On nonlocal Neumann boundary value problem for a second-order forward $(\alpha,\beta)$-difference equation

Abstract

In this paper, we present some properties of the forward $(\alpha,\beta )$-difference operators, and the existence results of two nonlocal boundary value problems for second-order forward $(\alpha,\beta)$-difference equations. The existence and uniqueness results are proved by using the Banach fixed point theorem, and the existence of at least one positive solution is established by using the Krasnoselskii’ fixed point theorem.

1 Introduction

The difference calculus is known as the calculus without considering limits and deals with sets of non-differentiable functions. The difference calculus appears in many applications such as statistics, computing, electrical circuit analysis, dynamical systems, economics, biology, and other fields; see for example [1–7] and the references therein. There are two types of difference operators, one based on the forward (delta) difference operator $\Delta _{h}f(t)=\frac{f(t+1)- f(t)}{h}$, and one on the backward (nabla) difference operator $\nabla_{h}f(t)=\frac{f(t)- f(t-h)}{h}$; if $h=1$, then Δ and ∇ are the standard difference operators. There is not much research involving the development of h-sums and h-difference operators (see [8–13]). Therefore, there is a gap in the literature as regards the details of this operation. In this paper, we aim to study the forward $(\alpha,\beta )$-difference of f defined by

$$ {_{h}}\Delta_{(\alpha,\beta)}f(t)=\frac{\beta f(t+h)-\alpha f(t)}{h},\quad\alpha, \beta,h>0, $$

where the coefficient of $f(t)$ and $f(t+h)$ can be chosen [14].

In particular, since the boundary value problem for forward $(\alpha ,\beta)$-difference equations has not been studied, we attempt to fill this gap by considering the existence and uniqueness result of the nonlocal Neumann boundary value problem for a second-order forward $(\alpha,\beta)$-difference equation,

$$\begin{aligned} \begin{aligned} &{_{h}}\Delta^{2}_{(\alpha,\beta)} u(t-h)+f \bigl(t ,u(t),(\Psi u) (t) \bigr)=0,\quad t\in (h{\mathbb{N}} ) _{h,Th}, \\ &{_{h}}\Delta_{(\alpha,\beta)} \biggl[ \biggl( \frac{\beta}{\alpha} \biggr)^{\frac{t}{h}}u(t) \biggr]_{t=0} = \phi_{1}(u),\qquad{_{h}}\Delta_{(\alpha ,\beta)} \biggl[ \biggl( \frac{\beta}{\alpha} \biggr)^{\frac {t}{h}}u(t) \biggr]_{t=Th} = \phi_{2}(u), \end{aligned} \end{aligned}$$

(1.1)

where $(h{\mathbb{N}} ) _{h,Th}:=\{h,2h,\ldots,Th\}$, $\alpha ,\beta,h>0$, $\alpha<\beta$, $f\in C ( (h{\mathbb{N}} ) _{0, (T+1 ) h} \times {\mathbb{R}}\times{\mathbb{R}},{\mathbb{R}} )$, $\phi_{1},\phi_{2}:C ( (h{\mathbb{N}} ) _{0, (T+1 ) h},{\mathbb{R}} )\rightarrow{\mathbb{R}}$, and for $\varphi\in ( (h{\mathbb {N}} ) _{0, (T+1 ) h} \times (h{\mathbb{N}} ) _{0, (T+1 ) h} ,[0,\infty) )$, $(\Psi u)(t) := ( {_{h}}\Delta_{(\alpha,\beta)}^{-1}\varphi u ) (t) = h \sum_{s=0}^{\frac{t}{h}-1}\varphi(t,s) u(hs)$.

Furthermore, we study the existence of at least one positive solution of boundary value problem for a second-order forward $(\alpha,\beta)$-difference equation of the form

$$\begin{aligned} \begin{aligned} &{_{h}}\Delta^{2}_{(\alpha,\beta)} u(t-h)+F \bigl(t,u(t)\bigr)=0,\quad t\in (h{\mathbb{N}} ) _{ (1+\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}, \\ &{_{h}}\Delta_{(\alpha,\beta)}u \biggl( \biggl( \frac{\beta}{\beta-\alpha } \biggr) h \biggr) =\frac{\beta}{h} u \biggl( \biggl( \frac{\beta}{\beta -\alpha}+1 \biggr) h \biggr) , \\ & \biggl[ {_{h}}\Delta_{(\alpha,\beta)} \biggl[ \biggl( \frac{\beta}{\alpha } \biggr)^{t}u( t) \biggr] \biggr]_{t= (T+\frac{\beta}{\beta-\alpha } ) h} = \theta(u), \end{aligned} \end{aligned}$$

(1.2)

where $F\in C ( (h{\mathbb{N}} ) _{ (1+\frac{\beta }{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h} \times[0,\infty),[0,\infty) )$ and $\theta:C ( (h{\mathbb {N}} ) _{ (1+\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}, [0,\infty) )\rightarrow [0,\infty)$.

In this paper, the plan is as follows. In Sect. 2 we recall some definitions and basic lemmas, and present some properties of the forward $(\alpha,\beta)$-difference operators. In this section, we also derive a representation for the solution to (1.1) and (1.2) by converting the problem to equivalent summation equations. In Sect. 3, we show the existence and uniqueness result of problem (1.1). In Sects. 4 and 5, we show the properties of Green’s function and the existence of at least one positive solution for problem (1.2) by using Krasnoselskii’s fixed point theorem in a cone. Finally, some examples are provided to illustrate our results in the last section.

Theorem 1.1

([15])

Let E be a Banach space, and $K\subset E$ be a cone. Assume $\Omega_{1}$, $\Omega_{2}$ are open subsets of E with $0\in \Omega_{1}$, $\overline{\Omega}_{1}\subset\Omega_{2}$, and let

$$A:K\cap(\overline{\Omega}_{2}\setminus\Omega_{1}) \longrightarrow K $$

be a completely continuous operator such that

• (i) $\|Au\|\leqslant\|u\|$, $u\in K\cap\partial\Omega_{1}$, and $\|Au\|\geqslant\|u\|$, $u\in K\cap\partial\Omega_{2}$, or

• (ii) $\|Au\|\geqslant\|u\|$, $u\in K\cap\partial\Omega_{1}$, and $\|Au\|\leqslant\|u\|$, $u\in K\cap\partial\Omega_{2}$.

Then A has a fixed point in $K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})$.

Lemma 1.1

([16] (Arzelá–Ascoli theorem))

A set of functions in $C[a,b]$ with the sup norm, is relatively compact if and only it is uniformly bounded and equicontinuous on $[a,b]$.

Lemma 1.2

([16])

If a set is closed and relatively compact then it is compact.

2 Preliminaries

In the following, there are notations, definitions, and lemmas which are used in the main results.

Let ${\mathbb{N}}_{a}:=\{a,a+1,a+2,\ldots\}$, ${\mathbb{N}}_{a,b}:=\{ a,a+1,a+2,\ldots,b=a+k\}$, $(h{\mathbb{N}})_{a}:=\{a,a+h,a+2h,\ldots\}$ and $(h{\mathbb{N}})_{a,b}:=\{a,a+h,a+2h,\ldots,b=a+kh\}$ for some $k \in{\mathbb{N}}$, $a\in{\mathbb{R}}$.

Definition 2.1

For $\alpha,\beta,h>0$ and f defined on $[0,\infty)$, the forward $(\alpha,\beta)$-difference of f is defined by

$$ {_{h}}\Delta_{(\alpha,\beta)}f(t)=\frac{\beta f(t+h)-\alpha f(t)}{h}. $$

Furthermore, we define the higher-order $(\alpha,\beta)$-difference by

$${_{h}}\Delta_{(\alpha,\beta)}^{n}f (t ):={_{h}} \Delta_{(\alpha,\beta )}{_{h}}\Delta_{(\alpha,\beta)}^{n-1} f (t )\quad \text{and}\quad {_{h}}\Delta_{(\alpha,\beta)}^{0}f(t):=f(t). $$

Example 2.1

${_{h}}\Delta_{(\alpha,\beta)}(1)=\frac{\beta-\alpha}{h}$ and ${_{h}}\Delta_{(\alpha,\beta)}(t)=\frac{\beta(t+h)-\alpha t }{h}= ( \frac{\beta-\alpha}{h} )t+\beta$.

Remark 2.1

• (i) If $\alpha=\beta=1$, the difference operator $_{h}\Delta _{(1,1)}$ becomes the forward h-difference operator

  $$\Delta_{h}f(t)=\frac{ f(t+h)- f(t)}{h}. $$

• (ii) Let $f(t)=\frac{1}{\alpha} ( \frac{\alpha}{\beta} )^{\frac{t}{h}}g(t)$ for $t \in(h{\mathbb{N}})_{a}$. Then

  $$\begin{aligned} {_{h}}\Delta_{(\alpha,\beta)}f(t)=\frac{1}{h} \biggl[ \frac{\beta}{\alpha} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}+1}g(t+h) - \biggl( \frac {\alpha}{\beta} \biggr)^{\frac{t}{h}}g(t) \biggr] = \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{t}{h}}\Delta_{h}g(t). \end{aligned}$$

The right inverse of the operator ${_{h}}\Delta_{(\alpha,\beta)}$ or the $(\alpha,\beta)$-sum operator is defined as follows.

Definition 2.2

Let $\alpha,\beta,h>0$ and $(h{\mathbb{N}})_{a,b} \subset(h{\mathbb{N}})_{c}$. Assuming that $f,g:(h{\mathbb{N}})_{c}\rightarrow\mathbb{R}$ and $f(t)=\frac {1}{\alpha} ( \frac{\alpha}{\beta} )^{\frac{t}{h}}g(t)$, we define the $(\alpha,\beta)$-sum of f by

$$ {_{h}}\Delta_{(\alpha,\beta)}^{-1}f(t):=h \sum _{k=\frac{c}{h}}^{\frac {t}{h}-1}f(hk)=h \sum_{k=\frac{c}{h}}^{\frac{t}{h}-1} \frac{\alpha ^{k-1}}{\beta^{k}}g(hk)\quad\text{for }t \in(h{\mathbb{N}})_{c+h}, $$

and for $a,b \in(h{\mathbb{N}})_{c+h}$, $a< b$, the $(\alpha,\beta)$-sum of f from a to b defined by

$$ \bigl[ {_{h}}\Delta_{(\alpha,\beta)}^{-1}f(t) \bigr]_{a}^{b}:=h \sum_{k=\frac {a}{h}}^{\frac{b}{h}-1}f(hk)=h \sum_{k=\frac{a}{h}}^{\frac{b}{h}-1}\frac {\alpha^{k-1}}{\beta^{k}}g(hk). $$

Example 2.2

• (i) If $f(t)=1$ for $t \in(h{\mathbb{N}})_{c}$, then we have $g(t)=\alpha ( \frac{\beta}{\alpha} )^{\frac{t}{h}}$ and

  $${_{h}}\Delta_{(\alpha,\beta)}^{-1}(1)=h \sum _{k=\frac{c}{h}}^{\frac {t}{h}-1}\frac{\alpha^{k-1}}{\beta^{k}} \biggl( \frac{\beta^{k}}{\alpha ^{k-1}} \biggr) =h \sum_{k=\frac{c}{h}}^{\frac{t}{h}-1}1=t-c. $$

• (ii) If $f(t)=t$ for $t \in(h{\mathbb{N}})_{c}$, then we have $g(t)=\alpha ( \frac{\beta}{\alpha} )^{\frac{t}{h}}t$ and

  $${_{h}}\Delta_{(\alpha,\beta)}^{-1}(t)=h \sum _{k=\frac{c}{h}}^{\frac {t}{h}-1}\frac{\alpha^{k-1}}{\beta^{k}} \biggl( \frac{\beta^{k}}{\alpha ^{k-1}}hk \biggr) =h^{2} \sum_{k=\frac{c}{h}}^{\frac{t}{h}-1}k= \frac {1}{2}(t-c) (t+c-h). $$

In the following lemma, we introduce the properties of forward $(\alpha ,\beta)$-operators.

Lemma 2.1

Letting $\alpha,\beta,h>0$ and $f: (h{\mathbb{N}})_{a}\rightarrow\mathbb{R}$,

$$\begin{aligned}& {_{h}}\Delta_{(\alpha,\beta)}{_{h}}\Delta_{(\alpha,\beta)}^{-1}f(t) = \beta f(t)+(\beta-\alpha) \sum_{k=\frac{a}{h}}^{\frac{t}{h}-1}f(kh), \\& {_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} \Delta_{(\alpha,\beta)}f(t) = \frac {\beta^{2}}{\alpha} \bigl[ f(t)-f(a) \bigr] + \biggl( \frac{\beta^{2}-\alpha ^{2}}{\alpha h} \biggr) {_{h}}\Delta_{(\alpha,\beta)}^{-1}f(t). \end{aligned}$$

Proof

Letting $f(t)=\frac{1}{\alpha} ( \frac{\alpha}{\beta } )^{\frac{t}{h}}g(t)$ for $t \in(h{\mathbb{N}})_{a}$, we have

$$\begin{aligned} {_{h}}\Delta_{(\alpha,\beta)}{_{h}}\Delta_{(\alpha,\beta)}^{-1}f(t) =& {_{h}}\Delta_{(\alpha,\beta)} \Biggl[ h\sum _{k=\frac{a}{h}}^{\frac{t}{h}-1}\frac{\alpha^{k-1}}{\beta ^{k}} g(hk) \Biggr] \\ =&\beta\sum_{k=\frac{a}{h}}^{\frac{t}{h}} \frac{\alpha^{k-1}}{\beta ^{k}}g(hk) - \alpha\sum_{k=\frac{a}{h}}^{\frac{t}{h}-1} \frac{\alpha ^{k-1}}{\beta^{k}}g(hk) \\ =&\beta \Biggl[ \frac{\alpha^{\frac{t}{h}-1}}{\beta^{\frac{t}{h}}}g(t)+ \sum _{k=\frac{a}{h}}^{\frac{t}{h}-1} \frac{\alpha^{k-1}}{\beta^{k}} g(hk) \Biggr] - \alpha \sum_{k=\frac{a}{h}}^{\frac{t}{h}-1}\frac{\alpha ^{k-1}}{\beta^{k}}g(hk) \\ =& \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-1} g(t)+(\beta - \alpha) \sum_{k=\frac{a}{h}}^{\frac{t}{h}-1} \frac{\alpha^{k-1}}{\beta^{k}} g(hk) \\ =& \beta f(t)+(\beta-\alpha) \sum_{k=\frac{a}{h}}^{\frac{t}{h}-1}f(kh). \end{aligned}$$

From Remark 2.1(ii), we obtain

$$\begin{aligned} {_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} \Delta_{(\alpha,\beta)}f(t) =&{_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} \biggl[ \biggl( \frac{\alpha}{\beta } \biggr)^{\frac{t}{h}} \Delta_{h}g(t) \biggr] \\ =&{_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} \biggl[ \biggl( \frac{\alpha}{\beta } \biggr)^{\frac{t}{h}} \biggl( \frac{g(t+h)-g(t)}{h} \biggr) \biggr] \\ =&\frac{\beta}{h}{_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} \biggl[ \frac {1}{\alpha} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}+1} g(t+h) \biggr] - \frac{\alpha}{h}{_{h}} \Delta_{(\alpha,\beta)}^{-1}{_{h}} \biggl[ \frac {1}{\alpha} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} g(t) \biggr] \\ =&\frac{\beta}{h}{_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} f(t+h) - \frac{\alpha}{h}{_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} f(t) \\ =&\beta\sum_{k=\frac{a}{h}}^{\frac{t}{h}-1} \frac{\alpha^{k-1}}{\beta ^{k}} g \bigl( (k+1)h \bigr) - \alpha\sum _{k=\frac{a}{h}}^{\frac{t}{h}-1} \frac{\alpha^{k-1}}{\beta^{k}} g ( kh ) \\ =&\beta\sum_{k=\frac{a}{h}+1}^{\frac{t}{h}} \frac{\alpha^{k-2}}{\beta ^{k-1}} g ( kh ) - \alpha\sum_{k=\frac{a}{h}}^{\frac{t}{h}-1} \frac{\alpha^{k-1}}{\beta^{k}} g ( kh ) \\ =&\beta \Biggl( \sum_{k=\frac{a}{h}}^{\frac{t}{h}-1} \frac{\alpha^{k-2}}{\beta^{k-1}} g(kh)+\frac{\alpha^{ \frac{t}{h}-2}}{\beta^{\frac{t}{h}-1}} g(t)-\frac {\alpha^{ \frac{a}{h}-2}}{\beta^{\frac{a}{h}-1}} g(a) \Biggr) - \alpha\sum_{k=\frac{a}{h}}^{\frac{t}{h}-1} \frac{\alpha^{k-1}}{\beta^{k}} g ( kh ) \\ =&\frac{\beta^{2}}{\alpha} \sum_{k=\frac{a}{h}}^{\frac{t}{h}-1} \frac{\alpha^{k-1}}{\beta^{k}} g(kh)+\frac{\alpha^{ \frac{t}{h}-2}}{\beta^{\frac{t}{h}-2}} g(t)-\frac {\alpha^{ \frac{a}{h}-2}}{\beta^{\frac{a}{h}-2}} g(a) -\alpha \sum_{k=\frac{a}{h}}^{\frac{t}{h}-1} \frac{\alpha^{k-1}}{\beta ^{k}} g ( kh ) \\ =&\frac{\beta^{2}}{\alpha} \bigl[ f(t)-f(a) \bigr] + \biggl( \frac{\beta ^{2}-\alpha^{2}}{\alpha h} \biggr) {_{h}}\Delta_{(\alpha,\beta)}^{-1}f(t). \end{aligned}$$

This completes the proof. □

Example 2.3

• (i) If $f(t)=1$ for $t \in(h{\mathbb{N}})_{a}$, then we have

  $$\begin{aligned}& {_{h}}\Delta_{(\alpha,\beta)}{_{h}}\Delta_{(\alpha,\beta)}^{-1}(1) = \beta+(\beta-\alpha) \sum_{k=\frac{a}{h}}^{\frac{t}{h}-1}(1)= \beta + \biggl( \frac{\beta-\alpha}{h} \biggr) (t-a), \\& {_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} \Delta_{(\alpha,\beta)}(1) = \beta ( 1-1 ) + \biggl( \frac{\beta-\alpha}{h} \biggr) {_{h}}\Delta_{(\alpha ,\beta)}^{-1}(1)= \biggl( \frac{\beta-\alpha}{h} \biggr) (t-a). \end{aligned}$$

  Note that if f is defined on $(h{\mathbb{N}})_{ (\frac{\beta }{\beta-\alpha} ) h}$, we obtain

  $$\begin{aligned}& {_{h}}\Delta_{(\alpha,\beta)}{_{h}}\Delta_{(\alpha,\beta)}^{-1}(1) = \biggl( \frac{\beta-\alpha}{h} \biggr)t, \\& {_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} \Delta_{(\alpha,\beta)}(1) = \biggl( \frac{\beta-\alpha}{h} \biggr)t-\beta. \end{aligned}$$

• (ii) If $f(t)=t$ for $t \in(h{\mathbb{N}})_{a}$, then we have

  $$\begin{aligned}& {_{h}}\Delta_{(\alpha,\beta)}{_{h}}\Delta_{(\alpha,\beta)}^{-1}(t)= \beta t+(\beta-\alpha) \sum_{k=\frac{a}{h}}^{\frac{t}{h}-1}(kh)= \beta t+ \biggl( \frac{\beta-\alpha}{2h} \biggr) (t-a) (t+a-h), \\& \begin{aligned} {_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} \Delta_{(\alpha,\beta)}(t)&=\beta ( t-a ) + \biggl( \frac{\beta-\alpha}{h} \biggr) {_{h}}\Delta_{(\alpha ,\beta)}^{-1}(t) \\ &=\beta ( t-a ) + \biggl( \frac{\beta-\alpha}{2h} \biggr) (t-a) (t+a-h). \end{aligned} \end{aligned}$$

  Note that if f is defined on $(h{\mathbb{N}})_{ (\frac{\beta }{\beta-\alpha} ) h}$, we obtain

  $$\begin{gathered} {_{h}}\Delta_{(\alpha,\beta)}{_{h}}\Delta_{(\alpha,\beta)}^{-1}(t) = \frac {1}{2} \biggl( \frac{\beta-\alpha}{h} \biggr) t^{2}+ \biggl( \alpha+\frac {\beta}{2} \biggr) t -\frac{1}{2} \biggl( \frac{\alpha}{\beta-\alpha} \biggr)h, \\ {_{h}}\Delta_{(\alpha,\beta)}^{-1}{_{h}} \Delta_{(\alpha,\beta)}(t) = \frac {1}{2} \biggl( \frac{\beta-\alpha}{h} \biggr) t^{2}+ \biggl( \alpha+\frac {\beta}{2} \biggr) t -\frac{1}{2} \biggl( \frac{2\beta^{2}+\alpha}{\beta -\alpha} \biggr)h. \end{gathered} $$

To study the solution of the boundary value problems (1.1), we need the following lemmas that deals with linear variant of the boundary value problems (1.1).

Lemma 2.2

Let $\alpha,\beta,h>0$, $\alpha<\beta$, function $x\in C ( (h{\mathbb{N}} ) _{0, (T+1 ) h} ,[0,\infty) )$ and functionals $\phi_{1},\phi_{2}:C ( (h{\mathbb {N}} ) _{0, (T+1 ) h},[0,\infty) )\rightarrow [0,\infty)$ be given. Then the problem

$$\begin{aligned} \begin{aligned} &{_{h}}\Delta^{2}_{(\alpha,\beta)} u(t-h)+x(t)=0,\quad t\in (h{\mathbb {N}} ) _{h,Th}, \\ &{_{h}}\Delta_{(\alpha,\beta)} \biggl[ \biggl( \frac{\beta}{\alpha} \biggr)^{\frac{t}{h}}u(t) \biggr]_{t=0} = \phi_{1}, \\ &{_{h}}\Delta_{(\alpha,\beta)} \biggl[ \biggl( \frac{\beta}{\alpha} \biggr)^{\frac{t}{h}}u( t) \biggr]_{t=Th} = \phi_{2}, \end{aligned} \end{aligned}$$

(2.1)

has the unique solution

$$\begin{aligned} u(t) &= \frac{\beta h}{(\beta-\alpha)^{2}T} \Biggl\{ \bigl[ \bigl[ (\beta- \alpha)T+\beta \bigr] \phi_{1}(u)-\beta\phi _{2}(u) \bigr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \\ &\quad {}-\frac{h}{\alpha\beta}\sum_{i=1}^{T} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-i} \bigl[ (T+1-i)\beta-(T-i)\alpha \bigr] x ( ih ) \Biggr\} -\frac{ t }{(\beta-\alpha)T } \\ &\quad {}\times \Biggl\{ \bigl[\phi_{1}(u)-\phi_{2}(u) \bigr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} -\frac{h}{\alpha\beta}\sum _{i=1}^{T} \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{t}{h}-i} \bigl[ (T+1-i)\beta-(T-i)\alpha \bigr] x ( ih ) \Biggr\} \\ &\quad {}-\frac{h^{2}}{\alpha\beta}\sum_{i=1}^{\frac{t}{h}-1} \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{t}{h}-i} \biggl[ \frac{t}{h}-i \biggr] x ( ih ),\quad t\in (h{\mathbb{N}} ) _{0, (T+1 ) h} . \end{aligned}$$

(2.2)

Proof

From ${_{h}}\Delta^{2}_{(\alpha,\beta)}u(t-h)=\frac{1}{h} [ \beta {_{h}}\Delta_{(\alpha,\beta)} u(t)-\alpha{_{h}}\Delta_{(\alpha,\beta)} u(t-h) ]$ and the first equation of (2.1), we create the system of n equations

$$\begin{aligned}& (A_{1})\quad\beta{_{h}}\Delta_{(\alpha,\beta)} u(t)- \alpha{_{h}}\Delta _{(\alpha,\beta)} u(t-h)=-hx(t), \\& (A_{2})\quad\beta{_{h}}\Delta_{(\alpha,\beta)} u(t-h)-\alpha{_{h}}\Delta _{(\alpha,\beta)} u(t-2h)=-hx(t-h), \\& (A_{3}) \quad\beta{_{h}}\Delta_{(\alpha,\beta)} u(t-2h)-\alpha{_{h}}\Delta _{(\alpha,\beta)} u(t-3h)=-hx(t-2h), \\& \ldots \\& (A_{n-1}) \quad\beta{_{h}}\Delta_{(\alpha,\beta)} u(2h)-\alpha {_{h}}\Delta_{(\alpha,\beta)} u(h)=-hx(2h), \\& (A_{n}) \quad\beta{_{h}}\Delta_{(\alpha,\beta)} u(h)-\alpha{_{h}}\Delta _{(\alpha,\beta)} u(0)=-hx(h). \end{aligned}$$

Considering $\beta\cdot(A_{1})+\alpha\cdot(A_{2})$, we obtain

$$\begin{aligned}& (A_{n+1}) \quad\beta^{2} {_{h}} \Delta_{(\alpha,\beta)} u(t)-\alpha^{2} {_{h}} \Delta_{(\alpha,\beta)} u(t-2h)=-h\sum_{k=0}^{1} \alpha^{k}\beta^{1-k}x(t-kh). \end{aligned}$$

For $\beta\cdot(A_{n+1})+\alpha^{2} \cdot(A_{3})$, we have the next equation:

$$\begin{aligned}& (A_{n+2}) \quad\beta^{3} {_{h}} \Delta_{(\alpha,\beta)} u(t)-\alpha^{3} {_{h}} \Delta_{(\alpha,\beta)} u(t-3h)=-h\sum_{k=0}^{2} \alpha^{k}\beta ^{2-k}x(t-kh). \\& \ldots \end{aligned}$$

Repeating the same process, we have the equation $(A_{2n-2})$

$$\begin{aligned}& (A_{2n-2}) \quad\beta^{n-1} {_{h}} \Delta_{(\alpha,\beta)} u(t)-\alpha ^{n-1} {_{h}} \Delta_{(\alpha,\beta)} u \bigl(t-(n-1)h \bigr) =-h\sum _{k=0}^{n-2}\alpha^{k}\beta^{n-2-k}x(t-kh). \end{aligned}$$

Finally, for $\beta\cdot(A_{2n-2})+\alpha^{n-1} \cdot(A_{n})$, we get the equation $(A_{2n-1})$

$$\begin{aligned}& (A_{2n-1}) \quad\beta^{n} {_{h}} \Delta_{(\alpha,\beta)} u(t)-\alpha^{n} {_{h}} \Delta_{(\alpha,\beta)} u(0)=-h\sum_{k=0}^{n-1} \alpha^{k}\beta^{n-1-k}x(t-kh). \end{aligned}$$

So, we obtain

$$\begin{aligned} {_{h}}\Delta_{(\alpha,\beta)} u(t) =& \biggl( \frac{\alpha}{\beta} \biggr) ^{n} {_{h}}\Delta_{(\alpha,\beta)} u(0)-h\sum_{k=0}^{n-1}\frac{\alpha ^{k}}{\beta^{k+1}}x(t-kh). \end{aligned}$$

(2.3)

For $t=nh \in(h{\mathbb{N}})_{0,Th}$, $n\in{\mathbb{N}}$, we can write (2.3) in the form

$$\begin{aligned}& {_{h}}\Delta_{(\alpha,\beta)} u(t)= \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} {_{h}}\Delta_{(\alpha,\beta)} u(0)-h\sum _{i=1}^{{\frac {t}{h}}}\frac{\alpha^{\frac{t}{h}-i}}{\beta^{{\frac {t}{h}}+1-i}}x(ih)\quad\text{and} \end{aligned}$$

(2.4)

$$\begin{aligned}& \beta u(t+h)-\alpha u(t)=h \biggl( \frac{\alpha}{\beta} \biggr)^{\frac {t}{h}} {_{h}}\Delta_{(\alpha,\beta)} u(0)-h^{2} \sum_{i=1}^{{\frac {t}{h}}}\frac{\alpha^{\frac{t}{h}-i}}{\beta^{{\frac{t}{h}}+1-i}}x(ih), \end{aligned}$$

(2.5)

where $\sum_{s=p}^{q} y(s)=0$; if $p< q$. By substituting $t=0,h,\ldots ,nh$, $n\in{\mathbb{N}}$ into (2.5), we have the system of n equations

$$\begin{aligned}& (B_{1}) \quad\beta u(h)-\alpha u(0)=h {_{h}} \Delta_{(\alpha,\beta)} u(0), \\& (B_{2}) \quad\beta u(2h)-\alpha u(h)=h \biggl( \frac{\alpha}{\beta} \biggr) {_{h}}\Delta_{(\alpha,\beta)} u(0)- \frac{h^{2}}{\beta}x(h), \\& (B_{3}) \quad\beta u(3h)-\alpha u(2h)=h \biggl( \frac{\alpha}{\beta} \biggr)^{2} {_{h}}\Delta_{(\alpha,\beta)} u(0)-h^{2}\sum_{i=1}^{2} \frac{\alpha ^{2-i}}{\beta^{{3-i}}}x(ih), \\& \ldots \\& (B_{n}) \quad\beta u(nh)-\alpha u \bigl((n-1)h \bigr)=h \biggl( \frac{\alpha }{\beta} \biggr)^{n-1} {_{h}}\Delta_{(\alpha,\beta)} u(0)-h^{2}\sum_{i=1}^{n-1} \frac{\alpha^{n-1-i}}{\beta^{n-i}}x(ih). \end{aligned}$$

Considering $\alpha\cdot(B_{1})+\beta\cdot(B_{2})$, we get

$$\begin{aligned}& (B_{n+1}) \quad\beta^{2} u(2h)-\alpha^{2} u(0)=2 \alpha h {_{h}}\Delta _{(\alpha,\beta)} u(0)-h^{2}x(h). \end{aligned}$$

Next, for $\alpha\cdot(B_{n+1})+\beta^{2} \cdot(B_{3})$, we get

$$\begin{aligned}& (B_{n+2}) \quad\beta^{3}u(3h)-\alpha^{3} u(0)=3 \alpha^{2} h {_{h}}\Delta _{(\alpha,\beta)} u(0) -h^{2} \sum_{i=1}^{2} (3-i) \alpha^{2-i}\beta^{i-1}x(ih). \\& \ldots \end{aligned}$$

Repeating this process, we have the equation $(B_{2n})$

$$\begin{aligned}& (B_{2n}) \quad\beta^{n}u (nh )-\alpha^{n} u(0)=n\alpha^{n-1} h {_{h}}\Delta_{(\alpha,\beta)} u(0) -h^{2} \sum_{i=1}^{n-1} (n-i) \alpha^{n-1-i}\beta^{i-1}x(ih). \end{aligned}$$

For $t=nh$, $n\in{\mathbb{N}}$, we can write $(B_{2n})$ in the form

$$\begin{aligned} u(t) =& \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}}u(0)+ \frac{t }{\alpha} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} {_{h}}\Delta_{(\alpha,\beta)}u(0) \\ &{}-\frac{h^{2}}{\alpha\beta}\sum_{i=1}^{{\frac{t}{h}}-1} \biggl[\frac {t}{h}-i \biggr] \biggl( \frac{\alpha}{\beta} \biggr) ^{\frac {t}{h}-i}x(ih),\quad t \in(h{\mathbb{N}})_{0,(T+1)h}. \end{aligned}$$

(2.6)

Hence

$$\begin{aligned} \biggl( \frac{\beta}{\alpha} \biggr)^{\frac{t}{h}}u(t) =&u(0)+ \frac{t }{\alpha} {_{h}}\Delta_{(\alpha,\beta)}u(0) -\frac{h^{2}}{\alpha\beta} \sum_{i=1}^{{\frac{t}{h}}-1} \biggl[\frac {t}{h}-i \biggr] \biggl( \frac{\beta}{\alpha} \biggr) ^{ i} x(ih) \\ =:&A+\frac{t }{\alpha} B-\frac{h^{2}}{\alpha\beta}\sum _{i=1}^{\frac {t}{h}-1} \biggl[ \frac{t}{h}-i \biggr] \biggl( \frac{\beta}{\alpha} \biggr) ^{ i} x ( ih ), \end{aligned}$$

(2.7)

for $t \in(h{\mathbb{N}})_{0,(T+1)h}$ and A, B are some constants.

Using the $(\alpha,\beta)$-difference for (2.7), we have

$$\begin{aligned} {_{h}}\Delta_{(\alpha,\beta)} \biggl[ \biggl( \frac{\beta}{\alpha} \biggr)^{\frac{t}{h}}u(t) \biggr] =&A \biggl( \frac{\beta-\alpha}{h} \biggr) +B \biggl[ \biggl( \frac{\beta -\alpha}{h} \biggr) \frac{t }{\alpha} +\frac{\beta}{\alpha} \biggr] \\ &{} -\frac{h}{\alpha\beta}\sum_{i=1}^{\frac{t}{h}} \biggl[ \biggl( \frac {t}{h}-i+1 \biggr)\beta- \biggl( \frac{t}{h}-i \biggr)\alpha \biggr] \biggl( \frac{\beta}{\alpha} \biggr)^{ i}x ( ih ). \end{aligned}$$

(2.8)

By using the boundary conditions of (2.1), we obtain

$$\begin{aligned}& \phi_{1}(u)= \biggl( \frac{\beta-\alpha}{h} \biggr)A + \biggl( \frac{\beta }{\alpha} \biggr) B, \end{aligned}$$

(2.9)

$$\begin{aligned}& \begin{aligned}[b] \phi_{2}(u)&= \biggl( \frac{\beta-\alpha}{h} \biggr)A + \biggl( \frac{(\beta -\alpha)T+\beta}{\alpha} \biggr) B \\ &\quad {} -\frac{h}{\alpha\beta}\sum_{i=1}^{T} \bigl[ ( T+1-i )\beta- ( T-i )\alpha \bigr] \biggl( \frac{\beta}{\alpha} \biggr)^{ i} x ( ih ).\end{aligned} \end{aligned}$$

(2.10)

The constants A and B can be obtained by solving the system of equations (2.9) and (2.10),

$$\begin{aligned}& \begin{aligned} A&= \frac{\beta h}{(\beta-\alpha)^{2}T} \Biggl\{ \bigl[ (\beta- \alpha)T+\beta \bigr] \phi_{1}(u)-\beta\phi_{2}(u)- \frac {h}{\alpha\beta}\sum_{i=1}^{T} \biggl( \frac{\beta}{\alpha} \biggr)^{i} \\ &\quad{}\times \bigl[ (T+1-i)\beta-(T-i)\alpha \bigr] x ( ih ) \Biggr\} , \end{aligned} \\& B=- \frac{ \alpha}{(\beta-\alpha)T } \Biggl\{ \phi_{1}(u)- \phi_{2}(u)-\frac{h}{\alpha\beta}\sum_{i=1}^{T} \biggl( \frac{\beta}{\alpha} \biggr)^{i} \bigl[ (T+1-i)\beta-(T-i)\alpha \bigr] x ( ih ) \Biggr\} . \end{aligned}$$

Substituting all constants A, B into (2.7), we obtain (2.2). This completes the proof. □

Next, we present the following lemma that deals with linear variant of the boundary value problem (1.2).

Lemma 2.3

Let $\alpha,\beta,h>0$, $\alpha<\beta$, function $y\in C ( (h{\mathbb{N}} ) _{ (\frac{\beta}{\beta -\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h} ,[0,\infty) )$ and functional $\theta:C ( (h{\mathbb{N}} ) _{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta }{\beta-\alpha} ) h},[0,\infty) )\rightarrow [0,\infty)$ be given. Then the problem

$$\begin{aligned} \begin{aligned} &{_{h}}\Delta^{2}_{(\alpha,\beta)} u(t-h)+y(t)=0,\quad t\in (h{\mathbb {N}} ) _{ (1+\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h} \\ &{_{h}}\Delta_{(\alpha,\beta)}u \biggl( \biggl( \frac{\beta}{\beta-\alpha } \biggr) h \biggr) =\frac{\beta}{h} u \biggl( \biggl( \frac{\beta}{\beta -\alpha}+1 \biggr) h \biggr) , \\ & \biggl[ {_{h}}\Delta_{(\alpha,\beta)} \biggl[ \biggl( \frac{\beta}{\alpha } \biggr)^{t}u( t) \biggr] \biggr]_{t= (T+\frac{\beta}{\beta-\alpha } ) h} = \theta(u), \end{aligned} \end{aligned}$$

(2.11)

has the unique solution

$$ \begin{aligned}[b] u(t)& = \biggl[ \frac{t}{(\beta-\alpha)T +2\beta} \biggr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \Biggl\{ \theta(u)+\frac{h}{\alpha\beta} \sum_{i=\frac{\beta}{\beta -\alpha}+1}^{T+\frac{\beta}{\beta-\alpha}} \biggl( \frac{\beta}{\alpha } \biggr)^{i} \\ &{}\quad \times \bigl[ (\beta-\alpha) (T-i)+2\beta \bigr] y ( ih ) \Biggr\} - \frac{h^{2}}{\alpha\beta}\sum_{i=\frac{\beta}{\beta-\alpha}+1}^{\frac {t}{h}-1} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-i} \biggl[ \frac {t}{h}-i \biggr] y ( ih ). \end{aligned} $$

(2.12)

Proof

With the same argument of Lemma 2.2, we obtain

$$\begin{aligned} \begin{aligned}[b] u(t)&= \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}}u \biggl( \biggl(\frac{\beta}{\beta-\alpha} \biggr) h \biggr)+ \frac{t }{\alpha} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} {_{h}}\Delta_{(\alpha,\beta)}u \biggl( \biggl(\frac{\beta}{\beta-\alpha } \biggr) h \biggr) \\ &\quad {}-\frac{h^{2}}{\alpha\beta}\sum_{i=\frac{\beta}{\beta-\alpha }+1}^{{\frac{t}{h}}-1} \biggl[\frac{t}{h}-i \biggr] \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{t}{h}-i} y(ih). \end{aligned} \end{aligned}$$

(2.13)

Using the first boundary condition of (2.11), we obtain $u ( (\frac{\beta}{\beta-\alpha} ) h )=0$.

So, (2.13) can be written in the form

$$\begin{aligned} \biggl( \frac{\beta}{\alpha} \biggr)^{\frac{t}{h}}u(t) =& \frac{t }{\alpha} {_{h}}\Delta_{(\alpha,\beta)}u \biggl( \biggl( \frac{\beta }{\beta-\alpha} \biggr) h \biggr) -\frac{h^{2}}{\alpha\beta}\sum _{i=\frac{\beta}{\beta-\alpha}+1}^{\frac {t}{h}-1} \biggl[ \frac{t}{h}-i \biggr] \biggl( \frac{\beta}{\alpha} \biggr) ^{ i}y ( ih ) \\ =:&\frac{t }{\alpha}C-\frac{h^{2}}{\alpha\beta}\sum _{i=\frac{\beta}{\beta -\alpha}+1}^{\frac{t}{h}-1} \biggl[ \frac{t}{h}-i \biggr] \biggl( \frac {\beta}{\alpha} \biggr) ^{ i} y ( ih ), \end{aligned}$$

(2.14)

for $t \in(h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}$; C is some constant.

Using the $(\alpha,\beta)$-difference for (2.14), we have

$$\begin{aligned} {_{h}}\Delta_{(\alpha,\beta)} \biggl[ \biggl( \frac{\beta}{\alpha} \biggr)^{\frac{t}{h}}u(t) \biggr] =&C \biggl[ \biggl( \frac{\beta-\alpha}{h} \biggr) \frac{t }{\alpha} +\frac{\beta}{\alpha} \biggr] \\ &{} -\frac{h}{\alpha\beta}\sum_{i=\frac{\beta}{\beta-\alpha}+1}^{\frac {t}{h}} \biggl[ \biggl( \frac{t}{h}-i+1 \biggr)\beta- \biggl( \frac {t}{h}-i \biggr)\alpha \biggr] \biggl( \frac{\beta}{\alpha} \biggr)^{ i} y ( ih ). \end{aligned}$$

(2.15)

By using the second boundary conditions of (2.11), we obtain

$$\begin{aligned} \theta(u) =&\frac{C}{\alpha} \bigl[(\beta-\alpha)T +2\beta \bigr] \\ &{} -\frac{h}{\alpha\beta}\sum_{i=\frac{\beta}{\beta-\alpha}+1}^{T+\frac {\beta}{\beta-\alpha}} \bigl[ ( T+2-i )\beta- ( T-i )\alpha \bigr] \biggl( \frac{\beta}{\alpha} \biggr)^{ i} y ( ih ). \end{aligned}$$

(2.16)

Solving the above equation, we have

$$\begin{aligned} C =& \frac{\alpha}{(\beta-\alpha) T+2\beta} \Biggl\{ \theta(u)+\frac{h}{\alpha\beta}\sum _{i=\frac{\beta}{\beta -\alpha}+1}^{T+\frac{\beta}{\beta-\alpha}} \biggl( \frac{\beta}{\alpha } \biggr)^{i} \bigl[ (\beta-\alpha) (T-i)+2\beta \bigr] y ( ih ) \Biggr\} . \end{aligned}$$

Substituting constants C into (2.14), we obtain (2.12). This completes the proof. □

Lemma 2.4

Problem (2.11) has the unique solution

$$ u(t)= {\mathcal{C}}(t)\theta(u)+\sum _{s=1}^{T}G \biggl( \frac {t}{h},s+ \frac{\beta}{\beta-\alpha} \biggr) y \biggl( \biggl(s+\frac{\beta }{\beta-\alpha} \biggr) h \biggr), $$

(2.17)

where

$$\begin{aligned} {\mathcal{C}}(t)= \biggl[ \frac{t}{(\beta-\alpha)T +2\beta} \biggr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}}, \end{aligned}$$

(2.18)

and

$$ G \biggl( \frac{t}{h},s \biggr)=\frac{h^{2}}{\alpha\beta} \textstyle\begin{cases} g_{1} ( \frac{t}{h},s ), & s\in{\mathbb{N}}_{\frac{\beta }{\beta-\alpha}+1,\frac{t}{h}-1} ,\\ g_{2} ( \frac{t}{h},s ), & s\in{\mathbb{N}}_{\frac {t}{h},T+\frac{\beta}{\beta-\alpha}}, \end{cases} $$

(2.19)

with

$$\begin{aligned}& g_{1} \biggl( \frac{t}{h},s \biggr) = \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-s} \biggl\{ \biggl(\frac{\frac{t}{h}}{(\beta-\alpha)T +2\beta} \biggr) \bigl[ (\beta-\alpha) (T-s)+2\beta \bigr]- \biggl[ \frac {t}{h}-s \biggr] \biggr\} , \\& g_{2} \biggl( \frac{t}{h},s \biggr) = \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-s} \biggl(\frac{\frac{t}{h}}{(\beta-\alpha)T +2\beta} \biggr) \bigl[ (\beta-\alpha) (T-s)+2\beta \bigr]. \end{aligned}$$

Proof

The unique solution of problem (2.1) can be written as

$$\begin{aligned} u(t) =& \biggl[ \frac{t}{(\beta-\alpha)T +2\beta} \biggr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \theta(u)+ \biggl[ \frac{t}{(\beta-\alpha)T +2\beta} \biggr] \frac{h}{\alpha\beta} \\ &{}\times \Biggl\{ \sum_{s=\frac{\beta}{\beta-\alpha}+1}^{\frac{t}{h}-1} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-s} \bigl[ (\beta-\alpha ) (T-s)+2 \beta \bigr] y ( sh ) +\sum_{s=\frac{t}{h}}^{T+\frac {\beta}{\beta-\alpha}} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac {t}{h}-s} \\ &{}\times \bigl[ (\beta-\alpha) (T-s)+2\beta \bigr] y ( sh ) \Biggr\} -\frac{h^{2}}{\alpha\beta}\sum_{s=\frac{\beta}{\beta-\alpha}+1}^{\frac {t}{h}-1} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-s} \biggl[ \frac {t}{h}-s \biggr] x ( sh ) \\ =& {\mathcal{C}}(t) \theta(u)+ \frac{h^{2}}{\alpha\beta}\sum _{s=\frac{\beta}{\beta-\alpha }+1}^{\frac{t}{h}-1} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-s} \Biggl\{ \biggl(\frac{\frac{t}{h}}{(\beta-\alpha)T +2\beta} \biggr) \\ &{}\times \bigl[ (\beta-\alpha) (T-s)+2\beta \bigr] -\sum _{s=\frac{\beta}{\beta -\alpha}+1}^{\frac{t}{h}-1} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac {t}{h}-s} \biggl[ \frac{t}{h}-s \biggr] \Biggr\} y ( ih ) \\ &{}+ \frac{h^{2}}{\alpha\beta}\sum_{s=\frac{t}{h}}^{T+\frac{\beta}{\beta -\alpha}} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-s} \biggl(\frac {\frac{t}{h}}{(\beta-\alpha)T +2\beta} \biggr) \bigl[ (\beta-\alpha) (T-s)+2\beta \bigr] y ( sh ) \\ =&{\mathcal{C}}(t)\theta(u)+\sum_{s=\frac{\beta}{\beta-\alpha }+1}^{T+\frac{\beta}{\beta-\alpha}} G \biggl( \frac{t}{h},s \biggr) y(sh) \\ =&{\mathcal{C}}(t)\theta(u)+\sum_{s=1}^{T}G \biggl( \frac{t}{h},s+\frac {\beta}{\beta-\alpha} \biggr) y \biggl( \biggl(s+ \frac{\beta}{\beta-\alpha } \biggr) h \biggr). \end{aligned}$$

This completes the proof. □

3 Existence and uniqueness of a solution for problem (1.1)

In this section, we present the existence and uniqueness result for problem (1.1). Let $E=C ((h{\mathbb{N}})_{0,(T+1)h}, {\mathbb{R}} )$ be a Banach space of all function u with the norm defined by

$$\Vert u \Vert =\max_{t\in(h{\mathbb{N}})_{0,(T+1)h}}\bigl\{ \bigl\vert u(t) \bigr\vert \bigr\} , $$

and also define an operator ${\mathcal{F}}:E\rightarrow E$

$$\begin{aligned} ({\mathcal{F}}u) (t) =& \frac{\beta h}{(\beta-\alpha)^{2}T} \Biggl\{ \bigl[ \bigl[ (\beta-\alpha)T+\beta \bigr] \phi_{1}(u)-\beta\phi _{2}(u) \bigr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \\ &{}-\frac{h}{\alpha\beta}\sum_{i=1}^{T} \biggl( \frac{\alpha}{\beta } \biggr)^{\frac{t}{h}-i} \bigl[ (T+1-i)\beta-(T-i)\alpha \bigr] f \bigl( ih,u(ih),(\Psi u) (ih) \bigr) \Biggr\} \\ &{}-\frac{ t }{(\beta-\alpha)T } \Biggl\{ \bigl[\phi_{1}(u)- \phi_{2}(u) \bigr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \\ &{}-\frac{h}{\alpha\beta}\sum_{i=1}^{T} \biggl( \frac{\alpha}{\beta } \biggr)^{\frac{t}{h}-i} \bigl[ (T+1-i)\beta-(T-i)\alpha \bigr] f \bigl( ih,u(ih),(\Psi u) (ih) \bigr) \Biggr\} \\ &{}-\frac{h^{2}}{\alpha\beta}\sum_{i=1}^{\frac{t}{h}-1} \biggl( \frac {\alpha}{\beta} \biggr)^{\frac{t}{h}-i} \biggl[ \frac{t}{h}-i \biggr] f \bigl( ih,u(ih),(\Psi u) (ih) \bigr). \end{aligned}$$

(3.1)

Obviously, problem (1.1) has solutions if and only if the operator $\mathcal{F}$ has fixed points.

Theorem 3.1

Assume that function $f\in C ( (h{\mathbb{N}} ) _{0, (T+1 ) h} \times[0,\infty),[0,\infty) )$, functionals $\phi_{1},\phi_{2}:C ( (h{\mathbb{N}} ) _{0, (T+1 ) h},[0,\infty) )\rightarrow [0,\infty)$, and for $\varphi\in ( (h{\mathbb{N}} ) _{0, (T+1 ) h} \times (h{\mathbb{N}} ) _{0, (T+1 ) h} ,[0,\infty) )$, with $\varphi_{0}=\max \{\varphi(t,s):(t,s)\in (h{\mathbb{N}} ) _{0, (T+1 ) h}\times (h{\mathbb{N}} ) _{0, (T+1 ) h} \}$. In addition, suppose that:

$(H_{1})$ :

There exist constants $\lambda_{1},\lambda_{2}>0$, such that

$$\begin{aligned} \bigl\vert f(t,u,\Psi u)-f(t,v,\Psi v) \bigr\vert \leq& \lambda_{1} \Vert u-v \Vert +\lambda_{2} \Vert \Psi u- \Psi v \Vert , \end{aligned}$$

for each $t\in(h{\mathbb{N}})_{0,(T+1)h}$ and $u, v\in E$.

$(H_{2})$ :

There exist constants $\ell_{1},\ell_{2}>0$, such that

$$\begin{aligned}& \bigl\vert \phi_{1}(u)-\phi_{1}(v) \bigr\vert \leq \ell_{1} \Vert u-v \Vert , \\& \bigl\vert \phi_{2}(u)-\phi_{2}(v) \bigr\vert \leq \ell_{2} \Vert u-v \Vert , \end{aligned}$$

for each $u, v\in E$.

$(H_{3})$ :

$\Theta:=\ell\Omega+ ( \lambda_{1}+\lambda_{2}\frac {h\varphi_{0}(T+1)}{\alpha} )\Phi < 1$, where

$$\begin{aligned}& \ell := \max\{\ell_{1},\ell_{2}\}, \end{aligned}$$

(3.2)

$$\begin{aligned}& \Omega := \frac{ h }{(\beta-\alpha) } \biggl\{ 1+ \beta+ \frac{2\beta^{2}}{(\beta-\alpha)T} \biggr\} , \end{aligned}$$

(3.3)

$$\begin{aligned}& \Phi := \frac{h^{2}T}{2\alpha\beta} \biggl( \frac{\beta}{\alpha} \biggr)^{T} \biggl\{ \biggl[T+\frac{\beta}{\beta-\alpha} \biggr] \biggl[1+ \frac{\beta }{\beta-\alpha} \biggr] +T+1 \biggr\} . \end{aligned}$$

(3.4)

Then problem (1.1) has a unique solution in $(h{\mathbb {N}})_{0,(T+1)h}$.

Proof

Denote

$${\mathcal{H}} \vert u-v \vert (t):=\bigl|f \bigl(t,u(t),(\Psi u) (t)\bigr)-f \bigl(t,v(t),(\Psi v) (t) \bigr)\bigr|. $$

For each $t\in(h{\mathbb{N}})_{0,(T+1)h}$ and $u,v\in{\mathcal{C}}$, we obtain

$$\begin{aligned} \bigl\vert (\Psi u) (t)-(\Psi v) (t) \bigr\vert \leq& h \varphi_{0} \sum_{s=0}^{\frac {t}{h}-1} \bigl\vert u(sh)-v(sh) \bigr\vert =\frac{h\varphi_{0}(T+1)}{\alpha} \Vert u-v \Vert \end{aligned}$$

and

$$\begin{aligned}& \bigl|({\mathcal{F}}u) (t)-({\mathcal{F}}v) (t)\bigr| \\& \quad\leq \frac{\beta h}{(\beta-\alpha)^{2}T} \Biggl\{ \bigl[ \bigl[ (\beta-\alpha)T+ \beta \bigr] \bigl\vert \phi_{1}(u)-\phi _{1}(v) \bigr\vert +\beta \bigl\vert \phi_{2}(u)-\phi_{2}(v) \bigr\vert \bigr] \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{t}{h}} \\& \qquad{}+\frac{h}{\alpha\beta}\sum_{i=1}^{T} \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{t}{h}-i} \bigl[ (T+1-i)\beta-(T-i)\alpha \bigr] {\mathcal{H}} \vert u-v \vert (ih) \Biggr\} \\& \qquad{}+\frac{ t }{(\beta-\alpha)T } \Biggl\{ \bigl[ \bigl\vert \phi_{1}(u)-\phi_{1}(v) \bigr\vert + \bigl\vert \phi_{2}(u)-\phi_{2}(v) \bigr\vert \bigr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \\& \qquad{}+\frac{h}{\alpha\beta}\sum_{i=1}^{T} \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{t}{h}-i} \bigl[ (T+1-i)\beta-(T-i)\alpha \bigr] {\mathcal{H}} \vert u-v \vert (ih) \Biggr\} \\& \qquad{}+\frac{h^{2}}{\alpha\beta}\sum_{i=1}^{\frac{t}{h}-1} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-i} \biggl[ \frac{t}{h}-i \biggr] {\mathcal{H}} \vert u-v \vert (ih) \\& \quad\leq \frac{\beta h}{(\beta-\alpha)^{2}T} \Biggl\{ \bigl[ \bigl[ (\beta-\alpha)T+ \beta \bigr] \ell_{1} \Vert u-v \Vert +\beta\ell _{2} \Vert u-v \Vert \bigr] \\& \qquad{}+\frac{h}{\alpha\beta} \bigl( \lambda_{1} \Vert u-v \Vert +\lambda _{2} \Vert \Psi u-\Psi v \Vert \bigr) \sum _{i=1}^{T} \biggl( \frac{\beta }{\alpha} \biggr)^{i} \bigl[ (T+1-i)\beta-(T-i)\alpha \bigr] \Biggr\} \\& \qquad{}+\frac{ h }{(\beta-\alpha)T} \Biggl\{ \bigl[\ell_{1} \Vert u-v \Vert +\ell_{2} \Vert u-v \Vert \bigr] \\& \qquad{}+\frac{h}{\alpha\beta} \bigl( \lambda_{1} \Vert u-v \Vert +\lambda _{2} \Vert \Psi u-\Psi v \Vert \bigr)\sum _{i=1}^{T} \biggl( \frac{\beta }{\alpha} \biggr)^{i} \bigl[ (T+1-i)\beta-(T-i)\alpha \bigr] \Biggr\} \\& \qquad{}+\frac{h^{2}}{\alpha\beta} \bigl( \lambda_{1} \Vert u-v \Vert +\lambda _{2} \Vert \Psi u-\Psi v \Vert \bigr)\sum _{i=1}^{T} \biggl( \frac{\beta }{\alpha} \biggr)^{i} [ T+1-i ] \\& \quad\leq \frac{\beta h}{(\beta-\alpha)^{2}T} \biggl\{ \ell \bigl[ (\beta-\alpha)T+2 \beta \bigr] \Vert u-v \Vert +\frac{h}{\alpha \beta} \biggl( \frac{\beta}{\alpha} \biggr)^{T} \\& \qquad{}\times \biggl( \lambda_{1}+\lambda_{2} \frac{h\varphi_{0}(T+1)}{\alpha } \biggr)\frac{T}{2} \bigl[ (\beta-\alpha)T+\beta \bigr] \Vert u-v \Vert \biggr\} \\& \qquad{}+\frac{ h }{(\beta-\alpha)T} \biggl\{ \ell \Vert u-v \Vert + \frac{h}{\alpha\beta} \biggl( \frac{\beta}{\alpha } \biggr)^{T} \biggl( \lambda_{1}+\lambda_{2}\frac{h\varphi_{0}(T+1)}{\alpha} \biggr) \\& \qquad{}\times\frac{T}{2} \bigl[ (\beta-\alpha)T+\beta \bigr] \Vert u-v \Vert \biggr\} \\& \qquad{}+\frac{h^{2}}{\alpha\beta} \biggl( \frac{\beta}{\alpha} \biggr)^{T} \biggl( \lambda_{1}+\lambda_{2} \frac{h\varphi_{0}(T+1)}{\alpha} \biggr)\frac {1}{2}T(T+1) \Vert u-v \Vert \\& \quad= \biggl[ \ell\Omega+ \biggl( \lambda_{1}+ \lambda_{2}\frac{h\varphi _{0}(T+1)}{\alpha} \biggr)\Phi \biggr] \Vert u-v \Vert \\& \quad = \Theta \Vert u-v \Vert . \end{aligned}$$

This implies that ${\mathcal{F}}$ is a contraction. Therefore, by using the Banach fixed point theorem, ${\mathcal{F}}$ has a fixed point which is a unique solution of problem (1.1) on $t\in(h{\mathbb {N}})_{0,(T+1)h}$. □

4 Properties of Green’s function for problem (1.2)

The necessary for considering the existence of a positive solution to problem (1.2) is to prove that Green’s function $G(t,s)$ in (2.19) satisfies a variety of properties. Firstly, we prove some necessary preliminary lemmas as follows.

Lemma 4.1

The coefficient function ${\mathcal{C}}(t)$ given in (2.18) is positive and strictly increasing in t, for $t\in(h{\mathbb {N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta }{\beta-\alpha} ) h}$. In addition,

$$\begin{aligned}& \min_{t\in (h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}}{\mathcal{C}}(t) \geq \biggl[ \frac{\beta h}{(\beta-\alpha) [ (\beta-\alpha)T +2\beta ] } \biggr] \biggl( \frac{\alpha}{\beta} \biggr)^{T+1+\frac{\beta }{\beta-\alpha}},\quad\textit{and} \\& \max_{t\in(h{\mathbb{N}})_{ (\frac{\beta}{\beta -\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}}{\mathcal{C}}(t) \leq \biggl[ \frac{ [ (\beta-\alpha)(T+1)+\beta ] h}{(\beta-\alpha) [ (\beta-\alpha)T +2\beta ] } \biggr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{\beta}{\beta-\alpha}}. \end{aligned}$$

Proof

It is clear that ${\mathcal{C}}(t) \geq0$ for $t\in (h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}$. Next, we prove that ${\mathcal{C}}(t)$ is strictly increasing in $t\in(h{\mathbb {N}})_{0,(T+1)h}$. Note that the forward $(\alpha,\beta)$-difference with respect to t for ${\mathcal{C}}(t)$ is

$$\begin{aligned} \Delta_{(\alpha,\beta)} {\mathcal{C}}(t) =& \biggl[ \frac{1}{(\beta -\alpha) T+ 2\beta} \biggr] \Delta_{(\alpha,\beta)} \biggl\{ t \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \biggr\} \\ =& \biggl[ \frac{1}{(\beta-\alpha) T+ 2\beta} \biggr] \alpha \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}}= \biggl( \frac {\alpha}{\beta} \biggr)^{\frac{t}{h}} \Delta_{h}{\mathcal{C}}(t). \end{aligned}$$

From $\alpha<\beta$, it implies that

$$\begin{aligned} \Delta_{h}{\mathcal{C}}(t) =&\frac{\alpha}{(\beta-\alpha)T+2\beta} > 0. \end{aligned}$$

Hence ${\mathcal{C}}(t)$ is strictly increasing in $t\in(h{\mathbb {N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta }{\beta-\alpha} ) h}$.

Finally, observe that

$$\begin{aligned}& \begin{aligned} \min_{t\in (h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}}{\mathcal{C}}(t)&=\min _{t\in (h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}} \biggl[ \frac{t}{(\beta -\alpha)T +2\beta} \biggr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac {t}{h}} \\ &\geq \biggl[ \frac{\beta h}{(\beta-\alpha) [ (\beta-\alpha)T +2\beta ] } \biggr] \biggl( \frac{\alpha}{\beta} \biggr)^{T+1+\frac {\beta}{\beta-\alpha}}, \quad\text{and}\end{aligned} \\& \begin{aligned} \max_{t\in (h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}}{ \mathcal{C}}(t)&=\max_{t\in (h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}} \biggl[ \frac{t}{(\beta-\alpha )T +2\beta} \biggr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \\ &\leq \biggl[ \frac{ [ (\beta-\alpha)(T+1)+\beta ] h}{(\beta -\alpha) [ (\beta-\alpha)T +2\beta ] } \biggr] \biggl( \frac {\alpha}{\beta} \biggr)^{\frac{\beta}{\beta-\alpha}}. \end{aligned} \end{aligned}$$

The proof is complete. □

Corollary 4.1

Let $I:= [ \frac{h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) , \frac{3h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) ]$. There is a constant $M_{{\mathcal{C}}} \in(0,1)$ such that $\min_{t \in I} {\mathcal{C}}(t)=M_{{\mathcal{C}}}\|{\mathcal {C}}\|$ where $\|\cdot\|$ is the usual maximum norm.

Proof

Since ${\mathcal{C}}(t)$ is strictly increasing in $t\in (h{\mathbb{N}})_{0,(T+1)h}$, it follows that there exists a positive constant $M_{{\mathcal{C}}}$, such that

$$\min_{t \in I} {\mathcal{C}}(t)=M_{{\mathcal{C}}} \biggl[ \frac{ [ (\beta-\alpha)(T+1)+\beta ] h}{(\beta-\alpha ) [ (\beta-\alpha)T +2\beta ] } \biggr] \biggl( \frac{\alpha }{\beta} \biggr)^{T+1+\frac{\beta}{\beta-\alpha}} =M_{{\mathcal{C}}} \Vert {\mathcal{C}} \Vert . $$

It is clear that $M_{{\mathcal{C}}} \in(0,1)$. The proof is complete. □

Lemma 4.2

Let $\alpha<\beta$ and $G ( \frac{t}{h},s )$ be Green’s function given in (2.19). Then $G ( \frac{t}{h},s ) \geq0$ for each $(t,s)\in (h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha } ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}\times{\mathbb {N}}_{\frac{\beta}{\beta-\alpha}+1,T+\frac{\beta}{\beta-\alpha}}$.

Proof

We aim to show that $g_{i} ( \frac{t}{h},s ) >0$, $i=1,2$, for each admissible pair $(t,s)$.

Firstly, we consider the function

$$g_{2} \biggl( \frac{t}{h},s \biggr) = \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \biggl(\frac{\frac{t}{h}}{(\beta-\alpha)T +2\beta} \biggr) \bigl[ (\beta-\alpha) (T-s)+2\beta \bigr],\quad s\in{\mathbb {N}}_{\frac{t}{h},T+\frac{\beta}{\beta-\alpha}} . $$

To guarantee that $g_{2} ( \frac{t}{h},s )>0$, it suffices to show that

$$\begin{aligned} (\beta-\alpha) (T-s)+2\beta \geq& (\beta-\alpha) \biggl[ T- \biggl(T+\frac{\beta}{\beta-\alpha} \biggr) \biggr] +2\beta=\beta > 0. \end{aligned}$$

(4.1)

Thus, we conclude that $g_{2} ( \frac{t}{h},s )>0$ for their respective domains.

Next, we consider the function $g_{1} ( \frac{t}{h},s )>0$ for $s\in{\mathbb{N}}_{\frac{\beta}{\beta-\alpha}+1,\frac{t}{h}-1}$ where

$$\begin{aligned} g_{1} \biggl( \frac{t}{h},s \biggr) =& \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \biggl\lbrace \biggl(\frac{\frac{t}{h}}{(\beta-\alpha)T +2\beta} \biggr) \bigl[ (\beta-\alpha) (T-s)+2\beta \bigr] - \biggl( \frac{t}{h}-s \biggr) \biggr\rbrace . \end{aligned}$$

To guarantee that $g_{1} ( \frac{t}{h},s )>0$, it suffices to show that

$$\begin{aligned}& \frac{\frac{t}{h} [ (\beta-\alpha)(T-s)+2\beta ]}{ ( \frac {t}{h}-s ) [ (\beta-\alpha)T+2\beta ]} \\& \quad = \frac{\frac{t}{h} [ (\beta-\alpha)(T-s)+2\beta ]}{\frac {t}{h} ( [ (\beta-\alpha)T+2\beta ]-s [ \frac{(\beta -\alpha)T+2\beta}{\frac{t}{h}} ] ) } \\& \quad = \frac{ [ (\beta-\alpha)T+2\beta ]-s(\beta-\alpha)}{ [ (\beta-\alpha)T+2\beta ]-s(\beta-\alpha) [ \frac{T+\frac{2\beta }{\beta-\alpha}}{\frac{t}{h}} ] } \\& \quad \geq \frac{(T-s) (\beta-\alpha)+2\beta}{ [ (\beta-\alpha)T+2\beta ]-s(\beta-\alpha) [ \frac{T+\frac{2\beta}{\beta-\alpha}}{ T+\frac {2\beta-\alpha}{\beta-\alpha} } ] } > 1. \end{aligned}$$

So, we conclude that $g_{1} ( \frac{t}{h},s )>0$ for their respective domains.

Consequently, it follows that $g_{i} ( \frac{t}{h},s )>0$ for each $i=1,2$. Therefore, $G ( \frac{t}{h},s )>0$. □

Lemma 4.3

Let $\alpha<\beta$ and $G ( \frac{t}{h},s )$ be Green’s function given in (2.19). Then it follows that

$$\begin{aligned} \max_{t,\in(h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}} G \biggl( \frac {t}{h},s \biggr)=G ( s,s ) \quad \textit{for each } s \in {\mathbb{N}}_{\frac{\beta}{\beta-\alpha}+1,T+\frac{\beta}{\beta-\alpha}}. \end{aligned}$$

(4.2)

Proof

Our aim is to show that the forward $(\alpha,\beta )$-difference with respect to t satisfies with

$${_{t,h}}\Delta_{(\alpha,\beta)} g_{1} \biggl( \frac{t}{h},s \biggr)= \biggl(\frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} {_{t}}\Delta_{h}g_{1} \biggl( \frac{t}{h},s \biggr) < 0 $$

and

$${_{t,h}}\Delta_{(\alpha,\beta)} g_{2} \biggl( \frac{t}{h},s \biggr)= \biggl(\frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} {_{t}}\Delta_{h}g_{2} \biggl( \frac{t}{h},s \biggr) >0. $$

This implies that $g_{1}$ is decreasing and $g_{2}$ is increasing in t. So, $G ( \frac{t}{h},s )\leq G ( s,s )$ for all $(t,s)\in{\mathbb{N}}_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h} \times{\mathbb{N}}_{\frac {\beta}{\beta-\alpha}+1,T+\frac{\beta}{\beta-\alpha}}$.

Firstly, for $g_{2} ( \frac{t}{h},s )$, we have

$$\begin{aligned} {_{t,h}}\Delta_{(\alpha,\beta)} g_{2} \biggl( \frac{t}{h},s \biggr) &= \biggl[\frac{ (\beta-\alpha)(T-s)+2\beta }{(\beta-\alpha)T +2\beta} \biggr] {_{t,h}}\Delta_{(\alpha,\beta)} \biggl\lbrace \biggl( \frac {\alpha}{\beta} \biggr)^{\frac{t}{h}}\frac{t}{h} \biggr\rbrace \\ &= \biggl[\frac{ (\beta-\alpha)(T-s)+2\beta }{(\beta-\alpha)T +2\beta} \biggr] \biggl\lbrace \biggl( \frac{\alpha}{\beta} \biggr)^{\frac {t}{h}}\frac{1}{h} \biggr\rbrace \\ &= \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}}{_{t}} \Delta_{h} g_{2} \biggl( \frac{t}{h},s \biggr) > 0, \end{aligned}$$

for all $(t,s)\in{\mathbb{N}}_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h} \times{\mathbb {N}}_{\frac{\beta}{\beta-\alpha}+1,T+\frac{\beta}{\beta-\alpha}}$.

Next, for $g_{1} ( \frac{t}{h},s )$, we note that

$$\begin{aligned} g_{1} \biggl( \frac{t}{h},s \biggr)= \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \biggl\lbrace \biggl(\frac{\frac{t}{h}}{(\beta-\alpha)T +2\beta} \biggr) \bigl[ (\beta-\alpha) (T-s)+2\beta \bigr] - \biggl( \frac{t}{h}-s \biggr) \biggr\rbrace . \end{aligned}$$

Here, we obtain

$$\begin{aligned} {_{t,h}}\Delta_{(\alpha,\beta)} g_{1} \biggl( \frac{t}{h},s \biggr) =& \biggl[\frac{ (\beta-\alpha)(T-s)+2\beta }{(\beta-\alpha)T +2\beta} \biggr] \biggl\lbrace \biggl( \frac{\alpha}{\beta} \biggr)^{\frac {t}{h}}\frac{1}{h} \biggr\rbrace - \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}}\frac{1}{h} \\ =& \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \frac{1}{h} \biggl[\frac{ (\beta-\alpha)(T-s)+2\beta }{(\beta-\alpha)T +2\beta }-1 \biggr] \\ =&- \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}} \frac{1}{h} \biggl[\frac{ (\beta-\alpha)s }{(\beta-\alpha)T +2\beta} \biggr] \\ =& \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}}{_{t}} \Delta_{h} g_{1} \biggl( \frac{t}{h},s \biggr) < 0, \end{aligned}$$

for all $(t,s)\in{\mathbb{N}}_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h} \times{\mathbb {N}}_{\frac{\beta}{\beta-\alpha}+1,T+\frac{\beta}{\beta-\alpha}}$.

Now, note that

$$\begin{aligned} G \biggl( \frac{\beta}{\beta-\alpha}+1,s \biggr) \leq G ( s,s ) \quad \text{for all } s\in {\mathbb{N}}_{\frac{\beta}{\beta-\alpha }+1,T+\frac{\beta}{\beta-\alpha}}. \end{aligned}$$

Consequently, this implies that

$$\begin{aligned} \max_{t,\in(h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}} G \biggl( \frac {t}{h},s \biggr)=G ( s,s ) \quad \text{for each } s \in {\mathbb{N}}_{\frac{\beta}{\beta-\alpha}+1,T+\frac{\beta}{\beta-\alpha}}. \end{aligned}$$

Observe that $G ( s,s )=\frac{h^{2}}{\alpha}g_{2} ( s,s )=\frac{h^{2}}{\alpha} ( \frac{\alpha}{\beta} )^{s} (\frac {s}{(\beta-\alpha)T +2\beta} ) [ (\beta-\alpha)(T-s)+2\beta ]$.

Thus, by the discussion in the first paragraph of this proof, we deduce that (4.2) holds. The proof is complete. □

Lemma 4.4

Let $\alpha<\beta$ and $G ( \frac{t}{h},s )$ be Green’s function given in (2.19). There exists a number $0<\sigma<1$ such that, for $s \in{\mathbb{N}}_{\frac{\beta}{\beta-\alpha}+1,T+\frac {\beta}{\beta-\alpha}}$,

$$\begin{aligned} \min_{ \frac{h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) \leq t\leq \frac{3h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) } G \biggl( \frac{t}{h},s \biggr) \geq& \sigma\max_{t\in(h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha } ) h}}G \biggl( \frac{t}{h},s \biggr) \\ =&\sigma G ( s,s ). \end{aligned}$$

(4.3)

Proof

We define the following notation:

$$\tilde{g}_{i} \biggl( \frac{t}{h},s \biggr):= \frac{g_{i} ( \frac {t}{h},s )}{g_{2} ( s,s )},\quad i=1,2. $$

For $t< sh$ and $\frac{h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) \leq t\leq \frac{3h}{4} ( T+1+\frac{\beta}{\beta-\alpha} )$, we find that

$$\begin{aligned} \tilde{g}_{2} \biggl( \frac{t}{h},s \biggr) =& \biggl( \frac{\alpha}{\beta } \biggr)^{\frac{t}{h}-s} \frac{t}{sh} \geq \frac{1}{4} \biggl( \frac{\beta}{\alpha} \biggr)^{\frac{1}{4} [ 3 ( T+\frac{\beta}{\beta-\alpha} )-1 ] } \biggl[ \frac{T+1+\frac{\beta}{\beta-\alpha}}{ T+\frac{\beta}{\beta -\alpha} } \biggr]:=\sigma_{1}. \end{aligned}$$

(4.4)

For $t>sh$, since $g_{1} ( \frac{t}{h},s )$ is decreasing with respect to t, we have

$$\begin{aligned} \tilde{g}_{1} \biggl( \frac{t}{h},s \biggr) =& \biggl( \frac{\alpha}{\beta } \biggr)^{\frac{t}{h}-s} \biggl\lbrace \frac{\frac{t}{h} [ (\beta-\alpha)(T-s)+2\beta ] - ( \frac{t}{h}-s ) [ (\beta-\alpha)T+2\beta ]}{ s \bigl[ (\beta-\alpha) (T-s)+2\beta \bigr]} \biggr\rbrace \\ =& \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t}{h}-s} \biggl\lbrace \frac{ (\beta-\alpha) ( T-\frac{t}{h} ) +2\beta}{ (\beta-\alpha) (T-s)+2\beta } \biggr\rbrace \\ \geq&\frac{1}{4} \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{1}{4} (T+ 1+\frac{\beta}{\beta-\alpha} )} \biggl[ \frac{ \alpha}{ (\beta -\alpha)(T-1)+\beta} \biggr]:=\sigma_{2}. \end{aligned}$$

(4.5)

Finally, since $\sigma_{1}>\frac{1}{4}$ and $\sigma_{2}<1$, it follows that

$$\begin{aligned} \sigma=\min\{\sigma_{1},\sigma_{2}\}< 1. \end{aligned}$$

(4.6)

We can conclude that (4.3) holds. □

Lemma 4.5

Let φ be a nonnegative functional. Then there is $\sigma^{*}\in (0,1)$ such that

$$\begin{aligned} &\min_{\frac{h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) \leq t\leq \frac{3h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) } \Biggl\{ {\mathcal{C}}(t)\theta(u)+\sum _{s=\frac{\beta}{\beta-\alpha }+1}^{T+\frac{\beta}{\beta-\alpha}}G \biggl( \frac{t}{h},s \biggr) F \bigl(sh,u(sh)\bigr) \Biggr\} \\ &\quad \geq\sigma^{*} \max_{t\in(h{\mathbb{N}})_{ (\frac{\beta }{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}} \Biggl\{ {\mathcal{C}}(t) \theta(u)+\sum_{s=\frac{\beta}{\beta-\alpha }+1}^{T+\frac{\beta}{\beta-\alpha}}G \biggl( \frac{t}{h},s \biggr) F\bigl(sh,u(sh)\bigr) \Biggr\} . \end{aligned}$$

(4.7)

Proof

By Lemma 4.4 and Corollary 4.1, we find that there exist constants $\sigma,M_{ {\mathcal{C}}}\in(0,1)$ such that

$$\begin{aligned}& \min_{ \frac{h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) \leq t\leq \frac{3h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) } G \biggl( \frac{t}{h},s \biggr) \geq \sigma\max_{t\in(h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha } ) h}}G \biggl( \frac{t}{h},s \biggr), \end{aligned}$$

(4.8)

$$\begin{aligned}& \min_{ \frac{h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) \leq t\leq \frac{3h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) } {\mathcal{C}}(t) \geq M_{ {\mathcal{C}}} \max_{t\in(h{\mathbb {N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta }{\beta-\alpha} ) h}} {\mathcal{C}}(t). \end{aligned}$$

(4.9)

We define

$$\begin{aligned} \sigma^{*}:=\min \{\sigma,M_{ {\mathcal{C}}} \}\in(0,1). \end{aligned}$$

(4.10)

Hence, we obtain (4.7). This completes the proof. □

5 Existence of positive solution to problem (1.2)

In this section, we consider the existence of at least one positive solution for problem (1.2) by using the Krasnoselskii fixed point theorem in a cone as mentioned in Sect. 1. Let ${\mathcal {E}}=C ((h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}, {\mathbb{R}} )$ be a Banach space of all function u with the norm defined by $\|u\|=\max_{t\in(h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha } ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}}\{|u(t)|\}$. Define the cone ${\mathcal{P}}\subseteq{\mathcal{E}}$ by

$$\begin{aligned} {\mathcal{P}}:= \Bigl\{ u\in{\mathcal{E}} : u(t)\geq0, \min_{\frac {h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) \leq t\leq\frac {3h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) }u (t)\geq\sigma^{*} \Vert u \Vert \text{ and } \theta(u)\geq0 \Bigr\} , \end{aligned}$$

where $\sigma^{*}$ is the number defined by (4.10).

From the nonlinear equation (1.2), we note that there exists a solution u of (1.2) if and only if u is a fixed point of the operator ${\mathcal{A}}:{\mathcal {P}}\rightarrow{\mathcal{P}}$ which is defined by

$$\begin{aligned} ({\mathcal{A}}u ) (t) :=&{\mathcal{C}}(t)\theta(u)+\sum _{s=\frac{\beta}{\beta-\alpha}+1}^{T+\frac{\beta}{\beta-\alpha}}G \biggl( \frac{t}{h},s \biggr) F \bigl(sh,u(sh)\bigr), \end{aligned}$$

(5.1)

where G is Green’s function for problem (1.2).

Lemma 5.1

Suppose that $F\in C ( (h{\mathbb{N}} ) _{ (1+\frac {\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha } ) h} \times[0,\infty),[0,\infty) )$ and $\theta: [4] C ( (h{\mathbb{N}} ) _{ (1+\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h},[0,\infty) )\rightarrow [0,\infty)$. Then the operator $\mathcal{A}:{\mathcal {P}}\rightarrow{\mathcal{P}}$ is completely continuous.

Proof

Since $G ( \frac{t}{h},s ) \geq0$ for all $(t,s)$, we have ${\mathcal{A}}\geq0$ for all $u\in{\mathcal{P}}$. For a constant $R>0$, we define

$$B_{R}=\bigl\{ u\in{\mathcal{P}} : \Vert u \Vert < R\bigr\} , $$

and let $M= \max_{(t,u)\in (h{\mathbb{N}} ) _{ (1+\frac {\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha } ) h}\times B_{R}} \vert F (t,u(t) ) \vert$, $N=\sup_{u\in B_{R}} |\theta(u)|$. Then, for $u\in B_{R}$, we obtain

$$\begin{aligned} \bigl|({\mathcal{A}}u) (t)\bigr| =& \Biggl\vert {\mathcal{C}}(t)\theta(u)+\sum _{s=\frac{\beta}{\beta-\alpha }+1}^{T+\frac{\beta}{\beta-\alpha}}G \biggl( \frac{t}{h},s \biggr) F\bigl(sh,u(sh)\bigr) \Biggr\vert \\ \leq&N \biggl[ \frac{ [ (\beta-\alpha)(T+1)+\beta ] h}{(\beta -\alpha) [ (\beta-\alpha)T +2\beta ] } \biggr] \biggl( \frac {\alpha}{\beta} \biggr)^{T+1+\frac{\beta}{\beta-\alpha}} +\frac {Mh^{2}}{\alpha\beta} \sum_{s=\frac{\beta}{\beta-\alpha}+1}^{T+\frac{\beta }{\beta-\alpha}}g_{2} ( s,s ) \\ =:&\mathcal{K}. \end{aligned}$$

Therefore, $\|{\mathcal{A}}u \|= \mathcal{K}$, and hence ${\mathcal{A}}(B_{R})$ is uniformly bounded.

We next prove that ${\mathcal{A}}(B_{R})$ is equicontinuous. For any $\epsilon>0$, there exists a positive constant $\delta^{*}=\max \lbrace\delta_{1},\delta_{2},\delta_{3} \rbrace$ such that, for $u\in B_{R}$ and $t_{1},t_{2}\in (h{\mathbb{N}} ) _{ (1+\frac {\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha } ) h}$ with $t_{1}< t_{2}$,

$$\begin{aligned}& \biggl\vert \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t_{2}}{h}}t_{2}- \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t_{1}}{h}}t_{1} \biggr\vert < \frac {\epsilon(\beta-\alpha)T+2\beta}{3N } \quad\text{whenever } \vert t_{2}-t_{1} \vert < \delta_{1}, \\& \bigl\vert t_{2}^{2}- t_{1}^{2} \bigr\vert < \frac{\epsilon\beta^{2}}{3M }\quad\text{whenever } \vert t_{2}-t_{1} \vert < \delta_{2}, \\& \vert t_{2}-t_{1} \vert < \frac{\epsilon\alpha\beta}{3hM ( T+2+\frac{\beta}{\beta-\alpha} ) }=\delta_{3}. \end{aligned}$$

Then we have

$$\begin{aligned}& \bigl\vert ({\mathcal{A}}u ) (t_{2})- ({\mathcal{A}}u ) (t_{1}) \bigr\vert \\& \quad\leq \bigl\vert {\mathcal{C}}(t_{2})-{ \mathcal{C}}(t_{1}) \bigr\vert N+M \Biggl\vert \sum _{s=\frac{\beta}{\beta-\alpha}+1}^{T+\frac{\beta}{\beta -\alpha}} \biggl[ G \biggl( \frac{t_{2}}{h},s \biggr)-G \biggl( \frac{t_{1}}{h},s \biggr) \biggr] \Biggr\vert \\& \quad\leq\frac{N}{(\beta-\alpha)T+2\beta} \biggl\vert \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{t_{2}}{h}}t_{2}- \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t_{1}}{h}}t_{1} \biggr\vert \\& \qquad{}+ \frac{h^{2}M}{\alpha\beta} \Biggl\vert \sum _{s=\frac{\beta}{\beta-\alpha}+1}^{\frac{t_{2}}{h}-1} g_{1} \biggl( \frac{t_{2}}{h},s \biggr) - \sum_{s=\frac{\beta}{\beta-\alpha}+1}^{\frac{t_{1}}{h}-1} g_{1} \biggl( \frac{t_{1}}{h},s \biggr) \Biggr\vert \\& \qquad{}+ \frac{h^{2}M}{\alpha\beta} \Biggl\vert \sum _{s=\frac{t_{2}}{h}}^{T+1+\frac{\beta}{\beta-\alpha}} g_{2} \biggl( \frac{t_{2}}{h},s \biggr) - \sum_{s=\frac{t_{1}}{h}}^{T+1+\frac{\beta}{\beta-\alpha}} g_{2} \biggl( \frac{t_{1}}{h},s \biggr) \Biggr\vert \\& \quad< \frac{N}{(\beta-\alpha)T+2\beta} \biggl\vert \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{t_{2}}{h}}t_{2}- \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t_{1}}{h}}t_{1} \biggr\vert \\& \qquad{}+ \frac{h^{2}M}{\alpha\beta} \Biggl\vert \biggl( \frac{\alpha}{\beta } \biggr)^{\frac{t_{2}}{h}}\sum_{s=\frac{\beta}{\beta-\alpha}+1}^{\frac{t_{2}}{h}-1} \biggl( \frac{\alpha}{\beta} \biggr)^{-s} \biggl[\frac{t_{2}}{h}\cdot \frac {(\beta-\alpha)(T-s)+2\beta}{(\beta-\alpha)T+2\beta}- \biggl( \frac {t_{2}}{h}-s \biggr) \biggr] \\& \qquad{}- \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t_{1}}{h}}\sum _{s=\frac{\beta}{\beta-\alpha}+1}^{\frac{t_{1}}{h}-1} \biggl( \frac{\alpha}{\beta} \biggr)^{-s} \biggl[\frac{t_{1}}{h}\cdot\frac {(\beta-\alpha)(T-s)+2\beta}{(\beta-\alpha)T+2\beta}- \biggl( \frac {t_{1}}{h}-s \biggr) \biggr] \Biggr\vert \\& \qquad {} +\frac{h^{2}M}{\alpha\beta} \Biggl\vert \biggl( \frac{\alpha}{\beta } \biggr)^{\frac{t_{2}}{h}}\frac{t_{2}}{h}\sum _{s=\frac{t_{2}}{h} }^{T+1+\frac {\beta}{\beta-\alpha}} \biggl( \frac{\alpha}{\beta} \biggr)^{-s} \biggl[ \frac{(\beta-\alpha )(T-s)+2\beta}{(\beta-\alpha)T+2\beta} \biggr] \\& \qquad{}- \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t_{1}}{h}} \frac {t_{1}}{h}\sum_{s=\frac{t_{1}}{h} }^{T+1+\frac{\beta}{\beta-\alpha}} \biggl( \frac{\alpha}{\beta} \biggr)^{-s} \biggl[ \frac{(\beta-\alpha )(T-s)+2\beta}{(\beta-\alpha)T+2\beta} \biggr] \Biggr\vert \\& \quad < \frac{N}{(\beta-\alpha)T+2\beta} \biggl\vert \biggl( \frac{\alpha}{\beta } \biggr)^{\frac{t_{2}}{h}}t_{2}- \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t_{1}}{h}}t_{1} \biggr\vert \\& \qquad{} + \frac{h^{2}M}{\alpha\beta} \biggl\vert \frac{\alpha}{\beta} \biggl( \frac{t_{2}}{h}-1 \biggr) \biggl( \frac{t_{2}}{h}-1 - \frac{\beta}{\beta-\alpha} \biggr) - \frac {\alpha}{\beta} \biggl( \frac{t_{1}}{h}-1 \biggr) \biggl( \frac{t_{1}}{h}-1 -\frac{\beta}{\beta-\alpha} \biggr) \biggr\vert \\& \qquad{} +\frac{h^{2}M}{\alpha\beta} \biggl\vert \frac{t_{2}}{h} \biggl( T+2+ \frac{\beta}{\beta-\alpha} \biggr) - \frac{t_{1}}{h} \biggl( T+2+ \frac{\beta}{\beta-\alpha} \biggr) \biggr\vert \\& \quad< \frac{N}{(\beta-\alpha)T+2\beta} \biggl\vert \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{t_{2}}{h}}t_{2}- \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{t_{1}}{h}}t_{1} \biggr\vert + \frac{M}{ \beta} \bigl\vert t_{2}^{2}- t_{1}^{2} \bigr\vert \\& \qquad{} +\frac{hM}{\alpha\beta} \biggl( T+2+\frac{\beta}{\beta-\alpha } \biggr) \vert t_{2}-t_{1} \vert \\& \quad < \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon }{3}= \epsilon. \end{aligned}$$

This implies that the set ${\mathcal{A}}(B_{R})$ is an equicontinuous set.

Finally, we apply Lemmas 4.2–4.4 to obtain

$$\begin{aligned} ({\mathcal{A}}x) (t)\geq0 \quad \text{for all } t\in (h{\mathbb {N}} ) _{ (1+\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}, \end{aligned}$$

(5.2)

and for $F\in\mathcal{P}$

$$\begin{aligned} \min_{t \in I} ({\mathcal{A}}u ) (t)& \geq \min _{t \in I} \Biggl( \sum_{s=\frac{\beta}{\beta-\alpha}+1}^{T+\frac {\beta}{\beta-\alpha}}G \biggl( \frac{t}{h},s \biggr) F\bigl(sh,u(sh)\bigr) \Biggr) +\min _{t \in I} {\mathcal{C}}(t) \theta (u ) \\ &\geq \sigma^{*} \max_{t \in(h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h}} \Biggl( \sum _{s=\frac {\beta}{\beta-\alpha}+1}^{T+\frac{\beta}{\beta-\alpha}}G \biggl( \frac {t}{h},s \biggr) F \bigl(sh,u(sh)\bigr) \Biggr)+\sigma^{*} \Vert {\mathcal{C}} \Vert \theta (u ) \\ &\geq \sigma^{*} \Vert {\mathcal{A}}u \Vert . \end{aligned}$$

(5.3)

So, ${\mathcal{A}}{\mathcal{P}}\subset{\mathcal{P}}$.

Consequently, from the Arzelá–Ascoli theorem, it follows that ${\mathcal{A}}: {\mathcal{P}}\rightarrow{\mathcal{P}}$ is the completely continuous operator. □

We next define

$$\begin{aligned}& {\mathcal{M}} := \Biggl[ \sum_{s=\frac{\beta}{\beta-\alpha}+1}^{T+\frac {\beta}{\beta-\alpha}}G ( s,s ) \Biggr] ^{-1}, \\& {\mathcal{N}} := \Biggl[ \sigma^{*} \sum_{s=\frac{\beta}{\beta-\alpha }+1}^{T+\frac{\beta}{\beta-\alpha}}G ( s,s ) \Biggr] ^{-1}, \\& \Upsilon_{1} := \biggl[ \frac{ [ (\beta-\alpha)(T+1)+\beta ] h}{ (\beta-\alpha) [ (\beta-\alpha)T +2\beta ] } \biggr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{\beta}{\beta-\alpha}}, \\& \Upsilon_{2} := \frac{1}{\sigma^{*}} \biggl[ \frac{ [ (\beta-\alpha)(T+1)+\beta ] h}{(\beta-\alpha ) [ (\beta-\alpha)T +2\beta ] } \biggr] \biggl( \frac{\alpha }{\beta} \biggr)^{\frac{\beta}{\beta-\alpha}}, \end{aligned}$$

and introduce some assumptions that will be used in the sequel.

$(X_{1})$ :

There exists a constant $r_{1}>0$ such that

$$\begin{aligned} F \bigl(t,u(t) \bigr)\leq\frac{1}{2} {\mathcal{M}} r_{1} \quad \text{whenever } 0\leq u \leq r_{1}. \end{aligned}$$

$(X_{2})$ :

There exists a constant $r_{2}>0$ with $r_{2}< r_{1}$ such that

$$\begin{aligned} F \bigl(t,u(t) \bigr)\geq\frac{1}{2}{\mathcal{N}} r_{2} \quad \text{whenever } \sigma^{*}r_{2} \leq u \leq r_{2}. \end{aligned}$$

$(X_{3})$ :

There exists a constant $r_{1}>0$ such that for each $u\in \mathcal{P}$ and $0\leq\|u\|\leq r_{1}$,

$$\begin{aligned} \theta(u)\leq\frac{1}{2}\Upsilon_{1} r_{1}. \end{aligned}$$

$(X_{4})$ :

There exists a constant $r_{2}>0$ such that for each $u\in \mathcal{P}$ and $\sigma^{*}r_{2}\leq\|u\|\leq r_{2}$,

$$\begin{aligned} \theta(u)\geq\frac{1}{2}\Upsilon_{2} r_{2}. \end{aligned}$$

The following theorem presents the proof of the existence result of at least one positive solution.

Theorem 5.1

Suppose that the conditions $(X_{1})$–$(X_{4})$ hold. Then problem (1.2) has at least one positive solution, $u^{*}$ where $r_{2}\leq\|u^{*}\|\leq r_{1}$.

Proof

Set $\Omega_{1}=\{u\in{\mathcal{E}} :\|u\|< r_{1} \}$. Then, for $u\in{\mathcal{P}}\cap\partial\Omega_{1}$, we have

$$\begin{aligned} \Vert {\mathcal{A}}u \Vert \leq& \max_{t\in (h{\mathbb{N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta}{\beta-\alpha} ) h} } \Biggl\{ { \mathcal{C}}(t)\theta(u)+\sum_{s=\frac{\beta}{\beta-\alpha }+1}^{T+\frac{\beta}{\beta-\alpha}}G \biggl( \frac{t}{h},s \biggr) F\bigl(sh,u(sh)\bigr) \Biggr\} \\ \leq& \biggl[ \frac{ [ (\beta-\alpha)(T+1)+\beta ] h}{(\beta -\alpha) [ (\beta-\alpha)T +2\beta ] } \biggr] \biggl( \frac {\alpha}{\beta} \biggr)^{\frac{\beta}{\beta-\alpha}}\cdot\frac {1}{2}\Upsilon_{1} r_{1} + \frac{1}{2} {\mathcal{M}} r_{1} \sum _{s=\frac{\beta }{\beta-\alpha}+1}^{T+\frac{\beta}{\beta-\alpha}}G ( s,s ) \\ =&\frac{r_{1}}{2}+\frac{r_{1}}{2}=r_{1}. \end{aligned}$$

Further, we let $\Omega_{2}=\{u \in{\mathcal{E}}:\|u\|< r_{2} \}$. Then, for $u\in {\mathcal{P}}\cap\partial\Omega_{2}$ and by using Lemma 4.4, we find that

$$\begin{aligned} ({\mathcal{A}}u) (t) \geq& \min_{t\in\frac{h}{4} ( T+1+\frac{\beta }{\beta-\alpha} ) \leq t\leq \frac{3h}{4} ( T+1+\frac{\beta }{\beta-\alpha} )} \bigl\{ { \mathcal{C}}(t)\theta(u) \bigr\} \\ &{}+\sum_{s=\frac{\beta}{\beta-\alpha}+1}^{T+\frac{\beta}{\beta-\alpha}} \biggl[ \min_{ \frac{h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) \leq t\leq \frac{3h}{4} ( T+1+\frac{\beta}{\beta-\alpha} ) } G \biggl( \frac{t}{h},s \biggr) F \bigl(sh,u(sh)\bigr) \biggr] \\ \geq& \frac{1}{2}\Upsilon_{2} r_{2} \sigma^{*} \max_{t\in (h{\mathbb {N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta }{\beta-\alpha} ) h}} \bigl\{ {\mathcal{C}}(t)\theta(u) \bigr\} \\ &{}+\frac{1}{2}{\mathcal{N}} r_{2}\sigma^{*} \max _{t\in (h{\mathbb {N}})_{ (\frac{\beta}{\beta-\alpha} ) h, (T+1+\frac{\beta }{\beta-\alpha} ) h} } \sum_{s=\frac{\beta}{\beta-\alpha}+1}^{T+\frac{\beta}{\beta-\alpha}} G \biggl( \frac{t}{h},s \biggr) \\ =&\frac{1}{2}\Upsilon_{2} r_{2}\sigma^{*} \biggl[ \frac{ [ (\beta-\alpha )(T+1)+\beta ] h}{(\beta-\alpha) [ (\beta-\alpha)T +2\beta ] } \biggr] \biggl( \frac{\alpha}{\beta} \biggr)^{\frac{\beta}{\beta -\alpha}}+ \frac{1}{2}{\mathcal{N}} r_{2}\sigma^{*} \sum_{s=\frac{\beta }{\beta-\alpha}+1}^{T+\frac{\beta}{\beta-\alpha}} G ( s,s ) \\ =&\frac{r_{2}}{2}+\frac{r_{2}}{2}=r_{2}. \end{aligned}$$

We can conclude by Theorem 1.1 that the operator $\mathcal{A}$ has a fixed point. This implies that problem (1.2) has a positive solution, $u^{*}$, where $r_{2}\leq\|u^{*}\| \leq r_{1}$. □

6 Some examples

In this section, we present some examples to illustrate our results.

Example 6.1

Consider the following boundary value problem for the second-order $(\alpha,\beta)$-difference equation:

$$\begin{aligned} \begin{gathered} {_{\frac{1}{2}}} \Delta^{2}_{(\frac{1}{3},\frac{3}{2})}u \biggl( t-\frac {1}{2} \biggr) + \frac{1}{ ( t+100 )^{10} } \biggl[ \frac{e^{-t/2} ( \vert u \vert +1 ) }{1+\cos^{2}u}+ ( \Psi u ) (t) \biggr] =0,\quad t \in \biggl( \frac{1}{2}{\mathbb{N}} \biggr)_{\frac{1}{2},10}, \\ {_{\frac{1}{2}}}\Delta_{(\frac{1}{3},\frac{3}{2})}u ( 0 )=\frac{ \vert u \vert }{(10e)^{2}} \sin^{2}(\pi u), \\ {_{\frac{1}{2}}}\Delta_{(\frac{1}{3},\frac{3}{2})} \biggl( \frac {9}{2} \biggr)^{20} u ( 10 )=\frac{ \vert u \vert }{(10 \pi)^{2}}\cos^{2}(\pi u), \end{gathered} \end{aligned}$$

(6.1)

where $(\Psi u)(t)=\frac{1}{2}\sum_{s=0}^{2t-1}\frac{e^{-4|s-t|}}{100 \sqrt{\pi}} u(sh)$.

Set $h=\frac{1}{2}$, $\alpha=\frac{1}{3}$, $\beta=\frac{3}{2}$, $T=20$, $\phi_{1}(u)=\frac{|u|}{(10e)^{2}}\sin^{2}(\pi u)$, $\phi_{2}(u)=\frac{|u|}{(10 \pi)^{2}}\cos^{2}(\pi u)$, $\varphi(t,s)=\frac {e^{-4|s-t|}}{100 \sqrt{\pi}}$ and

$$\begin{aligned} f \bigl( t,u(t),(\Psi u) (t) \bigr) = \frac{1}{ ( t+100 )^{10} } \Biggl[ \frac{e^{-t/2} ( \vert u \vert +1 ) }{1+\cos^{2}u}+\frac{1}{2}\sum_{s=0}^{2t-1} \frac{e^{-4 \vert s-t \vert }}{100 \sqrt{\pi}} u(sh) \Biggr]. \end{aligned}$$

We find that $\varphi_{0}=\frac{1}{100\sqrt{\pi}}=0.00564$, and

$$\begin{aligned} \bigl\vert f \bigl( t,u(t),(\Psi u) (t) \bigr) -f \bigl( t,v(t),(\Psi v) (t) \bigr) \bigr\vert \leq\frac{e^{-10}}{120^{10}} \vert u-v \vert + \frac{1}{120^{10}} \vert \Psi u-\Psi v \vert . \end{aligned}$$

Thus, $(H1)$ holds with $\lambda_{1}=7.332\times10^{-26}$ and $\lambda _{2}=1.615\times10^{-21}$. Since

$$\bigl\vert \phi_{1}(u)-\phi_{1}(v) \bigr\vert \leq \frac{1}{100e^{2}} \Vert u-v \Vert \quad \mbox{and}\quad \bigl\vert \phi_{2}(u)-\phi_{2}(v) \bigr\vert \leq\frac{1}{100\pi^{2}} \Vert u-v \Vert , $$

$(H2)$ holds with $\ell_{1}=0.00135$ and $\ell_{2}=0.00101$.

Clearly,

$$\ell=\max\{\ell_{1},\ell_{2}\}=0.00135, \qquad\Omega=1.1541 \quad\text{and} \quad\Phi=2.171\times10^{15}. $$

Then we find that

$$\Theta=\ell\Omega+ \biggl( \lambda_{1}+\lambda_{2} \frac{h\varphi_{0}(T+1)}{\alpha } \biggr)\Phi=0.00156 < 1. $$

Hence, by Theorem 3.1 problem (6.1) has a unique solution.

Example 6.2

Consider the following boundary value problem for the second-order $(\alpha,\beta)$-difference equation:

$$\begin{aligned} \begin{gathered} {_{\frac{3}{2}}} \Delta^{2}_{ ( \frac{1}{2},\frac{4}{3} ) }u \biggl( t-\frac{3}{2} \biggr) + \frac{e^{-3t} \vert u \vert }{(1+ \vert u \vert ) ( t+100 )^{5} }=0,\quad t \in \biggl( \frac{3}{2}{ \mathbb{N}} \biggr)_{\frac{39}{10},\frac{99}{10}}, \\ {_{\frac{3}{2}}}\Delta_{ ( \frac{1}{2},\frac{4}{3} ) }u \biggl( \frac{12}{5} \biggr) =\frac{8}{9}u \biggl(\frac{39}{10} \biggr), \\ {_{\frac{3}{2}}}\Delta_{ ( \frac{1}{2},\frac{4}{3} ) } \biggl( \biggl( \frac{8}{3} \biggr)^{\frac{99}{10}} u \biggl( \frac{99}{10} \biggr) \biggr)=\sum_{i=0}^{6}\frac{C_{i} \vert u(t_{i}) \vert }{1+ \vert u(t_{i}) \vert }, \quad t_{i}= \biggl( \frac{8}{5}+i \biggr)\frac{3}{2}, \end{gathered} \end{aligned}$$

(6.2)

where $C_{i}$ are given positive constants with $\frac{1}{\pi^{2}}\leq\sum_{i=0}^{6}C_{i}\leq\frac{3}{\pi^{2}}$.

Set $h=\frac{3}{2}$, $\alpha=\frac{1}{2}$, $\beta=\frac{4}{3}$, $T=5$, $\theta (u)=\sum_{i=0}^{6}\frac{C_{i}|u(t_{i})|}{1+|u(t_{i})|}$ and $F ( t,u(t) ) =\frac{e^{-3t}|u|}{(1+|u|) ( t+100 )^{5} }$.

We find

$$\begin{aligned}& \sigma_{1} = 0.00471, \qquad\sigma_{2}=0.1067, \qquad M_{\mathcal {C}}=0.000274, \\& \sigma = \min\{\sigma_{1},\sigma_{2}\}=0.00471, \qquad \sigma^{*} = \min \lbrace\sigma, M_{\mathcal{C}} \rbrace =0.000274, \\& {\mathcal{M}} = 3.972, \qquad{\mathcal{N}}=14495.004, \qquad \Upsilon_{1}= 0.2476, \qquad\Upsilon_{2}=903.832. \end{aligned}$$

Clearly,

$$\begin{aligned} \bigl\vert F \bigl( t,u(t) \bigr) \bigr\vert \leq9.3761\times10^{-6}=4.721 \times 10^{-6}\frac{{\mathcal{M}}}{2}, \end{aligned}$$

for $0\leq u\leq r_{1} \leq4.721\times10^{-6}$,

$$\begin{aligned} \bigl\vert F \bigl( t,u(t) \bigr) \bigr\vert \geq7.2879\times10^{-21}=1.006 \times 10^{-24}\frac{{\mathcal{N}}}{2}, \end{aligned}$$

for $2.755\times10^{-28}=\sigma^{*}r_{2}\leq u\leq r_{2} \leq 1.006\times 10^{-24}$,

$$\begin{aligned} \theta(u)\leq0.304=2.465\frac{\Upsilon_{1}}{2}\quad\text{for }0\leq u\leq r_{1} \leq2.465, \end{aligned}$$

and

$$\begin{aligned} \theta(u)\geq0.101=0.000223\frac{\Upsilon_{2}}{2}\quad\text{for }6.124 \times10^{-8}=\sigma^{*}r_{2}\leq u\leq r_{2} \leq0.000223. \end{aligned}$$

Therefore, conditions $(X_{1})$–$(X_{4})$ are satisfied. Consequently, by Theorem 5.1 problem (6.2) has at least one positive solution $u^{*}$ such that

$$r_{2}=6.124\times10^{-8} \leq \bigl\Vert u^{*} \bigr\Vert \leq 4.721\times10^{-6}=r_{1}. $$