ and $$\mathcal{K}_{n}$$:

$$\mathbb{R}_{+}^{n}=\{u=(u_{1},\ldots ,u_{n})^{T} \in \mathbb{R}^{n}: u_{i} \geq 0, \forall i={1,2,..,n}\}$$

and

$$\mathcal{K}_{n}^{+}=\{C=(c_{ij})\in \mathcal{K}_{n}: c_{ij}\geq 0, \forall i,j={1,2,..,n}\},$$

where $$\mathcal{K}_{n}$$ is the set of matrices of type $$n\times n$$. Also, a partial ordering “≤” concerning cone P is defined by $$u \leq v \Leftrightarrow v-u \in P$$. In this instance, the properties of a cone allow us to determine whether the properties of the standard ≤ for the real numbers, i.e., ≤ concerning $$\mathbb{R}_{+}$$, hold true for ≤ concerning any P (see [26]). Hence, we can consider here that

$$\forall u,v \in \mathbb{R}^{n}, u\leq v \text{ }(\text{or } v\geq u) \Leftrightarrow v-u \in \mathbb{R}_{+}^{n}$$

and

$$\forall C,D \in \mathcal{K}_{n}, C\leq D \text{ }(\text{or } D\geq C) \Leftrightarrow D-C \in \mathcal{K}_{n}^{+}.$$

Then, for each $$u\in \mathbb{R}^{n}$$, we write $$u>0$$ to show that $$u\geq 0$$ and $$u\neq 0$$; in a similar way, for each $$C\in \mathcal{K}_{n}$$, we write $$C>0 \Leftrightarrow C\geq 0 \text{ and }C\neq 0$$. For further studies on this subject, we refer readers to [2729].

Now, we give some basic concepts of q-calculus (see [14, 30, 31] and the references therein).

The q-analogue of any natural number n is defined by

$$[n]:=\frac{1-q^{n}}{1-q}.$$

The q-analogue of the factorial of natural number n is defined as

$$[n]!:= \textstyle\begin{cases} 1,&\quad n=0, \\ [1]\times [2]\times [3]\times \cdots\times [n],&\quad n\geq 1. \end{cases}$$

The q-form of the classical exponential function is as follows:

$$e_{q}^{t}=\sum _{n=0}^{\infty}\frac{t^{n}}{{[n]!}}=\frac{1}{(1-(1-q)t)_{q}^{\infty}}.$$

Another q-analogue of $$e^{t}$$ is given by

$$E_{q}^{t}=\sum _{n=0}^{\infty}q^{n.(n-1)/2}\frac{t^{n}}{[n]!}=(1+(1-q)t)_{q}^{ \infty}.$$

The connection between the exponential functions $$e_{q}^{t}$$ and $$E_{q}^{t}$$ is given by

$$e_{q}^{t}.E_{q}^{-t}=1\text{ and }e_{q}^{-t}.E_{q}^{t}=1.$$

The q-derivatives of exponential functions are $$D_{q} e_{q}^{t}=e_{q}^{t}$$ and $$D_{q}E_{q}^{t}=E_{q}^{qt}$$. Note that the q-derivative of $$E_{q}^{t}$$ is not exactly itself.

The q-integral of a function x on the interval $$[a,b]$$ is defined by

$$\int _{a}^{t}x(s) \,d_{q}s=\sum _{ n=0}^{\infty}(1-q).q^{n}.[t.x(tq^{n})-a.x(a q^{n})], \quad t\in [a,b],$$

and for $$a=0$$, we denote

$$\int _{0}^{t}x(s) \,d_{q}s=\sum _{n=0}^{\infty}t.(1-q).q^{n}.x(tq^{n}).$$

Here, the series on the right-hand side must be convergent. If $$a\in [0,b]$$ and the function x is defined in $$[0,b]$$, then the following property is valid:

$$\int _{a}^{b}x(t) \,d_{q}t=\int _{0}^{b}x(t) \,d_{q}t-\int _{0}^{a}x(t) \,d_{q}t.$$

The partial integration rule is formulated as follows:

$$\int _{0}^{s}y(t).D_{q}x(t) \,d_{q}t=[y(t).x(t)]_{0}^{s}-\int _{0}^{s}D_{q}y(t).x(qt) \,d_{q}t.$$

The change of order of integration is as follows:

$$\int _{0}^{t}\int _{0}^{s}x(r) \,d_{q}r \,d_{q}s=\int _{0}^{t}\int _{qr}^{t}x(r) \,d_{q}s \,d_{q}r.$$

Next, we give some fp theorems that will be the main tools for our results. For more details on fp theory, we refer the readers to [3236].

### Theorem 2.1

(Schaefer fp theorem) [37] Let X be a Banach space. Assume that $$\mathcal{U}:X \rightarrow X$$ is a completely continuous operator and the set $$\Delta =\{u\in X:u=\beta .\mathcal{U}(u),0<\beta <1 \}$$ is bounded. Then $$\mathcal{U}$$ has an fp in X.

### Theorem 2.2

(Krasnoselskii fp theorem) [38] Let M be a nonempty bounded closed convex subset of a Banach space X. Suppose that $$\mathcal{V}:M\rightarrow X$$ is a contraction operator and $$\Gamma :M\rightarrow X$$ is a compact operator such that

$$\mathcal{V}(u)+\Gamma (v)\in M, \quad \forall u, v \in M.$$

Then $$\mathcal{V} + \Gamma$$ has an fp in M.

### Theorem 2.3

(Banach fp theorem) [39] Let X be a Banach space. If $$\mathcal{U}:X \rightarrow X$$ is a contraction operator, that is, there exists $$\theta \in \ [0,1)$$ with

$$\| \mathcal{U}(u)-\mathcal{U}(v) \|\leq \theta . \|u-v\|$$

for any $$u,v \in X$$, then $$\mathcal{U}$$ has a unique fp in X.

## 3 Solutions of system (3)–(4)

First of all, we list the basic notations that will be used throughout this section as follows.

1. (i)

Let $$x_{0}\in \mathbb{R}^{n}$$, $$C_{j}\in \mathcal{K}_{n}$$ for $$j={1,..,N}$$ and $$0< T_{1}<\cdots <T_{N}=T$$ be constants.

2. (ii)

Let $$g,h\in C(S\times \mathbb{R}^{n}\times \mathbb{R}^{n};\mathbb{R}^{n})$$ and $$K:S\rightarrow \mathcal{K}_{n}$$ be continuous functions.

3. (iii)

Let $$L^{1}(S^{*})$$ be a Banach space with the norm $$\|L\|_{L^{1}}=\int _{0}^{T}L(t) \,d_{q}t$$.

4. (iv)

Let $$C(S;\mathbb{R}^{n})$$ and $$C'(S;\mathbb{R}^{n})$$ be Banach spaces with the usual norms

\begin{aligned}& \|x\|_{C(S;\mathbb{R}^{n})}=\max _{0\leq t \leq T}|x(t)|_{1},\\& \|x\|_{C'(S;\mathbb{R}^{n})}=\|x\|_{C(S;\mathbb{R}^{n})}+\|x'\|_{C(S; \mathbb{R}^{n})}, \end{aligned}

where $$|u|_{1}=|u_{1}|+\cdots+|u_{n}|$$ with $$u=(u_{1},\ldots,u_{n})^{T}\in \mathbb{R}^{n}$$, respectively.

5. (v)

The norm of matrices of type $$n\times n$$ for each $$C=(c_{ij})\in \mathcal{K}_{n}$$ is as follows:

$$\|C\|_{1}=\sup _{0\neq u \in \mathbb{R}^{n}} \frac{|Cu|_{1}}{{|u|_{1}}}=\max _{1\leq j \leq n}\sum _{i=1}^{n}|c_{ij}|.$$
6. (vi)

Additionally, we use the functions defined as follows:

$$\textstyle\begin{array}{rrc} g_{\gamma}(t,x,y)=g(t,x,y)+\gamma .x(qt),& \\ h_{\lambda}(t,x,y)=h(t,x,y)+\lambda .y(qt),& \end{array}$$

where $$\gamma ,\lambda \geq 0$$.

7. (vii)

Put

$$\sigma _{\lambda ,q}=I-\sum _{j=1}^{N}C_{j}.E_{q}^{-\lambda T_{j}}- \int _{0}^{T}K(\tau ).E_{q}^{-\lambda \tau} \,d_{q}\tau .$$

Now, we can give the following lemma, which states that the solution of system (3)–(4) is the solution of the system of integral equations.

### Lemma 3.1

Suppose that $$\det \sigma _{\lambda ,q}\neq 0$$. $$(x,y)\in C(S;\mathbb{R}^{n})\times C(S;\mathbb{R}^{n})$$ is a solution of equations (3)(4) if and only if $$(x,y)$$ is a solution of the below system of integral equations:

$$\left \{ \textstyle\begin{array}{llc} x(t)=&E_{q}^{-\gamma t}.x_{0}+\int _{0}^{t}E_{q}^{-\gamma t}.e_{q}^{- \gamma s}.g_{\gamma}(s,x(s),y(s)) \, d_{q}s, \\ y(t)=&\int _{0}^{t}E_{q}^{-\lambda t}.e_{q}^{\lambda s}.h_{\lambda}(s,x(s),y(s)) \,d_{q}s \\ &+E_{q}^{-\lambda t}.\sigma _{\lambda ,q}^{-1}\int _{0}^{T}\left ( \int _{qs}^{T}E_{q}^{-\lambda \tau}.e_{q}^{\lambda s}.K(\tau )d_{q} \tau \right ).h_{\lambda}(s,x(s),y(s)) \,d_{q}s \\ &+E_{q}^{-\lambda t}.\sigma _{\lambda ,q}^{-1}\sum _{j=1}^{N}C_{j} \int _{0}^{T_{j}}E_{q}^{-\lambda T_{j}}.e_{q}^{\lambda s}.h_{\lambda}(s,x(s),y(s)) \,d_{q}s. \end{array}\displaystyle \right .$$
(5)

### Proof

Let $$(x,y)\in C(S;\mathbb{R}^{n})\times C(S;\mathbb{R}^{n})$$ be a solution of problem (3)–(4). It is clear that $$(x,y)\in C'(S;\mathbb{R}^{n})\times C'(S;\mathbb{R}^{n})$$ and $$(x,y)$$ provides problem (3)–(4). For each $$\gamma , \lambda \geq 0$$, system (3) can be converted into an equivalent form as follows:

$$\left \{ \textstyle\begin{array}{llc} D_{q}x(t)+\gamma .x(qt)=g_{\gamma}(t,x,y),& & t\in S^{*}, \\ D_{q}y(t)+\lambda .y(qt)=h_{\lambda}(t,x,y), & & t\in S^{*}. \end{array}\displaystyle \right .$$
(6)

Multiplying the equations in (6) by $$e_{q}^{\gamma t}$$ and $$e_{q}^{\lambda t}$$, and taking q-integral from 0 to t, we get

$$x(t)=E_{q}^{-\gamma t}.x_{0}+\int _{0}^{t}E_{q}^{-\gamma t}.e_{q}^{ \gamma s}g_{\gamma}(s,x(s),y(s)) \,d_{q}s, \quad t\in S^{*},$$
(7)

and

$$y(t)=E_{q}^{-\lambda t}.y(0)+\int _{0}^{t}E_{q}^{-\lambda t}.e_{q}^{ \lambda s}h_{\lambda}(s,x(s),y(s)) \,d_{q}s, \quad t\in S^{*}.$$
(8)

It follows from (8) that

\begin{aligned} \int _{0}^{T}K(\tau ).y(\tau ) \,d_{q}\tau =&y(0)\int _{0}^{T}K(\tau ).E_{q}^{- \lambda \tau} \,d_{q}\tau \\ &+\int _{0}^{T}K(\tau )\left (\int _{0}^{\tau}E_{q}^{-\lambda \tau}.e_{q}^{ \lambda s}h_{\lambda}(s,x(s),y(s)) \,d_{q}s\right ) \,d_{q}\tau \\ =&y(0)\int _{0}^{T}K(\tau ).E_{q}^{-\lambda \tau} \,d_{q}\tau \\ &+\int _{0}^{T}\left (\int _{qs}^{T}E_{q}^{-\lambda \tau}.e_{q}^{ \lambda s}K(\tau ) \,d_{q}\tau \right )h_{\lambda}(s,x(s),y(s)) \,d_{q}s, \end{aligned}

and

$$\sum _{j=1}^{N}C_{j}.y(T_{j})-y(0)\sum _{j=1}^{N}C_{j}.E_{q}^{- \lambda T_{j}}=\sum _{j=1}^{N}C_{j}\int _{0}^{T_{j}}E_{q}^{-\lambda T_{j}}.e_{q}^{ \lambda s}h_{\lambda}(s,x(s),y(s)) \,d_{q}s.$$

It means that

\begin{aligned} y(0)=&\sum _{j=1}^{N}C_{j}.y(T_{j})+\int _{0}^{T}K(t).y(t) \,d_{q}t \\ =&y(0)\sum _{j=1}^{N}C_{j}.E_{q}^{-\lambda T_{j}}+\sum _{j=1}^{N}C_{j} \int _{0}^{T_{j}}E_{q}^{-\lambda T_{j}}.e_{q}^{\lambda s}h_{\lambda}(s,x(s),y(s)) \,d_{q}s \\ &+y(0)\int _{0}^{T}K(\tau )E_{q}^{-\lambda \tau} \,d_{q}\tau +\int _{0}^{T} \left (\int _{qs}^{T}E_{q}^{-\lambda \tau}.e_{q}^{\lambda s}.K(\tau ) \,d_{q}\tau \right ).h_{\lambda}(s,x(s),y(s)) \,d_{q}s. \end{aligned}

For this reason,

\begin{aligned} &y(0).\left (I-\sum _{j=1}^{N}C_{j}.E_{q}^{-\lambda T_{j}}-\int _{0}^{T}K( \tau ).E_{q}^{-\lambda \tau} \,d_{q}\tau \right ) \\ =&\sum _{j=1}^{N}C_{j}\int _{0}^{T_{j}}E_{q}^{-\lambda T_{j}}.e_{q}^{ \lambda s}.h_{\lambda}(s,x(s),y(s)) \,d_{q}s \\ &+\int _{0}^{T}\left (\int _{qs}^{T}E_{q}^{-\lambda \tau}.e_{q}^{ \lambda s}K(\tau ) \,d_{q}\tau \right ).h_{\lambda}(s,x(s),y(s)) \,d_{q}s, \end{aligned}

and so,

\begin{aligned} y(0)=&\sigma _{\lambda ,q}^{-1}\sum _{j=1}^{N}C_{j}\int _{0}^{T_{j}}E_{q}^{- \lambda T_{j}}.e_{q}^{\lambda s}.h_{\lambda}(s,x(s),y(s)) \,d_{q}s \\ &+\sigma _{\lambda ,q}^{-1}\int _{0}^{T}\left (\int _{qs}^{T}E_{q}^{- \lambda \tau}.e_{q}^{\lambda s}.K(\tau ) \,d_{q}\tau \right ).h_{ \lambda}(s,x(s),y(s)) \,d_{q}s. \end{aligned}
(9)

Using (7), (8), and (9), we deduce that $$(x,y)$$ provides system (5). Hence $$(x,y)$$ is a solution of system (5).

On the contrary, let $$(x,y)\in C(S;\mathbb{R}^{n})\times C(S;\mathbb{R}^{n})$$ be a solution of system (5). It is easy to prove that $$(x,y)\in C'(S;\mathbb{R}^{n})\times C'(S;\mathbb{R}^{n})$$ and $$(x,y)$$ ensure problem (3)–(4). Thus, the proof of Lemma 3.1 is completed. □

### Remark 3.1

Let $$S=[0,T]$$. If the classical exponential function is used instead of the q-exponential function in this lemma, Lemma 3.1 is reduced to Lemma 2.3 in [24] for $$q\rightarrow 1$$.

We can give the following result from Lemma 3.1.

### Corollary 3.1

The function $$y(t)$$ can be re-expressed as

$$y(t)=\int _{0}^{T}G(t,s;q).h_{\lambda}(s,x(s),y(s)) \,d_{q}s$$
(10)

with the help of the Green function, which can be defined as follows:

\begin{aligned} G(t,s;q)=&\left \{ \textstyle\begin{array}{llc} E_{q}^{-\lambda t}.e_{q}^{\lambda s}.I,& &0\leq s \leq t \leq T \quad \textit{for}\quad t\in S \\ 0,& &0\leq t \leq s \leq T\quad \textit{for}\quad t\in S \end{array}\displaystyle \right . \\ &+E_{q}^{-\lambda t}.e_{q}^{\lambda s}.\sigma _{\lambda ,q}^{-1}\int _{qs}^{T}E_{q}^{- \lambda \tau}.K(\tau ) \,d_{q}\tau \\ &+E_{q}^{-\lambda t}.e_{q}^{\lambda s}.\sigma _{\lambda ,q}^{-1} \left \{ \textstyle\begin{array}{llc} \sum _{j=1}^{N}E_{q}^{-\lambda T_{j}}.C_{j},& &0\leq s\leq T_{1}, \\ \quad \vdots & &\vdots \\ \sum _{j=k}^{N}E_{q}^{-\lambda T_{j}}.C_{j},& & T_{k-1}< s\leq T_{k}, \\ \quad \vdots & &\vdots \\ E_{q}^{-\lambda T}.C_{N},& & T_{N-1}< s\leq T_{N}=T. \end{array}\displaystyle \right . \end{aligned}
(11)

### Remark 3.2

Let $$C>0$$ and $$0,C \in \mathcal{K}_{n}$$. Then we have $$C \neq 0$$ and $$C-0=C \in \mathcal{K}_{n}^{+}$$. Also, let C be invertible. However, it does not emphasize that $$C^{-1}>0$$. We can give an instance

$$C= \begin{bmatrix} 1/2 & 1/6 \\ 0 & 1/3 \end{bmatrix} >0,\quad C^{-1}= \begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix}$$

definitely, it is not $$C^{-1}>0$$.

We can now give an important feature of the Green function.

### Lemma 3.2

Assume that $$K(t)\geq 0$$, $$\forall t\in S$$, and $$C_{j}\geq 0$$ for $$\forall j={1,..,N-1}$$, $$C_{N}>0$$ such that $$\det \sigma _{\lambda ,q}\neq 0$$, $$\sigma _{\lambda ,q}^{-1}>0$$, and $$\sigma _{\lambda ,q}^{-1}.C_{N}=d_{ij}$$ with $$d_{ij}>0$$ for $$\forall i,j={1,..,n}$$. Then

$$E_{q}^{-\lambda T}.\sigma _{\lambda ,q}^{-1}.C_{N}.E_{q}^{-\lambda t}e_{q}^{ \lambda s}\leq G(t,s;q)\leq \sigma _{\lambda ,q}^{-1}.E_{q}^{- \lambda t}.e_{q}^{\lambda s}\quad \textit{for}\quad \forall s,t\in S.$$

### Proof

With direct account, we have

$$G(t,s;q)\geq E_{q}^{-\lambda T}.\sigma _{\lambda ,q}^{-1}.C_{N}.E_{q}^{- \lambda t}.e_{q}^{\lambda s}$$
(12)

and

\begin{aligned} G(t,s;q) \leq &\left [I+\sigma _{\lambda ,q}^{-1}.\left (\int _{0}^{T}E_{q}^{- \lambda \tau}.K(\tau ) \,d_{q}\tau +\sum _{j=1}^{N}C_{j}.E_{q}^{- \lambda T_{j}}\right )\right ]E_{q}^{-\lambda t}.e_{q}^{\lambda s} \\ =&\left [I+\sigma _{\lambda ,q}^{-1}.\left (I-\sigma _{\lambda ,q} \right )\right ]E_{q}^{-\lambda t}.e_{q}^{\lambda s} \\ =&\sigma _{\lambda ,q}^{-1}.E_{q}^{-\lambda t}.e_{q}^{\lambda s}. \end{aligned}
(13)

From (12) and (13), the desired result is obtained. □

## 4 The existence and uniqueness of the solutions

In this part of the study, we prove the existence and uniqueness of the solutions of the equation given by (3)-(4). To do this, let us transform the problem into an fp problem by defining an operator $$\mathcal{U}$$ such that $$\mathcal{U}(x,y)=(x,y)$$. In this case, let $$\mathcal{A}=C(S;\mathbb{R}^{n})\times C(S;\mathbb{R}^{n})$$ be a Banach space defined by the norm

$$\|(x,y)\|_{\mathcal{A}}=\|x\|_{C(S;\mathbb{R}^{n})}+\|y\|_{C(S; \mathbb{R}^{n})}.$$
(14)

Also, $$\gamma =\lambda =0$$, we get

$$\sigma _{q}=I-\sum _{j=1}^{N}C_{j}-\int _{0}^{T}K(\tau ) \,d_{q}\tau$$

and

\begin{aligned} G(t,s;q)=&\left \{ \textstyle\begin{array}{llc} I,& & s\leq t\quad \text{for}\quad t\in S \\ 0,& & t\leq s\quad \text{for}\quad t\in S \end{array}\displaystyle \right .+\sigma _{q}^{-1}\int _{qs}^{T}K(\tau ) \,d_{q} \tau \\ &+\sigma _{q}^{-1}\left \{ \textstyle\begin{array}{llc} \sum _{j=1}^{N}C_{j},& & 0\leq s\leq T_{1}, \\ \quad \vdots & & \vdots \\ \sum _{j=k}^{N}C_{j},& & T_{k-1}< s\leq T_{k}, \\ \quad \vdots & &\vdots \\ C_{N},& & T_{N-1}< s\leq T_{N}=T. \end{array}\displaystyle \right . \end{aligned}
(15)

Thus, we can define an operator $$\mathcal{U}:\mathcal{A}\rightarrow \mathcal{A}$$ as follows:

$\begin{array}{rll}\mathcal{U}:& \mathcal{A}\to \mathcal{A}& \\ & \left(x,y\right)↦\left({\mathcal{U}}_{1}\left(x,y\right),{\mathcal{U}}_{2}\left(x,y\right)\right),\end{array}$

in which

\begin{aligned} &\mathcal{U}_{1}(x,y)(t)=x_{0}+\int _{0}^{t}g(s,x(s),y(s)) \,d_{q}s, \\ &\mathcal{U}_{2}(x,y)(t)=\int _{0}^{T}G(t,s;q).h(s,x(s),y(s)) \,d_{q}s. \end{aligned}
(16)

Our first result is based on the Schaefer fp theorem.

### Theorem 4.1

Let $$\mathcal{U}$$ be defined as in (16). Assume that the following hold.

1. (D1)

There exists a function $$k_{1}(t)\in C(S;\mathbb{R}^{+})$$ such that $$|g(t,x(t),y(t))|_{1}\leq k_{1}(t)$$ for all $$t \in S$$ and $$x(t),y(t)\in \mathbb{R}^{n}$$.

2. (D2)

There exists a function $$k_{2}(t)\in C(S;\mathbb{R}^{+})$$ such that $$|h(t,x(t),y(t))|_{1}\leq k_{2}(t)$$ for all $$t \in S$$ and $$x(t),y(t)\in \mathbb{R}^{n}$$.

Then problem (3)(4) has at least one solution on S.

### Proof

We divide the proof into four steps.

Step 1. We have to show that $$\mathcal{U}$$ is continuous. Let $$\{(x_{k},y_{k})\}$$ be a sequence such that $$(x_{k},y_{k}) \rightarrow (x,y)$$ in $$\mathcal{A}$$. By using the Lebesgue-dominated convergence theorem and the continuity of g and h, we get

$$\int _{0}^{t} |g(t,x_{k}(t),y_{k}(t))-g(t,x(t),y(t))|_{1} \,d_{q}t \rightarrow 0\quad \text{as} \quad k \rightarrow + \infty$$
(17)

and

$$\int _{0}^{T} |h(t,x_{k}(t),y_{k}(t))-h(t,x(t),y(t))|_{1} \,d_{q}t \rightarrow 0\quad \text{as} \quad k \rightarrow + \infty .$$
(18)

From (17) and (18), we have the following inequalities:

\begin{aligned} &|\mathcal{U}_{1}(x_{k},y_{k})(t)-\mathcal{U}_{1}(x,y)(t)|_{1} \\ \leq &\int _{0}^{t}|g(t,x_{k}(t),y_{k}(t))-g(t,x(t),y(t))|_{1} \,d_{q}t \rightarrow 0 \quad \text{as} \quad k \rightarrow + \infty \end{aligned}
(19)

and

\begin{aligned} &|\mathcal{U}_{2}(x_{k},y_{k})(t)-\mathcal{U}_{2}(x,y)(t)|_{1} \\ \leq &\|\sigma _{q}^{-1}\|_{1}\int _{0}^{T}|h(t,x_{k}(t),y_{k}(t))-h(t,x(t),y(t))|_{1} \,d_{q}t\rightarrow 0 \quad \text{as} \quad k \rightarrow + \infty . \end{aligned}
(20)

From (19) and (20), we infer $$\|\mathcal{U}(x_{k},y_{k})-\mathcal{U}(x,y)\|_{\mathcal{A}} \rightarrow 0$$ as $$k \rightarrow + \infty$$, emphasizing the continuity of $$\mathcal{U}$$.

Step 2. The operator $$\mathcal{U}$$ transforms the bounded sets in $$\mathcal{A}$$ into bounded sets in $$\mathcal{A}$$. For this, it is enough to prove that for any $$\rho >0$$ there exists a positive constant ω such that for each $$(x,y)\in B_{\rho}=\{(x,y)\in \mathcal{A}:\|(x,y)\|_{\mathcal{A}} \leq \rho \}$$ we have $$\|\mathcal{U}(x,y)\|_{\mathcal{A}}\leq \omega$$. With conditions (D1) and (D2), we obtain

\begin{aligned} \sup _{0\leq t \leq T}|\mathcal{U}_{1}(x,y)(t)|_{1}\leq |x_{0}|_{1}+T. \|k_{1}\|_{C(S;\mathbb{R}^{+})} \end{aligned}
(21)

and

\begin{aligned} \sup _{0\leq t \leq T}|\mathcal{U}_{2}(x,y)(t)|_{1}\leq \|\sigma _{q}^{-1} \|_{1}.T.\|k_{2}\|_{C(S;\mathbb{R}^{+})}. \end{aligned}
(22)

By using (21) and (22), we get

\begin{aligned} \|\mathcal{U}(x,y)\|_{\mathcal{A}}\leq |x_{0}|_{1}+T.[\|k_{1}\|_{C(S; \mathbb{R}^{+})}+\|\sigma _{q}^{-1}\|_{1}.\|k_{2}\|_{C(S;\mathbb{R}^{+})}]= \omega . \end{aligned}

Thus, step 2 is completed.

Step 3. The operator $$\mathcal{U}$$ transforms bounded sets into equicontinuous sets of $$\mathcal{A}$$. Let $$B_{\rho}$$ be a bounded set, as defined in step 2. Let $$t_{1},t_{2}\in S$$, $$t_{1}< t_{2}$$. Also, let $$(x,y)\in B_{\rho}$$. Then we have

\begin{aligned} |\mathcal{U}_{1}(x,y)(t_{2})-\mathcal{U}_{1}(x,y)(t_{1})|_{1}\leq \int _{t_{1}}^{t_{2}}|g(s,x(s),y(s))|_{1} \,d_{q}s\leq k_{1}|t_{2}-t_{1}| \end{aligned}
(23)

and

\begin{aligned} |\mathcal{U}_{2}(x,y)(t_{2})-\mathcal{U}_{2}(x,y)(t_{1})|_{1} \leq & \int _{0}^{T}|G(t_{2},s;q)-G(t_{1},s;q)|_{1}.|h(s,x(s),y(s))|_{1} \,d_{q}s \\ \leq &k_{2}|t_{2}-t_{1}|. \end{aligned}
(24)

As $$t_{1} \rightarrow t_{2}$$, the right-hand side of inequalities (23) and (24) tends to zero. Thus, step 3 is completed. Using the Arzela–Ascoli theorem together with steps 1–3, we deduce that operator U is completely continuous.

Step 4. Set $$\Delta =\{(x,y)\in \mathcal{A}:(x,y)=\beta .\mathcal{U}(x,y) \text{ for some }0<\beta <1 \}$$ is bounded. Let $$(x,y)\in \Delta$$. In this case $$(x,y)=\beta .\mathcal{U}(x,y)$$ for some $$0<\beta <1$$. Hence, for each $$t\in S$$, we have

$$x(t)=\beta .x_{0}+\beta .\int _{0}^{t}g(s,x(s),y(s)) \,d_{q}s$$

and

$$y(t)=\beta .\int _{0}^{T}G(t,s;q).h(s,x(s),y(s)) \,d_{q}s.$$

This emphasizes by (D1) and (D2) (as in step 2) that for each $$t\in S$$ we have

$$|\mathcal{U}_{1}(x,y)(t)|_{1}\leq |x_{0}|_{1}+T.\|k_{1}\|_{C(S; \mathbb{R}^{+})}$$

and

$$|\mathcal{U}_{2}(x,y)(t)|_{1}=\|\sigma _{q}^{-1}\|_{1}.T.\|k_{2}\|_{C(S; \mathbb{R}^{+})}.$$

Thus, for every $$t\in S$$, we get

$$\|(x,y)\|_{\mathcal{A}}\leq \omega .$$

This proves that Δ is a bounded set. Using the Schaefer fp theorem, we deduce that $$\mathcal{U}$$ has an fp, which is a solution of problem (3)–(4). □

### Remark 4.1

Let $$S=[0,T]$$. If the limit is taken when $$q\rightarrow 1$$, then problem (1)–(2) has at least one solution according to Schaefer fp theorem, which is new in the literature.

We describe operators $$\mathcal{V},\Gamma :\mathcal{A}\rightarrow \mathcal{A}$$ such that

$\begin{array}{rll}\mathcal{V}:& \mathcal{A}\to \mathcal{A}& \\ & \left(x,y\right)↦\left({\mathcal{U}}_{1}\left(x,y\right),0\right)\end{array}$

with

\begin{aligned} &\mathcal{U}_{1}(x,y)(t)=x_{0}+\int _{0}^{t}g(s,x(s),y(s)) \,d_{q}s, \end{aligned}

and

$\begin{array}{rll}\mathrm{\Gamma }:& \mathcal{A}\to \mathcal{A}& \\ & \left(x,y\right)↦\left(0,{\mathcal{U}}_{2}\left(x,y\right)\right)\end{array}$

with

\begin{aligned} &\mathcal{U}_{2}(x,y)(t)=\int _{0}^{T}G(t,s;q).h(s,x(s),y(s)) \,d_{q}s. \end{aligned}

Clearly, $$\mathcal{U}=\mathcal{V}+\Gamma$$.

Now, we can give the following theorem based on the Krasnoselskii fp theorem.

### Theorem 4.2

Suppose that the following conditions are provided.

1. (F1)

$$K\in C(S,\mathcal{K}_{n})$$ and $$C_{j}\in \mathcal{K}_{n}$$ ($$j={1,..,N}$$) such that $$0<\int _{0}^{T}\|K(t)\|_{1} \,d_{q}t+\sum _{j=1}^{N}\|C_{j}\|_{1}<1$$.

2. (F2)

Let $$L_{g}\in L^{1}(S^{*})$$ be a positive function such that

$$|g(t,x,y)-g(t,\bar{x},\bar{y})|_{1}\leq L_{g}(t)(|x-\bar{x}|_{1}+|y- \bar{y}|_{1})$$

for all $$(t,x,y),(t,\bar{x},\bar{y})\in S\times \mathbb{R}^{n}\times \mathbb{R}^{n}$$.

3. (F3)

Let $$h_{1},h_{2}\in L^{1}(S^{*})$$ be positive functions such that

$$|h(t,x,y)|_{1}\leq h_{1}(t)(|x|_{1}+|y|_{1})+h_{2}(t),\quad \forall (t,x,y)\in S\times \mathbb{R}^{n}\times \mathbb{R}^{n}.$$

$$L_{1}=\|L_{g}\|_{L^{1}(S^{*})}+\|\sigma _{q}^{-1}\|_{1}\|h_{1}\|_{L^{1}(S^{*})}< 1.$$
(25)

Then problem (3)(4) has at least one solution.

### Proof

We divide the proof into three steps.

Step 1. There exists a positive constant $$R>0$$ such that

$$\mathcal{V}(x,y)+\Gamma (\bar{x},\bar{y})\in B_{R}$$
(26)

for all $$(x,y),(\bar{x},\bar{y})\in B_{R}=\{(x,y)\in \mathcal{A}:\|(x,y)\|_{ \mathcal{A}}\leq R\}$$. To prove this, let $$(x,y)$$, $$(\bar{x},\bar{y})\in B_{R}$$. Then we get the following inequalities:

\begin{aligned} |\mathcal{U}_{1}(x,y)|_{1} \leq &|x_{0}|_{1}+\int _{0}^{t}|g(s,x(s),y(s))|_{1} \,d_{q}s \\ \leq &|x_{0}|_{1}+\|L_{g}\|_{L^{1}(S^{*})}R+g_{*}T, \end{aligned}
(27)
\begin{aligned} |\mathcal{U}_{2}(\bar{x},\bar{y})|_{1} \leq &\|\sigma _{q}^{-1}\|_{1} \int _{0}^{T}|h(s,\bar{x}(s),\bar{y}(s))|_{1} \,d_{q}s \\ \leq &\|\sigma _{q}^{-1}\|_{1}[\|h_{1}\|_{L^{1}(S^{*})}R+\|h_{2}\|_{L^{1}(S^{*})}] \end{aligned}
(28)

with the choice $$R>0$$ large enough such that

$$R> \frac{|x_{0}|_{1}+Tg_{*}+\|h_{2}\|_{L^{1}(S^{*})}\|\sigma _{q}^{-1}\|_{1}}{(1-L_{1})}.$$
(29)

By combining (27)–(29) and making some calculations, we reach the result in (26). Thus, step 1 is achieved.

Step 2. $$\mathcal{V}:\mathcal{A}\rightarrow \mathcal{A}$$ is a contraction operator. Let $$(x,y)$$ and $$(\bar{x}, \bar{y})$$ be arbitrary elements in $$\mathcal{A}$$. Then we obtain

\begin{aligned} |\mathcal{U}_{1}(x,y)(t)-\mathcal{U}_{1}(\bar{x},\bar{y})(t)|_{1} \leq &\int _{0}^{t}|g(s,x(s),y(s))-g(s,\bar{x}(s),\bar{y}(s))|_{1} \,d_{q}s \\ \leq &\|L_{g}\|_{L^{1}(S^{*})}\|(x,y)-(\bar{x},\bar{y})\|_{ \mathcal{A}}. \end{aligned}
(30)

Due to $$\|L_{g}\|_{L^{1}(S^{*})}\leq L_{1}<1$$, we mean that $$\mathcal{U}_{1}:\mathcal{A}\rightarrow C(S;\mathbb{R}^{n})$$ is a contraction mapping. In that case, the operator $$\mathcal{V}=(\mathcal{U}_{1},0):\mathcal{A}\rightarrow \mathcal{A}$$ is also a contraction mapping. So, step 2 is completed.

Step 3. $$\Gamma :B_{R}\rightarrow \mathcal{A}$$ is compact and continuous operator. Firstly, we show that $$\mathcal{U}_{2}$$ is continuous. Let $$\{(x_{k},y_{k})\} \subset B_{R}$$ and $$(x,y)\in B_{R}$$ such that

$$\|(x_{k},y_{k})-(x,y)\|_{\mathcal{A}}\rightarrow 0\quad \text{as} \quad k \rightarrow + \infty .$$

By using the Lebesgue dominated convergence theorem and the continuity of h, we obtain

$$\int _{0}^{T} |h(t,x_{k}(t),y_{k}(t))-h(t,x(t),y(t))|_{1} \,d_{q}t \rightarrow 0,\quad \text{as} \quad k \rightarrow + \infty .$$
(31)

From (31), we have

\begin{aligned} &\sup _{0\leq t \leq T}|\mathcal{U}_{2}(x_{k},y_{k})(t)-\mathcal{U}_{2}(x,y)(t)|_{1} \\ \leq &\|\sigma _{q}^{-1}\|\int _{0}^{T}|h(t,x_{k}(t),y_{k}(t))-h(t,x(t),y(t))|_{1} \,d_{q}t\rightarrow 0 \quad \text{as} \quad k \rightarrow + \infty . \end{aligned}

Now, we must show that $$\mathcal{U}_{2}(B_{R})$$ is relatively compact. Since h is continuous, there is $$k_{R}>0$$ so that $$|h(t,x(t),y(t))|_{1}\leq k_{R}$$ for all $$(x,y)\in B_{R}$$, $$\forall t\in S$$. Therefore, $$\mathcal{U}_{2}(B_{R})$$ is restricted in $$C(S;\mathbb{R}^{n})$$. Receiving arbitrary $$(x,y)\in B_{R}$$ and $$t_{1}, t_{2}\in S$$, $$t_{2}< t_{1}$$, we get

$$|\mathcal{U}_{2}(x,y)(t_{1})-\mathcal{U}_{2}(x,y)(t_{2})|_{1}=\left | \int _{t_{2}}^{t_{1}}h(s,x(s),y(s)) \,d_{q}s \right |_{1}\leq k_{R}|t_{1}-t_{2}|.$$

Thus $$\mathcal{U}_{2}(B_{R})$$ is equicontinuous. Hence, from the Arzela–Ascoli theorem, the $$\mathcal{U}_{2}(B_{R})$$ is relatively compact in $$C(S;\mathbb{R}^{n})$$.

We deduce from the Krasnoselskii fp theorem that $$\mathcal{U}=\mathcal{V}+\Gamma$$ has an fp. This ends the proof of the theorem. □

### Remark 4.2

Let $$S=[0,T]$$. Then Theorem 4.2 is reduced to Theorem 3.5 in [24] when $$q\rightarrow 1$$.

### Remark 4.3

From condition (F1), we have

$$\left \|\int _{0}^{T}K(\tau ) \,d_{q}\tau +\sum _{j=1}^{N}C_{j} \right \|_{1}\leq \int _{0}^{T}\|K(t)\|_{1} \,d_{q}t+\sum _{j=1}^{N} \|C_{j}\|_{1}< 1.$$

So, $$\sigma _{q}=I-\sum _{j=1}^{N}C_{j}-\int _{0}^{T}K(\tau ) \,d_{q} \tau$$ is invertible and

$$\|\sigma _{q}^{-1}\|_{1}\leq \frac{1}{1-\left \|\int _{0}^{T}K(\tau ) \,d_{q}\tau +\sum _{j=1}^{N}C_{j}\right \|_{1}} \leq \frac{1}{1-\int _{0}^{T}\|K(t)\|_{1} \,d_{q}t-\sum _{j=1}^{N}\|C_{j}\|_{1}}.$$

### Example 4.1

To explain Theorem 4.2, we examine the q-initial value problem as follows:

$$\textstyle\begin{cases} D_{q}x_{1}(t)=\frac{3}{100}E_{q}^{-t}|x(t)|_{1}, \\ D_{q}x_{2}(t)=\frac{7}{100}E_{q}^{-t}|y(t)|_{1}, \\ D_{q}y_{1}(t)=\frac{1}{136(E_{q}^{2})^{2}}E_{q}^{-t}\left [x_{1}(t). \cos{\left (\sqrt{y_{1}(t)}\right )}+y_{2}(t)\right ], \\ D_{q}y_{2}(t)=\frac{1}{272(E_{q}^{2})^{2}}E_{q}^{-t}\left [y_{1}(t). \sin{\left (\sqrt{y_{2}(t)}\right )}+x_{2}(t)\right ], \\ (x_{1}(0), x_{2}(0))^{T}=x_{0} \in \mathbb{R}^{2}, t \in S^{*}, \\ y_{1}(0)=\frac{1}{4}y_{1}(\frac{1}{2})+\frac{1}{8}y_{1}(1)+ \frac{1}{2}\int _{0}^{1}E_{q}^{-t}y_{1}(t) \,d_{q}t, \\ y_{2}(0)=\frac{1}{4}y_{1}(\frac{1}{2})+\frac{1}{8}y_{2}(\frac{1}{2})+ \frac{1}{8}y_{2}(1)+\frac{1}{4}\int _{0}^{1}E_{q}^{-t}y_{1}(t) \,d_{q}t+ \frac{1}{4}\int _{0}^{1}E_{q}^{-t}y_{2}(t) \,d_{q}t, \end{cases}$$
(32)

where $$t \in S^{*}$$.

Note that problem (32) is the form of problem (3)–(4) for $$n=2$$, $$g(t,x,y)=(g_{1}(t,x,y),g_{2}(t, x,y))^{T}$$, $$h(t,x,y)=(h_{1}(t,x,y),h_{2}(t,x,y))^{T}$$. Here,

\begin{aligned} &g_{1}(t,x,y)=\frac{3}{100}E_{q}^{-t}|x(t)|_{1}, \\ &g_{2}(t,x,y)=\frac{7}{100}E_{q}^{-t}|y(t)|_{1}, \\ &h_{1}(t,x,y)=\frac{1}{136(E_{q}^{2})^{2}}E_{q}^{-t}\left [x_{1}(t). \cos{\left (\sqrt{y_{1}(t)}\right )}+y_{2}(t)\right ], \\ &h_{2}(t,x,y)=\frac{1}{272(E_{q}^{2})^{2}}E_{q}^{-t}\left [y_{1}(t). \sin{\left (\sqrt{y_{2}(t)}\right )}+x_{2}(t)\right ], \end{aligned}

and $$N=2$$, $$T_{1}=\frac{1}{2}$$, $$T_{2}=T=1$$, $$q=\frac{1}{2}$$,

$$K(t)=\frac{1}{e_{q}^{t}} \begin{bmatrix} \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{4} \end{bmatrix} ,\quad C_{1}= \begin{bmatrix} \frac{1}{4} & 0 \\ \frac{1}{4} & \frac{1}{8} \end{bmatrix} ,\quad C_{2}= \begin{bmatrix} \frac{1}{8} & 0 \\ 0 & \frac{1}{16} \end{bmatrix} .$$

When the $$\|K(t)\|_{1}$$, $$\|C_{1}\|_{1}$$, and $$\|C_{2}\|_{1}$$ norms are calculated according to the given matrices, it is easily seen that

$$\int _{0}^{1}\|K(t)\|_{1} \,d_{q}t+\|C_{1}\|_{1}+\|C_{2}\|_{1}< 1.$$

In this case, condition (F1) is obtained.

Choosing $$L_{g}(t)=10^{-1}E_{q}^{-t}$$, $$h_{1}(t)=\max \left (\frac{1}{136(E_{q}^{2})^{2}}, \frac{1}{272(E_{q}^{2})^{2}}\right )E_{q}^{-t}$$ and $$h_{2}(t)=0$$, we get the following inequalities:

\begin{aligned} |g(t,x,y)-g(t,\bar{x},\bar{y})|_{1}\leq 10^{-1}E_{q}^{-t}(|x-\bar{x}|_{1}+|y- \bar{y}|_{1}) \end{aligned}

and

\begin{aligned} |h(t,x,y)|_{1}\leq \frac{1}{136(E_{q}^{2})^{2}}E_{q}^{-t}(|x|_{1}+|y|_{1}). \end{aligned}

Thus, conditions (F2) and (F3) are satisfied.

Also, we have

\begin{aligned} \sigma _{q}&\equiv I-\int _{0}^{1}K(t) \,d_{q}t-C_{1}-C_{2} \\ &=I-\frac{1}{2}(1-e_{q}^{-2}) \begin{bmatrix} \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{4} \end{bmatrix}- \begin{bmatrix} \frac{1}{4} & 0 \\ \frac{1}{4} & \frac{1}{8} \end{bmatrix}- \begin{bmatrix} \frac{1}{8} & 0 \\ 0 & \frac{1}{16} \end{bmatrix} \\ &= \begin{bmatrix} \sigma _{11}^{q} & 0 \\ \sigma _{21}^{q} & \sigma _{22}^{q} \end{bmatrix}, \end{aligned}

where

\begin{aligned} &\sigma _{11}^{q}=1-\frac{1-e_{q}^{-2}}{4}-\frac{1}{4}-\frac{1}{8}= \frac{1}{4E_{q}^{2}}+\frac{3}{8}>0, \\ &\sigma _{21}^{q}=-\frac{1-e_{q}^{-2}}{8}-\frac{1}{4}= \frac{1}{8E_{q}^{2}}-\frac{3}{8}< 0, \\ &\sigma _{22}^{q}=1-\frac{1-e_{q}^{-2}}{8}-\frac{1}{8}-\frac{1}{16}= \frac{1}{8E_{q}^{2}}+\frac{11}{16}>0. \end{aligned}

Then we get

$$\sigma _{q}^{-1}= \begin{bmatrix} \frac{8E_{q}^{2}}{2+3E_{q}^{2}} & 0 \\ \frac{16E_{q}^{2}(3E_{q}^{2}-1)}{(3E_{q}^{2}+2)(11E_{q}^{2}+2)} & \frac{16E_{q}^{2}}{11E_{q}^{2}+2} \end{bmatrix} >0.$$

Here, we obtain

\begin{aligned} \|\sigma _{q}^{-1}\|_{1}&=\text{max}\left \{{ \frac{8E_{q}^{2}}{2+3E_{q}^{2}}+ \frac{16E_{q}^{2}(3E_{q}^{2}-1)}{(3E_{q}^{2}+2)(11E_{q}^{2}+2)} , \frac{16E_{q}^{2}}{11E_{q}^{2}+2}}\right \}= \frac{136(E_{q}^{2})^{2}}{(3E_{q}^{2}+2)(11E_{q}^{2}+2)}. \end{aligned}

Since

\begin{aligned} L_{1}&=\|L_{g}\|_{L^{1}(S^{*})}+\|h_{1}\|_{L^{1}(S^{*})}\|\sigma _{q}^{-1} \|_{1} \\ &={10^{-1}}\frac{1}{2}\left (1-\frac{1}{E_{q}^{2}}\right )+ \frac{1}{136(E_{q}^{2})^{2}}.\frac{1}{2}\left (1-\frac{1}{E_{q}^{2}} \right ).\frac{136(E_{q}^{2})^{2}}{(3E_{q}^{2}+2)(11E_{q}^{2}+2)} \\ &=\frac{1}{2}\left (1-\frac{1}{E_{q}^{2}}\right )\left [10^{-1}+ \frac{1}{(3E_{q}^{2}+2)(11E_{q}^{2}+2)}\right ]< 1, \end{aligned}

we have (25).

From Theorem 4.2, we deduce that problem (32) has at least one solution.

By replacing condition (F3) with condition (), we get the following theorem proving the uniqueness of the solutions.

### Theorem 4.3

Assume that (F1), (F2) and the following condition are satisfied:

() Let $$L_{h}\in L^{1}(S^{*})$$ be a positive function such that

$$|h(t,x,y)-h(t,\bar{x},\bar{y})|_{1}\leq L_{h}(t)(|x-\bar{x}|_{1}+|y- \bar{y}|_{1})$$

for all $$(t,x,y),(t,\bar{x},\bar{y})\in S\times \mathbb{R}^{n}\times \mathbb{R}^{n}$$.

$$L=\|L_{g}\|_{L^{1}(S^{*})}+\|\sigma _{q}^{-1}\|_{1}.\|L_{h}\|_{L^{1}(S^{*})}< 1.$$
(33)

Then problem (3)(4) has a unique solution.

### Proof

We take

$$g_{*}=\max _{0\leq t \leq T}|g(t,0,0)|_{1}, \quad h_{*}=\max _{0 \leq t \leq T}|h(t,0,0)|_{1}$$

and pick sufficiently large $$R>0$$ such that

$$R>\frac{|x_{0}|_{1}+T(g_{*}+h_{*}\|\sigma _{q}^{-1}\|_{1})}{(1-L)}.$$
(34)

Primarily, we show that $$\mathcal{U}(B_{R})\subset B_{R}$$. Let $$B_{R}=\{(x,y)\in \mathcal{A}:\|(x,y)\|_{\mathcal{A}}\leq R\}$$. In fact, for $$(x,y)\in B_{R}$$ and all $$t\in S$$, we obtain the following inequalities:

\begin{aligned} &|\mathcal{U}_{1}(x,y)|_{1}\leq |x_{0}|_{1}+\int _{0}^{t}|g(s,x(s),y(s))-g(s,0,0)|_{1} \,d_{q}s+\int _{0}^{t}|g(s,0,0)|_{1} \,d_{q}s \\ &\hphantom{|\mathcal{U}_{1}(x,y)|_{1}}\leq |x_{0}|_{1}+\|L_{g}\|_{L^{1}(S^{*})}.R+g_{*}T, \end{aligned}
(35)
\begin{aligned} &|\mathcal{U}_{2}(x,y)|_{1}\leq \|\sigma _{q}^{-1}\|_{1}.\left [ \int _{0}^{T}|h(s,x(s),y(s))-h(s,0,0)|_{1} \,d_{q}s+\int _{0}^{T}|h(s,0,0)|_{1} \,d_{q}s\right ] \\ &\hphantom{|\mathcal{U}_{2}(x,y)|_{1}}\leq \|\sigma _{q}^{-1}\|_{1}.[\|L_{h}\|_{L^{1}(S^{*})}.R+h_{*}T]. \end{aligned}
(36)

By benefiting from inequalities (35)–(36) and with the pick of R as in (34), we deduce that $$\mathcal{U}(B_{R})\subset B_{R}$$. Then it indicates that $$\mathcal{U}:B_{R}\rightarrow B_{R}$$ is well defined.

Secondly, we demonstrate that $$\mathcal{U}$$ is a contraction operator. For $$(x,y),(\bar{x},\bar{y})\in B_{R}$$, we obtain the following inequalities:

\begin{aligned} |\mathcal{U}_{1}(x,y)(t)-\mathcal{U}_{1}(\bar{x},\bar{y})(t)|_{1} \leq &\int _{0}^{t}|g(s,x(s),y(s))-g(s,\bar{x}(s),\bar{y}(s))|_{1} \,d_{q}s \\ \leq &\|L_{g}\|_{L^{1}(S^{*})}.\|(x,y)-(\bar{x},\bar{y})\|_{ \mathcal{A}}, \end{aligned}
(37)
\begin{aligned} |\mathcal{U}_{2}(x,y)(t)-\mathcal{U}_{2}(\bar{x},\bar{y})(t)|_{1} \leq &\|\sigma _{q}^{-1}\|_{1}\int _{0}^{T}|h(s,x(s),y(s))-h(s, \bar{x}(s),\bar{y}(s))|_{1} \,d_{q}s \\ \leq &\|\sigma _{q}^{-1}\|_{1}.\|L_{h}\|_{L^{1}(S^{*})}.\|(x,y)-( \bar{x},\bar{y})\|_{\mathcal{A}}. \end{aligned}
(38)

Combining the achieved inequalities (37)–(38) and using assumption (33), we conclude that the operator $$\mathcal{U}:B_{R}\rightarrow B_{R}$$ is a contraction.

From the Banach fp theorem, problem (3)–(4) has only one solution $$(x,y)$$. Thus, the proof is completed. □

### Remark 4.4

Let $$S=[0,T]$$. Then Theorem 4.3 is reduced to Theorem 3.1 in [24] when $$q\rightarrow 1$$.

### Example 4.2

To explain Theorem 4.3, we examine the q-initial value problem as follows:

$$\textstyle\begin{cases} D_{q}x_{1}(t)=\frac{6}{100}E_{q}^{-t}[\sin (|x(t)|_{1})+\tan ^{-1}(|y(t)|_{1})], \\ D_{q}x_{2}(t)=\frac{4}{100}E_{q}^{-t}[\sin (|y(t)|_{1})+\tan ^{-1}(|x(t)|_{1})], \\ D_{q}y_{1}(t)=\frac{7}{2720(E_{q}^{2})^{2}}E_{q}^{-t}|x(t)|_{1}, \\ D_{q}y_{2}(t)=\frac{13}{2720(E_{q}^{2})^{2}}E_{q}^{-t}|y(t)|_{1}, \\ (x_{1}(0), x_{2}(0))^{T}=x_{0} \in \mathbb{R}^{2}, t \in S^{*}, \\ y_{1}(0)=\frac{1}{4}y_{1}(\frac{1}{2})+\frac{1}{8}y_{1}(1)+ \frac{1}{2}\int _{0}^{1}E_{q}^{-t}y_{1}(t) \,d_{q}t, \\ y_{2}(0)=\frac{1}{4}y_{1}(\frac{1}{2})+\frac{1}{8}y_{2}(\frac{1}{2})+ \frac{1}{8}y_{2}(1)+\frac{1}{4}\int _{0}^{1}E_{q}^{-t}y_{1}(t) \,d_{q}t+ \frac{1}{4}\int _{0}^{1}E_{q}^{-t}y_{2}(t) \,d_{q}t, \end{cases}$$
(39)

where $$t \in S^{*}$$.

Note that problem (39) is a form of problem (3)–(4) for $$n=2$$, $$g(t,x,y)=(g_{1}(t,x,y),g_{2}(t,x, y))^{T}$$, $$h(t,x,y)=(h_{1}(t,x,y),h_{2}(t,x,y))^{T}$$. Here,

\begin{aligned} &g_{1}(t,x,y)=\frac{6}{100}E_{q}^{-t}[\sin (|x(t)|_{1})+\tan ^{-1}(|y(t)|_{1})], \\ &g_{2}(t,x,y)=\frac{4}{100}E_{q}^{-t}[\sin (|y(t)|_{1})+\tan ^{-1}(|x(t)|_{1})], \\ &h_{1}(t,x,y)=\frac{7}{2720(E_{q}^{2})^{2}}E_{q}^{-t}|x(t)|_{1}, \\ &h_{2}(t,x,y)=\frac{13}{2720(E_{q}^{2})^{2}}E_{q}^{-t}|y(t)|_{1}, \end{aligned}

and $$N=2$$, $$T_{1}=\frac{1}{2}$$, $$T_{2}=T=1$$, $$q=\frac{1}{2}$$,

$$K(t)=\frac{1}{e_{q}^{t}} \begin{bmatrix} \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{4} \end{bmatrix} ,\quad C_{1}= \begin{bmatrix} \frac{1}{4} & 0 \\ \frac{1}{4} & \frac{1}{8} \end{bmatrix} ,\quad C_{2}= \begin{bmatrix} \frac{1}{8} & 0 \\ 0 & \frac{1}{16} \end{bmatrix} .$$

When the $$\|K(t)\|_{1}$$, $$\|C_{1}\|_{1}$$, and $$\|C_{2}\|_{1}$$ norms are calculated according to the given matrices, it is easily seen that

$$\int _{0}^{1}\|K(t)\|_{1} \, d_{q}t+\|C_{1}\|_{1}+\|C_{2}\|_{1}< 1.$$

In this case, condition (F1) is obtained.

On the other hand, choosing $$L_{g}(t)=10^{-1}E_{q}^{-t}$$ and $$L_{h}(t)=\frac{1}{136(E_{q}^{2})^{2}}E_{q}^{-t}$$, we get the following inequalities:

\begin{aligned} |g(t,x,y)-g(t,\bar{x},\bar{y})|_{1}\leq 10^{-1}E_{q}^{-t}(|x-\bar{x}|_{1}+|y- \bar{y}|_{1}) \end{aligned}

and

\begin{aligned} |h(t,x,y)-h(t,\bar{x},\bar{y})|_{1}\leq \frac{1}{136(E_{q}^{2})^{2}}E_{q}^{-t}(|x- \bar{x}|_{1}+|y-\bar{y}|_{1}). \end{aligned}

Thus, hypotheses (F2) and () are satisfied.

\begin{aligned} &\|L_{g}\|_{L^{1}(S^{*})}={20^{-1}}\left (1-\frac{1}{E_{q}^{2}} \right ), \\ &\|L_{h}\|_{L^{1}(S^{*})}=\frac{1}{272(E_{q}^{2})^{2}}.\left (1- \frac{1}{E_{q}^{2}}\right ), \\ &\|\sigma _{q}^{-1}\|_{1}= \frac{136(E_{q}^{2})^{2}}{(3E_{q}^{2}+2)(11E_{q}^{2}+2)}, \end{aligned}

we find that

\begin{aligned} L&=\|L_{g}\|_{L^{1}(S^{*})}+\|L_{h}\|_{L^{1}(S^{*})}\|\sigma _{q}^{-1} \|_{1} \\ &={20^{-1}}\left (1-\frac{1}{E_{q}^{2}}\right )+ \frac{1}{272(E_{q}^{2})^{2}}.\left (1-\frac{1}{E_{q}^{2}}\right ). \frac{136(E_{q}^{2})^{2}}{(3E_{q}^{2}+2)(11E_{q}^{2}+2)} \\ &=\frac{1}{2}\left (1-\frac{1}{E_{q}^{2}}\right )\left [10^{-1}+ \frac{1}{(3E_{q}^{2}+2)(11E_{q}^{2}+2)}\right ]< 1. \end{aligned}

All hypotheses of Theorem 4.3 are satisfied. So, there exists a unique solution of problem (39).

## 5 Conclusion

In the present paper, we have examined the existence and uniqueness of solutions to system (3)–(4) by using some fp theorems. Moreover, we have presented numerical examples to illustrate the main results.

In the future, a generalization of system (1)–(2) in the sense of $$(p,q)$$-calculus can be studied.

## Data Availability

No datasets were generated or analysed during the current study.

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## Acknowledgements

We would like to express our sincere gratitude to the anonymous referee for his/her helpful comments, helping to improve the quality of the manuscript.

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### Contributions

N.T.: conceptualization, methodology, writing—original draft, writing—review and editing; M.B.: methodology, supervision, writing—review and editing; A. Ş.: methodology, supervision, writing—review and editing. All authors have read and agreed to the published version of the manuscript.

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Correspondence to Nihan Turan.

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Turan, N., Başarır, M. & Şahin, A. On the solutions of a nonlinear system of q-difference equations. Bound Value Probl 2024, 92 (2024). https://doi.org/10.1186/s13661-024-01896-6