Existence of uncountably many bounded positive solutions to higher-order nonlinear neutral delay difference equations

Abstract

This paper studies higher-order nonlinear neutral delay difference equations of the form

$$\begin{aligned}& \varDelta \bigl(r^{m-1}_{n} \bigl(\varDelta \bigl(r^{m-2}_{n} \bigl(\cdots \bigl(\varDelta \bigl(r^{1}_{n} \bigl( \varDelta (x_{n}+p_{n}x_{n-\tau}) \bigr)^{\gamma_{1}} \bigr) \bigr)^{\gamma_{2}}\cdots \bigr)^{\gamma _{m-2}} \bigr) \bigr)^{\gamma_{m-1}} \bigr)\\& \quad = f(n,x_{n-\tau_{1}},\ldots,x_{n-\tau_{s}}). \end{aligned}$$

Using Krasnoselskii’s fixed point theorem, we obtain the existence of uncountably many bounded positive solutions to the considered problem.

1 Introduction and preliminaries

In mathematical models in diverse areas such as economy, biology, computer science, difference equations appear in a natural way; see, for example, [1, 2]. In the past thirty years, oscillation, nonoscillation, the asymptotic behavior and existence of bounded solutions to many types of difference equation have been widely examined. For the second order, see, for example, [3–9], and for higher orders, [10–15], and references therein.

Liu et al. [16] discussed the existence of uncountably many bounded positive solutions to

$$\varDelta \bigl( r_{n} \varDelta ( x_{n}+b_{n}x_{n-\tau}-c_{n} ) \bigr)+ f\bigl(n, x\bigl(f_{1}(n)\bigr),\ldots,x\bigl(f_{k}(n) \bigr)\bigr)=d_{n} $$

with respect $(b_{n})$. Using techniques of the measures of noncompactness, Galewski et al. [4] considered

$$\varDelta \bigl(r_{n} \bigl(\varDelta (x_{n}+p_{n}x_{n-\tau}) \bigr)^{\gamma}\bigr)+q_{n}x^{\alpha}_{n}+a_{n}f(x_{n+1})=0. $$

Migda and Schmeidel [12] studied the following equation:

$$\varDelta \bigl(r^{m-1}_{n}\varDelta \bigl(r^{m-2}_{n} \cdots \varDelta \bigl(r_{n}^{1}\varDelta (x_{n}+p_{n}x_{n-\tau}) \bigr)\cdots \bigr) \bigr)=a_{n}f(x_{n-\sigma})+b_{n}. $$

They established sufficient conditions under which for every real constant, there exists a solution to the studied problem convergent to this constant.

In this paper, we study higher-order nonlinear neutral delay difference equations of the form

$$\begin{aligned}& \varDelta \bigl(r^{m-1}_{n} \bigl(\varDelta \bigl(r^{m-2}_{n} \bigl(\cdots \bigl(\varDelta \bigl(r^{1}_{n} \bigl( \varDelta (x_{n}+p_{n}x_{n-\tau}) \bigr)^{\gamma_{1}} \bigr) \bigr)^{\gamma_{2}}\cdots \bigr)^{\gamma _{m-2}} \bigr) \bigr)^{\gamma_{m-1}} \bigr) \\ & \quad =f(n,x_{n-\tau_{1}},\ldots,x_{n-\tau_{s}}) \end{aligned}$$

(1)

under the following general settings:

(H₁):

$m\ge2$, $\gamma_{1},\ldots,\gamma_{m-1}\le1$ are ratios of odd positive integers, $\tau\in\mathbb{N}$, $\tau _{1},\ldots,\tau_{s}\in\mathbb{Z}$, $(p_{n})\subset\mathbb{R}$, $r^{i}=(r^{i}_{n})\subset\mathbb{R}\setminus\{0\}$, $i=1,\ldots,m-1$, and $f:\mathbb{N}\times\mathbb{R}^{s}\to\mathbb{R}$.

Additional conditions will be added to obtain the existence of uncountably many positive (nonoscillatory) solutions to equation (1). Krasnoselskii’s fixed point theorem will be used to prove our results. To illustrate them, three examples are included.

Throughout this paper, we assume that Δ is the forward difference operator. By a solution to equation (1) we mean a sequence $x:{\mathbb{N}}\rightarrow{\mathbb{R}}$ that satisfies (1) for every $n\ge k$ for some $k\geq\max\{\tau ,\tau_{1},\ldots,\tau_{s}\}$.

We consider the Banach space $l^{\infty}$ of all real bounded sequences $x:{\mathbb{N}}\rightarrow{\mathbb{R}}$ equipped with the standard supremum norm, that is, for $x=(x_{n})\in l^{\infty}$,

$$\Vert x\Vert=\sup_{n\in{\mathbb{N}}}|x_{n}|. $$

Definition 1

([17])

A subset A of $l^{\infty}$ is said to be uniformly Cauchy if for every $\varepsilon>0$, there exists $n_{0}\in\mathbb{N}$ such that $|x_{i}-x_{j}|<\varepsilon$ for any $i,j\geq n_{0}$ and $x=(x_{n})\in A$.

Theorem 1

([17])

A bounded, uniformly Cauchy subset of $l^{\infty}$ is relatively compact.

We shall use Krasnoselskii’s fixed point theorem in the following form.

Theorem 2

([18], 11.B, p.501)

Let X be a Banach space, B be a bounded closed convex subset of X, and $S,G:B\to X$ be mappings such that $Sx+Gy\in B$ for any $x,y\in B$. If S is a contraction and G is a compact, then the equation

$$Sx+Gx=x $$

has a solution in B.

2 Main results

For any nonnegative sequence $y=(y_{n})$ and $n\in\mathbb{N}$, we use the notation

$$\begin{aligned}& W_{1}(n,y)=\sum^{\infty}_{l_{1}=n}y_{l_{1}}; \\ & W_{2}(n,y)=\sum^{\infty}_{l_{2}=n} \Biggl(\biggl\vert \frac {1}{r^{m-1}_{l_{2}}}\biggr\vert \sum ^{\infty}_{l_{1}=l_{2}}y_{l_{1}} \Biggr)^{\gamma_{m-1}^{-1}}= \sum^{\infty}_{l_{2}=n} \biggl(\frac {W_{1}(l_{2},y)}{|r^{m-1}_{l_{2}}|} \biggr)^{\gamma_{m-1}^{-1}}; \\ & \ldots; \\& W_{k}(n,y)=\sum^{\infty}_{l_{k}=n} \biggl(\frac {W_{k-1}(l_{k},y)}{|r^{m-k+1}_{l_{k}}|} \biggr)^{\gamma_{m-k+1}^{-1}}, \quad k=2,\ldots,m. \end{aligned}$$

By $[0,M]^{s}$ we denote the set $[0,M]\times\cdots\times[0,M]\subset \mathbb{R}^{s}$.

Now we are in position to formulate and prove the main theorem.

Theorem 3

Suppose that (H₁) is satisfied. Assume further that

(H₂):

$\sup_{n\in\mathbb{N}}|p_{n}|=p^{\star}<1/4$;

(H₃):

there exists $M>0$ such that for any $n\in\mathbb {N}$, the function $f(n,\cdot)$ is a Lipschitz function on $[0,2M]^{s}$ with Lipschitz constant $P(n,M)$ satisfying

$$\sum^{\infty}_{l_{m}=1}\biggl\vert \frac{1}{r^{1}_{l_{m}}} \biggr\vert ^{\gamma_{1}^{-1}}\sum^{\infty}_{l_{m-1}=l_{m}} \biggl\vert \frac {1}{r^{2}_{l_{m-1}}}\biggr\vert ^{\gamma_{2}^{-1}}\cdots\sum ^{\infty}_{l_{2}=l_{3}}\biggl\vert \frac{1}{r^{m-1}_{l_{2}}}\biggr\vert ^{\gamma _{m-1}^{-1}}\sum^{\infty}_{l_{1}=l_{2}}P(l_{1},M)< \infty; $$

(H₄):

$W_{m}(1,|f(\cdot,0_{\mathbb{R}^{s}})|)<\infty$.

Then, equation (1) possesses uncountably many bounded positive solutions lying in $[M/2,2M]$.

Proof

Let $M>0$ be a constant fulfilling assumption (H₃). It is easy to see that (H₃) implies that

$$ \sum^{\infty}_{l_{1}=1}P(l_{1},M)< \infty $$

(2)

and

$$ \sum^{\infty}_{l_{k}=1}\biggl\vert \frac{1}{r^{m-k+1}_{l_{k}}} \biggr\vert ^{\gamma_{m-k+1}^{-1}}\cdots\sum ^{\infty}_{l_{2}=l_{3}} \biggl\vert \frac{1}{r^{m-1}_{l_{2}}}\biggr\vert ^{\gamma_{m-1}^{-1}}\sum^{\infty}_{l_{1}=l_{2}}P(l_{1},M)< \infty, \quad k=2,\ldots,m-1. $$

(3)

From (H₄) it is clear that

$$ \sum_{n=1}^{\infty}\bigl\vert f(n,0_{\mathbb{R}^{s}})\bigr\vert < \infty, W_{k}\bigl(1,\bigl\vert f(\cdot,0_{\mathbb{R}^{s}})\bigr\vert \bigr)< \infty,\quad k=2,\ldots,m. $$

(4)

Now we claim that (H₃) and (3) imply that

$$ W_{k}\bigl(1,P(\cdot,M)\bigr)< \infty, \quad k=2, \ldots,m. $$

(5)

Indeed, from (2) we get that there exists $n_{1}$ such that for any $n\ge n_{1}$, we have $\sum^{\infty}_{l_{1}=n}P(l_{1}, M)<1$; hence, since $\gamma_{m-1}\le1$, we get that for any $n\ge n_{1}$,

$$\Biggl(\sum^{\infty}_{l_{1}=n}P(l_{1},M) \Biggr)^{\gamma _{m-1}^{-1}}\le\sum^{\infty}_{l_{1}=n}P(l_{1},M). $$

Thus, for any $n\ge n_{1}$, we have

$$W_{2}\bigl(n,P(\cdot,M)\bigr)=\sum^{\infty}_{l_{2}=n} \Biggl\vert \frac {1}{r^{m-1}_{l_{2}}}\sum^{\infty}_{l_{1}=l_{2}}P(l_{1},M) \Biggr\vert ^{\gamma_{m-1}^{-1}}\le\sum^{\infty}_{l_{2}=n} \biggl\vert \frac {1}{r^{m-1}_{l_{2}}}\biggr\vert ^{\gamma_{m-1}^{-1}}\sum ^{\infty}_{l_{1}=l_{2}}P(l_{1},M). $$

To prove that $W_{2}(1,P(\cdot,M))<\infty$, we use the classical inequality

$$ (a+b )^{\alpha}\le2^{\alpha-1} \bigl(a^{\alpha }+b^{\alpha} \bigr)\quad \mbox{for } \alpha\ge1, a, b>0, $$

(6)

which gives

$$\begin{aligned} W_{2}\bigl(1,P(\cdot,M)\bigr) \le& W_{2} \bigl(n_{1},P(\cdot,M)\bigr)+\sum^{n_{1}-1}_{l_{2}=1} \biggl\vert \frac{W_{1}(l_{2},P(\cdot,M))}{r^{m-1}_{l_{2}}}\biggr\vert ^{\gamma _{m-1}^{-1}} \\ \le& W_{2}\bigl(n_{1},P(\cdot,M)\bigr) \\ &{}+\sum ^{n_{1}-1}_{l_{2}=1}\frac{2^{\gamma_{m-1}^{-1}-1}}{ \vert r^{m-1}_{l_{2}}\vert ^{\gamma^{-1}_{m-1}}} \Biggl( \Biggl(\sum ^{n_{1}-1}_{l_{1}=l_{2}}P(l_{1},M) \Biggr)^{\gamma_{m-1}^{-1}}+ \Biggl(\sum^{\infty}_{l_{1}=n_{1}}P(l_{1},M) \Biggr)^{\gamma_{m-1}^{-1}} \Biggr)< \infty. \end{aligned}$$

In an analogous way, we prove the remaining conditions in (5). We now claim that

$$ \sum_{n=1}^{\infty} \biggl( \frac{W_{k} (n,2M\sqrt {s}P(\cdot, M)+|f(\cdot,0_{\mathbb{R}^{s}})| )}{|r^{m-k}_{n}|} \biggr)^{\gamma_{m-k}^{-1}}< \infty,\quad k=1,\ldots,m-1. $$

(7)

We give the proof of (7) for the case $k=2$ and $m=3$; the other cases are analogous and are left to the reader. Indeed, using (6), we have

$$\begin{aligned}& \sum_{l_{3}=1}^{\infty} \biggl(\frac{W_{2} (l_{3},2M\sqrt {s}P(\cdot, M)+|f(\cdot,0_{\mathbb{R}^{s}})| )}{|r^{1}_{l_{3}}|} \biggr)^{\gamma_{1}^{-1}}\\& \quad =\sum^{\infty}_{l_{3}=1} \Biggl( \frac{1}{|r^{1}_{l_{3}}|}\sum^{\infty }_{l_{2}=l_{3}} \biggl( \frac{W_{1}(l_{2},2M\sqrt{s}P(\cdot,M))}{|r^{2}_{l_{2}}|} +\frac{W_{1}(l_{2},|f(\cdot,0_{\mathbb{R}^{s}})|)}{|r^{2}_{l_{2}}|} \biggr)^{\gamma_{2}^{-1}} \Biggr)^{\gamma_{1}^{-1}}\\& \quad \le2^{(\gamma_{2}^{-1}-1)\gamma_{1}^{-1}}\sum^{\infty}_{l_{3}=1} \Biggl(\frac{1}{|r^{1}_{l_{3}}|} \Biggl(\sum^{\infty}_{l_{2}=l_{3}} \biggl(\frac {W_{1}(l_{2},2M\sqrt{s}P(\cdot,M))}{|r^{2}_{l_{2}}|} \biggr)^{\gamma _{2}^{-1}} \\& \qquad {}+\sum ^{\infty}_{l_{2}=l_{3}} \biggl(\frac{W_{1}(l_{2},|f(\cdot ,0_{\mathbb{R}^{s}})|)}{|r^{2}_{l_{2}}|} \biggr)^{\gamma_{2}^{-1}} \Biggr) \Biggr)^{\gamma_{1}^{-1}}\\& \quad \le2^{(\gamma_{2}^{-1}-1)\gamma_{1}^{-1}+\gamma_{1}^{-1}-1}\cdot \Biggl(\sum^{\infty}_{l_{3}=1} \biggl(\frac{W_{2}(l_{3},2M\sqrt{s}P(\cdot ,M))}{|r^{1}_{l_{3}}|} \biggr)^{\gamma_{1}^{-1}} \\& \qquad {}+\sum ^{\infty}_{l_{3}=1} \biggl(\frac{W_{2}(l_{3},|f(\cdot,0_{\mathbb{R}^{s}})|)}{|r^{1}_{l_{3}}|} \biggr)^{\gamma_{1}^{-1}} \Biggr)\\& \quad =2^{(\gamma_{2}^{-1}-1)\gamma_{1}^{-1}+\gamma_{1}^{-1}-1}\cdot \bigl((2M\sqrt{s})^{\gamma_{1}^{-1}\gamma_{2}^{-1}}W_{3} \bigl(1,P(\cdot ,M)\bigr)+W_{3}\bigl(1,\bigl\vert f( \cdot,0_{\mathbb{R}^{s}})\bigr\vert \bigr) \bigr)< \infty. \end{aligned}$$

Once the claim proved, observe that we may find $n_{0}\ge\max\{\tau ,\tau_{1},\ldots,\tau_{s}\}$ such that

$$ W_{m}\bigl(n_{0},2M\sqrt{s}P(\cdot, M)+\bigl\vert f(\cdot,0_{\mathbb {R}^{s}})\bigr\vert \bigr)< 2M\biggl( \frac{1}{4}-p^{\star}\biggr). $$

(8)

We consider a subset of $l^{\infty}$ of the form

$$A_{n_{0}}= \bigl\{ x=(x_{n})\in l^{\infty}: x_{n}=3M/2, n< n_{0} \wedge \vert x_{n}-3M/2 \vert \leq M/2, n\geq n_{0} \bigr\} . $$

Observe that $A_{n_{0}}$ is a nonempty, bounded, convex, and closed subset of $l^{\infty}$.

Let us denote

$$u_{1}^{x}(n)=\sum^{\infty}_{l=n}f(l,x_{l-\tau_{1}}, \ldots,x_{l-\tau_{s}}), \quad x=(x_{n}), x_{n}\in[0,2M], n \ge\max\{\tau_{1},\ldots,\tau_{s}\}. $$

The following takes care of showing that $u_{1}^{x}$ is well defined and bounded above. By (H₃), for any $\mathbf{x}=(x_{1},\ldots,x_{s})\in [0,2M]^{s}$ and for any $n\in\mathbb{N}$, we have

$$ \bigl\vert f(n,\mathbf{x})\bigr\vert \le P(n,M)\Vert \mathbf{x}\Vert _{\mathbb{R}^{s}}+\bigl\vert f(n,0_{\mathbb{R}^{s}})\bigr\vert \le2M\sqrt {s}P(n,M)+\bigl\vert f(n,0_{\mathbb{R}^{s}})\bigr\vert , $$

(9)

where $\|\cdot\|_{\mathbb{R}^{s}}$ denotes the Euclidean norm in $\mathbb{R}^{s}$. Thus, for any $x=(x_{n})\in A_{n_{0}}$ and $n\ge\max\{ \tau_{1},\ldots,\tau_{s}\}$,

$$ \bigl\vert u_{1}^{x}(n)\bigr\vert \le W_{1}\bigl(n,2M\sqrt{s}P(\cdot,M)+\bigl\vert f(\cdot,0_{\mathbb{R}^{s}}) \bigr\vert \bigr). $$

(10)

Denote, for any $x=(x_{n})\in A_{n_{0}}$ and $n\ge\max\{\tau_{1},\ldots ,\tau_{s}\}$,

$$u_{2}^{x}(n)=\sum^{\infty}_{l=n} \biggl(\frac{u^{x}_{1}(l)}{r^{m-1}_{l}} \biggr)^{\gamma_{m-1}^{-1}}. $$

Thus, for any $x=(x_{n})\in A_{n_{0}}$ and $n\ge\max\{\tau_{1},\ldots ,\tau_{s}\}$,

$$ \bigl\vert u_{2}^{x}(n)\bigr\vert \le W_{2}\bigl(n,2M\sqrt{s}P(\cdot,M)+\bigl\vert f(\cdot,0_{\mathbb{R}^{s}}) \bigr\vert \bigr). $$

(11)

In an analogous way, for any $x=(x_{n})\in A_{n_{0}}$ and $n\ge\max\{\tau _{1},\ldots,\tau_{s}\}$, we denote

$$u_{k}^{x}(n)=\sum^{\infty}_{l=n} \biggl(\frac {u^{x}_{k-1}(l)}{r^{m-k+1}_{l}} \biggr)^{\gamma_{m-k+1}^{-1}},\quad k=2,\ldots,m.$$

Thus, for any $k=2,\ldots,m$, $x=(x_{n})\in A_{n_{0}}$, and $n\ge\max\{ \tau_{1},\ldots,\tau_{s}\}$,

$$ \bigl\vert u_{k}^{x}(n)\bigr\vert \le W_{k}\bigl(n,2M\sqrt{s}P(\cdot,M)+\bigl\vert f(\cdot,0_{\mathbb {R}^{s}}) \bigr\vert \bigr). $$

(12)

Define two mappings $T_{1},T_{2}: A_{n_{0}}\rightarrow l^{\infty}$ as follows:

$$\begin{aligned}& (T_{1}x)_{n}= \textstyle\begin{cases} 0 &\text{for } 1\leq n< n_{0}, \\ -p_{n}x_{n-\tau} &\text{for } n\geq n_{0}; \end{cases}\displaystyle \end{aligned}$$

(13)

$$\begin{aligned}& (T_{2}x)_{n}= \textstyle\begin{cases} 3M/2 &\text{for } 1\leq n< n_{0}, \\ 3M/2+(-1)^{m}u_{m}^{x}(n) &\text{for } n\geq n_{0}. \end{cases}\displaystyle \end{aligned}$$

(14)

Our next goal is to check the assumptions of Theorem 2 (Krasnoselskii’s fixed point theorem). Firstly, we show that $T_{1}x+T_{2}y\in A_{n_{0}}$ for $x,y\in A_{n_{0}}$. Let $x,y\in A_{n_{0}}$. For $n< n_{0}$, $(T_{1}x+T_{2}y)_{n}=3M/2$. For $n\geq n_{0}$, from assumption (H₂), (8), and (12) we get

$$\bigl\vert (T_{1}x+T_{2}y)_{n}-3M/2\bigr\vert \leq \vert p_{n}x_{n-\tau} \vert +\bigl\vert u_{m}^{x}(n)\bigr\vert \leq p^{\star}2M+2M \bigl(1/4-p^{\star}\bigr)=M/2. $$

It is easy to see that

$$\Vert T_{1}x-T_{1}y\Vert \leq p^{\star} \Vert x-y\Vert \quad \text{for } x,y\in A_{n_{0}}, $$

so that $T_{1}$ is a contraction.

To prove the continuity of $T_{2}$, notice that from (12) we get

$$\bigl\vert u^{x}_{m-1}(n)\bigr\vert \le W_{m-1} \bigl(1,2M\sqrt{s}P(\cdot,M)+\bigl\vert f(\cdot,0_{\mathbb {R}^{s}})\bigr\vert \bigr)=:d_{m-1} $$

for any $x=(x_{n})\in A_{n_{0}}$ and $n\ge\max\{\tau_{1},\ldots,\tau_{s}\}$. From the Lipschitz continuity of the function $x\mapsto x^{\gamma _{1}^{-1}}$ on $[0,d_{m-1}]$ with constant $L_{\gamma_{1}}$, say, we have

$$\bigl\vert (T_{2}x-T_{2}y)_{n}\bigr\vert \le \sum^{\infty}_{l_{m}=n}\biggl\vert \frac {1}{r^{1}_{l_{m}}}\biggr\vert ^{\gamma_{1}^{-1}}L_{\gamma_{1}}\bigl\vert u^{x}_{m-1}(n)-u^{y}_{m-1}(n)\bigr\vert $$

for any $x,y\in A_{n_{0}}$ and $n\ge n_{0}$. In an analogous way, by (12), for any $k=2,\ldots,m$, we get intervals $[0,d_{k}]$ on which the function $x\mapsto x^{\gamma_{k}^{-1}}$ is Lipschitz continuous, say, with constant $L_{\gamma_{k}}>0$. Hence, for any $x,y\in A_{n_{0}}$ and $n\ge n_{0}$, we have

$$\begin{aligned} \bigl\vert (T_{2}x-T_{2}y)_{n}\bigr\vert \le& L_{\gamma_{1}}\cdot\ldots\cdot L_{\gamma _{m-1}}\cdot\sum ^{\infty}_{l_{m}=n}\biggl\vert \frac{1}{r^{1}_{m}} \biggr\vert ^{\gamma_{1}^{-1}}\cdots\sum^{\infty}_{l_{k}=l_{k+1}} \biggl\vert \frac {1}{r^{m-k+1}_{l_{k}}}\biggr\vert ^{\gamma_{m-k+1}^{-1}}\cdots\sum ^{\infty}_{l_{2}=l_{3}}\biggl\vert \frac{1}{r^{m-1}_{l_{2}}} \biggr\vert ^{\gamma_{m-1}^{-1}} \\ &{}\times\sum^{\infty}_{l_{1}=l_{2}}\bigl\vert f(k,x_{k-\tau_{1}},\ldots ,x_{k-\tau_{s}})-f(k,y_{k-\tau_{1}}, \ldots,y_{k-\tau_{s}})\bigr\vert \\ \le&\sqrt{s}\cdot\prod^{m-1}_{j=1}L_{\gamma_{j}} \cdot \Biggl(\sum^{\infty}_{l_{m}=n}\biggl\vert \frac{1}{r^{1}_{m}}\biggr\vert ^{\gamma _{1}^{-1}}\cdots\sum ^{\infty}_{l_{2}=l_{3}}\biggl\vert \frac {1}{r^{m-1}_{l_{2}}}\biggr\vert ^{\gamma_{m-1}^{-1}}\sum^{\infty}_{l_{1}=l_{2}}P(l_{1},M) \Biggr)\|x-y\|, \end{aligned}$$

which, combined with (H₃), means that $T_{2}$ is continuous on $A_{n_{0}}$.

Now we show that $T_{2}(A_{n_{0}})$ is uniformly Cauchy. Let $\varepsilon >0$. From (7) we get the existence of $n_{\varepsilon}\in \mathbb{N}$ such that $n_{\varepsilon}\ge n_{0}$ and

$$2W_{m}\bigl(n_{\varepsilon},2M\sqrt{s}P(\cdot,M)+\bigl\vert f( \cdot,0_{\mathbb {R}^{s}})\bigr\vert \bigr)< \varepsilon. $$

From (12) we have, for $k, n\geq n_{\varepsilon}\geq n_{0}$ and for $x=(x_{n})\in A_{n_{0}}$,

$$\bigl\vert (T_{2}x)_{n}-(T_{2}x)_{k} \bigr\vert \le2\bigl\vert u^{x}_{m}(n_{\varepsilon}) \bigr\vert \le2W_{m}\bigl(n_{\varepsilon},2M\sqrt{s}P(\cdot,M)+ \bigl\vert f(\cdot,0_{\mathbb{R}^{s}})\bigr\vert \bigr)< \varepsilon. $$

Since $T_{2}(A_{n_{0}})$ is uniformly Cauchy and bounded, by Theorem 1, $T_{2}(A_{n_{0}})$ is relatively compact in $l^{\infty}$, which means that $T_{2}$ is a compact operator.

From Theorem 2 we get that there exists a fixed point $x=(x_{n})$ of $T_{1}+T_{2}$ on $A_{n_{0}}$. Hence,

$$x_{n}+p_{n}x_{n-\tau}=(-1)^{m}u_{m}^{x}(n)+3M/2 $$

for $n\ge n_{0}$. Applying the operator Δ to both sides of the last equation, raising to the power $\gamma_{1}$ (recalling that it is the ratio of odd positive integers), and multiplying by $r^{1}_{n}$, we get

$$r^{1}_{n} \bigl(\varDelta (x_{n}+p_{n}x_{n-\tau}) \bigr)^{\gamma _{1}}=(-1)^{m-1}u_{m-1}^{x}(n) $$

for $n\ge n_{0}$. Repeating this procedure $m-2$ times, we get that $x=(x_{n})$ is a solution to equation (1) for $n\geq n_{0}$ with $x_{n}\in[M, 2M]$.

Now we prove the existence of uncountably many solutions to (1) lying in $[M/2, 2M]$. Let $M_{1}$, $M_{2}$ be such that $M/2< M_{1}< M_{2}<M$. It is easy to see that the assumptions of the theorem are fulfilled for $M_{1}$, $M_{2}$. So there exist $n_{1},n_{2}\ge\max\{\tau,\tau _{1},\ldots,\tau_{s}\}$ and $x^{1}=(x^{1}_{n})$ and $x^{2}=(x^{2}_{n})$, each a fixed point of the operator $T^{i}_{1}+T^{i}_{2}$ in $A_{n_{i}}$, respectively, where

$$\begin{aligned}& \bigl(T_{1}^{i}x\bigr)_{n}= \textstyle\begin{cases} 0 &\text{for } 1\leq n< n_{i}, \\ -p_{n}x_{n-\tau} &\text{for } n\geq n_{i}; \end{cases}\displaystyle \\& \bigl(T^{i}_{2}x\bigr)_{n}= \textstyle\begin{cases} 3M_{i}/2 &\text{for } 1\leq n< n_{i}, \\ 3M_{i}/2+(-1)^{m}u_{m}^{x}(n) &\text{for } n\geq n_{i}. \end{cases}\displaystyle \end{aligned}$$

Thus, $x^{i}$ are solutions to (1) for $n\ge\max\{n_{1},n_{2}\}$. By (12) there exists $n_{3}\in\mathbb{N}$, $n_{3}\ge\max\{ n_{1},n_{2}\}$, such that

$$\bigl\vert u^{x^{1}}_{m}(n)\bigr\vert +\bigl\vert u^{x^{2}}_{m}(n)\bigr\vert \le3/4(M_{2}-M_{1}) \quad \mbox{for } n\ge n_{3}. $$

From this we get that, for $n\ge n_{3}$,

$$\bigl\vert x^{1}_{n}-x^{2}_{n}+p_{n} \bigl(x^{1}_{n-\tau}-x^{2}_{n-\tau}\bigr)\bigr\vert \ge 3/2(M_{2}-M_{1})-\bigl(\bigl\vert u^{x^{1}}_{m}(n)\bigr\vert +\bigl\vert u^{x^{2}}_{m}(n) \bigr\vert \bigr)>0, $$

which means that $x^{1}$ and $x^{2}$ are different solutions to (1) lying in $[M/2,2M]$. □

Remark 1

It is obvious that condition (H₄) in Theorem 3 can be replaced by the condition

($\mathrm{H}'_{4}$):

$W_{m}(1,|f(\cdot,\overline{\mathbf{x}})|)<\infty$ for some $\overline{\mathbf{x}}\in[0,2M]^{s}$.

3 Examples

Now, we present examples of equations for which our method can be applied.

Example 1

Let us consider the second-order nonlinear neutral delay difference equation

$$ \varDelta \bigl( \sqrt{n} \varDelta ( x_{n}+p_{n}x_{n-\tau} ) \bigr)=\frac{x^{2}_{n-\tau_{1}}}{4(2n-1)(2n+1)}, $$

(15)

where $\tau\in\mathbb{N}$, $\tau_{1}\in\mathbb{Z}$, $\gamma_{1}=1$, and $(p_{n})$ is any sequence of real numbers such that $\sup_{n\in \mathbb{N}}|p_{n}|<1/4$. Moreover, $r_{n}^{1}=\sqrt{n}$, and $f(n,u)=\frac{u^{2}}{4(2n-1)(2n+1)}$ for $n\in\mathbb{N}$ and $u\in \mathbb{R}$.

Since $f(n,\cdot)\in C^{1}(\mathbb{R})$ for any $n\in\mathbb{N}$, it follows that, for any $n\in\mathbb{N}$, $f(n,\cdot)$ is a locally Lipschitz function on $\mathbb{R}$. Hence, for any $n\in\mathbb{N}$, $f(n,\cdot)$ is a Lipschitz function on $[0,2M]$ for any $M>0$. It is easy to calculate that $P(n,M)=\frac{M}{(2n-1)(2n+1)}$ and

$$\sum^{\infty}_{l_{2}=1}\frac{1}{r^{1}_{l_{2}}}\sum ^{\infty }_{l_{1}=l_{2}}P(l_{1},M)=\sum ^{\infty}_{l_{2}=1}\frac{1}{\sqrt{l_{2}}}\sum ^{\infty}_{l_{1}=l_{2}}\frac{M}{(2l_{1}-1)(2l_{1}+1)}=\sum ^{\infty }_{l_{2}=1}\frac{M}{2\sqrt{l_{2}}(2l_{2}-1)}< \infty, $$

so that assumption (H₃) of Theorem 3 is satisfied. To see that assumption (H₄) of Theorem 3 is fulfilled, notice that $f(n,0)=0$, $n\in\mathbb{N}$. Hence, there exist uncountably many solutions to (15) in any interval $[M/2,2M]$ for any $M>0$. On the other hand, Theorem 3.1 in [16] is inapplicable because

$$\sum^{\infty}_{l_{2}=1}\frac{1}{r^{1}_{l_{2}}}\sum ^{l_{2}-1}_{l_{1}=1}P(l_{1},M)=\sum ^{\infty}_{l_{2}=1}\frac{1}{\sqrt {l_{2}}}\sum ^{l_{2}-1}_{l_{1}=1}\frac{M}{(2l_{1}-1)(2l_{1}+1)}=\sum ^{\infty }_{l_{2}=1}\frac{M(l_{2}-1)}{\sqrt{l_{2}}(2l_{2}-1)}=\infty. $$

Example 2

Consider the third-order nonlinear neutral delay difference equation

$$\begin{aligned}& \varDelta \biggl(n \biggl[\varDelta \biggl( \sqrt{n+2} \biggl[\varDelta \biggl( x_{n}+\frac{2+(-1)^{n}}{16}x_{n-\tau} \biggr) \biggr]^{1/3} \biggr) \biggr]^{1/5} \biggr) \\& \quad =\frac{(-1)^{n}\sin(x_{n-\tau_{1}})-n^{2}x^{6}_{n-\tau _{2}}}{(n^{5}+7n^{3}+1)(x^{4}_{n-\tau_{1}}+x^{2}_{n-\tau_{2}}+1)}, \end{aligned}$$

(16)

where $\tau\in\mathbb{N}$, $\tau_{1},\tau_{2}\in\mathbb{Z}$, $\gamma _{1}=1/3$, $\gamma_{2}=1/5$. Moreover, $p_{n}=\frac{2+(-1)^{n}}{16}$, $r_{n}^{1}=\sqrt{n+2}$, $r^{2}_{n}=n$, and $f(n,u,v)=\frac{(-1)^{n}\sin (u)-n^{2}v^{6}}{(n^{5}+7n^{3}+1)(u^{4}+v^{2}+1)}$ for any $n\in\mathbb{N}$ and $u,v\in\mathbb{R}$.

Because $f(n,\cdot)\in C^{1}(\mathbb{R}^{2})$ for any $n\in\mathbb{N}$, $f(n,\cdot)$ is a Lipschitz function on $[0,2M]^{2}$ for any $M>0$. It is easy to calculate that there exists $D(M)>0$ such that $P(n,M)\le \frac{D(M)n^{2}}{n^{5}+7n^{3}+1}$ for sufficiently large n and

$$\sum^{\infty}_{l_{1}=1}\frac{D(M)l_{1}^{2}}{l_{1}^{5}+7l_{1}^{3}+1}< \infty, \qquad \sum^{\infty}_{l_{2}=1}\frac{1}{l_{2}^{5}}\sum ^{\infty }_{l_{1}=l_{2}}\frac{D(M)l_{1}^{2}}{l_{1}^{5}+7l_{1}^{3}+1}< \infty, $$

and

$$\sum^{\infty}_{l_{3}=1}\frac{1}{\sqrt{l_{2}+3}^{3}}\sum ^{\infty }_{l_{2}=l_{3}}\frac{1}{l_{2}^{5}}\sum ^{\infty}_{l_{1}=l_{2}}\frac {D(M)l_{1}^{2}}{l_{1}^{5}+7l_{1}^{3}+1}< \infty. $$

Moreover, $f(n,0,0)=0$, $n\in\mathbb{N}$. This means that the assumptions of Theorem 3 are satisfied. Hence, there exist uncountably many solutions to (16) in any interval $[M/2,2M]$ for any $M>0$.

Example 3

Let us consider a nonlinear neutral delay difference equation of the form

$$ \varDelta \bigl( \bigl(\varDelta \bigl(\cdots \bigl(\varDelta \bigl(\varDelta (x_{n}+p_{n}x_{n-\tau}) \bigr)^{\gamma_{1}} \bigr)^{\gamma_{2}}\cdots \bigr)^{\gamma_{m-2}} \bigr)^{\gamma_{m-1}} \bigr)= \frac {x^{6}_{n-{\tau_{1}}}}{6^{n}}, $$

(17)

where $m\in\mathbb{N}$, $\tau\in\mathbb{N}$, $\tau_{1}\in\mathbb {Z}$, and $\gamma_{1},\ldots,\gamma_{m-1}$ are the ratios of odd positive integers. Moreover, $(p_{n})_{n\in\mathbb{N}}$ is any sequence of real numbers such that $\sup_{n\in\mathbb {N}}|p_{n}|<1/4$, $r_{n}^{1}=\cdots=r_{n}^{m-1}=1$, and $f(n,u)=\frac {u^{6}}{6^{n}}$ for any $n\in\mathbb{N}$ and $u\in\mathbb{R}$.

In an analogous way to Example 1, we have to check only assumption (H₄) of Theorem 3. We have that $f(n,\cdot)\in C^{1}(\mathbb{R})$ for any $n\in\mathbb{N}$, and it is easy to calculate that $P(n,M)=\frac{192M^{5}}{6^{n}}$ and

$$\sum^{\infty}_{l_{m}=1}\sum ^{\infty}_{l_{m-1}=l_{m}}\cdots\sum^{\infty}_{l_{2}=l_{3}} \sum^{\infty}_{l_{1}=l_{2}}\frac{192M^{5}}{6^{l_{1}}}=32M^{5} \biggl(\frac {6}{5} \biggr)^{m}< \infty. $$

Hence, there exist uncountably many solutions to (17) in any interval $[M/2,2M]$ for any $M>0$.