Theory and Modern Applications

# Parameter interval of positive solutions for a system of fractional difference equation

## Abstract

This paper deals with a typical system of Caputo fractional difference equations. Using the Guo–Krasnosel’skii fixed point theorem, we find a parameter interval for which at least one positive solution of the system exists. We give two examples to illustrate the results.

## Introduction

Fractional calculus, as a generalization of classical calculus, is one of those mathematical topics that received much attention. It has been shown for many years that the using of this emerging tool in modeling and design helps to improve the efficiency of various sciences. On the other hand, in recent years the fractional difference equations have been of great interest, there is much work focused on studying the existence and uniqueness of solutions [6, 19, 20]. The theory of discrete version of fractional calculus is very similar and parallel to the theory of continuous case.

Kutter was the first one studied the time differences of fractional order . Diaz and Osler introduced a discrete fractional difference operator defined as an infinite series . Grey and Zhang developed a fractional calculus for the discrete nabla difference operator . At the same time, Miller and Ross defined a fractional sum via the solution of a linear difference equation . Atici and Eloe introduced the Riemann–Liouville like fractional difference, and developed some of its properties that allow one to obtain solutions of certain fractional difference equations . Ferreira introduced the concept of left and right fractional sum/difference and started a fractional discrete-time theory of the calculus of variations [8, 9]. Holm developed and applied the tools of discrete fractional calculus to the area of fractional difference equations [14, 15]. Abdeljawad obtained dual identities in fractional difference calculus which they relate the delta and nabla and the left and right fractional sums and differences . Goodrich and Peterson develop basic theoretical results in the field of discrete fractional calculus . Moreover, Goodrich studied existence of positive solutions for discrete fractional systems and geometrical properties [10, 12].

Our objective is to explore the existence and uniqueness results for the following system of fractional difference equations:

$$\textstyle\begin{cases} -\Delta _{c}^{\alpha }y_{i}(t)=\lambda _{i} f_{i}( y_{1}(t+\alpha -1), y_{2}(t+\alpha -1),\ldots,y_{n}(t+\alpha -1) ), \\ y_{i}(\alpha -k)=a_{i}^{0}, \quad t\in [\alpha -(k-1), \alpha +T]_{ \mathbb{N}_{\alpha -(k-1)}},i=1,2,\ldots,n, \\ \Delta y_{i}(\alpha +T)=\Delta ^{j}y_{i}(\alpha -k)=0,\quad j=2,3,\ldots,k-1, \end{cases}$$
(1)

where $$\alpha \in (k-1,k]$$ and $$a_{i}^{0} \geq 0, T\geq k\geq 2$$ are real numbers, $$\Delta _{c}^{\alpha }$$ is the standard Caputo difference, $$f_{i} :[0, \infty )\times [0, \infty ) \times \cdots\times [0, \infty ) \rightarrow [0, \infty )$$ is continuous.

As we stated in the abstract, our objective is to use fixed point theory in special normed spaces to achieve an interval for parameter λ for which the problem (1) may or may not have a positive solution.

## Preliminaries and basic notations

Here, we give some basic definitions and properties of the discrete fractional calculus theory which can be found in .

### Definition 2.1

Let $$\mathbb{N}_{a}:=\{a, a+1,\ldots\}, a\in \mathbb{R}$$ and f: $$\mathbb{N}_{a}\rightarrow \mathbb{R}$$ be a real function. The difference operator Δ acts on f by

$$\Delta f(x):=f(x+1)- f(x), \quad x\in \mathbb{N}_{a}.$$

### Definition 2.2

The falling fractional power $$x^{\underline{\alpha }}$$ is given by

$$x^{\underline{\alpha }}=\frac{\varGamma (x+1)}{\varGamma (x+1-\alpha )}.$$

### Theorem 2.3

According to the definition of Δ and the falling fractional power we have

$$\Delta x^{\underline{\alpha }}= \alpha x^{\underline{\alpha -1}}.$$

### Definition 2.4

The fractional sum of order α for a given function h, for $$\alpha > 0$$, is defined by

\begin{aligned} \Delta ^{-\alpha }h(x) :=\frac{1}{\varGamma (\alpha )} \sum _{i=a}^{x- \alpha } \bigl(x-\sigma (i) \bigr)^{\underline{\alpha -1}}h(i), \end{aligned}

for $$x \in \{ \alpha +a, \alpha +a+1, \ldots\} := \mathbb{N}_{a+\alpha }$$ and $$\sigma (i)=i+1$$. The αth fractional difference for $$\alpha >0$$ is defined by $$\Delta ^{\alpha }h(x)=\Delta ^{n}\Delta ^{\alpha -n}h(x)$$, where $$x\in \mathbb{N}_{a+n-\alpha }$$ and $$n\in \mathbb{N}$$ such that $$0\leq n-1<\alpha \leq n$$.

Furthermore the Caputo fractional difference for $$\alpha > 0$$ is defined by

\begin{aligned} \Delta _{c}^{-\alpha }h(x) :=\Delta ^{-(n-\alpha )}\Delta ^{n}h(x)= \frac{1}{\varGamma (n-\alpha )} \sum _{i=a}^{x-(n-\alpha )} \bigl(x-\sigma (i) \bigr)^{ \underline{n-\alpha -1}} \Delta ^{n}h(i), \end{aligned}
(2)

where $$0\leq n-1<\alpha \leq n$$.

### Lemma 2.5

Let$$\alpha >0$$andhbe defined on$$\mathbb{N}_{a}$$, then

\begin{aligned} \Delta _{a+(n-\alpha )}^{-\alpha }\Delta _{c}^{\alpha }h(x) =h(x) - \sum_{i=0}^{n-1}c_{i}(x-a)^{\underline{i}}, \end{aligned}
(3)

where$$c_{i}\in \mathbb{R}, i=0,1,2,\ldots,n-1$$, and$$n-1<\alpha \leq n$$.

## Representation of the solution by Green’s function

Now we are ready to represent the solution of the problem (1) by Green’s function.

### Lemma 3.1

The discrete fractional boundary value problem

\begin{aligned} \textstyle\begin{cases} -\Delta _{c}^{\alpha }y(t)=\lambda h(t+\alpha -1), \\ y(\alpha -k)=a_{0}, \\ \Delta y(\alpha +T)=\Delta ^{j}y(\alpha -k)=0,\quad j=2,3,\ldots,k-1, \end{cases}\displaystyle \end{aligned}
(4)

has a unique solution given by

\begin{aligned} y(t)=\lambda \sum_{s=0}^{T+1}G(t,s)h(s+ \alpha -1)+a_{0}, \end{aligned}
(5)

where $$G(t,s)$$ is the Green’s function given by

\begin{aligned} G(t,s)=\frac{1}{\varGamma (\alpha )} \textstyle\begin{cases} (\alpha -1)(t-\alpha +k)(T+\alpha -s-1)^{\underline{\alpha -2}}-(t-s-1)^{ \underline{\alpha -1}} , \\ \quad 0\leq s< t-\alpha +1, \\ (\alpha -1)(t-\alpha +k)(T+\alpha -s-1)^{\underline{\alpha -2}} ,\quad 0 \leq t-\alpha +1\leq s . \end{cases}\displaystyle \end{aligned}

### Proof

Using Lemma 2.5

\begin{aligned} y(t)={}& {-}\frac{\lambda }{\varGamma (\alpha )}\sum_{s=0}^{t-\alpha }(t-s-1)^{ \underline{\alpha -1}}h \bigl(s+\alpha -1, y(s+\alpha -1) \bigr)+c_{0}+c_{1}t+c_{2}t^{ \underline{2}} \\ &{}+\cdots +c_{k-1}t^{\underline{k-1}}, \end{aligned}

for $$c_{i}\in \mathbb{R}, i=0,1,2,\ldots ,k-1$$.

Taking difference operator we find

\begin{aligned} &\Delta y(t)= -\frac{\lambda }{\varGamma (\alpha )}\sum_{s=0}^{t+1- \alpha }( \alpha -1) (t-s-1)^{\underline{\alpha -2}}h \bigl(s+\alpha -1, y(s+ \alpha -1) \bigr)+c_{1}+2c_{2}t \\ &\phantom{\Delta y(t)=}{}+\cdots +(k-1)c_{k-1}t^{\underline{k-2}} , \\ &\Delta ^{2}y(t)=-\frac{\lambda }{\varGamma (\alpha )}\sum _{s=0}^{t+2- \alpha }(\alpha -1) (\alpha -2) (t-s-1)^{\underline{\alpha -3}}h \bigl(s+ \alpha -1, y(s+\alpha -1) \bigr)+2c_{2} \\ &\phantom{\Delta ^{2}y(t)=}{}+\cdots +(k-1) (k-2)c_{k-1}t^{\underline{k-3}} , \\ &\vdots\\ &\Delta ^{k-1} y(t)= -\frac{\lambda }{\varGamma (\alpha )}\sum _{s=0}^{t+k-1- \alpha }(\alpha -1)\cdots (\alpha -k+1) (t-s-1)^{ \underline{\alpha -k}}h \bigl(s+\alpha -1, y(s+\alpha -1) \bigr) \\ &\phantom{\Delta ^{k-1} y(t)=}{}+(k-1) (k-2)\cdots c_{k-1}. \end{aligned}

From $$\Delta ^{j}y(\alpha -k)=0,j=2,3,\ldots ,k-1$$, we get $$c_{2}=c_{3}=\cdots =c_{k-1}=0$$, and by $$\Delta y(\alpha +T)=0,y(\alpha -k)=a_{0}$$, we have

\begin{aligned} c_{1}=\frac{\lambda }{\varGamma (\alpha )}\sum_{s=0}^{T+1}( \alpha -1) (T+ \alpha -s-1)^{\underline{\alpha -2}}h \bigl(s+\alpha -1, y(s+\alpha -1) \bigr), \end{aligned}

and $$c_{0}=-(\alpha -k)c_{1}+a_{0}$$, then we have

\begin{aligned} c_{0}=\frac{-(\alpha -k)\lambda }{\varGamma (\alpha )}\sum_{s=0}^{T+1}( \alpha -1) (T+\alpha -s-1)^{\underline{\alpha -2}}h \bigl(s+\alpha -1, y(s+ \alpha -1) \bigr)+a_{0}. \end{aligned}

Therefore, the solution of (1) is

\begin{aligned} y(t)={}& {-}\frac{\lambda }{\varGamma (\alpha )}\sum_{s=0}^{t-\alpha }(t-s-1)^{ \underline{\alpha -1}}h \bigl(s+\alpha -1, y(s+\alpha -1) \bigr) \\ &{}+\frac{-(\alpha -k)\lambda }{\varGamma (\alpha )}\sum_{s=0}^{T+1}( \alpha -1) (T+\alpha -s-1)^{\underline{\alpha -2}}h \bigl(s+\alpha -1, y(s+ \alpha -1) \bigr) \\ &{}+\frac{t\lambda }{\varGamma (\alpha )}\sum_{s=0}^{T+1}(\alpha -1) (T+ \alpha -s-1)^{\underline{\alpha -2}}h \bigl(s+\alpha -1, y(s+\alpha -1) \bigr)+a_{0} \\ = {}&\lambda \sum_{s=0}^{T+1}G(t,s)h \bigl(s+ \alpha -1, y(s+\alpha -1) \bigr)+a_{0}. \end{aligned}

□

### Lemma 3.2

The Green’s functionGgiven in Lemma 3.1satisfies in the following conditions:

1. (i)

$$G(t,s)>0, (t,s)\in [\alpha -(k-1), \alpha +T]_{\mathbb{N}_{\alpha -(k-1)}} \times [0,T+1]_{\mathbb{N}_{0}}$$.

2. (ii)

$$\max_{t\in [\alpha -(k-1), \alpha +T]_{\mathbb{N}_{\alpha -(k-1)}}} G(t,s)=G(T+ \alpha ,s)$$.

3. (iii)

$$\min_{t\in [\frac{(\alpha + T)}{4}, \frac{3(\alpha + T)}{4}]_{\mathbb{N}_{\alpha -(k-1)}}} G(t,s)\geq \frac{1}{4} \max_{t \in [\alpha -(k-1), \alpha +T]_{\mathbb{N}_{\alpha -(k-1)}}} G(t,s) =\frac{1}{4} G(T+\alpha ,s)$$.

### Proof

The proof is similar to the proof of Lemma 2.4 in . □

## Existence of positive solutions

In this section, we define the Banach space

\begin{aligned} \mathcal{E}_{i}={}& \bigl\{ y_{i} :[\alpha -k, \alpha +T]_{\mathbb{N}_{\alpha -k}} \rightarrow \mathbb{R} |y_{i}(\alpha -k)=a_{i}^{0}, \\ &\Delta y_{i}(\alpha +T)= \Delta ^{j}y_{i}(\alpha -k)=0,j=2,3,\ldots,k-1 \bigr\} , \end{aligned}

with norm

\begin{aligned} \Vert y_{i} \Vert _{\mathcal{E}_{i}} =\max \bigl\vert y_{i}(t) \bigr\vert ,\quad t \in [\alpha -k,\alpha +T]_{\mathbb{N}_{\alpha -k}}. \end{aligned}

It is clear that $$\mathcal{E}_{i}$$ is a Banach space. Put $$\mathcal{E}:=\mathcal{E}_{1}\times \mathcal{E}_{2}\times \cdots\times \mathcal{E}_{n}$$. By equipping $$\mathcal{E}$$ with the norm

\begin{aligned} \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert = \max \bigl\{ \Vert y_{1} \Vert _{\mathcal{E}_{1}}, \Vert y_{2} \Vert _{\mathcal{E}_{2}},\ldots, \Vert y_{n} \Vert _{\mathcal{E}_{n}} \bigr\} , \end{aligned}

it follows that $$(\mathcal{E},\| \cdot \|)$$ is a Banach space. It is clear that $$(y_{1},y_{2},\ldots,y_{n})$$ is solution of (1) if and only if $$y_{i}$$ satisfies

\begin{aligned} y_{i}(t)= \lambda _{i} \sum _{s=0}^{T+1}G(t,s)f_{i} \bigl(y_{1}(s+\alpha -1),y_{2}(s+ \alpha -1), \ldots, y_{n}(s+\alpha -1) \bigr)+a_{i}^{0}. \end{aligned}

Let $$\mathcal{T}_{i}: \mathcal{E}\rightarrow \mathcal{E}_{i}$$ be the operator defined by

\begin{aligned} & \mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) (t) \\ &\quad =\lambda _{i} \sum _{s=0}^{T+1}G(t,s)f_{i} \bigl(y_{1}(s+ \alpha -1),y_{2}(s+\alpha -1), \ldots, y_{n}(s+\alpha -1) \bigr)+a_{i}^{0}. \end{aligned}
(6)

Define the operator $$\mathcal{T} : \mathcal{E}\rightarrow \mathcal{E}$$ by

\begin{aligned} & \mathcal{T}(y_{1},y_{2}, \ldots, y_{n}) (t) \\ &\quad = \bigl(\mathcal{T}_{1}(y_{1},y_{2}, \ldots, y_{n}) (t), \mathcal{T}_{2}(y_{1},y_{2}, \ldots, y_{n}) (t),\ldots, \mathcal{T}_{n}(y_{1},y_{2}, \ldots, y_{n}) (t) \bigr). \end{aligned}
(7)

Let $$\mathcal{P}$$ be a cone defined by

\begin{aligned} \mathcal{P}&= \biggl\{ (y_{1},y_{2}, \ldots, y_{n}) \in \mathcal{E} : y_{i}(t) \geq 0,\min _{ t \in [\frac{\alpha +T}{4},\frac{3(\alpha +T)}{4}]}y_{i} \geq \frac{1}{4} \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert ,i=1,2, \ldots,n \biggr\} . \end{aligned}

### Lemma 4.1

Assume that$$\mathcal{T}$$is the operator defined in (7). Then$$\mathcal{T} : \mathcal{P}\rightarrow \mathcal{P}$$.

### Proof

By definition of $$\mathcal{T}_{i}$$, for $$(y_{1},y_{2}, \ldots, y_{n}) \in \mathcal{E}$$, we have

\begin{aligned} \mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) (t)\geq 0, \quad i=1,2,\ldots,n. \end{aligned}

We show that

\begin{aligned} \min_{ t \in [\frac{\alpha +T}{4},\frac{3(\alpha +T)}{4}]} \mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) (t)\geq \frac{1}{4} \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert , \end{aligned}

for $$(y_{1},y_{2}, \ldots, y_{n})\in \mathcal{E}$$. By Lemma 3.2(iii), we have

\begin{aligned} &\min _{ t \in [\frac{\alpha +T}{4},\frac{3(\alpha +T)}{4}]} \mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) (t) \\ &\quad= \min_{ t \in [\frac{\alpha +T}{4},\frac{3(\alpha +T)}{4}]} \lambda _{i} \sum _{s=0}^{T+1}G(t,s)f_{i}(y_{1}(s+ \alpha -1),y_{2} \bigl(s+ \alpha , \ldots, y_{n}(s+\alpha -1) \bigr)+a_{i}^{0} \\ &\quad\geq \frac{\lambda _{i}}{4}\max_{ t \in [\alpha -(k-1),\alpha +T]_{ \mathbb{N}_{\alpha -(k-1)}}}\sum _{s=0}^{T+1}G(t,s)f_{i} \bigl(y_{1}(s+ \alpha -1),y_{2}(s+ \alpha -1), \ldots, y_{n}(s+\alpha -1) \bigr) \\ &\qquad{}+\frac{a_{i}^{0}}{4} =\frac{1}{4} \bigl\Vert \mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert . \end{aligned}

This proves that $$\mathcal{T} : \mathcal{P}\rightarrow \mathcal{P}$$. □

### Theorem 4.2

Let$$f_{i} :[0,\infty )\times [0,\infty ) \times \cdots\times [0,\infty ) \rightarrow [0, \infty )$$be given for$$i=1,2,\ldots,n$$. If$$(y_{1},y_{2},\ldots,y_{n})\in \mathcal{E}$$is a fixed point of$$\mathcal{T}$$. Then$$(y_{1},y_{2},\ldots,y_{n})\in \mathcal{E}$$is a solution of (1).

### Proof

Let $$(y_{1},y_{2},\ldots,y_{n})\in \mathcal{E}$$ be a fixed point of $$\mathcal{T}$$, then we have

\begin{aligned} y_{i}(t)= \mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) (t)\geq 0,\quad i=1,2,\ldots,n, \end{aligned}

where $$\mathcal{T}_{i}$$ is defined as in (6). It is easy to see that

\begin{aligned} \mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) (\alpha -k)=a_{i}^{0} \end{aligned}

and

\begin{aligned} &\Delta \mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) (\alpha +T)= \mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) (\alpha +T+1)-\mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) (\alpha +T) \\ & \quad=\lambda _{i} \sum_{s=0}^{T+1}G( \alpha +T+1,s)f_{i} \bigl(y_{1}(s+ \alpha -1),y_{2}(s+\alpha -1), \ldots, y_{n}(s+\alpha -1) \bigr)+a_{i}^{0} \\ &\qquad{}- \lambda _{i} \sum_{s=0}^{T+1}G( \alpha +T,s)f_{i} \bigl(y_{1}(s+\alpha -1),y_{2}(s+ \alpha -1), \ldots, y_{n}(s+\alpha -1) \bigr)+a_{i}^{0} \\ &\quad= \lambda _{i} \sum_{s=0}^{T+1} \bigl[G(\alpha +T+1,s)-G(\alpha +T,s) \bigr]f_{i} \bigl(y_{1}(s+ \alpha -1),\ldots ,y_{n}(s+\alpha -1) \bigr) \\ &\quad= \frac{\lambda _{i}}{\varGamma (\alpha )}\sum_{s=0}^{T+1} \bigl[( \alpha -1) (\alpha +T+1-\alpha +k) (T+\alpha -s-1)^{ \underline{\alpha -2}} -(\alpha +T+1-s-1)^{\underline{\alpha -1}} \\ &\qquad{}-(\alpha -1) (\alpha +T-\alpha +k) (T+\alpha -s-1)^{ \underline{\alpha -2}}+(\alpha +T-s-1)^{\underline{\alpha -1}} \bigr] \\ &\qquad{}\times f_{i} \bigl(y_{1}(s+\alpha -1),y_{2}(s+ \alpha -1), \ldots, y_{n}(s+ \alpha -1) \bigr) \\ &\quad= \frac{\lambda _{i}}{\varGamma (\alpha )} \sum_{s=0}^{T+1} \biggl\{ ( \alpha -1) (T+\alpha -s-1)^{\underline{\alpha -2}} - \biggl[ \frac{\varGamma (\alpha +T-s+1)}{\varGamma (T-s+2)}- \frac{\varGamma (\alpha +T-s)}{\varGamma (T-s+1)} \biggr] \biggr\} \\ &\qquad{}\times f_{i} \bigl(y_{1}(s+\alpha -1),y_{2}(s+ \alpha -1), \ldots, y_{n}(s+ \alpha -1) \bigr) \\ &\quad= \frac{\lambda _{i}}{\varGamma (\alpha )} \sum_{s=0}^{T+1} \biggl\{ ( \alpha -1) (T+\alpha -s-1)^{\underline{\alpha -2}} - \biggl[ \frac{(\alpha +T-s)\varGamma (\alpha +T-s)}{(T-s+1)\varGamma (T-s+1)}- \frac{\varGamma (\alpha +T-s)}{\varGamma (T-s+1)} \biggr] \biggr\} \\ &\qquad{}\times f_{i} \bigl(y_{1}(s+\alpha -1),y_{2}(s+ \alpha -1), \ldots, y_{n}(s+ \alpha -1) \bigr) \\ &\quad= \frac{\lambda _{i}}{\varGamma (\alpha )} \sum_{s=0}^{T+1} \biggl\{ ( \alpha -1) (T+\alpha -s-1)^{\underline{\alpha -2}} - \biggl[ \frac{\varGamma (\alpha +T-s)}{\varGamma (T-s+1)} \biggl( \frac{\alpha +T-s}{T-s+1}-1 \biggr) \biggr] \biggr\} \\ &\qquad{}\times f_{i} \bigl(y_{1}(s+\alpha -1),y_{2}(s+ \alpha -1), \ldots, y_{n}(s+ \alpha -1) \bigr) \\ &\quad= \frac{\lambda _{i}}{\varGamma (\alpha )} \sum_{s=0}^{T+1} \biggl[( \alpha -1) (T+\alpha -s-1)^{\underline{\alpha -2}}- \frac{(\alpha -1)\varGamma (\alpha +T-s)}{(T-s+1)\varGamma (T-s+1)} \biggr] \\ &\qquad{}\times f_{i} \bigl(y_{1}(s+\alpha -1),y_{2}(s+ \alpha -1), \ldots, y_{n}(s+ \alpha -1) \bigr) \\ &\quad= \frac{\lambda _{i}}{\varGamma (\alpha )} \sum_{s=0}^{T+1} \biggl[( \alpha -1) (T+\alpha -s-1)^{\underline{\alpha -2}}- \frac{(\alpha -1)\varGamma (\alpha +T-s)}{\varGamma (T-s+2)} \biggr] \\ &\qquad{}\times f_{i} \bigl(y_{1}(s+\alpha -1),y_{2}(s+ \alpha -1), \ldots, y_{n}(s+ \alpha -1) \bigr) \\ &\quad= \frac{\lambda _{i}}{\varGamma (\alpha )} \sum_{s=0}^{T+1} \bigl[( \alpha -1) (T+\alpha -s-1)^{\underline{\alpha -2}}-(\alpha -1) (T+ \alpha -s-1)^{\underline{\alpha -2}} \bigr] \\ &\qquad{}\times f_{i} \bigl(y_{1}(s+\alpha -1),y_{2}(s+ \alpha -1), \ldots, y_{n}(s+ \alpha -1) \bigr) =0. \end{aligned}

Finally, when $$0< t-\alpha +1\leq s \leq T+1$$,

\begin{aligned} G(t,s)=(\alpha -1) (t-\alpha +k) (T+\alpha -s-1)^{\underline{\alpha -2}}, \end{aligned}

then

\begin{aligned} \Delta ^{j}G(t,s)=0,\quad j=2,\ldots,k-1. \end{aligned}

Therefore, we conclude

\begin{aligned} \Delta ^{j} \mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n}) (\alpha -k)=0, \quad j=2,\ldots,k-1, \end{aligned}

which completes the proof. □

### Theorem 4.3

()

Let$$\mathcal{E}$$be a Banach space, and let$$\mathcal{P}\subset \mathcal{E}$$be a cone in$$\mathcal{E}$$. Assume$$\varOmega _{1},\varOmega _{2}$$are open subsets of$$\mathcal{E}$$with$$0 \in \varOmega _{1}\subset \overline{\varOmega _{1}} \subset \varOmega _{2}$$and let$$S : \mathcal{P}\rightarrow \mathcal{P}$$be a completely continuous operator such that either

1. (A1)

$$\Vert \mathcal{T} w\Vert \leq \Vert w\Vert , w\in \mathcal{P} \cap \partial \varOmega _{1}, \Vert \mathcal{T}w\Vert \geq \Vert w \Vert , w \in \mathcal{P} \cap \partial \varOmega _{2}$$, or

2. (A2)

$$\Vert \mathcal{T}w\Vert \geq \Vert w\Vert , w\in \mathcal{P} \cap \partial \varOmega _{1}, \Vert \mathcal{T}w\Vert \leq \Vert w \Vert , w \in \mathcal{P} \cap \partial \varOmega _{2}$$.

Then$$\mathcal{T}$$has a fixed point in$$\mathcal{P}\cap ( \overline{\varOmega _{2}} \setminus \varOmega _{1})$$.

Now we find the parameter interval for which (1) has a positive solution. We use the following notations:

\begin{aligned} &F_{i}^{0}= \limsup_{y_{i}\rightarrow 0^{+}} \frac{f_{i}(y_{1},y_{2},\ldots,y_{n})}{y_{i}}, \\ &F_{i}^{\infty }= \limsup_{y_{i} \rightarrow +\infty } \frac{f_{i}(y_{1},y_{2},\ldots,y_{n})}{y_{i}}, \\ &f_{i}^{0 }= \liminf_{y_{i} \rightarrow 0^{+}} \frac{f_{i}(y_{1},y_{2},\ldots,y_{n})}{y_{i}}, \\ &f_{i}^{\infty }= \liminf_{y_{i} \rightarrow +\infty } \frac{f_{i}(y_{1},y_{2},\ldots,y_{n})}{y_{i}}, \\ &\varOmega _{r}= \bigl\{ (y_{1},y_{2}, \ldots, y_{n})\in \mathcal{E}: \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert < r \bigr\} , \\ &K= \max G(t,s), \quad\text{for } (t,s)\in \bigl[\alpha -(k-1), \alpha +T \bigr]_{ \mathbb{N}_{\alpha -(k-1)}}\times [0,T+1], \\ &l= \biggl(\frac{3(\alpha +T)}{4}-\alpha +1 \biggr)+1- \biggl( \frac{\alpha +T}{4}- \alpha +1 \biggr). \end{aligned}

In this section without loss of generality, we consider the operator $$\mathcal{T}_{i}$$, without $$a_{i}^{0}$$.

### Theorem 4.4

If$$\frac{1}{16}f_{i}^{\infty }Kl>F_{i}^{0 }K(T+1), f_{i}^{\infty }Kl \neq 0$$hold, then for each

\begin{aligned} \lambda _{i}\in \biggl( \biggl( \frac{1}{16}f_{i}^{\infty }Kl \biggr)^{-1}, \bigl(F_{i}^{0 }K(T+1) \bigr)^{-1} \biggr) \end{aligned}
(8)

the problem (1) has at least one positive solution. Note that we assume$$( f_{i}^{\infty }Kl)^{-1} =0$$if$$f_{i}^{\infty } =+\infty$$and$$(F_{i}^{0 }K(T+1))^{-1}=+ \infty$$if$$F_{i}^{0 }=0$$.

### Proof

If $$\lambda _{i}$$ satisfies in (8) and $$\varepsilon >0$$ is given such that

\begin{aligned} \biggl(\frac{1}{16} \bigl(f_{i}^{\infty }- \varepsilon \bigr)Kl \biggr)^{-1}\leq \lambda _{i} \leq \bigl( \bigl(F_{i}^{0 }+\varepsilon \bigr)K(T+1) \bigr)^{-1} , \end{aligned}
(9)

then, using the notation with $$F_{i}^{0 }$$, there exists $$r_{1} > 0$$ such that for $$(y_{1},y_{2}, \ldots, y_{n})\in \varOmega _{r_{1}}$$

\begin{aligned} f_{i}(y_{1},y_{2}, \ldots,y_{n})\leq \bigl(F_{i}^{0 }+\varepsilon \bigr)y_{i}. \end{aligned}
(10)

So if $$(y_{1},y_{2}, \ldots, y_{n})\in \partial \mathcal{P}$$ with $$\Vert (y_{1},y_{2}, \ldots, y_{n})\Vert =r_{1}$$ then, by (9) and (10), we have

\begin{aligned} \bigl\Vert \mathcal{T}_{i} (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert _{\mathcal{E}_{i}} \leq \lambda _{i} \bigl(F_{i}^{0}+\varepsilon \bigr) \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert K(T+1) \leq r_{1}= \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert . \end{aligned}

Hence for $$(y_{1},y_{2},\ldots,y_{n})\in \mathcal{P} \cap \partial \varOmega _{r_{1}}$$

\begin{aligned} \bigl\Vert \mathcal{T} (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert =\max \bigl\{ \bigl\Vert \mathcal{T}_{i} (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert _{\mathcal{E}_{i}} \bigr\} \leq \bigl\Vert (y_{1},y_{2},\ldots,y_{n}) \bigr\Vert . \end{aligned}
(11)

Let $$r_{3} > 0$$ be such that for $$y_{i}\geq r_{3}$$

\begin{aligned} f_{i}(y_{1},y_{2}, \ldots,y_{n})\geq \bigl(f_{i}^{\infty }-\varepsilon \bigr)y_{i}. \end{aligned}
(12)

If $$(y_{1},y_{2}, \ldots, y_{n})\in \partial \mathcal{P}$$ and $$\Vert (y_{1},y_{2}, \ldots, y_{n}) \Vert =r_{2}=\max \lbrace 2r_{1},r_{3} \rbrace$$ then using (9) and (12) implies

\begin{aligned} \bigl\Vert \mathcal{T}_{i} (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert _{{\mathcal{E}}_{i}}& \geq \mathcal{T}_{i} (y_{1},y_{2}, \ldots, y_{n}) = \lambda _{i}\sum_{s=0}^{T+1}G(t,s)f_{i}(y_{1},y_{2}, \ldots,y_{n}) \\ &\geq \lambda _{i} \bigl(f_{i}^{\infty }-\varepsilon \bigr)\sum_{s=( \frac{\alpha +T}{4}-\alpha +1)}^{(\frac{3(\alpha +T)}{4}-\alpha +1)}G(t,s)y_{i}(s+ \alpha -1) \\ &\geq \frac{\frac{1}{4}(f_{i}^{\infty }-\varepsilon ) \Vert (y_{1},y_{2}, \ldots, y_{n}) \Vert }{\frac{1}{16}(f_{i}^{\infty }-\varepsilon )Kl} \frac{1}{4}Kl \\ &= \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert . \end{aligned}

Then for $$(y_{1},y_{2},\ldots,y_{n}) \in \mathcal{P} \cap \partial \varOmega _{r_{2}}$$

\begin{aligned} \bigl\Vert \mathcal{T}(y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert =\max \bigl\{ \bigl\Vert \mathcal{T}_{i} (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert _{{\mathcal{E}}_{i}} \bigr\} \geq \bigl\Vert (y_{1},y_{2},\ldots,y_{n}) \bigr\Vert . \end{aligned}
(13)

Now, from (11), (13), and Theorem 4.3, we see that $$\mathcal{T}$$ has a fixed point $$(y_{1},y_{2},\ldots,y_{n}) \in \mathcal{P}\cap (\overline{\varOmega _{r_{2}}} \setminus \varOmega _{r_{1}})$$, where $$r_{1}\leq \Vert (y_{1},y_{2},\ldots,y_{n}) \Vert \leq r_{2}$$, and clearly $$(y_{1},y_{2},\ldots,y_{n})$$ is a positive solution of the problem (1). □

### Theorem 4.5

If$$\frac{1}{16}f_{i}^{0}Kl> F_{i}^{\infty }K(T+1) , f_{i}^{0}Kl\neq 0$$hold, then for each

\begin{aligned} \lambda _{i}\in \biggl( \biggl( \frac{1}{16}f_{i}^{0} Kl \biggr)^{-1}, \bigl( F_{i}^{\infty }K(T+1) \bigr)^{-1} \biggr) \end{aligned}
(14)

the problem (1) has at least one positive solution. Note that we assume$$( f_{i}^{0 }Kl)^{-1} =0$$if$$f_{i}^{0} =+\infty$$and$$(F_{i}^{\infty }K(T+1))^{-1}=+ \infty$$if$$F_{i}^{\infty }=0$$.

### Proof

Suppose $$\lambda _{i}$$ satisfies in (14) and $$\varepsilon >0$$ is such that

\begin{aligned} \biggl(\frac{1}{16} \bigl(f_{i}^{0}- \varepsilon \bigr)Kl \biggr)^{-1}\leq \lambda _{i}\leq \bigl( \bigl(F_{i}^{ \infty }+\varepsilon \bigr)K(T+1) \bigr)^{-1}. \end{aligned}
(15)

By using the notation of $$f_{i}^{0}$$, there exists $$r_{1} > 0$$ such that for $$(y_{1},y_{2},\ldots,y_{n})\in \varOmega _{ r_{1}}$$

\begin{aligned} f_{i} (y_{1},y_{2}, \ldots,y_{n}) \geq \bigl(f_{i}^{0}-\varepsilon \bigr)y_{i}. \end{aligned}
(16)

So if $$(y_{1},y_{2},\ldots,y_{n})\in \partial \mathcal{P}$$ with $$\Vert (y_{1},y_{2},\ldots,y_{n})\Vert =r_{1}$$ then analogous to Theorem 4.4, we deduce

\begin{aligned} \bigl\Vert \mathcal{T}_{i}(y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert _{{\mathcal{E}}_{i}} \geq \bigl\Vert (y_{1},y_{2},\ldots,y_{n}) \bigr\Vert . \end{aligned}

Hence, for $$(y_{1},y_{2},\ldots,y_{n}) \in \mathcal{P}\cap \partial \varOmega _{r_{1}}$$

\begin{aligned} \bigl\Vert \mathcal{T}(y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert \geq \bigl\Vert (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert . \end{aligned}
(17)

Next, we may choose $$R_{1} > 0$$ such that for $$y_{i}\geq R_{1}$$

\begin{aligned} f_{i}(y_{1},y_{2}, \ldots,y_{n})\leq \bigl(F_{i}^{\infty }+\varepsilon \bigr)y_{i}. \end{aligned}
(18)

Case 1. If $$f_{i}$$ is bounded, then, for some $$N_{i} > 0$$, we have

\begin{aligned} f_{i}(y_{1},y_{2},\ldots,y_{n})\leq N_{i} \quad\text{for } y_{i} \in [0,+\infty ). \end{aligned}

Now let $$r_{3}=\max \lbrace 2r_{1},\lambda N_{i} K(T+1)\rbrace$$ and $$(y_{1},y_{2},\ldots,y_{n})\in \mathcal{P}$$ with $$\Vert (y_{1},y_{2},\ldots,y_{n}) \Vert =r_{3}$$, thus

\begin{aligned} \bigl\Vert \mathcal{T}_{i}(y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert _{{\mathcal{E}}_{i}} &\leq \lambda _{i}\sum_{s=0}^{T+1}G(\alpha +T,s)f_{i}(y_{1},y_{2},\ldots,y_{n}) \\ &\leq \lambda _{i} N_{i}\sum _{s=0}^{T+1}G(\alpha +T,s)= \lambda _{i} K(T+1)N_{i} \leq r_{3}\\ &= \bigl\Vert (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert . \end{aligned}

Hence, for $$(y_{1},y_{2},\ldots,y_{n}) \in \partial \varOmega _{r_{3}}$$,

\begin{aligned} \bigl\Vert \mathcal{T} (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert \leq \bigl\Vert (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert . \end{aligned}
(19)

Case 2. If $$f_{i}$$ is not bounded. Then for some $$r_{4}>\max \lbrace 2r_{1}, R_{1}\rbrace$$ we have

$$f_{i}(y_{1},y_{2},\ldots,y_{n})\geq f_{i}(r_{4}, r_{4}, \ldots,r_{4}) \quad\text{for } (y_{1},y_{2},\ldots,y_{n})\in \varOmega _{r_{4}}.$$

If $$(y_{1},y_{2},\ldots,y_{n}) \in \mathcal{P}$$ with $$\Vert (y_{1},y_{2},\ldots,y_{n})\Vert =r_{4}$$, then, by (15) and (18), we have

\begin{aligned} \bigl\Vert \mathcal{T}_{i}(y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert _{{\mathcal{E}}_{i}} &\leq \lambda _{i}\sum_{s=0}^{T+1}G(\alpha +T,s) \bigl(F_{i}^{\infty }+ \varepsilon \bigr) y_{i}(s+ \alpha -1) \\ &\leq \lambda _{i} \bigl(F_{i}^{\infty }+ \varepsilon \bigr) \bigl\Vert (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert K(T+1) \\ &= \lambda _{i} K(T+1) \bigl(F_{i}^{\infty }+ \varepsilon \bigr) \bigl\Vert (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert \\ &\leq \bigl\Vert (y_{1},y_{2},\ldots,y_{n}) \bigr\Vert . \end{aligned}

Thus (19) holds.

For $$(y_{1},y_{2},\ldots,y_{n}) \in \mathcal{P} \cap \partial \varOmega _{r_{2}}$$, we have

\begin{aligned} \bigl\Vert \mathcal{T}(y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert \leq \bigl\Vert (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert . \end{aligned}
(20)

Theorem 4.3 implies that $$\mathcal{T}$$ has a fixed point $$(y_{1},y_{2},\ldots,y_{n}) \in \mathcal{P} \cap (\overline{\varOmega _{r_{2}}} \setminus \varOmega _{r_{1}})$$, where $$r_{1}\leq \Vert (y_{1},y_{2},\ldots,y_{n}) \Vert \leq r_{2}$$, and easily $$(y_{1},y_{2},\ldots,y_{n})$$ is a positive solution of (1). □

### Theorem 4.6

Suppose there exist$$r_{2} > r_{1} > 0$$such that, for$$\lambda _{i}>0$$,

\begin{aligned} \max_{0 \leq y_{i}\leq r_{2}} f_{i}(y_{1},y_{2}, \ldots,y_{n})\leq \frac{r_{2}}{\lambda _{i} K(T+1)}, \qquad\min_{0\leq y_{i} \leq r_{1}} f_{i}(y_{1},y_{2},\ldots,y_{n})\geq \frac{r_{1}}{\frac{1}{4}\lambda _{i} Kl}. \end{aligned}

Then (1) has a positive solution$$(y_{1},y_{2},\ldots,y_{n})\in \mathcal{P}$$, where$$r_{1} \leq \Vert (y_{1},y_{2},\ldots,y_{n})\Vert \leq r_{2}$$.

### Proof

If $$(y_{1},y_{2},\ldots,y_{n})\in \mathcal{P} \cap \partial \varOmega _{r_{1}}$$, we have

\begin{aligned} \bigl\Vert \mathcal{T}_{i}(y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert _{{\mathcal{E}}_{i}} &\geq \mathcal{T}_{i}(y_{1},y_{2}, \ldots,y_{n})= \lambda _{i} \sum _{s=0}^{T+1}G(t,s) f_{i}(y_{1},y_{2}, \ldots,y_{n}) \\ &\geq \lambda _{i} \sum_{s=(\frac{\alpha +T}{4}-\alpha +1)}^{( \frac{3(\alpha +T)}{4}-\alpha +1)}G( \alpha +T,s)\min_{0\leq y_{i} \leq r_{1}} f_{i}(y_{1},y_{2}, \ldots,y_{n}) \\ &\geq \lambda _{i}\frac{1}{4}Kl \frac{r_{1}}{\lambda _{i} \frac{1}{4}Kl}=r_{1}= \bigl\Vert (y_{1},y_{2},\ldots,y_{n}) \bigr\Vert . \end{aligned}

That is, for $$(y_{1},y_{2},\ldots,y_{n})\in \mathcal{P} \cap \partial \varOmega _{r_{1}}$$,

\begin{aligned} \bigl\Vert \mathcal{T}(y_{1},y_{2},\ldots,y_{n}) \bigr\Vert \geq \bigl\Vert (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert . \end{aligned}

Also, for $$(y_{1},y_{2},\ldots,y_{n}) \in \mathcal{P} \cap \partial \varOmega _{r_{2}}$$, we have

\begin{aligned} \bigl\Vert \mathcal{T}_{i}(y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert _{{\mathcal{E}}_{i}} &\leq \lambda _{i} \sum_{s=0}^{T+1}G(\alpha +T,s) \max_{0 \leq y_{i} \leq r_{2}}f_{i}(y_{1},y_{2}, \ldots,y_{n}) \\ &\leq \lambda _{i} K(T+1)\frac{r_{2}}{\lambda _{i} K(T+1)}=r_{2}= \bigl\Vert (y_{1},y_{2},\ldots,y_{n}) \bigr\Vert . \end{aligned}

That is, for $$(y_{1},y_{2},\ldots,y_{n})\in \mathcal{P} \cap \partial \varOmega _{r_{2}}$$,

\begin{aligned} \bigl\Vert \mathcal{T}(y_{1},y_{2},\ldots,y_{n}) \bigr\Vert \leq \bigl\Vert (y_{1},y_{2}, \ldots,y_{n}) \bigr\Vert . \end{aligned}

Thus, by Theorem 4.3, the problem (1) has a positive solution $$(y_{1},y_{2},\ldots,y_{n}) \in \mathcal{P}$$, where $$r_{1}\leq \Vert (y_{1},y_{2},\ldots,y_{n})\Vert \leq r_{2}$$. □

## Nonexistence results

In order to find some nonexistence results for problem (1), we consider the following condition:

1. (H)

$$\sup_{r>0}\min_{y_{i} \in (0, r)}f_{i}(y_{1},y_{2},\ldots,y_{n})>0$$.

### Theorem 5.1

Assume the condition$$\mathrm{(H)}$$is true. If$$F_{i}^{0}<+\infty$$and$$F_{i}^{\infty } <+\infty$$, then there is a positive real number$$\lambda _{i}^{0}>0$$such that for$$0< \lambda _{i} <\lambda _{i}^{0}$$the problem (1) does not have a positive solution.

### Proof

Since $$F_{i}^{0}, F_{i}^{\infty }$$ are finite, we can find positive real numbers $$l_{i}^{1}, l_{i}^{2}, r_{1},r_{2}$$, where $$r_{1} < r_{2}$$ and

\begin{aligned} &f_{i}(y_{1},y_{2}, \ldots,y_{n})\leq l_{i}^{1} y_{i},\quad \text{for } y_{i}\in [0, r_{1}], \\ &f_{i}(y_{1},y_{2},\ldots,y_{n})\leq l_{i}^{2} y_{i}, \quad\text{for } y_{i} \in [r_{2}, +\infty ). \end{aligned}

Let $$L_{i}= \max \lbrace l_{i}^{1}, l_{i}^{2}, \max_{r_{1}\leq y_{i} \leq r_{2}} \lbrace \frac{f_{i}(y_{1},y_{2},\ldots,y_{n})}{y_{i}} \rbrace \rbrace$$. Then we have

\begin{aligned} f_{i}(y_{1},y_{2}, \ldots,y_{n})\leq L_{i} y_{i}, \quad\text{for } y_{i} \in [0,+\infty ). \end{aligned}

Assume $$(w_{1},w_{2},\ldots,w_{n})(t)$$ is a positive solution of (1). We find a contradiction for $$0 <\lambda _{i} < \lambda _{i}^{0}:= (L_{i}K(T+1))^{-1}$$. Since $$\mathcal{T}(w_{1},w_{2},\ldots,w_{n})(t) = (w_{1},w_{2},\ldots,w_{n})(t)$$ for $$t \in [\alpha -k,\alpha +T]_{\mathbb{Z}_{\alpha -k}}$$,

\begin{aligned} \bigl\Vert (w_{1},w_{2}, \ldots,w_{n}) \bigr\Vert &= \bigl\Vert \mathcal{T} (w_{1},w_{2},\ldots,w_{n}) (t) \bigr\Vert =\max \bigl\{ \bigl\Vert \mathcal{T}_{i} (w_{1},w_{2}, \ldots,w_{n}) \bigr\Vert _{{ \mathcal{E}}_{i}} \bigr\} \\ &\leq \lambda _{i}K(T+1)L_{i} \bigl\Vert (w_{1},w_{2},\ldots,w_{n}) \bigr\Vert < \bigl\Vert (w_{1},w_{2},\ldots,w_{n}) \bigr\Vert , \end{aligned}

### Theorem 5.2

Suppose$$\mathrm{(H)}$$holds. If$$f_{i}^{0}>0$$and$$f_{i}^{\infty } >0$$, then there is a real number$$\lambda _{i}^{0}>0$$such that for$$\lambda _{i} >\lambda _{i}^{0}$$the problem (1) does not have a positive solution.

### Proof

Since $$f_{i}^{0}, f_{i}^{\infty }$$ are positive, there exist $$\gamma _{i}^{1},\gamma _{i}^{2}, r_{1},r_{2}>0$$, where $$r_{1} < r_{2}$$, and

\begin{aligned} &f_{i}(y_{1},y_{2}, \ldots,y_{n})\geq \gamma _{i}^{1} y_{i}, \quad\text{for } y_{i}\in [0, r_{1}], \\ &f_{i}(y_{1},y_{2},\ldots,y_{n})\geq \gamma _{i}^{2} y_{i}, \quad\text{for } y_{i}\in [r_{2}, +\infty ). \end{aligned}

Let $$\gamma _{i} = \min \lbrace \gamma _{i}^{1}, \gamma _{i}^{2}, \min_{r_{1}\leq y_{i} \leq r_{2}} \lbrace \frac{f_{i}(y_{1},y_{2},\ldots,y_{n})}{ y_{i}}\rbrace \rbrace >0$$. Then we get

\begin{aligned} f_{i}(y_{1},y_{2}, \ldots,y_{n})\geq \gamma _{i} y_{i}, \quad\text{for } y_{i} \in [0,+\infty ). \end{aligned}

Assume $$(w_{1},w_{2},\ldots,w_{n})$$ is a positive solution of (1). We find a contradiction for $$\lambda _{i} >\lambda _{i}^{0}:= ( \frac{1}{16}\gamma _{i} Kl)^{-1}$$. Since $$\mathcal{T} (w_{1},w_{2},\ldots,w_{n})(t) = (w_{1},w_{2},\ldots,w_{n})(t)$$ for $$t \in [\alpha -k,\alpha +T]_{\mathbb{Z}_{\alpha -k}}$$,

\begin{aligned} \bigl\Vert (w_{1},w_{2}, \ldots,w_{n}) \bigr\Vert &= \bigl\Vert \mathcal{T} (w_{1},w_{2},\ldots,w_{n}) (t) \bigr\Vert =\max \bigl\{ \bigl\Vert \mathcal{T}_{i} (w_{1},w_{2}, \ldots,w_{n}) \bigr\Vert _{{ \mathcal{E}}_{i}} \bigr\} \\ &\geq \lambda _{i}\gamma _{i}\frac{1}{4} \bigl\Vert (w_{1},w_{2},\ldots,w_{n}) \bigr\Vert \frac{1}{4} Kl> \bigl\Vert (w_{1},w_{2}, \ldots,w_{n}) \bigr\Vert , \end{aligned}

## Uniqueness

### Theorem 6.1

Assume that the Banach space$$\mathcal{E}:=\mathcal{E}_{1}\times \mathcal{E}_{2}\times \cdots\times \mathcal{E}_{n}$$is endowed with the norm

\begin{aligned} \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) \bigr\Vert = \max \bigl\{ \Vert y_{1} \Vert _{\mathcal{E}_{1}}, \Vert y_{2} \Vert _{\mathcal{E}_{2}},\ldots, \Vert y_{n} \Vert _{\mathcal{E}_{n}} \bigr\} , \end{aligned}

and$$f_{i}$$satisfying the Lipschitz condition

\begin{aligned} \bigl\Vert \bigl(f_{i}(y_{1},y_{2}, \ldots, y_{n}) - f_{i}(w_{1},w_{2}, \ldots,w_{n}) \bigr) \bigr\Vert \leq L_{i} \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) -(w_{1},w_{2}, \ldots,w_{n}) \bigr\Vert , \end{aligned}
(21)

where$$y_{i} ,w_{i}\in \mathcal{E}_{i} , L_{i}>0$$. Then the problem (1) has exactly one positive solution provided

$$L(T+1) K \bigl\Vert (\lambda _{1},\ldots,\lambda _{n}) \bigr\Vert < 1,$$

where$$L=\max \{L_{i},i=0,1,2,\ldots,n\}$$, $$K=\max G(t,s)$$and$$\Vert (\lambda _{1},\ldots,\lambda _{n}) \Vert =\max \{|\lambda _{i} |,i=0,1,2,\ldots,n\}$$.

### Proof

For any $$(y_{1},y_{2}, \ldots, y_{n}), (w_{1},w_{2},\ldots,w_{n}) \in \mathcal{E}$$, using the assumption (21), we have

\begin{aligned} & \bigl\Vert \bigl(\mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n})-\mathcal{T}_{i}(w_{1},w_{2}, \ldots,w_{n}) \bigr) \bigr\Vert _{\mathcal{E}_{i}} \\ &\quad\leq \lambda _{i}\sum_{s=0}^{T+1} G(\alpha +T,s) \bigl\Vert \bigl(f_{i}(y_{1},y_{2}, \ldots, y_{n}) - f_{i}(w_{1},w_{2}, \ldots,w_{n}) \bigr) \bigr\Vert \\ &\quad\leq L_{i}\lambda _{i} (T+1) K \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) -(w_{1},w_{2}, \ldots,w_{n}) \bigr\Vert \\ &\quad\leq L(T+1) K \bigl\Vert (\lambda _{1},\ldots,\lambda _{n}) \bigr\Vert \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) -(w_{1},w_{2},\ldots,w_{n}) \bigr\Vert , \end{aligned}

$$i=1,2,\ldots ,n$$. That is,

\begin{aligned} &\bigl\Vert \mathcal{T}(y_{1},y_{2}, \ldots,y_{n}) - \mathcal{T}(w_{1},w_{2}, \ldots,w_{n}) \bigr\Vert \\ &\quad= \bigl\Vert \bigl(\mathcal{T}_{1}(y_{1},y_{2}, \ldots, y_{n}),\ldots, \mathcal{T}_{n}(y_{1},y_{2}, \ldots, y_{n}) \bigr) \\ &\qquad{}- \bigl(\mathcal{T}_{1}(w_{1},w_{2}, \ldots,w_{n}),\ldots, \mathcal{T}_{n}(w_{1},w_{2}, \ldots,w_{n}) \bigr) \bigr\Vert \\ &\quad= \bigl\Vert \bigl(\mathcal{T}_{1}(y_{1},y_{2}, \ldots, y_{n})-\mathcal{T}_{1}(w_{1},w_{2}, \ldots,w_{n}) \bigr),\ldots, \\ &\qquad \bigl(\mathcal{T}_{n}(y_{1},y_{2}, \ldots, y_{n})-\mathcal{T}_{n}(w_{1},w_{2}, \ldots,w_{n}) \bigr) \bigr\Vert \\ &\quad= \max \bigl\{ \bigl\Vert \bigl(\mathcal{T}_{i}(y_{1},y_{2}, \ldots, y_{n})-\mathcal{T}_{i}(w_{1},w_{2}, \ldots,w_{n}) \bigr) \bigr\Vert _{\mathcal{E}_{i}}, i=0,1,\ldots,n \bigr\} \\ &\quad\leq L(T+1) K \bigl\Vert (\lambda _{1},\ldots,\lambda _{n}) \bigr\Vert \bigl\Vert (y_{1},y_{2}, \ldots, y_{n}) -(w_{1},w_{2},\ldots,w_{n}) \bigr\Vert . \end{aligned}

Since $$L=\max \{L_{i},i=0,1,2,\ldots,n\}$$ and $$\|(\lambda _{1},\ldots,\lambda _{n})\|=\max \{|\lambda _{i}|,i=0,1,2,\ldots,n \}$$, for $$L(T+1) K \Vert (\lambda _{1},\ldots,\lambda _{n}) \Vert <1$$, the operator $$\mathcal{T}$$ is the contraction mapping. Therefore, the problem (1) has exactly one positive solution. □

## Example

### Example 7.1

Suppose that $$\alpha =1.8, T=10,\lambda _{i}>0$$, and $$a_{i}^{0}>0$$ is an integer and

\begin{aligned} \textstyle\begin{cases} \Delta _{c} ^{\alpha } y_{1}(t)+\lambda _{1}f_{1}(y_{1},y_{2})=0, \qquad y_{1}(-0.2)=a_{1}^{0},\qquad\Delta y_{1}(11.8)=0, \\ \Delta _{c}^{\alpha } y_{2}(t)+\lambda _{2} f_{2}(y_{1},y_{2})=0, \qquad y_{2}(-0.2)=a_{2}^{0},\qquad\Delta y_{2}(11.8)=0, \end{cases}\displaystyle \end{aligned}
(22)

where

\begin{aligned} &f_{1}(y_{1},y_{2})= \frac{(y_{1}^{2}+y_{2}^{2}+y_{1})(4+\cos (y_{1}y_{2}))}{300y_{1}+12y_{2}+1}, \\ &f_{2}(y_{1},y_{2})= \frac{(y_{1}^{2}+y_{2}^{2}+y_{2})(4+\sin (y_{1}y_{2}))}{20y_{1}+200y_{2}+1}. \end{aligned}

Clearly $$f_{i}(y_{1},y_{2}): \mathbb{R}^{+}\times \mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$$ is continuous. Moreover, it is easy to prove that

\begin{aligned} &F_{1}^{\infty }=\limsup_{y_{1}\rightarrow \infty } \frac{f_{1}(y_{1},y_{2})}{y_{1}}=\frac{1}{60},\qquad F_{2}^{\infty }= \limsup _{y_{2}\rightarrow \infty }\frac{f_{2}(y_{1},y_{2})}{y_{2}}= \frac{1}{40}, \\ &f_{1}^{0}=\liminf_{y_{1}\rightarrow 0^{+}} \frac{f_{1}(y_{1},y_{2})}{y_{1}}=4,\qquad f_{2}^{0}=\liminf_{y_{2} \rightarrow 0^{+}} \frac{f_{2}(y_{1},y_{2})}{y_{2}}=4. \end{aligned}

Therefore, from Theorem 4.5, for each

$$\lambda _{1}\in (0.366, 3.4441),\qquad \lambda _{2}\in (0.366, 2.2961),$$

the boundary value problem (22) has at least one positive solution.

### Example 7.2

In this example we focus on the linearized system as follows:

\begin{aligned} \textstyle\begin{cases} \Delta _{c} ^{\alpha } y_{1}(t)+\frac{\lambda _{1}}{1000}y_{1}=0, \qquad y_{1}(-0.02)=a_{1}^{0},\\ \Delta ^{2} y_{1}(-0.02)=\Delta y_{1}(12.98)=0, \\ \Delta _{c} ^{\alpha } y_{2}(t)+\frac{\lambda _{2}}{1000} y_{2}=0, \qquad y_{2}(-0.02)=a_{2}^{0}, \\\Delta ^{2}y_{2}(-0.02)=\Delta y_{2}(12.98)=0, \\ \Delta _{c} ^{\alpha } y_{3}(t)+\frac{\lambda _{3}}{1000}y_{3}=0,\\ y_{3}(-0.02)=a_{3}^{0},\qquad\Delta ^{2} y_{3}(-0.02)=\Delta y_{3}(12.98)=0, \\ \Delta _{c} ^{\alpha } y_{4}(t)+\frac{\lambda _{4}}{1000}y_{4}=0,\\ y_{4}(-0.02)=a_{4}^{0},\qquad\Delta ^{2} y_{4}(-0.02)=\Delta y_{4}(12.98)=0, \end{cases}\displaystyle \end{aligned}
(23)

where $$\alpha =2.98, T=10,\lambda _{i}>0$$, and $$a_{i}^{0}>0$$ is an integer and

\begin{aligned} &f_{1}(y_{1},y_{2},y_{3},y_{4})= \frac{1}{1000}y_{1},\qquad f_{2}(y_{1},y_{2},y_{3},y_{4})= \frac{1}{1000}y_{2}, \\ &f_{3}(y_{1},y_{2},y_{3},y_{4})= \frac{1}{1000}y_{3},\qquad f_{4}(y_{1},y_{2},y_{3},y_{4})= \frac{1}{1000}y_{4}. \end{aligned}

Clearly $$f_{i}(y_{1},y_{2},y_{3},y_{4}): \mathbb{R}^{+}\times \mathbb{R}^{+}\times \mathbb{R}^{+}\times \mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$$ is continuous. Take $$l_{i}^{1}=l_{i}^{2}=l_{i}^{3}=l_{i}^{4}=\frac{1}{1000}$$. It is easy to prove that

\begin{aligned} &F_{1}^{\infty }=\limsup_{y_{1}\rightarrow \infty } \frac{f_{1}(y_{1},y_{2},y_{3},y_{4})}{y_{1}}=\frac{1}{1000},\qquad F_{2}^{ \infty }=\limsup _{y_{2}\rightarrow \infty } \frac{f_{2}(y_{1},y_{2},y_{3},y_{4})}{y_{2}}=\frac{1}{1000}, \\ &F_{3}^{\infty }=\limsup_{y_{3}\rightarrow \infty } \frac{f_{3}(y_{1},y_{2},y_{3},y_{4})}{y_{3}}=\frac{1}{1000},\qquad F_{4}^{ \infty }=\limsup _{y_{4}\rightarrow \infty } \frac{f_{4}(y_{1},y_{2},y_{3},y_{4})}{y_{4}}=\frac{1}{1000}, \\ &F_{1}^{0}=\limsup_{y_{1}\rightarrow 0^{+}} \frac{f_{1}(y_{1},y_{2},y_{3},y_{4})}{y_{1}}=\frac{1}{1000},\qquad F_{2}^{0}= \limsup _{y_{2}\rightarrow 0^{+}} \frac{f_{2}(y_{1},y_{2},y_{3},y_{4})}{y_{2}}=\frac{1}{1000}, \\ &F_{3}^{0}=\limsup_{y_{3}\rightarrow 0^{+}} \frac{f_{3}(y_{1},y_{2},y_{3},y_{4})}{y_{3}}=\frac{1}{1000},\qquad F_{4}^{0}= \limsup _{y_{4}\rightarrow 0^{+}} \frac{f_{4}(y_{1},y_{2},y_{3},y_{4})}{y_{4}}=\frac{1}{1000}, \\ &L_{i}=\max \biggl\{ l_{i}^{1}, l_{i}^{2},l_{i}^{3}, l_{i}^{4},\max_{r_{1} \leq y\leq r_{2}}\frac{f_{i}(y_{1},y_{2},y_{3},y_{4})}{y_{i}} \biggr\} = \frac{1}{1000},\qquad \lambda _{i}^{0}\approx 0.6078. \end{aligned}

Thus, the conditions of Theorem 5.1 are satisfied. Therefore the problem (23) does not have a positive solution for $$0< \lambda _{i}<\lambda _{i}^{0}$$. Moreover, we have

\begin{aligned} &f_{1}^{\infty }=\liminf_{y_{1}\rightarrow \infty } \frac{f_{1}(y_{1},y_{2},y_{3},y_{4})}{y_{1}}=\frac{1}{1000},\qquad f_{2}^{ \infty }=\liminf _{y_{2}\rightarrow \infty } \frac{f_{2}(y_{1},y_{2},y_{3},y_{4})}{y_{2}}=\frac{1}{1000}, \\ &f_{3}^{\infty }=\liminf_{y_{3}\rightarrow \infty } \frac{f_{3}(y_{1},y_{2},y_{3},y_{4})}{y_{3}}=\frac{1}{1000},\qquad f_{4}^{ \infty }=\liminf _{y_{4}\rightarrow \infty } \frac{f_{4}(y_{1},y_{2},y_{3},y_{4})}{y_{4}}=\frac{1}{1000}, \\ &f_{1}^{0}=\liminf_{y_{1}\rightarrow 0^{+}} \frac{f_{1}(y_{1},y_{2},y_{3},y_{4})}{y_{1}}=\frac{1}{1000},\qquad f_{2}^{0}= \liminf _{y_{2}\rightarrow 0^{+}} \frac{f_{2}(y_{1},y_{2},y_{3},y_{4})}{y_{2}}=\frac{1}{1000}, \\ &f_{3}^{0}=\liminf_{y_{3}\rightarrow 0^{+}} \frac{f_{3}(y_{1},y_{2},y_{3},y_{4})}{y_{3}}=\frac{1}{1000},\qquad f_{3}^{0}= \liminf _{y_{3}\rightarrow 0^{+}} \frac{f_{4}(y_{1},y_{2},y_{3},y_{4})}{y_{4}}=\frac{1}{1000}, \\ &l=6.4,\qquad \gamma _{i}=\min \biggl\{ \gamma _{i}^{1}, \gamma _{i}^{2},\gamma _{i}^{3}, \gamma _{i}^{4},\min_{r_{1}\leq y\leq r_{2}} \frac{f_{i}(t,y_{1},y_{2},y_{3},y_{4})}{y_{i}} \biggr\} =\frac{1}{1000},\\ & \lambda _{i}^{0} \approx 16.4842. \end{aligned}

Thus, the conditions of Theorem 5.2 are satisfied. Therefore the problem (23) does not have a positive solution for $$\lambda _{i}>\lambda _{i}^{0}$$.

## Conclusion

In this research we consider a typical system of Caputo fractional difference equations of the form (1). Using the Guo–Krasnosel’skii fixed point theorem, we find a parameter interval for existence and nonexistence of positive solutions dependent on the parameter λ and two examples are given to illustrate the main results.

In this paper we used Caputo discrete fractional operators on the time scale $$\mathbb{Z}$$. It could be interesting to extend this work to the time scale $$h\mathbb{Z}$$. Working on $$h\mathbb{Z}, 0< h<1$$ rather than on $$\mathbb{Z}$$ makes it possible to guarantee the convergence of solutions for a larger class of fractional difference initial value problems [4, 18].

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## Funding

This research has been supported by Sahand University of Technology under the title of “Ph.D. Thesis”.

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### Contributions

The contributions of both authors have been joint discussion, proving the results and providing the numerical examples. All authors read and approved the final manuscript.

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Correspondence to Kazem Ghanbari.

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