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

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.

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, [1–7]). The readers can find more details as regards elementary notions and definitions of fractional finite difference equations in [8–15]. Also, much attention was devoted to the fractional differential inclusions (see for example, [9, 10, 16–24]). To the best of our knowledge, there is no published research work about fractional finite difference inclusions.

In 2011, Goodrich [25] 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

[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 [12].

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 $z\in Ty$ [26]. 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 [26]).

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

[26]

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.