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Existence and attractivity of solutions for fractional difference equations

Advances in Difference Equations20182018:191

https://doi.org/10.1186/s13662-018-1637-4

  • Received: 31 January 2018
  • Accepted: 9 May 2018
  • Published:

Abstract

In this paper, we study a kind of difference equations with Riemann–Liouville-like fractional difference. The results on existence and attractivity are obtained by using the Picard iteration method and Schauder’s fixed point theorem. Examples are provided to illustrate the main results.

Keywords

  • Fractional difference equations
  • Existence
  • Attractivity

MSC

  • 39A06
  • 39A13

1 Introduction

In this paper, we study the fractional difference equation
$$ \textstyle\begin{cases} \Delta^{\alpha}_{\alpha-1} u(t)=f(t+\alpha,u(t+\alpha)), \quad t\in\mathbb {N}_{0}, \alpha\in(0,1], \\ \Delta^{\alpha-1}_{\alpha-1} u(t)|_{t={0}}=u_{0}, \end{cases} $$
(1.1)
where \(\Delta^{\alpha}\) is the Riemann–Liouville-like fractional difference, which is defined in later sections, \(f:\mathbb{N}_{\alpha}\times\mathbb{R}\to\mathbb{R}\), \(f(t,u)\) is continuous with respect to t and u, and \(\mathbb{N}_{\alpha}=\{\alpha ,\alpha+1,\alpha+2,\ldots\}\).

In the last decade, fractional differential equations have been recognized as valuable tools to describe many phenomena in various fields of engineering, physics, science, and so on. A huge number of results focused on fractional differential equations; see the monographs of Kilbas et al. [12] and Zhou [17, 18], the papers [1, 10], and the references therein. Within the past ten years, however, there has been more interest in developing discrete fractional equations, that is, fractional difference equations. This development has demonstrated that fractional difference equations have a number of unexpected difficulties and technical complications [26, 9, 1315].

Motivated by the works mentioned, in this paper, we investigate the existence and attractivity of solutions for fractional difference equations. In Sect. 2, we describe the discrete fractional difference calculus and some properties. The main results are obtained in Sect. 3. Using the Picard iteration method, we prove the existence of equation (1.1) in Sect. 3.1. In Sect. 3.2, we also obtain the existence of attractive solutions for fractional difference equations. Section 3.3 is devoted to inducing the existence and attractivity of equation (1.1) by Schauder’s fixed point theorem according to the introduced weighted space. Finally, we provide three examples to illustrate our results.

2 Preliminaries

This section is devoted to some preliminary facts.

Definition 2.1

(see [5, 11])

Let \(\nu>0\). The νth fractional sum is defined by
$$ \Delta^{-\nu}_{a} f(t)=\frac{1}{\Gamma(\nu)}\sum ^{t-\nu}_{s=a}\bigl(t-\sigma (s) \bigr)^{(\nu-1)}f(s), $$
(2.1)
where f is defined for \(s=a\mod(1)\), and \(\Delta^{-\nu}_{a} f\) is defined for \(t=a+\nu\mod(1)\), \(t^{(\nu)}=\frac{\Gamma(t+1)}{\Gamma (t-\nu+1)}\), \(\sigma(s)=s+1\), and Γ is the gamma function. The fractional sum \(\Delta^{-\nu}\) maps functions defined on \(\mathbb{N}_{a}\) to functions defined on \(\mathbb{N}_{a+\nu}\).

Definition 2.2

(see [5, 11])

Let \(\mu>0\) and \(m-1\le\mu\le m\), where m is a positive integer. The μth fractional difference is defined as
$$ \Delta^{\mu}_{a} f(t)=\Delta^{m} \bigl(\Delta^{-(m-\mu)}_{a} f(t) \bigr). $$
(2.2)

Lemma 2.1

(see [4])

Assume that \(\mu+1\) is not a nonpositive integer. Then
$$ \Delta^{-\nu}_{a} t^{(\mu)}= \frac{\Gamma(\mu+1)}{\Gamma(\mu+\nu +1)}t^{(\mu+\nu)}. $$
(2.3)

Lemma 2.2

(see [4])

Assume that the following factorial functions are well defined. Then:
  1. (i)

    \(\Delta t^{(\mu)}=\mu t^{(\mu-1)}\);

     
  2. (ii)

    \((t-\mu)t^{(\mu)}=t^{(\mu+1)}\);

     
  3. (iii)

    \(\mu^{(\mu)}=\Gamma(\mu+1)\);

     
  4. (iv)

    \(t^{(\mu+\nu)}=(t-\nu)^{(\mu)}t^{(\nu)}\).

     
We indicate the following properties of the gamma function (see [7, 12]):
  1. (i)
    for all \(x\in\mathbb{R}\), excluding \(x=0,-1,-2,\ldots \) ,
    $$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}; $$
     
  2. (ii)

    the gamma function is logarithmically convex function on the positive real axis, that is, \(\log\Gamma(x)\) is convex.

     
By the log-convexity property of the gamma function we have that
$$\Gamma\bigl(\lambda x+(1-\lambda)y\bigr)\le\Gamma^{\lambda}( x) \Gamma^{1-\lambda }(y), \quad \lambda\in(0,1),x,y>0. $$

Lemma 2.3

  1. (i)

    If \(0<\alpha<1\), then \((t^{(\nu)} )^{\alpha}\le t^{(\alpha\nu)}\);

     
  2. (ii)
    If \(\alpha>1\) and \(\nu>0\), then
    $$\bigl(t^{(-\nu)} \bigr)^{\alpha}\le\frac{\Gamma(t-[t]+\alpha\nu )}{\Gamma^{\alpha}(t-[t]+\nu)}t^{(-\alpha\nu)} \quad \textit{for }t\in\mathbb{R}^{+}. $$
    Here \([t]\) means the integer part of t.
     

Proof

The proof of (i) is given in [4]. Chen [8] proved (ii) for \(t\in\mathbb{N}_{1}\). We prove (ii) for all \(t\in\mathbb{R}^{+}\):
$$\begin{aligned} &\Gamma^{\alpha-1}(t+1)\Gamma(t+\alpha\nu+1)\Gamma^{\alpha}\bigl(t-[t]+ \nu \bigr) \\ &\quad =t^{\alpha-1}(t-1)^{\alpha-1}\cdots\bigl(t-[t]\bigr)^{\alpha-1}(t+ \alpha\nu ) (t-1+\alpha\nu)\cdots\bigl(t-[t]+\alpha\nu\bigr) \\ &\qquad {}\times\Gamma\bigl(t-[t]+\alpha\nu\bigr)\Gamma^{\alpha} \bigl(t-[t]+\nu\bigr) \\ &\quad = \biggl[t^{\alpha} \biggl(1+\frac{\alpha\nu}{t} \biggr) \biggr] \biggl[(t-1)^{\alpha} \biggl(1+\frac{\alpha\nu}{t-1} \biggr) \biggr] \cdots \biggl[\bigl(t-[t]\bigr)^{\alpha} \biggl(1+\frac{\alpha\nu}{(t-[t])} \biggr) \biggr] \\ &\qquad {}\times\Gamma\bigl(t-[t]+\alpha\nu\bigr)\Gamma^{\alpha} \bigl(t-[t]+\nu\bigr) \\ &\quad < \biggl[t^{\alpha} \biggl(1+\frac{\nu}{t} \biggr)^{\alpha}\biggr] \biggl[(t-1)^{\alpha} \biggl(1+ \frac{\nu}{t-1} \biggr)^{\alpha}\biggr] \cdots \biggl[\bigl(t-[t] \bigr)^{\alpha} \biggl(1+\frac{\nu}{(t-[t])} \biggr)^{\alpha}\biggr] \\ &\qquad {}\times\Gamma\bigl(t-[t]+\alpha\nu\bigr)\Gamma^{\alpha} \bigl(t-[t]+\nu\bigr) \\ &\quad =(t+\nu)^{\alpha}(t-1+\nu)^{\alpha}\cdots\bigl(t-[t]+\nu \bigr)^{\alpha}\Gamma ^{\alpha}\bigl(t-[t]+\nu\bigr)\Gamma\bigl(t-[t]+ \alpha\nu\bigr) \\ &\quad =\Gamma^{\alpha}(t+\nu+1)\Gamma\bigl(t-[t]+\alpha\nu\bigr), \end{aligned}$$
where the inequality \(1+\frac{\alpha\nu}{t}<(1+\frac{\nu}{t})^{\alpha}\) is used. Then
$$ \frac{\Gamma^{\alpha-1}(t+1)}{\Gamma^{\alpha}(t+\nu+1)}\le\frac{\Gamma (t-[t]+\alpha\nu)}{\Gamma^{\alpha}(t-[t]+\nu)} \cdot\frac{1}{\Gamma(t+\alpha\nu+1)}. $$
Hence
$$ \bigl(t^{(-\nu)} \bigr)^{\alpha}=\frac{\Gamma^{\alpha}(t+1)}{\Gamma ^{\alpha}(t+\nu+1)} \le \frac{\Gamma(t-[t]+\alpha\nu)}{\Gamma^{\alpha}(t-[t]+\nu)} \cdot\frac{\Gamma(t+1)}{\Gamma(t+\alpha\nu+1)} =\frac{\Gamma(t-[t]+\alpha\nu)}{\Gamma^{\alpha}(t-[t]+\nu)}t^{(-\alpha \nu)} $$
for \(t\in\mathbb{R}^{+}\). The proof is completed. □

Remark 2.1

In (ii), if \(t-[t]+\alpha\nu\in(0,1)\), then we also have that
$$\bigl(t^{(-\nu)} \bigr)^{\alpha}\le t^{(-\alpha\nu)}. $$

Lemma 2.4

(see [12, (1.5.15)])

The quotient expansion of two gamma functions at infinity is
$$ \frac{\Gamma(z+a)}{\Gamma(z+b)}=z^{a-b} \biggl(1+O \biggl( \frac{1}{z} \biggr) \biggr) \quad \bigl( \bigl\vert \arg(z+a) \bigr\vert < \pi, \vert z \vert \to\infty\bigr). $$
(2.4)
By Lemma 2.4 we can easily get that
$$ t^{(-\nu)}\to0 \quad \mbox{as }t\to\infty\ (\nu>0) $$
(2.5)
and
$$ \frac{t^{(a)}}{t^{(b)}}\to0 \quad \mbox{as }t\to\infty\ (a< b). $$
(2.6)

Lemma 2.5

  1. (i)

    If \(-1< t< r\), then \(t^{(-\nu)}\ge r^{(-\nu)}\) for \(\nu>0\);

     
  2. (ii)

    If \(\nu-1< t< r\), then \(t^{(\nu)}\le r^{(\nu)}\) for \(\nu>0\);

     
  3. (iii)

    If \(\alpha,\beta\ge0\), then \(t^{(\beta)}t^{(-\alpha)}\le t^{(\beta-\alpha)}\) for \(t>\beta-1\).

     

Proof

(i) Since
$$\begin{aligned} \frac{t^{(-\nu)}}{r^{(-\nu)}}&=\frac{\Gamma(t+1)\Gamma(r+\nu+1)}{\Gamma (t+\nu+1)\Gamma(r+1)} \\ &=\frac{\Gamma(t+1)\Gamma(r+\nu+1)}{ \Gamma(\lambda(r+\nu+1)+(1-\lambda)(t+1))\Gamma((1-\lambda)(r+\nu +1)+\lambda(t+1))}, \end{aligned}$$
where \(\lambda=\frac{\nu}{r-t+\nu}\in(0,1)\), by the log-convexity property of the gamma function, we get that
$$\begin{aligned} \frac{t^{(-\nu)}}{r^{(-\nu)}}&=\frac{\Gamma(t+1)\Gamma(r+\nu+1)}{ \Gamma(\lambda(r+\nu+1)+(1-\lambda)(t+1))\Gamma((1-\lambda)(r+\nu +1)+\lambda(t+1))} \\ &\ge\frac{\Gamma(t+1)\Gamma(r+\nu+1)}{(\Gamma(r+\nu+1))^{\lambda +1-\lambda}(\Gamma(t+1))^{1-\lambda+\lambda}} \\ &=1. \end{aligned}$$
Then \(t^{(-\nu)}\ge r^{(-\nu)}\) for \(-1< t< r\) and \(\nu>0\).
(ii) Since
$$ \frac{t^{(\nu)}}{r^{(\nu)}} =\frac{\Gamma(t+1)\Gamma(r-\nu+1)}{\Gamma(t-\nu+1)\Gamma(r+1)}, $$
similarly to (i), we can easily get that \(t^{(\nu)}\le r^{(\nu)}\) for \(\nu-1< t< r\) and \(\nu>0\).
(iii) By (i) we know that
$$t^{(-\alpha)}\le(t-\beta)^{(-\alpha)}\quad \mbox{for } t>\beta-1. $$
According to (iv) of Lemma 2.2, it is easy to get that
$$t^{(\beta)}t^{(-\alpha)}\le t^{(\beta)}(t-\beta)^{(-\alpha)}=t^{(\beta -\alpha)} \quad \mbox{for } t>\beta-1. $$

The proof is completed. □

Remark 2.2

Chen and Liu [9] obtained (i) of Lemma 2.5. However, the proof holds in the case that \(t\to\infty\). Then we give the proof as before.

Lemma 2.6

(see [3])

Let \(a,\nu\in\mathbb{R}\), \(a>\nu>-1\), and \(a\le b\). Then
$$ \sum^{b}_{s=a}s^{(\nu)}= \frac{(b+1)^{(\nu+1)}-a^{(\nu+1)}}{\nu+1}. $$
(2.7)
In particular, if \(a=\nu\), then
$$ \sum^{b}_{s=\nu}s^{(\nu)}= \frac{(b+1)^{(\nu+1)}}{\nu+1}. $$
(2.8)

Remark 2.3

Let \(\nu>0\) be noninteger, and let \(m-1\le\mu\le m\), where m is a positive integer. Set \(\nu=m-\mu\). Then we have
$$ \sum^{t-\mu}_{s=a+\nu}\bigl(t- \sigma(s)\bigr)^{(\mu-1)}=\frac{(t-a-\nu)^{(\mu )}}{\mu}. $$
(2.9)

Lemma 2.7

(see [4])

Let f be a real-value function defined on \(\mathbb{N}_{a}\), and let \(\mu,\nu>0\). Then we have the following equalities:
  1. (i)

    \(\Delta^{-\nu}_{a+\mu} (\Delta^{-\mu}_{a} f(t) )=\Delta^{-(\mu+\nu)}_{a} f(t) =\Delta^{-\mu}_{a+\nu} (\Delta^{-\nu}_{a} f(t) )\);

     
  2. (ii)

    \(\Delta^{-\nu}_{a}\Delta f(t)=\Delta\Delta^{-\nu}_{a} f(t)-\frac{(t-a)^{(\nu-1)}}{\Gamma(\nu)}f(a)\).

     

Definition 2.3

The discrete Mittag-Leffler function is defined by
$$ F_{\alpha,\beta}(\lambda,t)=\sum^{\infty}_{n=0} \lambda^{n} \frac{t^{(n\alpha)}}{\Gamma(n\alpha+\beta)}\quad \bigl( \vert \lambda \vert < 1 \bigr), $$
where \(\alpha,\beta\in\mathbb{R}^{+}\) and \(\lambda\in\mathbb{C}\).
Notice that the series is absolutely convergent for \(|\lambda|<1\). In fact, since
$$\begin{aligned} \biggl\vert \frac{t^{(n\alpha)}}{\Gamma(n\alpha+\beta)} \biggr\vert &= \biggl\vert \frac {\Gamma(t+1)}{\Gamma(t-n\alpha+1)\Gamma(n\alpha+\beta)} \biggr\vert \\ &= \biggl\vert \frac{\Gamma(t+1)\Gamma(n\alpha-t)\sin(\pi(t-n\alpha+1))}{\pi \Gamma(n\alpha+\beta)} \biggr\vert \\ &\le\frac{1}{\pi}\frac{\Gamma(t+1)\Gamma(n\alpha-t)}{\Gamma(n\alpha +\beta)} \\ &\le\frac{1}{\pi}\frac{\Gamma(t+1)(\Gamma(t+1))^{\eta}(\Gamma(n\alpha +\beta))^{1-\eta}}{\Gamma(n\alpha+\beta)} \\ &=\frac{1}{\pi}\frac{(\Gamma(t+1))^{1+\eta}}{(\Gamma(n\alpha+\beta ))^{\eta}} \\ &\le\frac{(\Gamma(t+1))^{1+\eta}}{\pi} \end{aligned}$$
for \(\eta=\frac{t+1}{n\alpha+\beta-t-1}\) and \(n>\frac{2t+1}{\alpha}\), we have
$$ \sum^{\infty}_{n>\frac{2t+1}{\alpha}} \biggl\vert \lambda^{n}\frac{t^{(\alpha )}}{\Gamma(n\alpha+\beta)} \biggr\vert \le \frac{(\Gamma(t+1))^{1+\eta}}{\pi}\sum ^{\infty}_{n>\frac {2t+1}{\alpha}}|\lambda|^{n}. $$
It is easy to see that the series \(\sum^{\infty}_{n=0}\lambda^{n} \frac{t^{(\alpha)}}{\Gamma(n\alpha+\beta)}\) (\(|\lambda|<1\)) is absolutely convergent.

Definition 2.4

A solution u of the fractional difference equation (1.1) is said to be attractive if \(u(t)\to0\) as \(t\to\infty\).

The space \(\ell^{\infty}_{n_{0}}\) is the set of real sequences defined on the set of positive integers where any individual sequence is bounded with respect to the usual supremum norm. It is well known that, under the supremum norm, \(\ell^{\infty}_{n_{0}}\) is a Banach space [16].

Definition 2.5

(see [16])

A set Ω of sequences in \(\ell^{\infty}_{n_{0}}\) is uniformly Cauchy (or equi-Cauchy) if for every \(\varepsilon>0\), there exists an integer N such that \(|x(i)-x(j)|<\varepsilon\) whenever \(i, j >N\) for any \(x=\{x(n)\}\) in Ω.

Theorem 2.1

(see [16, Discrete Arzelà–Ascoli’s theorem])

A bounded, uniformly Cauchy subset Ω of \(\ell^{\infty}_{n_{0}}\) is relatively compact.

Definition 2.6

Let \((X,d)\) be a metric space. An operator \(T:X\to X\) is a Picard operator if there exists \(x^{*}\in X\) such that \(\operatorname{Fix} T=\{x^{*}\}\) and the sequence \(\{T^{n}(x_{0})\}_{n\in\mathbb{N}}\) converges to \(x^{*}\) for all \(x_{0}\in X\).

Theorem 2.2

(Schauder’s fixed point theorem)

Let Ω be a closed, convex, and nonempty subset of a Banach space X. Let \(T : \Omega\to\Omega\) be a continuous operator such that TΩ is a relatively compact subset of X. Then T has at least one fixed point in Ω.

According to Definitions 2.12.2, it is suitable to rewrite the fractional difference equations (1.1) in the equivalent summation equation
$$ u(t)=\frac{u_{0}}{\Gamma(\alpha)}t^{(\alpha-1)}+\frac{1}{\Gamma(\alpha)} \sum ^{t-\alpha}_{s=0}\bigl(t-\sigma(s) \bigr)^{(\alpha-1)}f\bigl(s+\alpha,u(s+\alpha )\bigr),\quad t\in \mathbb{N}_{\alpha}. $$
(2.10)

3 Main results

3.1 Results via Picard iteration method

Theorem 3.1

Assume that
(H1): 
\(f(t,u)\) satisfies the Lipschitz condition
$$ \bigl\vert f(t,u_{1})-f(t,u_{2}) \bigr\vert \le A \vert u_{1}-u_{2} \vert , $$
(3.1)
where \(0< A<1\) is independent of t;
(H2): 

there exists a constant \(M>0\) such that \(|f(t,u(t))|\le M\) for \(t\in\mathbb{N}_{\alpha}\).

Then the fractional difference equation (1.1) has at least one solution.

Proof

Define the sequence \(\{g_{n}(\cdot): n\in\mathbb{N}_{0}\}\) as follows:
$$\begin{aligned} &g_{0}(t)=\frac{u_{0}}{\Gamma(\alpha)}t^{(\alpha-1)}, \quad t\in\mathbb {N}_{\alpha}, \end{aligned}$$
(3.2)
$$\begin{aligned} &g_{n}(t)=g_{0}(t)+\frac{1}{\Gamma(\alpha)}\sum ^{t-\alpha}_{s=0}\bigl(t-\sigma (s)\bigr)^{(\alpha-1)} f \bigl(s+\alpha,g_{n-1}(s+\alpha)\bigr),\quad t\in\mathbb{N}_{\alpha}, n\in\mathbb{N}_{0}. \end{aligned}$$
(3.3)
Clearly, by induction we have
$$ \bigl\vert g_{n}(t)-g_{n-1}(t) \bigr\vert \le MA^{n-1}\frac{t^{(n\alpha)}}{\Gamma(n\alpha+1)}. $$
In fact, for \(n=1\), by condition (H1) we can conclude that
$$\begin{aligned} \begin{aligned} \bigl\vert g_{1}(t)-g_{0}(t) \bigr\vert &\le \frac{1}{\Gamma(\alpha)} \sum^{t-\alpha}_{s=0}\bigl(t- \sigma(s)\bigr)^{(\alpha-1)}M \\ &=M\frac{t^{(\alpha)}}{\Gamma(\alpha+1)}. \end{aligned} \end{aligned}$$
Without loss of generality, we assume that
$$ \bigl\vert g_{n-1}(t)-g_{n-2}(t) \bigr\vert \le MA^{n-2}\frac{t^{((n-1)\alpha)}}{\Gamma ((n-1)\alpha+1)}. $$
Then
$$\begin{aligned} \bigl\vert g_{n}(t)-g_{n-1}(t) \bigr\vert &\le A \frac{1}{\Gamma(\alpha)}\sum^{t-\alpha }_{s=0}\bigl(t- \sigma(s)\bigr)^{(\alpha-1)} {MA^{n-2}}\frac{(s+\alpha)^{((n-1)\alpha)}}{\Gamma((n-1)\alpha+1)} \\ &=\frac{MA^{n-1}}{\Gamma((n-1)\alpha+1)}\Delta^{-\alpha}(t+\alpha )^{((n-1)\alpha)} \\ &={MA^{n-1}}\frac{t^{(n\alpha)}}{\Gamma(n\alpha+1)}. \end{aligned}$$
Set
$$ g(t)=\lim_{n\to\infty}g_{n}(t)=\lim_{n\to\infty } \bigl(g_{n}(t)-g_{0}(t)\bigr)+g_{0}(t)=\sum ^{\infty}_{k=1}\bigl(g_{k}(t)-g_{k-1}(t) \bigr)+g_{0}(t). $$
Since the series \(\frac{M}{A}\sum^{\infty}_{k=1}{A^{k}}\frac{t^{(k\alpha )}}{\Gamma(k\alpha+1)}=\frac{M}{A}F_{\alpha,\beta}(A,t)\) is absolutely convergent for \(0< A<1\), the existence of the solution for the fractional difference equation (1.1) is proved. The proof is completed. □

3.2 Results via Schauder’s fixed point theorem

In this section, we deal with the existence and attractivity of the solution for fractional difference equations by a fixed point theorem. First, for any \(u\in\ell^{\infty}_{\alpha}\), we define the operator T as follows:
$$ Tu(t)=\frac{u_{0}}{\Gamma(\alpha)}t^{(\alpha-1)}+\frac{1}{\Gamma(\alpha)} \sum ^{t-\alpha}_{s=0}\bigl(t-\sigma(s) \bigr)^{(\alpha-1)}f\bigl(s+\alpha,u(s+\alpha )\bigr),\quad t\in \mathbb{N}_{\alpha}. $$
(3.4)
Then the existence of the solution of the fractional difference equation (1.1) is equivalent to that T has a fixed point.

Theorem 3.2

Assume that
(H3): 
there exist constants \(\beta_{1}\in(\alpha,1)\), \(\delta>0\), and \(L_{1}\ge0\) such that
$$ \bigl\vert f(t,u) \bigr\vert \le L_{1}(t+1)^{(-\beta_{1})} \vert u \vert ^{\delta}\quad \textit{for }t\in\mathbb {N}_{\alpha} \textit{ and }u\in\mathbb{R}. $$
(3.5)
Furthermore, suppose that
$$ \frac{|u_{0}|}{\Gamma(\alpha)}\alpha^{(\alpha-1)} +\frac{L_{1}\Gamma(1-\beta_{1})}{\Gamma(1+\alpha-\beta_{1})} \alpha^{(\alpha-\beta_{1})}\le1. $$
Then the fractional difference equation (1.1) has at least one solution in \(S_{1}\). Further, the solutions of (1.1) are attractive.

Proof

Choose \(\gamma>0\) sufficiently small such that
$$ \alpha+\gamma-1< 0,\qquad 1-\beta_{1}-\gamma\delta>0,\qquad \alpha+ \gamma-\beta _{1}-\gamma\delta< 0, $$
and
$$\frac{|u_{0}|}{\Gamma(\alpha)}(\alpha+\gamma)^{(\alpha+\gamma-1)} +\frac{L_{1}\Gamma(1-\beta_{1}-\gamma\delta)}{\Gamma(1+\alpha-\beta _{1}-\gamma\delta)} (\alpha+ \gamma)^{(\alpha-\beta_{1}+\gamma-\gamma\delta)}\le1. $$
For \(t\in\mathbb{N}_{\alpha}\), define the closed subset \(S_{1}\subset\ell^{\infty}_{\alpha}\) as follows:
$$ S_{1}= \bigl\{ u\in\ell^{\infty}_{\alpha}: \bigl\vert u(t) \bigr\vert \le t^{(-\gamma)}\mbox{ for }t\in\mathbb{N}_{\alpha} \bigr\} . $$
It is easy to see that \(S_{1}\) is a closed, bounded, and convex subset of the Banach space \(\ell^{\infty}_{\alpha}\).
In the case \(\delta>1\), we can get the following inequality by (ii) of Lemma 2.3 and Remark 2.1:
$$ \bigl(t^{(-\gamma)} \bigr)^{\delta}\le\frac{\Gamma(\alpha+\gamma\delta)}{ \Gamma^{\delta}(\alpha+\gamma)}t^{(-\gamma\delta)} \le t^{(-\gamma \delta)}\quad \mbox{for }t\in\mathbb{N}_{\alpha}. $$
Combining this with (i) of Lemma 2.3, we have
$$ \bigl(t^{(-\gamma)} \bigr)^{\delta}\le t^{(-\gamma\delta)} \quad \mbox{for }\delta>0, t\in\mathbb{N}_{\alpha}. $$
(3.6)
We first show that T is continuous in \(S_{1}\) and maps \(S_{1}\) into \(S_{1}\). Applying (H3), (3.4), and (3.6) to any \(u\in S_{1}\), we have
$$\begin{aligned} \bigl\vert Tu(t) \bigr\vert \le&\frac{ \vert u_{0} \vert }{\Gamma(\alpha)}t^{(\alpha-1)}+ \frac{1}{\Gamma (\alpha)} \sum^{t-\alpha}_{s=0}\bigl(t- \sigma(s)\bigr)^{(\alpha-1)} \bigl\vert f\bigl(s+\alpha,u(s+\alpha )\bigr) \bigr\vert \\ \le&\frac{ \vert u_{0} \vert }{\Gamma(\alpha)}t^{(\alpha-1)}+\frac{1}{\Gamma(\alpha)} \sum ^{t-\alpha}_{s=0}\bigl(t-\sigma(s)\bigr)^{(\alpha-1)}L_{1}(s+ \alpha+1)^{(-\beta _{1})} \bigl\vert u(s+\alpha) \bigr\vert ^{\delta} \\ \le&\frac{ \vert u_{0} \vert }{\Gamma(\alpha)}t^{(\alpha-1)}+\frac{1}{\Gamma(\alpha)} \sum ^{t-\alpha}_{s=0}\bigl(t-\sigma(s)\bigr)^{(\alpha-1)}L_{1}(s+ \alpha+\gamma \delta)^{(-\beta_{1})} \bigl\vert u(s+\alpha) \bigr\vert ^{\delta} \\ \le&\frac{ \vert u_{0} \vert }{\Gamma(\alpha)}t^{(\alpha-1)}+\frac{L_{1}}{\Gamma(\alpha)} \sum ^{t-\alpha}_{s=0}\bigl(t-\sigma(s)\bigr)^{(\alpha-1)}(s+ \alpha+\gamma\delta )^{(-\beta_{1})} \bigl((s+\alpha)^{(-\gamma)} \bigr)^{\delta} \\ \le&\frac{ \vert u_{0} \vert }{\Gamma(\alpha)}t^{(\alpha-1)}+\frac{L_{1}}{\Gamma(\alpha)} \sum ^{t-\alpha}_{s=0}\bigl(t-\sigma(s)\bigr)^{(\alpha-1)}(s+ \alpha+\gamma\delta )^{(-\beta_{1})}(s+\alpha)^{(-\gamma\delta)} \\ \le&\frac{ \vert u_{0} \vert }{\Gamma(\alpha)}t^{(\alpha-1)}+\frac{L_{1}}{\Gamma(\alpha)} \sum ^{t-\alpha}_{s=0}\bigl(t-\sigma(s)\bigr)^{(\alpha-1)}(s+ \alpha)^{(-\beta _{1}-\gamma\delta)} \\ =&\frac{ \vert u_{0} \vert }{\Gamma(\alpha)}t^{(\alpha-1)}+\frac{L_{1}\Gamma(1-\beta _{1}-\gamma\delta)}{\Gamma(1+\alpha-\beta_{1}-\gamma\delta)}(t+\alpha )^{(\alpha-\beta_{1}-\gamma\delta)} \\ \le&\frac{ \vert u_{0} \vert }{\Gamma(\alpha)}t^{(\alpha-1)}+\frac{L_{1}\Gamma(1-\beta _{1}-\gamma\delta)}{\Gamma(1+\alpha-\beta_{1}-\gamma\delta)}t^{(\alpha-\beta _{1}-\gamma\delta)} \\ =& \biggl(\frac{ \vert u_{0} \vert }{\Gamma(\alpha)}(t+\gamma)^{(\alpha+\gamma -1)}+\frac{L_{1}\Gamma(1-\beta_{1}-\gamma\delta)}{\Gamma(1+\alpha-\beta _{1}-\gamma\delta)} (t+ \gamma)^{(\alpha-\beta_{1}-\gamma\delta+\gamma)} \biggr)t^{(-\gamma)} \\ \le& \biggl(\frac{ \vert u_{0} \vert }{\Gamma(\alpha)}(\alpha+\gamma)^{(\alpha+\gamma -1)} \\ &{}+\frac{L_{1}\Gamma(1-\beta_{1}-\gamma\delta)}{\Gamma(1+\alpha-\beta _{1}-\gamma\delta)} (\alpha+\gamma)^{(\alpha-\beta_{1}-\gamma\delta+\gamma)} \biggr)t^{(-\gamma )} \\ \le&t^{(-\gamma)}, \end{aligned}$$
which implies that T maps \(S_{1}\) into \(S_{1}\).
By (2.5) it is easy to see that for all \(\varepsilon>0\), there exists \(K_{1}\in\mathbb{N}_{+}\) large enough such that
$$ \frac{L_{1}\Gamma(1-\beta_{1}-\gamma\delta)}{\Gamma(1+\alpha-\beta_{1}-\gamma \delta)}t^{(\alpha-\beta_{1}-\gamma\delta)}\le\frac{\varepsilon}{2}\quad \mbox{and}\quad \frac{u_{0}}{\Gamma(\alpha)}t^{(\alpha-1)}\le\frac{\varepsilon}{2}\quad \mbox{for } t\in \mathbb{N}_{\alpha+K_{1}}. $$
(3.7)
Let \(u_{n},u\in S_{1}\), \(n=1,2,\ldots \) , and \(\lim_{n\to\infty}u_{n}=u\). Applying (2.9), for \(t\in\{\alpha,\alpha+1,\ldots,\alpha+K_{1}-1\} \), we have
$$\begin{aligned} & \bigl\vert Tu_{n}(t)-Tu(t) \bigr\vert \\ &\quad \le\frac{1}{\Gamma(\alpha)}\sum^{t-\alpha}_{s=0} \bigl(t-\sigma(s)\bigr)^{(\alpha-1)} \bigl\vert f\bigl(s+\alpha,u_{n}(s+ \alpha)\bigr)-f\bigl(s+\alpha,u(s+\alpha)\bigr) \bigr\vert \\ &\quad \le\frac{1}{\Gamma(\alpha)}\sum^{t-\alpha}_{s=0} \bigl(t-\sigma(s)\bigr)^{(\alpha-1)} \max_{s\in[0,1,\ldots,K_{1}-1]} \bigl\vert f\bigl(s+\alpha,u_{n}(s+\alpha)\bigr)-f\bigl(s+\alpha ,u(s+\alpha) \bigr) \bigr\vert \\ &\quad =\frac{t^{(\alpha)}}{\Gamma(\alpha+1)} \max_{s\in[0,1,\ldots,K_{1}-1]} \bigl\vert f\bigl(s+ \alpha,u_{n}(s+\alpha)\bigr)-f\bigl(s+\alpha,u(s+\alpha)\bigr) \bigr\vert \\ &\quad \le\frac{(\alpha+K_{1}-1)^{(\alpha)}}{\Gamma(\alpha+1)} \max_{s\in [0,1,\ldots,K_{1}-1]} \bigl\vert f \bigl(s+\alpha,u_{n}(s+\alpha)\bigr)-f\bigl(s+\alpha,u(s+\alpha)\bigr) \bigr\vert \\ &\quad \to 0\quad \mbox{as }n\to\infty. \end{aligned}$$
For \(t\in\mathbb{N}_{\alpha+K_{1}}\), by (3.6) we also have
$$\begin{aligned}& \bigl\vert Tu_{n}(t)-Tu(t) \bigr\vert \\& \quad \le \frac{1}{\Gamma(\alpha)}\sum^{t-\alpha}_{s=0} \bigl(t-\sigma(s)\bigr)^{(\alpha-1)} \bigl( \bigl\vert f\bigl(s+ \alpha,u_{n}(s+\alpha)\bigr) \bigr\vert + \bigl\vert f\bigl(s+ \alpha,u(s+\alpha)\bigr) \bigr\vert \bigr) \\& \quad \le \frac{2L_{1}}{\Gamma(\alpha)}\sum^{t-\alpha}_{s=0} \bigl(t-\sigma (s)\bigr)^{(\alpha-1)}(s+\alpha+\gamma\delta)^{(-\beta_{1})} \bigl\vert u(s+\alpha ) \bigr\vert ^{\delta} \\& \quad \le \frac{2L_{1}}{\Gamma(\alpha)}\sum^{t-\alpha}_{s=0} \bigl(t-\sigma (s)\bigr)^{(\alpha-1)}(s+\alpha+\gamma\delta)^{(-\beta_{1})} \bigl((s+\alpha )^{(-\gamma)} \bigr)^{\delta} \\& \quad \le \frac{2L_{1}}{\Gamma(\alpha)}\sum^{t-\alpha}_{s=0} \bigl(t-\sigma (s)\bigr)^{(\alpha-1)}(s+\alpha+\gamma\delta)^{(-\beta_{1})}(s+ \alpha )^{(-\gamma\delta)} \\& \quad \le \frac{2L_{1}}{\Gamma(\alpha)}\sum^{t-\alpha}_{s=0} \bigl(t-\sigma (s)\bigr)^{(\alpha-1)}(s+\alpha)^{(-\beta_{1}-\gamma\delta)} \\& \quad = \frac{2L_{1}\Gamma(1-\beta_{1}-\gamma\delta)}{\Gamma(1+\alpha-\beta _{1}-\gamma\delta)}(t+\alpha)^{(\alpha-\beta_{1}-\gamma\delta)} \\& \quad \le \frac{2L_{1}\Gamma(1-\beta_{1}-\gamma\delta)}{\Gamma(1+\alpha-\beta _{1}-\gamma\delta)}t^{(\alpha-\beta_{1}-\gamma\delta)} \\& \quad \le \varepsilon. \end{aligned}$$
Hence we can get that
$$ \|Tu_{n}-Tu\|=\sup_{t\in\mathbb{N}_{\alpha}} \bigl\vert Tu_{n}(t)-Tu(t) \bigr\vert \to0\quad \mbox{as }n\to\infty. $$
Thus T is continuous in \(S_{1}\).
In the following, we prove that \(TS_{1}\) is relatively compact. For \(t_{1},t_{2}\in\mathbb{N}_{\alpha+K_{1}}\), we have
$$\begin{aligned} \bigl\vert Tu(t_{2})-Tu(t_{1}) \bigr\vert \le& \biggl\vert \frac{u_{0}}{\Gamma(\alpha)}t^{(\alpha -1)}_{2}-\frac{u_{0}}{\Gamma(\alpha)}t^{(\alpha-1)}_{1} \biggr\vert \\ &{}+ \Biggl\vert \frac{1}{\Gamma(\alpha)}\sum^{t_{2}-\alpha}_{s=0} \bigl(t_{2}-\sigma (s)\bigr)^{(\alpha-1)} f\bigl(s+\alpha,u(s+\alpha) \bigr) \\ &{}-\frac{1}{\Gamma(\alpha)}\sum^{t_{1}-\alpha}_{s=0} \bigl(t_{1}-\sigma (s)\bigr)^{(\alpha-1)} f\bigl(s+\alpha,u(s+\alpha) \bigr) \Biggr\vert \\ \le&\frac{u_{0}}{\Gamma(\alpha)}t^{(\alpha-1)}_{2}+\frac{u_{0}}{\Gamma(\alpha )}t^{(\alpha-1)}_{1} \\ &{}+ \Biggl\vert \frac{1}{\Gamma(\alpha)}\sum^{t_{2}-\alpha}_{s=0} \bigl(t_{2}-\sigma (s)\bigr)^{(\alpha-1)} f\bigl(s+\alpha,u(s+\alpha) \bigr) \Biggr\vert \\ &{}+ \Biggl\vert \frac{1}{\Gamma(\alpha)}\sum^{t_{1}-\alpha}_{s=0} \bigl(t_{1}-\sigma (s)\bigr)^{(\alpha-1)} f\bigl(s+\alpha,u(s+\alpha) \bigr) \Biggr\vert \\ \le&\frac{u_{0}}{\Gamma(\alpha)}t^{(\alpha-1)}_{2}+\frac{u_{0}}{\Gamma(\alpha )}t^{(\alpha-1)}_{1} +\frac{L_{1}\Gamma(1-\beta_{1}-\gamma\delta)}{\Gamma(1+\alpha-\beta _{1}-\gamma\delta)}t_{2}^{(\alpha-\beta_{1}-\gamma\delta)} \\ &{}+\frac{L_{1}\Gamma(1-\beta_{1}-\gamma\delta)}{\Gamma(1+\alpha-\beta _{1}-\gamma\delta)}t_{1}^{(\alpha-\beta_{1}-\gamma\delta)} \\ \le&2\varepsilon. \end{aligned}$$
Therefore \(\{Tu:u\in S_{1}\}\) is a bounded and uniformly Cauchy subset by Definition 2.5. Hence \(TS_{1}\) is relatively compact by Theorem 2.1. Due to Schauder’s fixed point theorem, T has a fixed point. All functions in \(S_{1}\) tend to 0 as \(t\to\infty\), and hence the solution of (1.1) is attractive. The proof is completed. □

3.3 Results in a weighted space

Let \(\ell^{\infty}_{\alpha}\) be the set of all real sequences \(u=\{u(t)\}^{\infty}_{t=\alpha}\) with the norm \(\|u\|=\sup_{t\in\mathbb{N}_{\alpha}}|u(t)|\). Then \((\ell^{\infty}_{\alpha},\|\cdot\|)\) is a Banach space. For any \(u\in\ell^{\infty}_{\alpha}\), we introduce a new norm as follows:
$$ \|u\|_{\beta_{2}}=\sup_{t\in\mathbb{N}_{\alpha}} \biggl\{ \frac {|u(t)|}{t^{(\beta_{2})}} \biggr\} , $$
where \(\beta_{2}\in(\alpha,1)\). It is easy to see that \((\ell^{\infty }_{\alpha},\|\cdot\|_{\beta_{2}})\) is also a Banach space.
Define the closed subset
$$S_{\beta_{2}}= \biggl\{ u\in\ell^{\infty}_{\alpha}:\|u \|_{\beta_{2}}\le \frac{(\alpha|u_{0}|+M)\Gamma(\alpha+1-\beta_{2})}{\Gamma(\alpha+1)} \biggr\} . $$
Obviously, \(S_{\beta_{2}}\) is a bounded, closed, and convex subset of the Banach space \((\ell^{\infty}_{\alpha},\|\cdot\|_{\beta_{2}})\).

Definition 3.1

A function \(f:[0,\infty)\times\mathbb{R}\to\mathbb{R}\) is \(\beta_{2}\)-continuous if for any \(\varepsilon>0\), there exists \(\delta=\delta_{\varepsilon}>0\) such that for all \((t,u_{1}),(t,u_{2})\in[0,\infty)\times\mathbb{R}\), we have
$$ \bigl\vert f(t,u_{1})-f(t,u_{2}) \bigr\vert < \varepsilon, $$
provided that \(\frac{|u_{1}-u_{2}|}{t^{(\beta_{2})}}<\delta\).

Obviously, u is a solution of (1.1) if and only if the operator T has a fixed point.

Theorem 3.3

Assume that (H2) holds and f is \(\beta_{2}\)-continuous. Then the fractional difference equation (1.1) has at least one solution in \(S_{\beta_{2}}\). Further, the solutions of (1.1) are attractive.

Proof

We first show that T is continuous in \(S_{\beta_{2}}\) and maps \(S_{\beta_{2}}\) into \(S_{\beta_{2}}\). Applying (3.4), for any \(u\in S_{\beta_{2}}\), we have
$$\begin{aligned} \frac{|Tu(t)|}{t^{(\beta_{2})}} \le&\frac{|u_{0}|t^{(\alpha-1)}}{\Gamma (\alpha)t^{(\beta_{2})}}+\frac{1}{\Gamma(\alpha)t^{(\beta_{2})}} \sum ^{t-\alpha}_{s=0}\bigl(t-\sigma(s)\bigr)^{(\alpha-1)} \bigl\vert f\bigl(s+\alpha,u(s+\alpha )\bigr) \bigr\vert \\ \le&\frac{|u_{0}|t^{(\alpha-1)}}{\Gamma(\alpha)t^{(\beta_{2})}} +\frac{M}{\Gamma(\alpha)t^{(\beta_{2})}}\sum^{t-\alpha}_{s=0} \bigl(t-\sigma (s)\bigr)^{(\alpha-1)} \\ =&\frac{|u_{0}|t^{(\alpha-1)}}{\Gamma(\alpha)t^{(\beta_{2})}}+\frac {Mt^{(\alpha)}}{\Gamma(\alpha+1)t^{(\beta_{2})}} \\ =&\frac{|u_{0}|\Gamma(t-\beta_{2}+1)}{\Gamma(\alpha)\Gamma(t-\alpha +2)}+\frac{M\Gamma(t-\beta_{2}+1)}{\Gamma(\alpha+1)\Gamma(t-\alpha+1)} \\ \le&\frac{(\alpha|u_{0}|+M)\Gamma(\alpha+1-\beta_{2})}{\Gamma(\alpha+1)}, \end{aligned}$$
which implies that T maps \(S_{\beta_{2}}\) into \(S_{\beta_{2}}\).
Let \(u_{n},u\in S_{\beta_{2}}\), \(n=1,2,\ldots \) , and \(\lim_{n\to\infty}u_{n}=u\), which means that \(\|u_{n}-u\|_{\beta_{2}}\to0\) as \(n\to\infty\), that is,
$$ \sup_{t\in\mathbb{N}_{\alpha}}\frac{|u_{n}(t)-u(t)|}{t^{(\beta_{2})}}\to 0 \quad \mbox{as }n\to\infty. $$
Combining this with (2.9), we have
$$\begin{aligned}& \frac{|Tu_{n}(t)-Tu(t)|}{t^{(\beta_{2})}} \\& \quad \le \frac{1}{\Gamma(\alpha)t^{(\beta_{2})}}\sum^{t-\alpha }_{s=0} \bigl(t-\sigma(s)\bigr)^{(\alpha-1)} \bigl\vert f\bigl(s+\alpha,u_{n}(s+ \alpha)\bigr)-f\bigl(s+\alpha,u(s+\alpha)\bigr) \bigr\vert \\& \quad \le \frac{\varepsilon}{\Gamma(\alpha)t^{(\beta_{2})}}\sum^{t-\alpha }_{s=0} \bigl(t-\sigma(s)\bigr)^{(\alpha-1)} \\& \quad = \frac{\varepsilon t^{(\alpha)}}{\Gamma(\alpha+1)t^{(\beta_{2})}} \\& \quad \le \varepsilon. \end{aligned}$$
Hence we get that
$$ \|Tu_{n}-Tu\|_{\beta_{2}}=\sup_{t\in\mathbb{N}_{\alpha}} \biggl\{ \frac {|Tu_{n}(t)-Tu(t)|}{t^{(\beta_{2})}} \biggr\} \to0\quad \mbox{as }n\to\infty, $$
and thus T is continuous in \(S_{\beta_{2}}\).
In the following, we prove that \(TS_{\beta_{2}}\) is relatively compact. To this end, we define
$$ Hu(t)=\frac{Tu(t)}{t^{(\beta_{2})}} \quad \mbox{for }t\in\mathbb{N}_{\alpha }, u\in S_{\beta_{2}}. $$
According to this process, it is easy to see that \(\{Hu(\cdot):u\in S_{\beta_{2}}\}\) is uniformly bounded. Next, we prove that \(\{Hu(\cdot):u\in S_{\beta_{2}}\}\) is uniformly Cauchy. For \(t_{1},t_{2}\in\mathbb{N}_{\alpha}\) such that \(t_{1}< t_{2}\), we have
$$\begin{aligned}& \bigl\vert Hu(t_{2})-Hu(t_{1}) \bigr\vert \\& \quad = \biggl\vert \frac{Tu(t_{2})}{t^{(\beta_{2})}_{2}}-\frac {Tu(t_{1})}{t^{(\beta_{2})}_{1}} \biggr\vert \\& \quad \le \biggl\vert \frac{u_{0}t^{(\alpha-1)}_{2}}{\Gamma(\alpha)t^{(\beta _{2})}_{2}}-\frac{u_{0} t^{(\alpha-1)}_{1}}{\Gamma(\alpha)t^{(\beta _{2})}_{1}} \biggr\vert + \Biggl\vert \frac{1}{\Gamma(\alpha)t^{(\beta_{2})}_{2}}\sum^{t_{2}-\alpha}_{s=0} \bigl(t_{2}-\sigma(s)\bigr)^{(\alpha-1)} f\bigl(s+\alpha,u(s+\alpha) \bigr) \\& \qquad {} -\frac{1}{\Gamma(\alpha)t^{(\beta_{2})}_{1}}\sum^{t_{1}-\alpha }_{s=0} \bigl(t_{1}-\sigma(s)\bigr)^{(\alpha-1)} f\bigl(s+\alpha,u(s+\alpha) \bigr) \Biggr\vert \\& \quad \le \frac{ \vert u_{0} \vert }{\Gamma(\alpha)}\biggl(\frac{\Gamma(t_{2}-\beta _{2}+1)}{\Gamma(t_{2}-\alpha+2)} +\frac{\Gamma(t_{1}-\beta_{2}+1)}{\Gamma(t_{1}-\alpha+2)}\biggr) \\& \qquad {} + \Biggl\vert \frac{1}{\Gamma(\alpha)t^{(\beta_{2})}_{2}}\sum^{t_{2}-\alpha }_{s=0} \bigl(t_{2}-\sigma(s)\bigr)^{(\alpha-1)} f\bigl(s+\alpha,u(s+\alpha) \bigr) \Biggr\vert \\& \qquad {} + \Biggl\vert \frac{1}{\Gamma(\alpha)t^{(\beta_{2})}_{1}}\sum^{t_{1}-\alpha }_{s=0} \bigl(t_{1}-\sigma(s)\bigr)^{(\alpha-1)} f\bigl(s+\alpha,u(s+\alpha) \bigr) \Biggr\vert \\& \quad \le \frac{ \vert u_{0} \vert }{\Gamma(\alpha)}\biggl(\frac{\Gamma(t_{2}-\beta _{2}+1)}{\Gamma(t_{2}-\alpha+2)} +\frac{\Gamma(t_{1}-\beta_{2}+1)}{\Gamma(t_{1}-\alpha+2)}\biggr) \\& \qquad {} +\frac{M}{\Gamma(\alpha)t^{(\beta_{2})}_{2}}\sum^{t_{2}-\alpha }_{s=0} \bigl(t_{2}-\sigma(s)\bigr)^{(\alpha-1)} +\frac{M}{\Gamma(\alpha)t^{(\beta_{2})}_{1}}\sum ^{t_{1}-\alpha }_{s=0}\bigl(t_{1}-\sigma(s) \bigr)^{(\alpha-1)} \\& \quad \le \frac{ \vert u_{0} \vert }{\Gamma(\alpha)}\biggl(\frac{\Gamma(t_{2}-\beta _{2}+1)}{\Gamma(t_{2}-\alpha+2)} +\frac{\Gamma(t_{1}-\beta_{2}+1)}{\Gamma(t_{1}-\alpha+2)} \biggr)+\frac {Mt^{(\alpha)}_{2}}{\Gamma(\alpha+1)t^{(\beta_{2})}_{2}} +\frac{Mt^{(\alpha)}_{1}}{\Gamma(\alpha+1)t^{(\beta_{2})}_{1}} \\& \quad \le \frac{ \vert u_{0} \vert }{\Gamma(\alpha)}\biggl(\frac{\Gamma(t_{2}-\beta _{2}+1)}{\Gamma(t_{2}-\alpha+2)} +\frac{\Gamma(t_{1}-\beta_{2}+1)}{\Gamma(t_{1}-\alpha+2)} \biggr) \\& \qquad {}+\frac {M}{\Gamma(\alpha+1)}\biggl(\frac{\Gamma(t_{2}-\beta_{2}+1)}{\Gamma (t_{2}-\alpha+1)} +\frac{\Gamma(t_{1}-\beta_{2}+1)}{\Gamma(t_{1}-\alpha+1)}\biggr) \\& \quad = \frac{ \vert u_{0} \vert }{\Gamma(\alpha)}\biggl[t_{2}^{\alpha-\beta_{2}-1} \biggl(1+O \biggl(\frac{1}{t_{2}}\biggr)\biggr) +t_{1}^{\alpha-\beta_{2}-1} \biggl(1+O\biggl(\frac{1}{t_{1}}\biggr)\biggr) \biggr] \\& \qquad {} +\frac{M}{\Gamma(\alpha+1)}\biggl[t_{2}^{\alpha-\beta_{2}}\biggl(1+O \biggl(\frac{1}{t_{2}}\biggr)\biggr) +t_{1}^{\alpha-\beta_{2}} \biggl(1+O\biggl(\frac{1}{t_{1}}\biggr)\biggr)\biggr] \\& \quad \to 0\quad \mbox{as }t_{1}\to\infty,t_{2}\to\infty. \end{aligned}$$
Therefore, \(\{Hu:u\in S_{\beta_{2}}\}\) is a bounded and uniformly Cauchy subset by Definition 2.5. Thus \(\{\frac{Tu(t)}{t^{(\beta_{2})}}:u\in S_{\beta_{2}}\}\) is relatively compact by Theorem 2.1. Hence \(T(S_{\beta_{2}})\) is relatively compact in \(S_{\beta_{2}}\). Due to Schauder’s fixed point theorem, T has a fixed point. All functions in \(S_{\beta_{2}}\) tend to 0 as \(t\to\infty\), and hence the solution of (1.1) is attractive. The proof is completed. □

4 Example

As applications of our main results, we consider the following examples.

Example 4.1

Consider the equation
$$ \textstyle\begin{cases} \Delta^{0.5}_{-0.5} u(t)=(t+0.7)^{(-0.7)}\sin u(t+0.5), \quad t\in\mathbb {N}_{0}, \\ \Delta^{-0.5}_{-0.5} u(t)|_{t={0}}=0, \end{cases} $$
(4.1)
where \(f(t,u(t))=(t+0.2)^{(-0.7)}\sin u(t)\), \(t\in\mathbb{N}_{0.5}\). Since
$$\begin{aligned} \bigl\vert f(t,u_{1})-f(t,u_{2}) \bigr\vert &= \bigl\vert (t+0.2)^{(-0.7)}(\sin u_{1}-\sin u_{2}) \bigr\vert \end{aligned}$$
(4.2)
$$\begin{aligned} &\le \frac{\Gamma(1.7)}{\Gamma(2.4)} \vert u_{1}-u_{2} \vert \end{aligned}$$
(4.3)
and
$$ \bigl\vert f(t,u) \bigr\vert = \bigl\vert (t+0.2)^{(-0.7)}\sin u \bigr\vert \le\frac{\Gamma (1.7)}{\Gamma(2.4)}\quad \mbox{for }t\in\mathbb{N}_{0.5} \mbox{ and }u\in \ell^{\infty}_{0.5}, $$
(4.4)
(H1) and (H2) hold. By Theorem 3.1 we get that (4.1) has at least one attractive solution.

Example 4.2

Consider
$$ \textstyle\begin{cases} \Delta^{0.5}_{-0.5} u(t)=(t+0.7)^{(-0.7)}u^{1.2}(t+0.5), \quad t\in\mathbb {N}_{0}, \\ \Delta^{-0.5}_{-0.5} u(t)|_{t={0}}=0, \end{cases} $$
(4.5)
where \(f(t,u(t))=(t+0.2)^{(-0.7)}u^{1.2}(t)\), \(t\in\mathbb{N}_{0.5}\). Since \(\alpha=0.5\), \(\beta_{1}=0.7\), \(\delta=1.2\), \(\gamma=0.2\), we have that
$$ \alpha+\gamma-1< 0,\qquad 1-\beta_{1}-\gamma\delta>0,\qquad \alpha+ \gamma-\beta _{1}-\gamma\delta< 0, $$
(4.6)
and
$$ \bigl\vert f(t,u) \bigr\vert \le L_{1}(t+\gamma \delta)^{(-\beta_{1})} \vert u \vert ^{\delta}\quad \mbox{for }t\in \mathbb{N}_{0.5}\mbox{ and }u\in\ell^{\infty}_{0.5}. $$
(4.7)
Then (H3) holds. By Theorem 3.2 we get that (4.5) has at least one attractive solution.

Example 4.3

Consider
$$ \textstyle\begin{cases} \Delta^{0.5}_{-0.5} u(t)=(t+0.7)^{(-0.7)}\sin ((t+0.5)^{(-0.6)}u(t+0.5) ),\quad t\in\mathbb{N}_{0}, \\ \Delta^{-0.5}_{-0.5} u(t)|_{t={0}}=0, \end{cases} $$
(4.8)
where \(f(t,u(t))=(t+0.2)^{(-0.7)}\sin (t^{(-0.6)}u(t) )\), \(t\in\mathbb{N}_{0.5}\). Let \(\beta_{2}=0.6\in(0.5,1)\). Since
$$\begin{aligned} \bigl\vert f(t,u_{1})-f(t,u_{2}) \bigr\vert &= \bigl\vert (t+0.2)^{(-0.7)} \bigl(\sin \bigl(t^{(-0.6)}u_{1} \bigr)-\sin \bigl(t^{(-0.6)}u_{2} \bigr) \bigr) \bigr\vert \\ &\le\frac{\Gamma(1.7)}{\Gamma(2.4)}t^{(-0.6)}t^{(0.6)}\frac {|u_{1}-u_{2}|}{t^{(0.6)}} \\ &\le\frac{\Gamma(1.7)}{\Gamma(2.4)}\frac{|u_{1}-u_{2}|}{t^{(0.6)}} \end{aligned}$$
(4.9)
and
$$ \bigl\vert f(t,u) \bigr\vert = \bigl\vert (t+0.2)^{(-0.7)}\sin \bigl(t^{(-0.6)}u(t) \bigr) \bigr\vert \le\frac{\Gamma(1.7)}{\Gamma(2.4)}\quad \mbox{for }t\in\mathbb {N}_{0.5}\mbox{ and }u\in\ell^{\infty}_{0.5}. $$
(4.10)
Then f is \(\beta_{2}\)-continuous, and (H2) holds. By Theorem 3.3 we get that (4.8) has at least one attractive solution.

Declarations

Acknowledgements

The authors are very grateful to the reviewers for their valuable comments. The work was supported by the National Natural Science Foundation of China (No. 11671339).

Authors’ contributions

Both authors LZ and YZ contributed equally to each part of this work. Both authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Faculty of Mathematics and Computational Science, Xiangtan University, Xiangtan, P.R. China
(2)
Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Faculty of Science, King Abdulaziz University, Jeddah, Saudi Arabia

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