Anti-periodic boundary value problem for first order impulsive delay difference equations

Abstract

In this paper, we investigate the anti-periodic boundary value problem for first order impulsive delay difference equations. To begin with, we establish two comparison theorems. Then, by using these theorems, we prove the existence and uniqueness of solutions for the linear problem. Finally, by using the method of upper and lower solutions coupled with the monotone iterative technique, we obtain the new existence results of extremal solutions. Meanwhile, an example is given to illustrate the results obtained.

1 Introduction

Impulsive differential equations are recognized as important models which describe many evolution processes that abruptly change their state at a certain moment. Such equations have extensive application in economics, physics, chemical technology, medicine, dynamic systems, optimal control, population dynamics and many other fields. The theory of impulsive differential equations has drawn much attention in recent years and is much richer than the corresponding theory of differential equations. For more information about the theory of important differential equations, see [1–4] and the references therein.

Anti-periodic boundary value problem is an important branch of boundary value problem, and it has recently become an interesting area of investigation. The existence and uniqueness of solutions for such a problem have received a great deal of attention, we refer the readers to [5–12] and the references therein. For the case of differential equations, Chen et al. [5] investigated the anti-periodic solutions for first order differential equations, Aftabizadeh et al. [7] discussed the anti-periodic boundary value problem for second order differential equations, Wang and Zhang [8] considered the anti-periodic problem for impulsive differential equations. Ahmad and Nieto [9] studied anti-periodic problem for impulsive functional differential equations. Moreover, for difference equations, a lot of results have been investigated in the literature [13–17]. For example, Liu [14] studied higher order functional difference equations with p-Laplacian. Immediately after this, he [15] studied higher order nonlinear periodic difference equations. However, we noticed that all these known results are related to anti-periodic problem for differential equations and to difference equations. Motivated by some recent work on anti-periodic problems and difference equations with impulse (see [18–23]), in this paper, we attempt to propose some results concerning the impulsive delay difference equations with anti-periodic boundary conditions

$$\begin{aligned} \left \{ \begin{array}{@{}l} \Delta u(n)=f(n,u(n),u(\theta(n))),\quad {n}\neq {n}_{k}, n\in J,\\ \Delta u({n_{k}})=I_{k}(u({n_{k}})), \quad k=1,2,\ldots,p,\\ u(0)=-u(T), \end{array} \right . \end{aligned}$$

(1)

where $\Delta u(n)=u(n+1)-u(n)$, $f\in C(J\times\mathbb{R}^{2},\mathbb {R})$, $\theta\in C[J,\mathbb{Z}]$, $0\leq\theta(n)\leq n$, $J=[0,T]=\{ 0,1,\ldots,T\}$, $I_{k}\in C(\mathbb{R},\mathbb{R})$ ($k=1,2,\ldots,p$), $0< n_{1}<n_{2}<\cdots<n_{p}<T$, T is a positive integer.

In [19], He and Zhang investigated first order impulsive difference equations with periodic boundary conditions. Wang and Wang [18] analyzed first order impulsive difference equations with linear boundary conditions. Zhang et al. [12] investigated impulsive anti-periodic boundary value problems for nonlinear $q_{k}$-difference equations. To the best of our knowledge, there are few results on the anti-periodic boundary value problem for impulsive delay difference equations. Hence, we are concerned with the existence of solutions for anti-periodic boundary value problem (1). In Section 2, we introduce the concept of upper and lower solutions and establish two comparison principles. In Section 3, we discuss the existence of solutions and uniqueness for the linear anti-periodic boundary value problem. Moreover, by using the monotone iterative technique and the method of upper and lower solutions, we obtain the existence theorem of extremal solutions for problem (1). Finally, an example is worked out to demonstrate the obtained results.

2 Comparison results

In this section, we introduce relative notation and some lemmas. Throughout this paper, let ℕ denote the set of all natural numbers and let Ω denote the set of real-valued functions defined on J with the norm $\|u\|= \max_{n\in J}|u(n)|$ for $u\in\Omega$. For $\alpha,\beta\in\Omega$, we write $\alpha\leq\beta$ if $\alpha(n)\leq\beta(n)$ for all $n\in J$.

A function $u\in\Omega$ is said to be a solution of problem (1) if it satisfies (1).

Definition 2.1

Functions $\alpha, \beta\in \Omega$ are called lower and upper solutions of problem (1) if

$$\left \{ \begin{array}{@{}l} \Delta\alpha(n)\leq f(n,\alpha(n),\alpha(\theta(n))),\quad {n}\neq{n}_{k}, n\in J, \\ \Delta\alpha({n_{k}})\leq I_{k}(\alpha({n_{k}})),\quad k=1,2,\ldots,p,\\ \alpha(0)\leq-\alpha(T), \end{array} \right . $$

and

$$\left \{ \begin{array}{@{}l} \Delta\beta(n)\geq f(n,\beta(n),\beta(\theta(n))),\quad {n}\neq{n}_{k}, n\in J, \\ \Delta\beta({n_{k}})\geq I_{k}(\beta({n_{k}})),\quad k=1,2,\ldots,p,\\ \beta(0)\geq-\beta(T). \end{array} \right . $$

Lemma 2.1

([19])

Assume that

(i) the sequence ${n_{k}}$ satisfies $0\leq n_{0}< n_{1}<\cdots<n_{k}<\cdots$ with $\lim_{k\to\infty}n_{k}=\infty$;

(ii) for $k\in\mathbb{N}$, $n\geq n_{0}$,

$$\left \{ \begin{array}{@{}l} \Delta m(n)\leq l_{n}m(n)+q_{n},\quad n\neq n_{k},\\ m(n_{k}+1)\leq b_{k}m(n_{k})+e_{k}, \end{array} \right . $$

where $\{l_{n}\}$ and $\{q_{n}\}$ are two real-valued consequences and $l_{n}>-1$, $b_{k}$ and $e_{k}$ are constants, and $b_{k}\geq0$. Then

$$\begin{aligned} m(n)\leq{}& m(n_{0}) \prod_{n_{0}< n_{k}<n}b_{k} \prod_{n_{0}<i<n,i\neq n_{k},k\in \mathbb{N}}(1+l_{i})+ \sum_{i=n_{0},i\neq n_{k}}^{n-1} \prod _{i<n_{k}<n}b_{k}\prod_{i<s<n,s\neq n_{k}}(1+l_{s})q_{i} \\ &{}+ \sum_{n_{0}<n_{k}<n}e_{k} \prod _{n_{k}<n_{j}<n}b_{j}\prod_{n_{k}<i<n,i\neq n_{j},j\in\mathbb{N}}(1+l_{i}). \end{aligned}$$

Remark 2.2

When k is finite, Lemma 2.1 also holds. In this paper, we only consider this case.

Next, we will establish two new comparison results which play an important role in the monotone iterative technique.

Lemma 2.2

Let $m\in\Omega$ be such that

$$\left \{ \begin{array}{@{}l} \Delta m(n)+Mm(n)+Nm(\theta(n))\leq0,\quad n\neq n_{k}, n\in J,\\ \Delta m(n_{k})\leq-L_{k}m(n_{k}),\quad k=1,2,\ldots,p,\\ m(0)\leq0, \end{array} \right . $$

where $N\geq0$, $0\leq L_{k}< M<1$ for $k=1,2,\ldots,p$, and

$$\begin{aligned} N \sum_{i=0,i\neq n_{k}}^{T} \prod _{i< n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-1}- \prod _{k=1}^{p}(1-L_{k})<M-1. \end{aligned}$$

(2)

Then $m(n)\leq0$ on J.

Proof

Let $v(n)=(1-M)^{-n}m(n)$, $n\in[0,T]$,

$$\begin{aligned}& \Delta v(n)\leq-N(1-M)^{\theta(n)-n-1}v\bigl(\theta(n)\bigr), \end{aligned}$$

(3)

$$\begin{aligned}& v(n_{k}+1)\leq(1-M)^{-1}(1-L_{k})v(n_{k}). \end{aligned}$$

(4)

Obviously, $v(n)\leq0$ implies $m(n)\leq0$. So it suffices to show $v(n)\leq0$ on J. Suppose on the contrary that there exists $n^{*}\in J$ such that $v(n^{*})>0$. Since $v(0)=m(0)\leq0$, then $n^{*}\in(0,T]$. Let $\bar{n}\in[0,n^{*})$ such that $v(\bar{n})=\mbox{inf}_{n\in [0,n^{*})}v(n)=-\lambda\leq0$. We suppose that $\bar{n}\neq n_{k}$ (if $\bar{n}=n_{k}$, the proof is similar). From (3) we have

$$\begin{aligned} \Delta v(n) \leq\lambda N(1-M)^{\theta(n)-n-1},\quad n\in \bigl[0,n^{*}\bigr]. \end{aligned}$$

(5)

By (4) and (5), using Lemma 2.1, we have for $n\in [\bar{n},n^{*}]$

$$\begin{aligned}[b] v(n)&\leq v(\bar{n}) \prod_{\bar{n}< n_{k}<n}(1-M)^{-1}(1-L_{k}) +\lambda N \sum_{i=\bar{n},i\neq n_{k}}^{n-1} \prod _{i<n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-2} \\ &\leq-\lambda \prod_{k=1}^{p}(1-M)^{-1}(1-L_{k}) +\lambda N \sum_{i=0,i\neq n_{k}}^{T} \prod _{i<n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-2}. \end{aligned} $$

Let $n=n^{*}$, we can get

$$\begin{aligned} v\bigl(n^{*}\bigr)&\leq\lambda \Biggl\{ N \sum_{i=0,i\neq n_{k}}^{T} \prod_{i< n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-2} - \prod_{k=1}^{p}(1-M)^{-1}(1-L_{k}) \Biggr\} \\ &=\lambda(1-M)^{-1} \Biggl\{ N \sum_{i=0,i\neq n_{k}}^{T} \prod_{i<n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-1} - \prod_{k=1}^{p}(1-L_{k}) \Biggr\} \\ &<0, \end{aligned}$$

which is a contradiction. Hence $v(n)\leq0$ on J, this completes the proof. □

Lemma 2.3

Let (2) hold and $m\in \Omega$,

$$\begin{aligned} \left \{ \begin{array}{@{}l} \Delta m(n)+Mm(n)+Nm(\theta(n))\leq0,\quad n\neq n_{k}, n\in J,\\ \Delta m(n_{k})\leq-L_{k}m(n_{k}),\quad k=1,2,\ldots,p,\\ m(0)\leq-m(T), \end{array} \right . \end{aligned}$$

(6)

where $N\geq0$, $0\leq L_{k}< M<1$ for $k=1,2,\ldots,p$.

Then $m(n)\leq0$ on J.

Proof

Suppose on the contrary that $m(n)>0$ for some $n\in J$. Then there are two cases as follows.

Case 1: There exists $n^{*}\in J$ such that $m(n^{*})>0$ and $m(n)\geq0$ for all $n\in J$. In this case, (6) implies $\Delta m(n)\leq0$ and $m(n_{k}+1)\leq(1-L_{k})m(n_{k})\leq m(n_{k})$, $k=1,2,\ldots,p$. So $m(n)$ is a nonincreasing function. Then we have $m(0)\geq m(T)$. Since $m(0)\leq-m(T)$ and $m(n)\geq0$ for all $n\in J$, we get $m(0)=m(T)=0$, then $m(n)\equiv0$, which is a contradiction with $m(n^{*})>0$.

Case 2: There exist $n^{*}$ and $n_{*}$ such that $m(n^{*})>0$ and $m(n_{*})<0$. The proof demonstrates that $m(0)\leq0$, so that we can apply Lemma 2.2 and affirm that $m(n)\leq0$ on J. If $m(0)>0$, then $m(T)<0$. Let $v(n)=(1-M)^{-n}m(n)$ on J, we get $v(0)>0$, $v(T)<0$, $v(n^{*})>0$, $v(n_{*})<0$. Set $\min_{n\in J}v(n)=-\lambda$, then $\lambda>0$. Without loss of generality, we shall suppose $v(n_{*})=-\lambda$ (if for some k such that $v(n_{k})=-\lambda$, the proof is similar), we have

$$\begin{aligned} \left \{ \begin{array}{@{}l} \Delta v(n)\leq -N(1-M)^{\theta(n)-n-1}v(\theta(n)),\quad n\neq n_{k}, n\in J,\\ v(n_{k}+1)\leq(1-M)^{-1}(1-L_{k})v(n_{k}),\quad k=1,2,\ldots,p,\\ v(0)\leq-v(T)(1-M)^{T}. \end{array} \right . \end{aligned}$$

(7)

If $n_{*}< n^{*}$, using (7) and Lemma 2.1, for $n\in[n_{*},n^{*}]$, we obtain

$$\begin{aligned}[b] v(n)&\leq v(n_{*}) \prod_{n_{*}< n_{k}<n}(1-M)^{-1}(1-L_{k}) +\lambda N \sum_{i=n_{*},i\neq n_{k}}^{n-1} \prod _{i<n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-2} \\ &\leq-\lambda \prod_{k=1}^{p}(1-M)^{-1}(1-L_{k}) +\lambda N \sum_{i=0,i\neq n_{k}}^{T} \prod _{i<n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-2}. \end{aligned} $$

Let $n=n^{*}$, we get

$$\begin{aligned} v\bigl(n^{*}\bigr)&\leq\lambda \Biggl\{ N \sum_{i=0,i\neq n_{k}}^{T} \prod_{i< n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-2} - \prod_{k=1}^{p}(1-M)^{-1}(1-L_{k}) \Biggr\} \\ &=\lambda(1-M)^{-1} \Biggl\{ N \sum_{i=0,i\neq n_{k}}^{T} \prod_{i<n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-1} - \prod_{k=1}^{p}(1-L_{k}) \Biggr\} \\ &<0, \end{aligned}$$

which is a contradiction.

If $n_{*}>n^{*}$, using (7) and Lemma 2.1, for $n\in[n_{*},T]$, we get

$$v(n)\leq v(n_{*}) \prod_{0< n_{k}<n}(1-M)^{-1}(1-L_{k}) +\lambda N \sum_{i=0,i\neq n_{k}}^{n-1} \prod _{i<n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-2}. $$

Let $n=T$, it then follows from (2) that

$$\begin{aligned} v(T)&\leq v(n_{*}) \prod_{0< n_{k}<n}(1-M)^{-1}(1-L_{k}) +\lambda N \sum_{i=0,i\neq n_{k}}^{n-1} \prod _{i<n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-2} \\ &\leq\lambda \Biggl\{ N \sum_{i=0,i\neq n_{k}}^{T} \prod _{i<n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-2} - \prod_{k=1}^{p}(1-M)^{-1}(1-L_{k}) \Biggr\} \\ &=\lambda(1-M)^{-1} \Biggl\{ N \sum_{i=0,i\neq n_{k}}^{T} \prod_{i<n_{k}<n}(1-L_{k}) (1-M)^{\theta(i)-i-1} - \prod_{k=1}^{p}(1-L_{k}) \Biggr\} \\ &<-\lambda=v(n_{*}), \end{aligned}$$

which is a contradiction with the definition of $v(n_{*})$. So $m(0)\leq 0$. By Lemma 2.2, we get $m(n)\leq0$ on J, the proof is complete. □

3 Main results

Let us consider the linear problem of (1) as follows:

$$\begin{aligned} \left \{ \begin{array}{@{}l} \Delta u(n)+Mu(n)+Nu(\theta(n))=\sigma(n),\quad n\neq n_{k}, n\in J,\\ \Delta u(n_{k})= -L_{k}u(n_{k})+\gamma_{k}, \quad k=1,2,\ldots,p,\\ u(0)=-u(T), \end{array} \right . \end{aligned}$$

(8)

where $N\geq0$, $0\leq L_{k}< M<1$, $\gamma_{k}\in\mathbb{R}$, $\sigma\in C(J,\mathbb{R})$.

Lemma 3.1

A function $u\in\Omega$ is a solution of (8) if and only if u is a solution of the following impulsive summation equation:

$$\begin{aligned} u(n) =& \sum_{j=0,j\neq n_{k}}^{T-1}G(n,j) \bigl(\sigma(j)-Nu\bigl(\theta(j)\bigr) \bigr) + \sum_{0< n_{k}\leq T-1}G(n,n_{k}) \bigl[(M-L_{k})u(n_{k}) +\gamma_{k} \bigr], \end{aligned}$$

(9)

where

$$G(n,j)=\frac{1}{1+(1-M)^{T}}\left \{ \begin{array}{@{}l@{\quad}l} \frac{(1-M)^{n}}{(1-M)^{j+1}}, &0\leq j\leq n-1, \\ \frac{-(1-M)^{T+n}}{(1-M)^{j+1}}, &n\leq j\leq T-1. \end{array} \right . $$

Proof

Assume that $u\in\Omega$ is a solution of (8). Set $v(n)= \frac{u(n)}{(1-M)^{n}}$, $n\in J$. From (8), we see that $v(n)$ satisfies

$$\begin{aligned} \left \{ \begin{array}{@{}l} v(n+1)=v(n)+ \frac{\sigma(n)-Nv(\theta(n))(1-M)^{\theta (n)}}{(1-M)^{n+1}},\quad n\neq n_{k}, n\in J, \\ \Delta v(n_{k})= \frac{M-L_{k}}{1-M}v(n_{k})+\frac{\gamma _{k}}{(1-M)^{n_{k}+1}}, \quad k=1,2,\ldots,p, \\ v(0)=-v(T)(1-M)^{T}. \end{array} \right . \end{aligned}$$

(10)

By using (10), we have

$$\begin{aligned} v(n) =&v(0)+ \sum_{j=0,j\neq n_{k}}^{n-1} \frac{\sigma (j)-Nv(\theta(j))(1-M)^{\theta(j)}}{(1-M)^{j+1}} \\ &{}+ \sum_{0< n_{k}\leq n-1} \biggl(\frac{M-L_{k}}{1-M}v(n_{k})+ \frac {\gamma_{k}}{(1-M)^{n_{k}+1}} \biggr). \end{aligned}$$

(11)

If we set $n=T$ in (11), then we get

$$\begin{aligned} v(T) =&v(0)+ \sum_{j=0,j\neq n_{k}}^{T-1} \frac{\sigma(j)-Nv(\theta(j))(1-M)^{\theta(j)}}{(1-M)^{j+1}} \\ &{}+ \sum_{0< n_{k}\leq T-1} \biggl(\frac{M-L_{k}}{1-M}v(n_{k})+ \frac{\gamma_{k}}{(1-M)^{n_{k}+1}} \biggr). \end{aligned}$$

(12)

From the boundary condition $v(T)= \frac{-v(0)}{(1-M)^{T}}$, we obtain

$$\begin{aligned} v(0) =& -\frac{(1-M)^{T}}{1+(1-M)^{T}} \Biggl[ \sum _{j=0,j\neq n_{k}}^{T-1}\frac{\sigma(j)-Nv(\theta(j))(1-M)^{\theta (j)}}{(1-M)^{j+1}} \\ &{}+ \sum_{0< n_{k}\leq T-1} \biggl(\frac {M-L_{k}}{1-M}v(n_{k})+ \frac{\gamma_{k}}{(1-M)^{n_{k}+1}} \biggr) \Biggr]. \end{aligned}$$

(13)

Substituting (13) into (11) and using $v(n)= \frac{u(n)}{(1-M)^{n}}$, $n\in J$, we have

$$\begin{aligned} \frac{u(n)}{(1-M)^{n}} =& \frac{1}{1+(1-M)^{T}} \Biggl( \sum _{j=0,j\neq n_{k}}^{n-1}\frac{\sigma(j)-Nu(\theta(j))}{(1-M)^{j+1}}+ \sum _{0< n_{k}\leq n-1}\frac{(M-L_{k})u(n_{k}) +\gamma_{k}}{(1-M)^{n_{k}+1}} \Biggr) \\ &{}- \frac{(1-M)^{T}}{1+(1-M)^{T}} \Biggl( \sum_{j=n,j\neq n_{k}}^{T-1} \frac{\sigma(j)-Nu(\theta(j))}{(1-M)^{j+1}}+ \sum_{n<n_{k}\leq T-1}\frac{(M-L_{k})u(n_{k}) +\gamma_{k}}{(1-M)^{n_{k}+1}} \Biggr). \end{aligned}$$

Let

$$G(n,j)=\frac{1}{1+(1-M)^{T}} \left \{ \begin{array}{@{}l@{\quad}l} \frac{(1-M)^{n}}{(1-M)^{j+1}}, &0\leq j\leq n-1, \\ \frac{-(1-M)^{T+n}}{(1-M)^{j+1}}, &n\leq j\leq T-1, \end{array} \right . $$

we see that u is a solution of (9). The proof is complete. □

Lemma 3.2

Assume that constants $0< M<1$, $N\geq0$, $0\leq L_{k}<1$, $\gamma_{k}\in\mathbb{R}$, $\sigma\in C(J,\mathbb{R})$, and

$$\begin{aligned} \frac{1}{1+(1-M)^{T}} \Biggl(TN+ \sum_{k=1}^{p}|M-L_{k}| \Biggr)< 1. \end{aligned}$$

(14)

Then (8) has a unique solution.

Proof

Define an operator $F:\Omega\to\Omega$ by

$$\begin{aligned} (Fu) (n)={}& \sum_{j=0,j\neq n_{k}}^{T-1}G(n,j) \bigl( \sigma(j)-Nu\bigl(\theta(j)\bigr) \bigr) + \sum_{0< n_{k}\leq T-1}G(n,n_{k}) \bigl[(M-L_{k})u(n_{k}) +\gamma_{k} \bigr]. \end{aligned}$$

For any $u_{1}, u_{2}\in\Omega$, we have

$$\begin{aligned} \bigl|Fu_{1}(n)-Fu_{2}(n)\bigr|\leq{}& \sum _{j=0,j\neq n_{k}}^{T-1} \bigl|G(n,j) \bigr| \bigl|N \bigl(u_{1}\bigl( \theta(j)\bigr)-u_{2}\bigl(\theta(j)\bigr) \bigr) \bigr| \\ &{} + \sum_{0< n_{k}\leq T-1} \bigl|G(n,n_{k}) \bigr| \bigl|(M-L_{k}) \bigl(u_{1}(n_{k})-u_{2}(n_{k}) \bigr) \bigr| \\ \leq{}& \frac{1}{1+(1-M)^{T}} \Biggl(TN+ \sum_{k=1}^{p}|M-L_{k}| \Biggr)\|u_{1}-u_{2}\|. \end{aligned}$$

Hence, $\|Fu_{1}(n)-Fu_{2}(n)\|= \max_{n\in J}|Fu_{1}(n)-Fu_{2}(n)|=\tau\|u_{1}-u_{2}\|$, where

$$\tau=\frac{1}{1+(1-M)^{T}} \Biggl(TN+ \sum_{k=1}^{p}|M-L_{k}| \Biggr)< 1. $$

By the Banach contraction principle, F has a unique fixed point. The proof is complete. □

Theorem 3.1

Let the following conditions hold:

(A₀):

Functions $\alpha,\beta\in\Omega$ are lower and upper solutions for (1) with $\alpha\leq\beta$.

(A₁):

There exist $N\geq0$, $0\leq L_{k}< M<1$ for $k=1,2,\ldots,p$ such that the function $f\in C(J\times\mathbb{R}^{2},\mathbb{R})$ satisfies

$$f(n,x,y)-f(n,u,v)\geq-M(x-u)-N(y-v), $$

where $\alpha(n)\leq u\leq x\leq\beta(n)$, $\alpha(\theta(n))\leq v(\theta(n))\leq y(\theta(n))\leq\beta(\theta(n))$, $n\in J$.

(A₂):

The functions $I_{k}\in C(\mathbb{R},\mathbb{R})$ satisfy

$$I_{k}(x)-I_{k}(y)\geq-L_{k}(x-y), $$

where $\alpha(n_{k})\leq y\leq x\leq\beta(n_{k})$, $0\leq L_{k}<1$, $k=1,2,\ldots,p$.

(A₃):

$$N \sum_{i=0,i\neq n_{k}}^{T} \prod _{i< n_{k}<T}(1-L_{k}) (1-M)^{\theta(i)-i-1} - \prod _{k=1}^{p}(1-L_{k})<M-1. $$

(A₄):

$$\frac{1}{1+(1-M)^{T}} \Biggl(TN+ \sum_{k=1}^{p}|M-L_{k}| \Biggr)< 1. $$

Then there exists a solution u of problem (1) such that $\alpha(n)\leq u(n)\leq\beta(n)$ on J.

Proof

We consider the following modified problem:

$$\begin{aligned} \left \{ \begin{array}{@{}l} \Delta u(n)+Mu(n)+Nu(\theta(n))=\sigma_{q}(n),\quad n\neq n_{k},~n\in J,\\ \Delta u(n_{k})= -L_{k}u(n_{k})+d_{k},\quad k=1,2,\ldots,p,\\ u(0)=-u(T), \end{array} \right . \end{aligned}$$

(15)

where

$$\begin{aligned}& \sigma_{q}(n)=f\bigl(n,q\bigl(n,u(n)\bigr),q\bigl(\theta(n),u\bigl(\theta (n)\bigr)\bigr)\bigr)+Mq\bigl(n,u(n)\bigr)+Nq\bigl(\theta(n),u\bigl(\theta(n)\bigr)\bigr),\\& d_{k}=I_{k}\bigl(q\bigl(n_{k},u(n_{k})\bigr)\bigr)+L_{k}q\bigl(n_{k},u(n_{k})\bigr),\\& q\bigl(n,u(n)\bigr)=\max\bigl\{ \alpha (n),\min\bigl\{ u,\beta(n)\bigr\} \bigr\} \quad \mbox{for }n\in J. \end{aligned}$$

We can easily see that if $\alpha(n)\leq u(n)\leq\beta(n)$ on J, then u is a solution of (1) if and only if u is a solution of (15). Indeed, suppose that $u\in\Omega$ is a solution of (15), we now prove that $\alpha(n)\leq u(n)$ on J. Set $m(n)=\alpha (n)-u(n)$, $n\in J$. Owing to (A₀)-(A₂), we acquire

$$\begin{aligned}& \begin{aligned}[b] \Delta m(n)={}&\Delta\alpha(n)-\Delta u(n) \\ \leq{}& f\bigl(n,\alpha(n),\alpha\bigl(\theta(n)\bigr)\bigr)-\bigl[-Mu(n)-Nu \bigl(\theta(n)\bigr) \\ &{} +f \bigl(n,q\bigl(n,u(n)\bigr),q\bigl(\theta(n),u\bigl(\theta(n)\bigr)\bigr) \bigr)+Mq\bigl(n,u(n)\bigr)+Nq\bigl(\theta(n),u\bigl(\theta(n)\bigr)\bigr)\bigr] \\ \leq{}&{-}Mm(n)-Nm\bigl(\theta(n)\bigr), \end{aligned} \\& \begin{aligned}[b] \Delta m(n_{k})&=\Delta\alpha(n_{k})- \Delta u(n_{k}) \\ &\leq I_{k}\bigl(\alpha(n_{k})\bigr)-I_{k} \bigl(q\bigl(n_{k},u(n_{k})\bigr) \bigr)-L_{k}q \bigl(n_{k},u(n_{k})\bigr)+L_{k}u(n_{k}) \\ &\leq-L_{k}m(n_{k}), \end{aligned} \end{aligned}$$

and

$$m(0)=\alpha(0)-u(0)\leq-\alpha(T)+u(T)=-m(T). $$

By Lemma 2.3, we have $m(n)\leq0$ on J, i.e., $\alpha (n)\leq u(n)$ on J. Similarly, we can show that $u(n)\leq\beta(n)$ on J.

Next, we need to prove that problem (15) has a solution. To do this, we write problem (15) in the following way by Lemma 3.1:

$$u(n)= \sum_{j=0,j\neq n_{k}}^{T-1}G(n,j) \bigl(\sigma (j)-Nu\bigl(\theta(j)\bigr) \bigr)+ \sum_{0< n_{k}\leq T-1}G(n,n_{k}) \bigl[(M-L_{k})u(n_{k})+d_{k} \bigr]. $$

We define the continuous and compact operator $\mathcal{A}:\Omega\to \Omega$ by

$$\begin{aligned}{} [\mathcal{A}u](n) =& \sum_{j=0,j\neq n_{k}}^{T-1}G(n,j) \bigl(\sigma(j)-Nu\bigl(\theta(j)\bigr) \bigr) \\ &{} + \sum_{0< n_{k}\leq T-1}G(n,n_{k}) \bigl[(M-L_{k})u(n_{k}) +d_{k} \bigr],\quad n\in J. \end{aligned}$$

The continuity of f and the definition of q imply that $\sigma_{q}$ and $d_{k}$ are bounded. We can choose constants $h>0$ and $w>0$ such that $|\sigma_{q}|\leq h$, $|d_{k}|\leq w$. For $\lambda\in(0,1)$, we find that any solution of $u=\lambda\mathcal{A}u$ satisfies

$$\begin{aligned} \|u\| =&\lambda\|\mathcal{A}u\| \\ \leq& \max_{n\in J}\sum_{j=0,j\neq n_{k}}^{T-1} \bigl|G(n,j) \bigr| \bigl[ \bigl|\sigma_{q}(j) \bigr|+N \bigl|u\bigl(\theta(j)\bigr) \bigr| \bigr] \\ &{} + \max_{n\in J}\sum_{0< n_{k}\leq T-1} \bigl|G(n,n_{k}) \bigr| \bigl[ |M-L_{k} | \bigl|u(n_{k}) \bigr|+ |d_{k} | \bigr] \\ \leq& \frac{hT}{1+(1-M)^{T}}+ \frac {TN}{1+(1-M)^{T}}\|u\|+ \frac{ \sum_{k=1}^{p}|M-L_{k}|}{1+(1-M)^{T}}\|u\|+ \frac{pw}{1+(1-M)^{T}}. \end{aligned}$$

Then by (A₄) we have

$$\|u\|\leq \Biggl(1-\frac{1}{1+(1-M)^{T}} \Biggl(TN+ \sum _{k=1}^{p}|M-L_{k}| \Biggr) \Biggr)^{-1} \biggl( \frac {hT+pw}{1+(1-M)^{T}} \biggr). $$

From the Schaefer fixed-point theorem, $\mathcal{A}$ has at least a fixed point. It is clear that this fixed point is the solution of (15). Such a solution lies between α and β and in consequence is a solution of (1). The proof is complete. □

Theorem 3.2

Let all assumptions of Theorem  3.1 hold. Then there exist monotone sequences $\{\alpha_{j}(n)\}$, $\{\beta_{j}(n)\}$ with $\alpha_{0}=\alpha$, $\beta_{0}=\beta$ such that $\lim_{j\to\infty}\alpha_{j}(n)=\rho(n)$, $\lim_{j\to\infty}\beta_{j}(n)=r(n)$ uniformly on J, and $\rho(n)$, $r(n)$ are the extremal solutions of problem (1).

Proof

Let

$$\begin{aligned} \left \{ \begin{array}{@{}l} \Delta \alpha_{j}(n)+M\alpha_{j}(n)+N\alpha_{j}(\theta(n))\\ \quad=f(n,\alpha_{j-1}(n),\alpha _{j-1}(\theta(n))) +M\alpha_{j-1}(n)+N\alpha_{j-1}(\theta(n)),\quad n\neq n_{k},n\in J,\\ \Delta\alpha_{j}(n_{k})+L_{k}\alpha_{j}(n_{k})=I_{k}(\alpha_{j-1}(n_{k}))+L_{k}\alpha _{j-1}(n_{k}),\quad k=1,2,\ldots,p,\\ \alpha_{j}(0)=-\alpha_{j}(T), \end{array} \right . \end{aligned}$$

(16)

and

$$\begin{aligned} \left \{ \begin{array}{@{}l} \Delta \beta_{j}(n)+M\beta_{j}(n)+N\beta_{j}(\theta(n))\\ \quad=f(n,\beta_{j-1}(n),\beta _{j-1}(\theta(n))) +M\beta_{j-1}(n)+N\beta_{j-1}(\theta(n)),\quad n\neq n_{k},n\in J,\\ \Delta\beta_{j}(n_{k})+L_{k}\beta_{j}(n_{k})=I_{k}(\beta_{j-1}(n_{k}))+L_{k}\beta _{j-1}(n_{k}),\quad k=1,2,\ldots,p,\\ \beta_{j}(0)=-\beta_{j}(T), \end{array} \right . \end{aligned}$$

(17)

for $j=1,2,\ldots$ , where $\alpha_{0}=\alpha$, $\beta_{0}=\beta$. It follows from Lemma 3.2 that problems (16) and (17) have a unique solution, respectively.

First, we show that $\alpha_{0}\leq\alpha_{1}\leq\beta_{1}\leq\beta_{0}$.

Let $m=\alpha_{0}-\alpha_{1}$, then owing to (A₀) and $\alpha _{1}(0)=-\alpha_{1}(T)$, we obtain

$$\begin{aligned}& \begin{aligned}[b] \Delta m(n)={}&\Delta\alpha_{0}(n)-\Delta \alpha_{1}(n)\\ \leq{}& f\bigl(n,\alpha_{0}(n),\alpha _{0}\bigl(\theta(n)\bigr)\bigr)-\bigl[-M\alpha_{1}(n)-N \alpha_{1}\bigl(\theta(n)\bigr) \\ &{} +f\bigl(n,\alpha_{0}(n),\alpha_{0}\bigl(\theta(n) \bigr)\bigr)+M\alpha_{0}(n)+N\alpha _{0}\bigl(\theta(n) \bigr)\bigr] \\ \leq{}&{-}Mm(n)-Nm\bigl(\theta(n)\bigr), \end{aligned} \\& \begin{aligned}[b] \Delta m(n_{k})&=\Delta\alpha_{0}(n_{k})- \Delta\alpha_{1}(n_{k}) \\ &\leq I_{k}\bigl(\alpha_{0}(n_{k})\bigr)- \bigl[I_{k}\bigl(\alpha_{0}(n_{k}) \bigr)+L_{k}\alpha_{0}(n_{k})-L_{k} \alpha _{1}(n_{k})\bigr] \\ &\leq-L_{k}m(n_{k}), \end{aligned} \end{aligned}$$

and

$$m(0)=\alpha_{0}(0)-\alpha_{1}(0)\leq-\alpha_{0}(T)+ \alpha_{1}(T)=-m(T). $$

It then follows from Lemma 2.3 that we have $m(n)\leq0$ on J, which implies $\alpha_{0}(n)\leq\alpha_{1}(n)$ for $n\in J$. Similarly, we get $\beta_{1}(n)\leq\beta_{0}(n)$ on J.

Next, take $m=\alpha_{1}-\beta_{1}$, by using (A₀), (A₁), we have

$$\begin{aligned} &\Delta m(n)+Mm(n)+Nm\bigl(\theta(n)\bigr) \\ &\quad=\Delta\alpha_{1}(n)+M\alpha_{1}(n)+N \alpha_{1}\bigl(\theta(n)\bigr)-\Delta\beta _{1}(n)-M \beta_{1}(n)-N\beta_{1}\bigl(\theta(n)\bigr) \\ &\quad\leq f\bigl(n,\alpha_{0}(n),\alpha_{0}\bigl( \theta(n)\bigr)\bigr)+M\alpha_{0}(n)+N\alpha_{0}\bigl(\theta (n)\bigr)-f\bigl(n,\beta_{0}(n),\beta_{0}\bigl(\theta(n) \bigr)\bigr) \\ &\qquad{}-M\beta_{1}(n)-N\beta_{1}\bigl(\theta(n)\bigr) \leq0. \end{aligned}$$

Noticing $\alpha_{0}\leq\beta_{0}$ and (A₂), we get

$$\begin{aligned}& \begin{aligned}[b] \Delta m(n_{k})&=-L_{k} \alpha_{1}(n_{k})+I_{k}\bigl( \alpha_{0}(n_{k})\bigr)+L_{k} \alpha_{0}(n_{k})-I_{k}\bigl(\beta _{0}(n_{k})\bigr)-L_{k}\beta_{0}(n_{k}) \\ &\leq-L_{k}m(n_{k}), \end{aligned} \\& m(0)=\alpha_{1}(0)-\beta_{1}(0)=-\alpha_{1}(T)+ \beta_{1}(T)=-m(T). \end{aligned}$$

Again by Lemma 2.3 we get $m(n)\leq0$ on J, that is, $\alpha_{1}(n)\leq \beta_{1}(n)$ for all $n\in J$. Thus we get $\alpha_{0}\leq\alpha_{1}\leq\beta _{1}\leq\beta_{0}$. Continuing this process, by induction, we can get the sequences $\{\alpha_{j}(n)\}$ and $\{\beta_{j}(n)\}$ such that

$$\alpha_{0}\leq\alpha_{1}\leq\cdots\leq \alpha_{j}\leq\cdots \leq\beta_{j}\leq\cdots\leq \beta_{1}\leq\beta_{0} \quad\mbox{on } J. $$

Clearly, the sequences $\{\alpha_{j}(n)\}$, $\{\beta_{j}(n)\}$ are uniformly bounded and equi-continuous. Since they are monotone sequences, by the Ascoli-Arzela theorem, we can get that the entire sequences $\{\alpha_{j}(n)\}$ and $\{\beta_{j}(n)\}$ converge uniformly and monotonically on J with $\lim_{j\to\infty}\alpha _{j}(n)=\rho(n)$ and $\lim_{j\to\infty}\beta_{j}(n)=r(n)$.

Obviously ρ, r are the solutions of (1). Next we prove that ρ, r are extremal solutions of (1), let $u\in \Omega$ be any solutions of problem (1) such that $\alpha _{0}\leq u\leq\beta_{0}$. Suppose that there exists a positive integer j such that $\alpha_{j}\leq u\leq{\beta}_{j}$ on J. Setting $m=\alpha_{j+1}-u$, we have

$$\begin{aligned}& \begin{aligned}[b] \Delta m(n)={}&\Delta{\alpha}_{j+1}(n)-\Delta u(n) \\ \leq{}& f\bigl(n,{\alpha}_{j}(n),{\alpha}_{j}\bigl( \theta(n)\bigr)\bigr)-M{\alpha}_{j+1}(n)-N{\alpha }_{j+1}\bigl( \theta(n)\bigr) \\ &{}-f\bigl(n,u(n),u\bigl(\theta(n)\bigr)\bigr)+M{\alpha}_{j}(n)+N \alpha_{j}\bigl(\theta(n)\bigr) \\ \leq{}&-M\bigl({\alpha}_{j+1}(n)-u(n)\bigr)-N\bigl({ \alpha}_{j+1}\bigl(\theta(n)\bigr)-u\bigl(\theta (n)\bigr)\bigr) \\ ={}&-Mm(n)-Nm\bigl(\theta(n)\bigr),\quad n\neq n_{k}, n\in J, \end{aligned} \\& \begin{aligned}[b] \Delta m(n_{k})&=\Delta{\alpha}_{j+1}(n_{k})- \Delta u(n_{k}) \\ &=I_{k}\bigl({\alpha}_{j}(n_{k}) \bigr)-I_{k}\bigl(u(n_{k})\bigr)-L_{k}{ \alpha}_{j+1}(n_{k})+L_{k}{\alpha }_{j}(n_{k}) \\ &\leq-L_{k}m(n_{k}),\quad k=1,2,\ldots,p, \end{aligned} \\& m(0)=\alpha_{j+1}(0)-u(0)=-\alpha_{j+1}(T)+u(T)=-m(T). \end{aligned}$$

By Lemma 2.3, $m(n)\leq0$ on J, i.e., $\alpha_{j+1}(n)\leq u(n)$ $n\in J$. Similarly, one derives $\alpha_{j+1}(n)\leq u(n)\leq \beta_{j+1}(n)$ on J. Since $\alpha_{0}(n)\leq u(n)\leq\beta_{0}(n)$ on J, by induction we see that $\alpha_{j}(n)\leq u(n)\leq\beta_{j}(n)$ on J for every j. Taking the limit as $j\to\infty$, we conclude $\rho(n)\leq u(n)\leq r(n)$ on J. The proof is then finished. □

Example 3.1

Consider the equations

$$\begin{aligned} \left \{ \begin{array}{@{}l} \Delta u(n)=f(n,u(n),u(\theta(n)))=-u^{2}(n)- \frac{1}{200}u ( \frac{1}{2}n )+ \frac{2^{-n}}{25},\quad n\in Z[0,3],n\neq n_{1}, \\ \Delta u({n_{k}})=- \frac{1}{2}u(n_{1}),\quad n_{1}=2, \\ u(0)=-u(3). \end{array} \right . \end{aligned}$$

(18)

It is easy to verify that $\alpha=- \frac{1}{201}$ is a lower solution and $\beta(n)= \frac{1}{12}(4-2^{-n})$ is an upper solution for (18). Indeed,

$$\begin{aligned}& \Delta\beta(n)= \frac{1}{12}\bigl(2^{-n}-2^{-n-1}\bigr) \geq -\biggl[ \frac{1}{12}\bigl(4-2^{-n}\bigr)\biggr]^{2}- \frac{1}{200}\cdot \frac{1}{12}\bigl(4-2^{-\frac{1}{2}n}\bigr)+ \frac {1}{25}\cdot2^{-n}, \\& \Delta\beta(2)= \frac{1}{12}\bigl(2^{-2}-2^{-3} \bigr)>- \frac{1}{2}\cdot \frac{1}{12}\bigl(4-2^{-2}\bigr), \\& \beta(0)= \frac{1}{12}(4-1)>- \frac {1}{12}\bigl(4-2^{-3} \bigr)=-\beta(3), \\& f(n,x,y)-f(n,u,v)=-\bigl(x^{2}-u^{2}\bigr)- \frac{1}{200}(y-v)\geq - \frac{2}{3}(x-u)- \frac{1}{200}(y-v), \end{aligned}$$

for $\alpha\leq u\leq x\leq\beta$. Taking $M= \frac {2}{3}$, $N= \frac{1}{200}$, $L_{k}= \frac {1}{2}$, we get

$$\begin{aligned}& \prod_{k=1}^{p}(1-L_{k})+M-1= \frac {1}{2}+ \frac{2}{3}-1= \frac{1}{6}, \\& \begin{aligned}[b] & \frac{1}{200} \sum_{i=0,i\neq n_{k}}^{3} \prod_{i< n_{k}<3}(1-L_{k}) (1-M)^{-\frac{1}{2}i-1} = \frac{1}{200} \sum_{i=0,i\neq n_{k}}^{3} \prod _{i<n_{k}<3} \frac{1}{2}\cdot \biggl( \frac{1}{3} \biggr)^{-\frac{1}{2}i-1} \\ &\quad= \frac{1}{200} \sum_{i=0}^{1} \biggl( \frac{1}{3} \biggr)^{-\frac{1}{2}i-1} + \frac{1}{200} \sum _{i=2}^{3} \frac {1}{2}\cdot \biggl( \frac{1}{3} \biggr)^{-\frac {1}{2}i-1}< \frac{1}{6}, \end{aligned} \end{aligned}$$

which shows that condition (A₃) is satisfied, and

$$\frac{1}{1+ (1- \frac{2}{3} )^{3}} \Biggl(3\cdot \frac{1}{200}+ \sum _{k=1}^{p} \biggl( \frac{2}{3}- \frac{1}{2} \biggr) \Biggr)< 1, $$

which shows that condition (A₄) is satisfied. Hence, by Theorem 3.2, we obtain the existence of monotone sequences that converge to the extremal solutions of (18) in a function interval contained in $[\alpha,\beta]$.