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# On the existence of solutions for a fractional finite difference inclusion via three points boundary conditions

Advances in Difference Equations20152015:242

https://doi.org/10.1186/s13662-015-0559-7

• Received: 14 April 2015
• Accepted: 30 June 2015
• Published:

## Abstract

In this paper, we discussed the existence of solutions for the fractional finite difference inclusion $$\Delta^{\nu}x(t)\in F(t,x(t),\Delta x(t),\Delta^{2} x(t))$$ via the boundary value conditions $$\xi x(\nu-3)+\beta\Delta x(\nu -3)=0$$, $$x(\eta)=0$$, and $$\gamma x(b+\nu)+\delta\Delta x(b+\nu)=0$$, where $$\eta\in\mathbb{N}_{\nu-2}^{b+\nu-1}$$, $$2<\nu<3$$, and $$F:\mathbb {N}_{\nu-3}^{b+\nu+1}\times\mathbb{R}\times\mathbb{R}\times\mathbb {R}\to2^{\mathbb{R}}$$ is a compact valued multifunction.

## Keywords

• fixed point
• fractional finite difference inclusion
• three points boundary conditions

## 1 Introduction

There are many works concerned with the existence of solutions for some fractional finite difference equations from different views by using the fixed point theory techniques (see for example, ). The readers can find more details as regards elementary notions and definitions of fractional finite difference equations in . Also, much attention was devoted to the fractional differential inclusions (see for example, [9, 10, 1624]). To the best of our knowledge, there is no published research work about fractional finite difference inclusions.

In 2011, Goodrich  investigated the general discrete fractional boundary problem, namely
$$\left \{\textstyle\begin{array}{l} -\Delta^{\nu}y(t)=f(t+\nu-1,y(t+\nu-1)), \\ \alpha y(\nu-2)-\beta\Delta y(\nu-2)=0, \\ \gamma y(\nu+b)-\delta\Delta y(\nu+b)=0, \end{array}\displaystyle \right .$$
where $$t\in[0,b]_{\mathbb{N}_{0}}$$, $$\nu\in(1,2]$$, and $$\alpha\gamma+\alpha \delta+\beta\gamma\neq0$$ with $$\alpha,\beta,\gamma,\delta\geq0$$. In this paper, with this thought and motivation in our minds, we investigate the existence of solution for the fractional finite difference inclusion
$$\left \{\textstyle\begin{array}{l} \Delta^{\nu}x(t)\in F(t,x(t),\Delta x(t),\Delta^{2} x(t)), \\ \xi x(\nu-3)+\beta\Delta x(\nu-3)=0, \\ x(\eta)=0, \\ \gamma x(b+\nu)+\delta\Delta x(b+\nu)=0, \end{array}\displaystyle \right .$$
where $$\eta\in\mathbb{N}_{\nu-2}^{b+\nu-1}$$, $$2<\nu<3$$ and $$F:\mathbb {N}_{\nu-3}^{b+\nu+1}\times\mathbb{R}\times\mathbb{R}\times\mathbb {R}\to2^{\mathbb{R}}$$ is a compact valued multifunction.

## 2 Preliminaries

As is well known, the Gamma function has some properties as $$\Gamma (z+1)=z\Gamma(z)$$ and $$\Gamma(n)=(n-1)!$$ for all $$n\in\mathbb{N}$$. Define
$$t^{\underline{\nu}}=\frac{\Gamma(t+1)}{\Gamma(t+1-\nu)}$$
for all $$t,\nu\in\mathbb{R}$$ whenever the right-hand side is defined. If $$t+1-\nu$$ is a pole of the gamma function and $$t+1$$ is not a pole, then we define $$t^{\underline{\nu}}=0$$. One can verify that $$\nu ^{\underline{\nu}}=\nu^{\underline{\nu-1}}=\Gamma(\nu+1)$$ and $$t^{\underline{\nu+1}}=(t-\nu)t^{\underline{\nu}}$$. We use the notations $$\mathbb{N}_{a}=\{a, a+1, a+2, \ldots\}$$ for all $$a\in\mathbb {R}$$ and $$\mathbb{N}^{b}_{a}=\{a, a+1, a+2, \ldots, b\}$$ for all real numbers a and b whenever $$b-a$$ is a natural number.
Let $$\nu>0$$ be such that $$m-1<\nu\leq m$$ for some natural number m. Then the νth fractional sum of f based at a is defined by
$$\Delta^{-\nu}_{a}f(t)=\frac{1}{\Gamma(\nu)}\sum ^{t-\nu}_{k=a}\bigl(t-\sigma (k)\bigr)^{\underline{\nu-1}}f(k)$$
for all $$t\in\mathbb{N}_{a+\nu}$$. Similarly, we define
$$\Delta^{\nu}_{a}f(t)=\frac{1}{\Gamma(-\nu)}\sum ^{t+\nu}_{k=a}\bigl(t-\sigma (k)\bigr)^{\underline{-\nu-1}}f(k)$$
for all $$t\in\mathbb{N}_{a+m-\nu}$$.

### Lemma 2.1



Let $$h:\mathbb{N}_{\nu-3}^{b+\nu+1}\to\mathbb{R}$$ be a mapping and $$2<\nu\leq3$$. The general solution of the equation $$\Delta^{\nu}_{\nu -3}x(t)=h(t)$$ is given by
$$x(t)=\sum^{3}_{i=1}c_{i}t^{\underline{\nu-i}}+ \frac{1}{\Gamma(\nu)}\sum^{t-\nu}_{s=0}\bigl(t- \sigma(s)\bigr)^{\underline{\nu-1}}h(s),$$
(1)
where $$c_{1}$$, $$c_{2}$$, $$c_{3}$$ are arbitrary constants.
Since $$\Delta t^{\underline{\mu}}=\mu t^{\underline{\mu-1}}$$, we have
$$\Delta x(t)=\sum^{3}_{i=1}c_{i}( \nu-i)t^{\underline{\nu-i-1}}+\frac {1}{\Gamma(\nu-1)}\sum^{t-\nu+1}_{s=0} \bigl(t-\sigma(s)\bigr)^{\underline{\nu-2}}h(s)$$
(2)
for more information see .

Let $$(X,d)$$ be a metric space. Denote by $$2^{X}$$, $$\mathit{CB}(X)$$, and $$P_{\mathrm{cp}}(X)$$ the class of all nonempty subsets, the class of all closed and bounded subsets, and the class of all compact subsets of X, respectively. A mapping $$Q: X\to2^{X}$$ is called a multifunction on X and $$u\in X$$ is called a fixed point of Q whenever $$u\in Qu$$.

Consider the Hausdorff metric $$H_{d}: 2^{X}\times2^{X}\to[0,\infty)$$ by
$$H_{d}(A,B)=\max\Bigl\{ \sup_{a\in A}d(a,B), \sup _{b\in B}d(A,b)\Bigr\} ,$$
where $$d(A,b)=\inf_{a\in A}d(a, b)$$. Let $$(X,d)$$ be a metric space, $$\alpha: X\times X\to[0,\infty)$$ a map, and $$T:X\to2^{X}$$ a multifunction.

We say that X obeys the condition ($$C_{\alpha}$$) whenever for each sequence $$\{x_{n}\}$$ in X with $$\alpha(x_{n}, x_{n+1})\geq1$$ for all n and $$x_{n}\to x$$, there exists a subsequence $$\{x_{n_{k}}\}$$ of $$\{x_{n}\}$$ such that $$\alpha(x_{n_{k}},x)\geq1$$ for all k. The map T is said to be α-admissible whenever for each $$x\in X$$ and $$y\in Tx$$ with $$\alpha(x, y)\geq1$$, we have $$\alpha(y, z)\geq1$$ for all $$z\in Ty$$ . Suppose that Ψ is the family of nondecreasing functions $$\psi:[0,\infty)\to[0,\infty)$$ such that $$\sum_{n=1}^{\infty }\psi^{n}(t)<\infty$$ for all $$t>0$$ (for more on this please see ).

By using the following fixed point result, we review the existence of solutions for the fractional finite difference inclusion
$$\Delta_{\nu-3}^{\nu}x(t)\in F\bigl(t,x(t),\Delta x(t), \Delta^{2} x(t)\bigr)$$
via the boundary conditions $$\xi x(\nu-3)+\beta\Delta x(\nu-3)=0$$, $$\gamma x(b+\nu)+\delta\Delta x(b+\nu)=0$$, and $$x(\eta)=0$$, where $$\eta \in\mathbb{N}_{\nu-2}^{b+\nu-1}$$, $$2<\nu<3$$, and $$F:\mathbb{N}_{\nu -3}^{b+\nu}\times\mathbb{R}\times\mathbb{R}\times\mathbb{R}\to 2^{\mathbb{R}}$$ is a compact valued multifunction.

### Lemma 2.2



Let $$(X,d)$$ be a complete metric space, $$\psi\in\Psi$$ a strictly increasing map, $$\alpha: X\times X\to[0,\infty)$$ a map and $$T:X\to \mathit{CB}(X)$$ an α-admissible multifunction such that $$\alpha (x,y)H(Tx,Ty)\leq\psi(d(x,y))$$ for all $$x,y \in X$$ and there exist $$x_{0}\in X$$ and $$x_{1}\in Tx_{0}$$ with $$\alpha(x_{0},x_{1})\geq1$$. If X obeys the condition ($$C_{\alpha}$$), then T has a fixed point.

## 3 Main result

In this section, we consider the fractional finite difference inclusion
$$\Delta_{\nu-3}^{\nu}x(t)\in F\bigl(t,x(t),\Delta x(t),\Delta^{2} x(t)\bigr)$$
(3)
via the boundary value conditions $$\xi x(\nu-3)+\beta\Delta x(\nu -3)=0$$, $$\gamma x(b+\nu)+\delta\Delta x(b+\nu)=0$$, and $$x(\eta)=0$$, where ξ, β, γ, δ are non-zero numbers, $$\eta\in \mathbb{N}_{\nu-2}^{b+\nu-1}$$, $$2<\nu<3$$, $$x:\mathbb{N}_{\nu-3}^{b+\nu +1}\to\mathbb{R}$$ and $$F:\mathbb{N}_{\nu-3}^{b+\nu+1}\times\mathbb {R}\times\mathbb{R}\times\mathbb{R}\to2^{\mathbb{R}}$$ is a compact valued multifunction.

### Lemma 3.1

Let $$y:\mathbb{N}_{0}^{b+1}\to\mathbb{R}$$ and $$2<\nu< 3$$. Then $$x_{0}$$ is a solution for the fractional finite difference equation $$\Delta_{\nu-3}^{\nu}x(t)=y(t)$$ via the boundary conditions $$\xi x(\nu -3)+\beta\Delta x(\nu-3)=0$$, $$x(\eta)=0$$, and $$\gamma x(b+\nu)+\delta \Delta x(b+\nu)=0$$ if and only if $$x_{0}$$ is a solution of the fractional sum equation $$x(t)=\sum_{s=0}^{b+1}G(t,s,\eta)y(s)$$, where
\begin{aligned} G(t,s,\eta) =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}} -\theta[\gamma+\delta(\nu-1)]t^{\underline{\nu-1}}}{\theta\beta_{0}\mu \Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)] [\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu -2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu -4}}} \biggr] \\ &{}\times(b-s+2) \bigl(b+\nu-\sigma(s) \bigr)^{\underline{\nu-2}} + \biggl[\frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\ &{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)} \biggr] \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}} \\ &{}+ \frac{(t-\sigma(s))^{\underline{\nu -1}}}{\Gamma(\nu)}, \end{aligned}
whenever $$0\leq s\leq t-\nu\leq b+1$$ and $$0\leq s\leq\eta-\nu\leq b+1$$,
\begin{aligned} G(t,s,\eta) =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)] t^{\underline{\nu-1}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline {\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)] [\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu -2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu -4}}} \biggr] \\ &{}\times(b-s+2) \bigl(b+\nu-\sigma(s) \bigr)^{\underline{\nu-2}} + \biggl[\frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\ &{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)} \biggr] \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}, \end{aligned}
whenever $$0\leq t-\nu< s\leq\eta-\nu\leq b+1$$,
\begin{aligned} G(t,s,\eta) =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)] t^{\underline{\nu-1}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline {\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu )(b+\nu)^{\underline{\nu-4}}} \biggr] \\ &{}\times(b-s+2) \bigl(b+\nu-\sigma(s) \bigr)^{\underline{\nu-2}} +\frac{(t-\sigma(s))^{\underline{\nu-1}}}{\Gamma(\nu)}, \end{aligned}
whenever $$0\leq\eta-\nu< s\leq t-\nu\leq b+1$$ and
\begin{aligned} G(t,s,\eta) =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)]t^{\underline{\nu-1}}}{ \theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu )(b+\nu)^{\underline{\nu-4}}} \biggr] \\ &{}\times(b-s+2) \bigl(b+\nu-\sigma(s) \bigr)^{\underline{\nu-2}}, \end{aligned}
whenever $$0\leq t-\nu< s\leq b+1$$ and $$0\leq\eta-\nu< s\leq b+1$$. Here,
\begin{aligned}& \theta=\frac{\eta\beta\nu-\eta\xi-3\eta\beta-2\xi+\xi\nu-\beta\nu ^{2}+6\beta\nu-8\beta}{\beta(\nu-2)}, \\& \mu=\frac{b\xi\delta\nu-2b\delta\xi+\gamma\xi b^{2}+3b\gamma\xi+\beta b\nu ^{2}\delta+\delta b^{2}\beta\nu+\beta b\delta\nu-6\beta\delta b+3\beta \delta b^{2}+4\xi\delta\nu}{\beta(\nu-2)} \\& \hphantom{\mu=}{}+\frac{-8\delta\xi+4\gamma\xi b+12\gamma\xi+4\beta\nu^{2}\delta+7\gamma \beta\nu b+12\gamma\beta\nu+4\beta\delta\nu-24\beta\delta+21\beta\gamma b+36\beta\gamma}{\beta(\nu-2)} \end{aligned}
and
$$\beta_{0}=\frac{\theta[\delta(\nu-1)+\gamma(b+2)](b+3)(b+4)+\mu(\eta+2-\nu )(\eta+3-\nu)}{\theta\mu}.$$

### Proof

Let $$x_{0}$$ be a solution for the equation $$\Delta_{\nu-3}^{\nu}x(t)=y(t)$$ via the boundary conditions $$\xi x(\nu -3)+\beta\Delta x(\nu-3)=0$$, $$x(\eta)=0$$, and $$\gamma x(b+\nu)+\delta \Delta x(b+\nu)=0$$. Then by using (2) and Lemma 2.1, we get
$$x_{0}(t)=c_{1}t^{\underline{\nu-1}}+c_{2}t^{\underline{\nu -2}}+c_{3}t^{\underline{\nu-3}} +\frac{1}{\Gamma(\nu)}\sum_{s=0}^{t-\nu}\bigl(t- \sigma(s)\bigr)^{\underline{\nu-1}}y(s)$$
and
\begin{aligned} \Delta x_{0}(t) =&c_{1}(\nu-1)t^{\underline{\nu-2}}+c_{2}( \nu-2)t^{\underline {\nu-3}}+c_{3}(\nu-3)t^{\underline{\nu-4}} \\ &{}+\frac{1}{\Gamma(\nu-1)} \sum_{s=0}^{t-\nu+1}\bigl(t-\sigma(s) \bigr)^{\underline {\nu-2}}y(s), \end{aligned}
where $$c_{1},c_{2},c_{3}\in\mathbb{R}$$ are arbitrary constants. Now, by using the boundary condition
$$\xi x(\nu-3)+\beta\Delta x(\nu-3)=0,$$
we get $$\xi c_{3} +\beta[c_{2}(\nu-2)+c_{3}(\nu-3)]=0$$. Also, by using the condition $$x(\eta)=0$$ we obtain
\begin{aligned} c_{3} =&-(\eta+2-\nu) (\eta+3-\nu)c_{1}-(\eta+2- \nu)c_{2} \\ &{}-\frac{1}{\eta^{\underline{\nu-3}}\Gamma(\nu)}\sum_{s=0}^{\eta-\nu} \bigl(\eta -\sigma(s)\bigr)^{\underline{\nu-1}}y(s). \end{aligned}
Moreover, by using the boundary condition $$\gamma x(b+\nu)+\delta \Delta x(b+\nu)=0$$, we get
\begin{aligned}& c_{1}\bigl[\delta(\nu-1)+\gamma(b+2)\bigr](b+\nu)^{\underline{\nu-2}}+c_{2} \bigl[\delta(\nu -2)+\gamma(b+3)\bigr](b+\nu)^{\underline{\nu-3}} \\& \qquad {}+c_{3} \bigl[\delta(\nu-3)+\gamma(b+4)\bigr] (b+\nu)^{\underline{\nu-4}} \\& \quad =-\frac{\delta}{\Gamma(\nu-1)}\sum_{s=0}^{b+1} \bigl(b+\nu-\sigma (s)\bigr)^{\underline{\nu-2}}y(s)-\frac{\gamma}{\Gamma(\nu)}\sum _{s=0}^{b} \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-1}}y(s). \end{aligned}
Thus, by using a simple calculation, we get
\begin{aligned}& c_{1}=-\frac{1}{\beta_{0}\theta\eta^{\underline{\nu-3}}\Gamma(\nu)}\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y(s) \\& \hphantom{c_{1}={}}{}-\frac{\gamma+\delta(\nu-1)}{\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline {\nu-4}}}\sum _{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s) \bigr)^{\underline{\nu-2}}y(s), \\& c_{2}=\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline{\nu-3}}\Gamma(\nu)}\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y(s) \\& \hphantom{c_{2}={}}{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)]}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu -4}}} \\& \hphantom{c_{2}={}}{}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}y(s) \end{aligned}
and
\begin{aligned} c_{3} =&\frac{(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}}{\theta^{2}\beta_{0}\eta ^{\underline{\nu-3}}\Gamma(\nu)}\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y(s) \\ &{}+\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu)]}{\theta\beta _{0}\mu\Gamma(\nu) (b+\nu)^{\underline{\nu-4}}}\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma (s)\bigr)^{\underline{\nu-2}}y(s). \end{aligned}
Hence,
\begin{aligned} x_{0}(t) =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)]t^{\underline{\nu-1}}}{ \theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu )(b+\nu)^{\underline{\nu-4}}} \biggr] \\ &{}\times \sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}y(s) \\ &{}+ \biggl[\frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\ &{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)} \biggr] \sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y(s) \\ &{}+\sum_{s=0}^{t-\nu}\frac{(t-\sigma(s))^{\underline{\nu-1}}}{\Gamma(\nu)}y(s) =\sum_{s=0}^{b+1}G(s,t,\eta)y(s). \end{aligned}
Now, let $$x_{0}$$ be a solution for the equation $$x(t)=\sum_{s=0}^{b+1}G(s,t,\eta)y(s)$$. Then we have
\begin{aligned} x_{0}(t) =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)]t^{\underline{\nu-1}}}{ \theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)] t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu)(b+\nu )^{\underline{\nu-4}}} \biggr] \\ &{}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}y(s) \\ &{}+ \biggl[\frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\ &{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)} \biggr] \\ &{}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y(s) +\sum_{s=0}^{t-\nu}\frac{(t-\sigma(s))^{\underline{\nu-1}}}{\Gamma(\nu)}y(s). \end{aligned}
Since $$(\nu-3)^{\underline{\nu-1}}=(\nu-3)^{\underline{\nu-2}}=0$$, $$(\nu -3)^{\underline{\nu-3}}=(\nu-3)^{\underline{\nu-4}}=\Gamma(\nu-2)$$, and
$$\sum_{s=0}^{-3}\bigl(\nu-3-\sigma(s) \bigr)^{\underline{\nu-1}}y(s)=\sum_{s=0}^{-2} \bigl(\nu-3-\sigma(s)\bigr)^{\underline{\nu-2}}y(s)=0,$$
we get $$\xi x_{0}(\nu-3)+\beta\Delta x_{0}(\nu-3)=0$$. A simple calculation shows us $$\gamma x_{0}(b+\nu)+\delta\Delta x_{0}(b+\nu)=0$$ and $$x_{0}(\eta)=0$$. On the other hand,
$$x_{0}(t)=c_{1}t^{\underline{\nu-1}}+c_{2}t^{\underline{\nu-2}} +c_{3}t^{\underline{\nu-3}}+\frac{1}{\Gamma(\nu)}\sum _{s=0}^{t-\nu }\bigl(t-\sigma(s)\bigr)^{\underline{\nu-1}}y(s)$$
is a solution for the equation $$\Delta^{\nu}_{\nu-3} x(t)=y(t)$$ and so $$\Delta^{\nu}_{\nu-3} x_{0}(t)=y(t)$$. □
A function $$x:\mathbb{N}_{\nu-3}^{b+\nu+1}\to\mathbb{R}$$ is a solution of the problem (3) whenever it satisfies the boundary conditions and there exists a function $$y:\mathbb{N}_{0}^{b+1}\to\mathbb {R}$$ such that
$$y(t)\in F\bigl(t, x(t),\Delta x(t),\Delta^{2} x(t)\bigr)$$
for all $$t\in\mathbb{N}_{0}^{b+1}$$ and
\begin{aligned} x(t) =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)]t^{\underline{\nu-1}}}{ \theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu )(b+\nu)^{\underline{\nu-4}}} \biggr] \\ &{}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}y(s) \\ &{}+ \biggl[\frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\ &{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)} \biggr] \\ &{}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y(s) +\sum_{s=0}^{t-\nu}\frac{(t-\sigma(s))^{\underline{\nu-1}}}{\Gamma(\nu)}y(s). \end{aligned}
Let $$\mathcal{X}$$ be the set of all functions $$x:\mathbb{N}_{\nu -3}^{b+\nu+1}\to\mathbb{R}$$ endowed with the norm
$$\|x\|=\max_{t\in\mathbb{N}_{\nu-3}^{b+\nu+1}}\bigl\vert x(t)\bigr\vert +\max _{t\in\mathbb {N}_{\nu-3}^{b+\nu+1}}\bigl\vert \Delta x(t)\bigr\vert +\max _{t\in\mathbb{N}_{\nu-3}^{b+\nu +1}}\bigl\vert \Delta^{2} x(t)\bigr\vert .$$
We show that $$(\mathcal{X},\|\cdot\|)$$ is a Banach space. Let $$\{x_{n}\}$$ be a Cauchy sequence in $$\mathcal{X}$$ and $$\epsilon>0$$ be given. Choose a natural number N such that $$\|x_{n}-x_{m}\|<\epsilon$$ for all $$m,n>N$$. This implies that $$\max_{t\in\mathbb{N}_{\nu-3}^{b+\nu+1}}|x_{n}(t)-x_{m}(t)|<\epsilon$$, $$\max_{t\in\mathbb{N}_{\nu-3}^{b+\nu+1}}|\Delta x_{n}(t)-\Delta x_{m}(t)|<\epsilon$$ and
$$\max_{t\in\mathbb{N}_{\nu-3}^{b+\nu+1}}\bigl\vert \Delta^{2} x_{n}(t)-\Delta^{2} x_{m}(t)\bigr\vert < \epsilon.$$
Choose $$x(t), z(t), w(t)\in\mathbb{R}$$ such that $$x_{n}(t)\to x(t)$$, $$\Delta x_{n}(t)\to z(t)$$, and $$\Delta^{2} x_{n}(t)\to w(t)$$ for all $$t\in \mathbb{N}_{\nu-3}^{b+\nu+1}$$. Note that $$\Delta x_{n}(t)=x_{n}(t+1)-x_{n}(t)$$ and so $$\Delta x(t)=x(t+1)-x(t)=z(t)$$. Similarly, we get $$\Delta^{2} x(t)=w(t)$$. This implies that $$|x_{n}(t)-x(t)|<\frac{\epsilon}{3}$$, $$|\Delta x_{n}(t)-\Delta x(t)|<\frac {\epsilon}{3}$$, and $$|\Delta^{2} x_{n}(t)-\Delta^{2} x(t)|<\frac{\epsilon }{3}$$ for all $$t\in\mathbb{N}_{\nu-3}^{b+\nu+1}$$ and $$n>M$$ for some natural number M. Thus,
$$\|x_{n}-x\|=\max_{t\in\mathbb{N}_{\nu-3}^{b+\nu+1}}\bigl\vert x_{n}(t)-x(t)\bigr\vert +\max_{t\in\mathbb{N}_{\nu-3}^{b+\nu+1}}\bigl\vert \Delta x_{n}(t)-\Delta x(t)\bigr\vert +\max_{t\in\mathbb{N}_{\nu-3}^{b+\nu+1}} \bigl\vert \Delta^{2} x(t)-\Delta^{2} x(t)\bigr\vert < \epsilon.$$
Hence, $$(\mathcal{X},\|\cdot\|)$$ is a Banach space.
Let $$x\in\mathcal {X}$$. Define the set of selections of F by
$$S_{F,x}=\bigl\{ y:\mathbb{N}_{0}^{b+1}\to\mathbb{R} \mid y(t)\in F\bigl(t, x(t),\Delta x(t),\Delta^{2} x(t)\bigr) \mbox{ for all } t \in\mathbb {N}_{0}^{b+1}\bigr\} .$$
Since $$F(t, x(t),\Delta x(t),\Delta^{2} x(t))\neq\emptyset$$, the selection principle implies that $$S_{F,x}$$ is nonempty.

### Theorem 3.2

Suppose that $$\psi\in\Psi$$ and $$F: \mathbb{N}_{\nu-3}^{b+\nu+1}\times \mathbb{R} \times\mathbb{R}\times\mathbb{R}\to P_{\mathrm{cp}}(\mathbb{R})$$ is a multifunction such that
$$H_{d}\bigl(F(t,x_{1},x_{2},x_{3})-F(t,z_{1},z_{2},z_{3}) \bigr)\leq\psi\bigl(\vert x_{1}-z_{1}\vert +\vert x_{2}-z_{2}\vert +\vert x_{3}-z_{3} \vert \bigr)$$
for all $$t\in\mathbb{N}_{\nu-3}^{b+\nu+1}$$ and $$x_{1},x_{2},x_{3},z_{1},z_{2},z_{3}\in\mathbb{R}$$. Then the boundary value inclusion (3) has a solution.

### Proof

Choose $$y\in S_{F,x}$$ and put $$h(t)=\sum_{s=0}^{b+1}G(t,s,\eta)y(s)$$ for all $$t\in\mathbb{N}_{\nu-3}^{\nu+b+1}$$. Then $$h\in\mathcal{X}$$ and so the set
$$\Biggl\{ h\in\mathcal{X}: \mbox{there exists } y\in S_{F,x} \mbox{ such that } h(t)=\sum_{s=0}^{b+1}G(t,s,\eta)y(s) \mbox{ for all } t\in\mathbb {N}_{\nu-3}^{b+\nu+1} \Biggr\}$$
is nonempty. Now define $$\mathcal{F}: \mathcal{X}\to2^{\mathcal{X}}$$ by
\begin{aligned} \mathcal{F}(x) =& \Biggl\{ h\in\mathcal{X}: \mbox{there exists } y\in S_{F,x} \mbox{ such that } h(t)=\sum_{s=0}^{b+1}G(t,s, \eta)y(s) \\ &\mbox{for all } t\in\mathbb{N}_{\nu-3}^{b+\nu+1} \Biggr\} . \end{aligned}
We show that the multifunction $$\mathcal{F}$$ has a fixed point. First, we show that $$\mathcal{F}(x)$$ is closed subset of $$\mathcal{X}$$ for all $$x\in\mathcal{X}$$. Let $$x\in\mathcal{X}$$ and $$\{u_{n}\}_{n\geq1}$$ be a sequence in $$\mathcal{F}(x)$$ with $$u_{n}\to u$$. For each n, choose $$y_{n} \in S_{F,x}$$ such that
\begin{aligned} u_{n}(t) =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)] t^{\underline{\nu-1}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline {\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)] [(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta _{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \biggr] \\ &{}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}y_{n}(s) \\ &{}+ \biggl[\frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\ &{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)} \biggr] \\ &{}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y_{n}(s) +\sum_{s=0}^{t-\nu}\frac{(t-\sigma(s))^{\underline{\nu-1}}}{\Gamma(\nu)}y_{n}(s) \end{aligned}
for all $$t\in\mathbb{N}_{\nu-3}^{b+\nu+1}$$. Since F has compact values, $$\{y_{n}\}_{n\geq1}$$ has a subsequence which converges to some $$y\in S_{F,x}$$. We denote this subsequence again by $$\{y_{n}\}_{n\geq 1}$$. So
\begin{aligned} u_{n}(t) \to& u(t) \\ =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)] t^{\underline{\nu-1}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline {\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu )(b+\nu)^{\underline{\nu-4}}} \biggr] \\ &{}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}y(s) \\ &{}+ \biggl[\frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\ &{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)} \biggr] \\ &{}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y(s) +\sum_{s=0}^{t-\nu}\frac{(t-\sigma(s))^{\underline{\nu-1}}}{\Gamma(\nu)}y(s) \end{aligned}
for all $$t\in\mathbb{N}_{\nu-3}^{b+\nu+1}$$. This implies that $$u\in \mathcal{F}(x)$$. Thus, the multifunction $$\mathcal{F}$$ has closed values. Since F is a compact multifunction, it is easy to check that $$\mathcal {F}(x)$$ is bounded set in $$\mathcal{X}$$ for all $$x\in\mathcal{X}$$. Let $$x,z\in\mathcal{X}$$, $$h_{1}\in\mathcal{F}(x)$$, and $$h_{2}\in\mathcal {F}(z)$$. Choose $$y_{1}\in S_{F,x}$$ and $$y_{2}\in S_{F,z}$$ such that
\begin{aligned} h_{1}(t) =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)] t^{\underline{\nu-1}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline {\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)] [(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta _{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \biggr] \\ &{}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}y_{1}(s) \\ &{}+ \biggl[\frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\ &{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)} \biggr] \\ &{}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y_{1}(s) +\sum_{s=0}^{t-\nu}\frac{(t-\sigma(s))^{\underline{\nu-1}}}{\Gamma(\nu)}y_{1}(s) \end{aligned}
and
\begin{aligned} h_{2}(t) =& \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)] t^{\underline{\nu-1}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline {\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)] [(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta _{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \biggr] \\ &{}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}y_{2}(s) \\ &{}+ \biggl[\frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\ &{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)} \biggr] \\ &{}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y_{2}(s) +\sum_{s=0}^{t-\nu}\frac{(t-\sigma(s))^{\underline{\nu-1}}}{\Gamma(\nu)}y_{2}(s) \end{aligned}
for all $$t\in\mathbb{N}_{\nu-3}^{b+\nu+1}$$. Since
\begin{aligned}& H_{d}\bigl(F\bigl(t,x(t),\Delta x(t),\Delta^{2} x(t) \bigr)-F\bigl(t,z(t),\Delta z(t),\Delta^{2} z(t)\bigr)\bigr) \\& \quad \leq \psi\bigl(\bigl\vert x(t)-z(t)\bigr\vert +\bigl\vert \Delta x(t)-\Delta z(t)\bigr\vert +\bigl\vert \Delta^{2} x(t)- \Delta^{2} z(t)\bigr\vert \bigr) \end{aligned}
for all $$x,z\in\mathcal{X}$$ and $$t\in\mathbb{N}_{\nu-3}^{b+\nu+1}$$, we get
$$\bigl\vert y_{1}(t)-y_{2}(t)\bigr\vert \leq\psi\bigl( \bigl\vert x(t)-z(t)\bigr\vert +\bigl\vert \Delta x(t)-\Delta z(t)\bigr\vert +\bigl\vert \Delta^{2} x(t)-\Delta^{2} z(t)\bigr\vert \bigr).$$
Now, put
\begin{aligned}& G_{1}=\max_{t\in\mathbb{N}_{\nu-3}^{b+1+\nu}} \Biggl\{ \biggl\vert \frac{[\gamma +\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu-3}} -\theta[\gamma+\delta(\nu-1)]t^{\underline{\nu-1}}}{\theta\beta_{0}\mu \Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\& \hphantom{G_{1}={}}{}-\frac{[\xi-\beta(\nu-3)] [\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu -2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu -4}}}\biggr\vert \\& \hphantom{G_{1}={}}{}\times\sum _{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s) \bigr)^{\underline{\nu-2}} +\biggl\vert \frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\& \hphantom{G_{1}={}}{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)}\biggr\vert \\& \hphantom{G_{1}={}}{}\times\sum _{s=0}^{\eta-\nu}\bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}+ \sum_{s=0}^{t-\nu}\frac{(t-\sigma(s))^{\underline{\nu-1}}}{\Gamma(\nu)} \Biggr\} , \\& G_{2}=\max_{t\in\mathbb{N}_{\nu-3}^{b+1+\nu}} \Biggl\{ \biggl\vert \frac{(\nu -3)[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu-4}}}{\theta\beta _{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\& \hphantom{G_{2}={}}{}-\frac{(\nu-1)\theta[\gamma+\delta(\nu-1)]t^{\underline{\nu-2}}}{\theta\beta _{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\& \hphantom{G_{2}={}}{}-\frac{[\xi-\beta(\nu-3)] [\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu -3}}}{\beta\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}}\biggr\vert \\& \hphantom{G_{2}={}}{}\times\sum _{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s) \bigr)^{\underline{\nu-2}} \\& \hphantom{G_{2}={}}{}+\biggl\vert \frac{(\nu-3)[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-4}}-\theta(\nu-1) t^{\underline{\nu-2}}}{\beta _{0}\theta^{2}\eta^{\underline{\nu-3}}\Gamma(\nu)} \\& \hphantom{G_{2}={}}{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-3}}}{\beta\theta^{2}\beta_{0}\eta^{\underline{\nu -3}}\Gamma(\nu)}\biggr\vert \\& \hphantom{G_{2}={}}{}\times\sum _{s=0}^{\eta-\nu}\bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}+ \sum_{s=0}^{t-\nu+1}\frac{(t-\sigma(s))^{\underline{\nu-2}}}{\Gamma(\nu -1)} \Biggr\} \end{aligned}
and
\begin{aligned} G_{3} =&\max_{t\in\mathbb{N}_{\nu-3}^{b+1+\nu}} \Biggl\{ \biggl\vert \frac{(\nu-3)(\nu-4)[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu)] t^{\underline{\nu-5}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu )^{\underline{\nu-4}}} \\ &{}-\frac{(\nu-1)(\nu-2)\theta[\gamma+\delta(\nu -1)]t^{\underline{\nu-3}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu )^{\underline{\nu-4}}} \\ &{}-\frac{(\nu-3)[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu )(\eta+3-\nu)]t^{\underline{\nu-4}}}{\beta\theta\beta_{0}\mu\Gamma(\nu) (b+\nu)^{\underline{\nu-4}}}\biggr\vert \\ &{}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma (s)\bigr)^{\underline{\nu-2}} \\ &{}+\biggl\vert \frac{(\nu-3)(\nu-4)[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-5}} -\theta(\nu-1)(\nu-2) t^{\underline{\nu-3}}}{\beta_{0}\theta^{2}\eta ^{\underline{\nu-3}}\Gamma(\nu)} \\ &{}+\frac{(\nu-3)[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-4}}}{\beta\theta^{2}\beta_{0}\eta^{\underline{\nu -3}}\Gamma(\nu)}\biggr\vert \\ &{}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}} +\sum_{s=0}^{t-\nu+2}\frac{(t-\sigma(s))^{\underline{\nu-3}}}{\Gamma(\nu -2)} \Biggr\} . \end{aligned}
Then we have
\begin{aligned}& \bigl\vert h_{1}(t)-h_{2}(t)\bigr\vert \\& \quad = \Biggl\vert \biggl[\frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu )(\eta+3-\nu)]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)] t^{\underline{\nu-1}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline {\nu-4}}} \\& \qquad {}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)] [(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta _{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \biggr] \\& \qquad {}\times \sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}(y_{1}-y_{2}) (s) \\& \qquad {}+ \biggl[\frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\& \qquad {}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)} \biggr] \\& \qquad {}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}(y_{1}-y_{2}) (s) +\sum_{s=0}^{t-\nu} \frac{(t-\sigma(s))^{\underline{\nu-1}}}{ \Gamma(\nu)}(y_{1}-y_{2}) (s)\Biggr\vert \\& \quad \leq \biggl\vert \frac{[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu-1)]t^{\underline{\nu-1}}}{ \theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\& \qquad {}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)] t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu)(b+\nu )^{\underline{\nu-4}}}\biggr\vert \\& \qquad {}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}\bigl\vert y_{1}(s)-y_{2}(s) \bigr\vert \\& \qquad {}+\biggl\vert \frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\& \qquad {}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)}\biggr\vert \\& \qquad {}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}} \bigl\vert y_{1}(s)-y_{2}(s) \bigr\vert +\sum_{s=0}^{t-\nu}\frac{(t-\sigma(s))^{\underline{\nu-1}}}{ \Gamma(\nu)} \bigl\vert y_{1}(s)-y_{2}(s)\bigr\vert \\& \quad \leq \max_{t\in\mathbb{N}_{0}^{b+1}}\bigl\vert y_{1}(t)-y_{2}(t) \bigr\vert \\& \qquad {}\times\max_{t\in\mathbb{N}_{\nu-3}^{b+1+\nu}} \Biggl\{ \biggl\vert \frac{[\gamma+\delta (\nu-1)][(\eta+2-\nu) (\eta+3-\nu)]t^{\underline{\nu-3}}-\theta[\gamma+\delta(\nu -1)]t^{\underline{\nu-1}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu )^{\underline{\nu-4}}} \\& \qquad {}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta\beta_{0}\mu\Gamma(\nu) (b+\nu)^{\underline{\nu-4}}}\biggr\vert \\& \qquad {}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma (s)\bigr)^{\underline{\nu-2}} +\biggl\vert \frac{[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta_{0}]t^{\underline{\nu -3}}-\theta t^{\underline{\nu-1}}}{\beta_{0}\theta^{2}\eta^{\underline{\nu -3}}\Gamma(\nu)} \\& \qquad {}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-2}}}{\beta(\nu-2)\theta^{2}\beta_{0}\eta^{\underline {\nu-3}}\Gamma(\nu)}\biggr\vert \\ & \qquad {}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}+\sum_{s=0}^{t-\nu} \frac{(t-\sigma(s))^{\underline{\nu-1}}}{\Gamma(\nu)} \Biggr\} \\& \quad \leq \psi\bigl(\bigl\vert x(t)-z(t)\bigr\vert +\bigl\vert \Delta x(t)-\Delta z(t)\bigr\vert +\bigl\vert \Delta^{2} x(t)- \Delta^{2} z(t)\bigr\vert \bigr)\times G_{1}. \end{aligned}
Since
\begin{aligned} \Delta h_{1}(t) =& \biggl[\frac{(\nu-3)[\gamma+\delta(\nu-1)][(\eta+2-\nu )(\eta+3-\nu)]t^{\underline{\nu-4}}-(\nu-1)\theta[\gamma+\delta(\nu-1)] t^{\underline{\nu-2}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline {\nu-4}}} \\ &{}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)] [(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu-3}}}{\beta\theta\beta_{0}\mu \Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \biggr] \\ &{}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}y_{1}(s) \\ &{}+ \biggl[\frac{(\nu-3)[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-4}}-\theta(\nu-1) t^{\underline{\nu-2}}}{\beta _{0}\theta^{2}\eta^{\underline{\nu-3}}\Gamma(\nu)} \\ &{}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-3}}}{\beta\theta^{2}\beta_{0}\eta^{\underline{\nu -3}}\Gamma(\nu)} \biggr] \\ &{}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}y_{1}(s) +\sum_{s=0}^{t-\nu+1}\frac{(t-\sigma(s))^{\underline{\nu-2}}}{\Gamma(\nu -1)}y_{1}(s), \end{aligned}
we get
\begin{aligned}& \bigl\vert \Delta h_{1}(t)-\Delta h_{2}(t)\bigr\vert \\& \quad \leq\biggl\vert \frac{(\nu-3)[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-4}}-(\nu-1)\theta[\gamma+\delta(\nu-1)] t^{\underline{\nu-2}}}{\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline {\nu-4}}} \\& \qquad {}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu) (\eta+3-\nu)]t^{\underline{\nu-3}}}{\beta\theta\beta_{0}\mu\Gamma(\nu )(b+\nu)^{\underline{\nu-4}}}\biggr\vert \\& \qquad {}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}}\bigl\vert y_{1}(s)-y_{2}(s) \bigr\vert \\& \qquad {}+\biggl\vert \frac{(\nu-3)[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-4}}-\theta(\nu-1) t^{\underline{\nu-2}}}{\beta _{0}\theta^{2}\eta^{\underline{\nu-3}}\Gamma(\nu)} \\& \qquad {}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-3}}}{\beta\theta^{2}\beta_{0}\eta^{\underline{\nu -3}}\Gamma(\nu)}\biggr\vert \\& \qquad {}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}} \bigl\vert y_{1}(s)-y_{2}(s) \bigr\vert +\sum_{s=0}^{t-\nu+1}\frac{(t-\sigma(s))^{\underline{\nu-2}}}{ \Gamma(\nu-1)} \bigl\vert y_{1}(s)-y_{2}(s)\bigr\vert \\& \quad \leq\max_{t\in\mathbb{N}_{0}^{b+1}}\bigl\vert y_{1}(t)-y_{2}(t) \bigr\vert \\ & \qquad {}\times \max_{t\in\mathbb{N}_{\nu-3}^{b+1+\nu}} \Biggl\{ \biggl\vert \frac{(\nu-3)[\gamma +\delta(\nu-1)] [(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu-4}}}{\theta\beta_{0}\mu\Gamma(\nu) (b+\nu)^{\underline{\nu-4}}} \\& \qquad {}-\frac{(\nu-1)\theta[\gamma +\delta(\nu-1)]t^{\underline{\nu-2}}}{\theta\beta_{0}\mu\Gamma(\nu) (b+\nu)^{\underline{\nu-4}}} \\& \qquad {}-\frac{[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta +3-\nu)]t^{\underline{\nu-3}}}{ \beta\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}}\biggr\vert \\& \qquad {}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}} \\& \qquad {}+\biggl\vert \frac{(\nu-3)[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-4}}-\theta(\nu-1) t^{\underline{\nu-2}}}{\beta _{0}\theta^{2}\eta^{\underline{\nu-3}}\Gamma(\nu)} \\& \qquad {}+\frac{[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-3}}}{\beta\theta^{2}\beta_{0}\eta^{\underline{\nu -3}}\Gamma(\nu)}\biggr\vert \\& \qquad {}\times \sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}}+\sum_{s=0}^{t-\nu+1} \frac{(t-\sigma(s))^{\underline{\nu-2}}}{\Gamma(\nu -1)} \Biggr\} \\& \quad \leq\psi\bigl(\bigl\vert x(t)-z(t)\bigr\vert +\bigl\vert \Delta x(t)-\Delta z(t)\bigr\vert +\bigl\vert \Delta^{2} x(t)- \Delta^{2} z(t)\bigr\vert \bigr)\times G_{2}. \end{aligned}
Also, we have
\begin{aligned}& \bigl\vert \Delta^{2} h_{1}(t)-\Delta^{2} h_{2}(t)\bigr\vert \\& \quad \leq\biggl\vert \frac{(\nu-3)(\nu-4)[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu )]t^{\underline{\nu-5}}}{\theta\beta _{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\& \qquad {}-\frac{(\nu-1) (\nu-2)\theta[\gamma+\delta(\nu-1)]t^{\underline{\nu-3}}}{\theta\beta _{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\& \qquad {}-\frac{(\nu-3)[\xi-\beta(\nu-3)] [\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu -4}}}{\beta\theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}}\biggr\vert \\& \qquad {}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma(s)\bigr)^{\underline{\nu-2}} \bigl\vert y_{1}(s)-y_{2}(s) \bigr\vert \\& \qquad {}+\biggl\vert \frac{(\nu-3)(\nu-4)[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-5}} -\theta(\nu-1)(\nu-2) t^{\underline{\nu-3}}}{\beta_{0}\theta^{2}\eta ^{\underline{\nu-3}}\Gamma(\nu)} \\& \qquad {}+\frac{(\nu-3)[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-4}}}{\beta\theta^{2}\beta_{0}\eta^{\underline{\nu -3}}\Gamma(\nu)}\biggr\vert \\& \qquad {}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}} \bigl\vert y_{1}(s)-y_{2}(s) \bigr\vert +\sum_{s=0}^{t-\nu+2}\frac{(t-\sigma(s))^{\underline{\nu-3}}}{ \Gamma(\nu-2)} \bigl\vert y_{1}(s)-y_{2}(s)\bigr\vert \\& \quad \leq\max_{t\in\mathbb{N}_{0}^{b+1}}\bigl\vert y_{1}(t)-y_{2}(t) \bigr\vert \\ & \qquad {}\times \max_{t\in \mathbb{N}_{\nu-3}^{b+1+\nu}} \Biggl\{ \biggl\vert \frac{(\nu-3) (\nu-4)[\gamma+\delta(\nu-1)][(\eta+2-\nu)(\eta+3-\nu)]t^{\underline{\nu -5}}}{ \theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\& \qquad {}-\frac{(\nu-1)(\nu-2)\theta[\gamma+\delta(\nu-1)]t^{\underline{\nu-3}}}{ \theta\beta_{0}\mu\Gamma(\nu)(b+\nu)^{\underline{\nu-4}}} \\& \qquad {}-\frac{(\nu-3)[\xi-\beta(\nu-3)][\gamma+\delta(\nu-1)][(\eta+2-\nu )(\eta+3-\nu)] t^{\underline{\nu-4}}}{\beta\theta\beta_{0}\mu\Gamma(\nu)(b+\nu )^{\underline{\nu-4}}}\biggr\vert \\& \qquad {}\times\sum_{s=0}^{b+1}(b-s+2) \bigl(b+\nu-\sigma (s)\bigr)^{\underline{\nu-2}} \\& \qquad {}+\biggl\vert \frac{(\nu-3)(\nu-4)[(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-5}} -\theta(\nu-1)(\nu-2) t^{\underline{\nu-3}}}{\beta_{0}\theta^{2}\eta ^{\underline{\nu-3}}\Gamma(\nu)} \\& \qquad {}+\frac{(\nu-3)[-\xi+\beta(\nu-3)][(\eta+2-\nu)(\eta+3-\nu)-\theta\beta _{0}]t^{\underline{\nu-4}}}{\beta\theta^{2}\beta_{0}\eta^{\underline{\nu -3}}\Gamma(\nu)}\biggr\vert \\& \qquad {}\times\sum_{s=0}^{\eta-\nu} \bigl(\eta-\sigma(s)\bigr)^{\underline{\nu-1}} +\sum_{s=0}^{t-\nu+2}\frac{(t-\sigma(s))^{\underline{\nu-3}}}{\Gamma(\nu -2)} \Biggr\} \\& \quad \leq\psi\bigl(\bigl\vert x(t)-z(t)\bigr\vert +\bigl\vert \Delta x(t)-\Delta z(t)\bigr\vert +\bigl\vert \Delta^{2} x(t)- \Delta^{2} z(t)\bigr\vert \bigr)\times G_{3}. \end{aligned}
Hence, we obtain
\begin{aligned} \|h_{1}-h_{2}\| =&\max_{t\in\mathbb{N}_{\nu-3}^{b+1+\nu}}\bigl\vert h_{1}(t)-h_{2}(t)\bigr\vert + \max_{t\in\mathbb{N}_{\nu-3}^{b+1+\nu}} \bigl\vert \Delta h_{1}(t)-\Delta h_{2}(t)\bigr\vert \\ &{}+ \max_{t\in\mathbb{N}_{\nu-3}^{b+1+\nu}}\bigl\vert \Delta^{2} h_{1}(t)- \Delta^{2} h_{2}(t)\bigr\vert \\ \leq&\psi\bigl(\bigl\vert x(t)-z(t)\bigr\vert +\bigl\vert \Delta x(t)- \Delta z(t)\bigr\vert +\bigl\vert \Delta^{2} x(t)- \Delta^{2} z(t)\bigr\vert \bigr) (G_{1}+G_{2}+G_{3}) \\ \leq&(G_{1}+G_{2}+G_{3})\psi\bigl(\Vert x-z \Vert \bigr) \end{aligned}
for all $$x,z\in\mathcal{X}$$, $$h_{1}\in\mathcal{F}(x)$$, and $$h_{2}\in \mathcal{F}(z)$$. So $$H_{d}(\mathcal{F}(x),\mathcal{F}(z))\leq(G_{1}+G_{2}+G_{3})\psi(\|x-z\|)$$ for all $$x,z\in\mathcal{X}$$.
Define the function α on $$\mathcal{X}\times\mathcal{X}$$ by $$\alpha(x,z)=1$$ whenever $$G_{1}+G_{2}+G_{3}< 1$$ and $$\alpha(x,z)=\frac{1}{G_{1}+G_{2}+G_{3}}$$ otherwise. Thus,
$$\alpha(x,z) H_{d}\bigl(\mathcal{F}(x),\mathcal{F}(z)\bigr)\leq\psi \bigl(\Vert x-z\Vert \bigr)$$
for all $$x,z\in\mathcal{X}$$. Let $$\{x_{n}\}$$ be a sequence in $$\mathcal {X}$$ with $$\alpha(x_{n}, x_{n+1})\geq1$$ for all n and $$x_{n}\to x$$. Then it is easy to check that there exists a subsequence $$\{x_{n_{k}}\}$$ of $$\{ x_{n}\}$$ such that $$\alpha(x_{n_{k}},x)\geq1$$ for all k. This implies that $$\mathcal{X}$$ obeys the condition ($$C_{\alpha}$$). If $$x\in\mathcal {X}$$ and $$y\in\mathcal{F}(x)$$ with $$\alpha(x, y)\geq1$$, then it is easy to see that $$\alpha(y, z)\geq1$$ for all $$z\in\mathcal{F}(y)$$. Thus, $$\mathcal{F}$$ is an α-admissible α-ψ-contractive multifunction. Hence by using Theorem 2.2, there exists $$x^{*}\in\mathcal{X}$$ such that $$x^{*}\in\mathcal{F}(x^{*})$$. One can check that $$x^{*}$$ is a solution for the problem (3). □

### Example 3.1

Consider the fractional finite difference inclusion
$$\Delta^{2.5}_{-0.5}x(t)\in \biggl[1 , e^{t^{2}}+2+\frac{\sin x(t)}{e^{2|t|}}+\sinh^{2} t+\frac{|\Delta x(t)|}{4|t|}+ \frac {3}{6t^{2}-1}+\frac{|\Delta^{2}x(t)|}{\cosh|3t|} \biggr]$$
(4)
via the boundary value conditions $$\xi x(-0.5)+\beta\Delta x(-0.5)=0$$, $$\gamma x(6.5)+\delta\Delta x(6.5)=0$$, and $$x(3.5)=0$$, where ξ, β, γ, δ are non-zero numbers. In fact, this problem is a special case of the problem (3), where $$\nu=2.5$$, $$\eta =3.5$$, $$b=4$$, and
$$F(t,x_{1},x_{2},x_{3})= \biggl[1 , e^{t^{2}}+2+\frac{\sin x_{1}}{e^{2|t|}}+\sinh^{2} t+\frac{|x_{2}|}{4|t|}+ \frac{3}{6t^{2}-1}+\frac {|x_{3}|}{\cosh|3t|} \biggr].$$
Note that $$e^{t^{2}}+2+\frac{\sin x_{1}}{e^{2|t|}}+\sinh^{2} t+\frac {|x_{2}|}{4|t|}+\frac{3}{6t^{2}-1}+\frac{|x_{3}|}{\cosh|3t|}>1$$ for all $$t\in \mathbb{N}_{-0.5}^{7.5}$$ and $$x_{1},x_{2},x_{3}\in\mathbb{R}$$. Also, $$e^{2|t|}\geq2$$, $$4|t|\geq2$$, and $$\cosh|3t|\geq2$$ for all $$t\in \mathbb{N}_{-0.5}^{7.5}$$ and F is a compact valued multifunction on $$\mathbb{N}_{-0.5}^{7.5}\times\mathbb{R}\times\mathbb{R}\times\mathbb {R}$$. Now, define $$\psi\in\Psi$$ by $$\psi(z)=\frac{z}{2}$$ for all $$z\geq0$$. Since
\begin{aligned}& H_{d}\bigl(F(t,x_{1},x_{2},x_{3}),F(t,z_{1},z_{2},z_{3}) \bigr) \\& \quad \leq\biggl\vert \frac{\sin x_{1}}{e^{2|t|}}-\frac{x_{2}}{4|t|}+ \frac{x_{3}}{\cosh|3t|}-\frac{\sin z_{1}}{e^{2|t|}}+\frac{z_{2}}{4|t|}-\frac{z_{3}}{\cosh|3t|}\biggr\vert \\& \quad \leq\frac{|x_{1}-z_{1}|+|x_{2}-z_{2}|+|x_{3}-z_{3}|}{2} \\& \quad =\psi\bigl(\vert x_{1}-z_{1} \vert +\vert x_{2}-z_{2}\vert +\vert x_{3}-z_{3}\vert \bigr) \end{aligned}
for all $$t\in\mathbb{N}_{-0.5}^{7.5}$$ and $$x_{1},x_{2},x_{3},z_{1},z_{2},z_{3}\in \mathbb{R}$$, by using Theorem 3.2 the problem (4) has at least one solution.

## 4 Conclusions

In this manuscript, based on a fixed point theorem, we provided the existence result for a fractional finite difference inclusion in the presence of the general boundary conditions. An example illustrates our result.

## Declarations

### Acknowledgements

The research of the second and third authors was supported by Azarbaijan Shahid Madani University.

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)
Department of Mathematics, Cankaya University, Ogretmenler Cad. 14, Balgat, Ankara, 06530, Turkey
(2)
Institute of Space Sciences, Magurele, Bucharest, Romania
(3)
Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran

## References

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