On a close to symmetric system of difference equations of second order

Abstract

Closed form formulas of the solutions to the following system of difference equations:

$$x_{n}=\frac{y_{n-1}y_{n-2}}{x_{n-1}(a_{n}+b_{n}y_{n-1}y_{n-2})},\qquad y_{n}=\frac{x_{n-1}x_{n-2}}{y_{n-1}(\alpha _{n}+\beta _{n}x_{n-1}x_{n-2})},\quad n\in \mathbb {N}_{0}, $$

where $a_{n}$, $b_{n}$, $\alpha _{n}$, $\beta _{n}$, $n\in \mathbb {N}_{0}$, and initial values $x_{-i}$, $y_{-i}$, $i\in\{1,2\}$ are real numbers, are found. The domain of undefinable solutions to the system is described. The long-term behavior of its solutions is studied in detail for the case of constant $a_{n}$, $b_{n}$, $\alpha _{n}$ and $\beta _{n}$, $n\in \mathbb {N}_{0}$.

1 Introduction

Studying concrete nonlinear difference equations and systems is a topic of a great recent interest (see, e.g., [1–46] and the references therein). Studying systems of difference equations, especially symmetric and close to symmetric ones, is a topic of considerable interest (see, e.g., [2, 6, 7, 10, 12–16, 18, 19, 23, 24, 26–29, 31–38, 40, 41, 44, 46]). Another topic of interest is solvable difference equations and systems and their applications (see, e.g., [1–5, 7, 17, 20, 21, 23–27, 29–37, 39–46]). Renewed interest in the area started after the publication of [20] where a formula for a solution of a difference equation was theoretically explained. The most interesting thing in [20] was a change of variables which reduced the equation to a linear one with constant coefficients. Related ideas were later used, e.g., in [1, 4, 7, 17, 21, 23–27, 29–37, 39–45].

Quite recently in [2] the following systems of difference equations were presented:

$$ \begin{aligned} &x_{n}=\frac{y_{n-1}y_{n-2}}{x_{n-1}(\pm1\pm y_{n-1}y_{n-2})},\\ &y_{n}= \frac{x_{n-1}x_{n-2}}{y_{n-1}(\pm1\pm x_{n-1}x_{n-2})},\quad n\in \mathbb {N}_{0}, \end{aligned} $$

(1)

where $x_{-i}$, $y_{-i}$, $i\in\{1,2\}$ are real numbers, and some formulas for their solutions are given, some of which are proved by induction.

The next system of difference equations

$$ \begin{aligned} &x_{n}=\frac{y_{n-1}y_{n-2}}{x_{n-1}(a_{n}+b_{n}y_{n-1}y_{n-2})},\\ &y_{n}=\frac{x_{n-1}x_{n-2}}{y_{n-1}(\alpha _{n}+\beta _{n}x_{n-1}x_{n-2})},\quad n\in \mathbb {N}_{0}, \end{aligned} $$

(2)

where $a_{n}$, $b_{n}$, $\alpha _{n}$, $\beta _{n}$, $n\in \mathbb {N}_{0}$, and initial values $x_{-i}$, $y_{-i}$, $i\in\{1,2\}$, are real numbers, is a generalization of the system in (1). Our aim is to show that more general system (2) is solvable by giving a natural method for getting its solutions. The domain of undefinable solutions to the system is also described. For the case when $a_{n}$, $b_{n}$, $\alpha _{n}$, $\beta _{n}$, $n\in \mathbb {N}_{0}$, are constant, the long-term behavior of its solutions is investigated in detail.

A solution $(x_{n}, y_{n})_{n\ge-2}$ of system (2) is called periodic, or eventually periodic, with period p if there is $n_{0}\ge-2$ such that

$$x_{n+p}=x_{n}\quad \mbox{and}\quad y_{n+p}=y_{n} \quad \mbox{for } n\ge n_{0}. $$

For some results in the area, see, e.g., [6, 9–11, 19, 21, 22, 28].

2 Solutions to system (2) in closed form

Assume first that $x_{-i}\ne0$, $y_{-i}\ne0$, $i\in\{1,2\}$. Then, by the method of induction and the equations in (2), it follows that for every well-defined solution to system (2), $x_{n}\ne0$ and $y_{n}\ne0$, for every $n\in \mathbb {N}_{0}$. On the other hand, if $x_{n_{0}}=0$ for some $n_{0}\in \mathbb {N}$, then the first equation in (2) implies that $y_{n_{0}-1}=0$ or $y_{n_{0}-2}=0$. If $y_{n_{0}-1}=0$, then $x_{n_{0}-2}=0$ or $x_{n_{0}-3}=0$, while if $y_{n_{0}-2}=0$, then $x_{n_{0}-3}=0$ or $x_{n_{0}-4}=0$. Repeating this procedure, we get that $x_{-i}=0$ or $y_{-i}=0$ for some $i\in\{1,2\}$. Similarly, if $y_{n_{1}}=0$ for some $n_{1}\in \mathbb {N}$, we get $x_{-i}=0$ or $y_{-i}=0$ for some $i\in\{1,2\}$. Hence, for a well-defined solution $(x_{n},y_{n})_{n\ge-2}$ of system (2), we have that

$$\begin{aligned} x_{n}y_{n}\ne0,\quad n\ge-2 \end{aligned}$$

(3)

if and only if $x_{-i}y_{-i}\ne0$, $i\in\{1,2\}$.

Assume now that $(x_{n},y_{n})_{n\ge-2}$ is a solution to system (2) such that (3) holds. Then, by multiplying the first equation in (2) by $x_{n-1}$ and the second one by $y_{n-1}$, and using the following changes of variables

$$\begin{aligned} u_{n}=\frac{1}{x_{n}x_{n-1}},\qquad v_{n}= \frac{1}{y_{n}y_{n-1}}, \end{aligned}$$

(4)

$n\ge-1$, system (2) is transformed in the following one:

$$\begin{aligned} u_{n}=a_{n}v_{n-1}+b_{n}, \qquad v_{n}=\alpha _{n}u_{n-1}+\beta _{n},\quad n \in \mathbb {N}_{0}. \end{aligned}$$

(5)

From (5) it follows that

$$\begin{aligned} &u_{n}=a_{n}\alpha _{n-1}u_{n-2}+a_{n} \beta _{n-1}+b_{n}, \end{aligned}$$

(6)

$$\begin{aligned} &v_{n}=\alpha _{n}a_{n-1}v_{n-2}+ \alpha _{n}b_{n-1}+\beta _{n},\quad n\in \mathbb {N}. \end{aligned}$$

(7)

This means that $(u_{2n})_{n\in \mathbb {N}_{0}}$, $(u_{2n-1})_{n\in \mathbb {N}_{0}}$, $(v_{2n})_{n\in \mathbb {N}_{0}}$, and $(v_{2n-1})_{n\in \mathbb {N}_{0}}$ are solutions to two linear first-order difference equations, which are solvable.

Solving these equations, we get

$$\begin{aligned}& u_{2n} =u_{0}\prod_{j=1}^{n}a_{2j} \alpha _{2j-1} +\sum_{i=1}^{n} (a_{2i}\beta _{2i-1}+b_{2i} ) \prod _{s=i+1}^{n}a_{2s}\alpha _{2s-1}, \end{aligned}$$

(8)

$$\begin{aligned}& u_{2n-1} =u_{-1}\prod_{j=1}^{n}a_{2j-1} \alpha _{2j-2} +\sum_{i=1}^{n} (a_{2i-1}\beta _{2i-2}+b_{2i-1} ) \prod _{s=i+1}^{n}a_{2s-1}\alpha _{2s-2}, \end{aligned}$$

(9)

$$\begin{aligned}& v_{2n} =v_{0}\prod_{j=1}^{n} \alpha _{2j}a_{2j-1} +\sum_{i=1}^{n} (\alpha _{2i}b_{2i-1}+\beta _{2i} ) \prod _{s=i+1}^{n}\alpha _{2s}a_{2s-1}, \end{aligned}$$

(10)

$$\begin{aligned}& v_{2n-1} =v_{-1}\prod_{j=1}^{n} \alpha _{2j-1}a_{2j-2} +\sum_{i=1}^{n} (\alpha _{2i-1}b_{2i-2}+\beta _{2i-1} ) \prod _{s=i+1}^{n}\alpha _{2s-1}a_{2s-2}. \end{aligned}$$

(11)

Using (4) we obtain

$$\begin{aligned} x_{2n+i}=\frac{1}{u_{2n+i}x_{2n+i-1}}=\frac {u_{2n+i-1}}{u_{2n+i}}x_{2(n-1)+i},\quad i\in \{0,1\}, \end{aligned}$$

and

$$\begin{aligned} y_{2n+i}=\frac{1}{v_{2n+i}y_{2n+i-1}}=\frac {v_{2n+i-1}}{v_{2n+i}}y_{2(n-1)+i},\quad i\in \{0,1\}, \end{aligned}$$

for $2n+i\ge0$, from which it follows that

$$\begin{aligned}& x_{2m+i}=x_{i-2}\prod_{j=0}^{m} \frac{u_{2j+i-1}}{u_{2j+i}}, \end{aligned}$$

(12)

$$\begin{aligned}& y_{2m+i}=y_{i-2}\prod_{j=0}^{m} \frac{v_{2j+i-1}}{v_{2j+i}} \end{aligned}$$

(13)

for every $m\in \mathbb {N}_{0}$, $i\in\{0,1\}$.

3 Case of constant coefficients

In this section we consider the case when all the coefficients in system (2) are constant, that is, when

$$a_{n}=a, \qquad b_{n}=b,\qquad \alpha _{n}=\alpha ,\qquad \beta _{n}=\beta , \quad n\in \mathbb {N}_{0}. $$

Then (2) is

$$ \begin{aligned} &x_{n}=\frac{y_{n-1}y_{n-2}}{x_{n-1}(a+by_{n-1}y_{n-2})},\\ & y_{n}=\frac{x_{n-1}x_{n-2}}{y_{n-1}(\alpha +\beta x_{n-1}x_{n-2})}, \quad n\in \mathbb {N}_{0}. \end{aligned} $$

(14)

Assume that $(x_{n},y_{n})_{n\ge-2}$ is a solution to system (2) such that (3) holds. Then we have

$$\begin{aligned} u_{n}=av_{n-1}+b, \qquad v_{n}=\alpha u_{n-1}+\beta ,\quad n\in \mathbb {N}_{0}, \end{aligned}$$

(15)

and

$$\begin{aligned}& u_{n}=a\alpha u_{n-2}+a\beta +b, \end{aligned}$$

(16)

$$\begin{aligned}& v_{n}=a\alpha v_{n-2}+\alpha b+\beta ,\quad n\in \mathbb {N}. \end{aligned}$$

(17)

From (8)-(11), we obtain

$$\begin{aligned} u_{2n-l}&=u_{-l}(a\alpha )^{n}+(a\beta +b) \frac{1-(a\alpha )^{n}}{1-a\alpha } \\ &=\frac{a\beta +b+(a\alpha )^{n}(u_{-l}(1-a\alpha )-a\beta -b)}{1-a\alpha } \end{aligned}$$

(18)

for $n\in \mathbb {N}_{0}$, $l\in\{0,1\}$ when $a\alpha \ne 1$, while if $a\alpha =1$, we have

$$\begin{aligned} u_{2n-l}=u_{-l}+(a\beta +b)n,\quad n\in \mathbb {N}_{0}, l\in\{0,1\}, \end{aligned}$$

(19)

and we also have

$$\begin{aligned} v_{2n-l}&=v_{-l}(a\alpha )^{n}+(\alpha b+\beta ) \frac{1-(a\alpha )^{n}}{1-a\alpha } \\ &=\frac{\alpha b+\beta +(a\alpha )^{n}(v_{-l}(1-a\alpha )-\alpha b-\beta )}{1-a\alpha }, \end{aligned}$$

(20)

$n\in \mathbb {N}_{0}$, $l\in\{0,1\}$ if $a\alpha \ne1$, while if $a\alpha =1$, we have

$$\begin{aligned} v_{2n-l}=v_{-l}+(\alpha b+\beta )n,\quad n\in \mathbb {N}_{0}, l\in \{0,1\}. \end{aligned}$$

(21)

Now we present formulae for solutions to system (14).

Case $a\alpha \ne1$. We have

$$\begin{aligned}& x_{2m}= x_{-2}\prod_{j=0}^{m} \frac{u_{2j-1}}{u_{2j}}=x_{-2}\prod_{j=0}^{m} \frac{a\beta +b+(a\alpha )^{j}(u_{-1}(1-a\alpha )-a\beta -b)}{a\beta +b+(a\alpha )^{j}(u_{0}(1-a\alpha )-a\beta -b)}, \end{aligned}$$

(22)

$$\begin{aligned}& x_{2m+1} =x_{-1}\prod_{j=0}^{m} \frac{u_{2j}}{u_{2j+1}}=x_{-1}\prod_{j=0}^{m} \frac{a\beta +b+(a\alpha )^{j}(u_{0}(1-a\alpha )-a\beta -b)}{a\beta +b+(a\alpha )^{j+1}(u_{-1}(1-a\alpha )-a\beta -b)}, \end{aligned}$$

(23)

$$\begin{aligned}& y_{2m} =y_{-2}\prod_{j=0}^{m} \frac{v_{2j-1}}{v_{2j}}=y_{-2}\prod_{j=0}^{m} \frac{\alpha b+\beta +(a\alpha )^{j}(v_{-1}(1-a\alpha )-\alpha b-\beta )}{\alpha b+\beta +(a\alpha )^{j}(v_{0}(1-a\alpha )-\alpha b-\beta )}, \end{aligned}$$

(24)

$$\begin{aligned}& y_{2m+1} =y_{-1}\prod_{j=0}^{m} \frac{v_{2j}}{v_{2j+1}}=y_{-1}\prod_{j=0}^{m} \frac{\alpha b+\beta +(a\alpha )^{j}(v_{0}(1-a\alpha )-\alpha b-\beta )}{\alpha b+\beta +(a\alpha )^{j+1}(v_{-1}(1-a\alpha )-\alpha b-\beta )} \end{aligned}$$

(25)

for every $m\in \mathbb {N}_{0}$.

Case $a\alpha =1$. We have

$$\begin{aligned}& x_{2m} =x_{-2}\prod_{j=0}^{m} \frac{u_{2j-1}}{u_{2j}}=x_{-2}\prod_{j=0}^{m} \frac{u_{-1}+(a\beta +b)j}{u_{0}+(a\beta +b)j}, \end{aligned}$$

(26)

$$\begin{aligned}& x_{2m+1} =x_{-1}\prod_{j=0}^{m} \frac{u_{2j}}{u_{2j+1}}=x_{-1}\prod_{j=0}^{m} \frac{u_{0}+(a\beta +b)j}{u_{-1}+(a\beta +b)(j+1)}, \end{aligned}$$

(27)

$$\begin{aligned}& y_{2m} =y_{-2}\prod_{j=0}^{m} \frac{v_{2j-1}}{v_{2j}}=y_{-2}\prod_{j=0}^{m} \frac{v_{-1}+(\alpha b+\beta )j}{v_{0}+(\alpha b+\beta )j}, \end{aligned}$$

(28)

$$\begin{aligned}& y_{2m+1}=y_{-1}\prod_{j=0}^{m} \frac{v_{2j}}{v_{2j+1}}=y_{-1}\prod_{j=0}^{m} \frac{v_{0}+(\alpha b+\beta )j}{v_{-1}+(\alpha b+\beta )(j+1)} \end{aligned}$$

(29)

for every $m\in \mathbb {N}_{0}$.

4 Long-term behavior of solutions to system (14)

Before we formulate and prove the main results regarding the long-term behavior of well-defined solutions to system (14), we quote the following well-known asymptotic formula which will be used in the proofs of the main results:

$$\begin{aligned} (1+x)^{-1}=1-x+O\bigl(x^{2}\bigr),\quad \mbox{as } x\to 0. \end{aligned}$$

(30)

We also define the following quantities:

$$\begin{aligned}& L_{1}:=\frac{u_{-1}(1-a\alpha )-a\beta -b}{u_{0}(1-a\alpha )-a\beta -b},\qquad L_{2}:=\frac {u_{0}(1-a\alpha )-a\beta -b}{a\alpha (u_{-1}(1-a\alpha )-a\beta -b)}, \\& L_{3}:=\frac{v_{-1}(1-a\alpha )-\alpha b-\beta }{v_{0}(1-a\alpha )-\alpha b-\beta }, \qquad L_{4}:= \frac {v_{0}(1-a\alpha )-\alpha b-\beta }{a\alpha (v_{-1}(1-a\alpha )-\alpha b-\beta )}. \end{aligned}$$

Finally, we give another auxiliary result.

Lemma 1

If $a\alpha \ne1$, $a\beta +b\ne0\ne \alpha b+\beta$. Then system (14) has two-periodic solutions.

Proof

The equilibrium solution to system (15) is

$$\begin{aligned} u_{n}=\bar{u}=\frac{a\beta +b}{1-a\alpha }\ne0,\qquad v_{n}=\bar{v}= \frac{\alpha b+\beta }{1-a\alpha }\ne0,\quad n\in \mathbb {N}_{0}. \end{aligned}$$

(31)

From (4) and (31) it follows that

$$\begin{aligned} x_{n}=\frac{1-a\alpha }{(a\beta +b)x_{n-1}}=x_{n-2},\quad n\in \mathbb {N}_{0}, \end{aligned}$$

(32)

and

$$\begin{aligned} y_{n}=\frac{1-a\alpha }{(\alpha b+\beta )y_{n-1}}=y_{n-2},\quad n\in \mathbb {N}_{0}, \end{aligned}$$

(33)

as desired. □

The next three results are devoted to the long-term behavior of well-defined solutions to system (14).

Theorem 1

Assume that $\vert a\alpha \vert \ne1$ and $(x_{n}, y_{n})_{n\ge-2}$ is a well-defined solution to system (14). Then the following statements are true.

• (a) If $a\beta +b\ne0\ne \alpha b+\beta$ and $\vert a\alpha \vert <1$, then $(x_{n},y_{n})$ converges to a, not necessarily prime, two-periodic solution.

• (b) If $u_{-1}=u_{0}=(a\beta +b)/(1-a\alpha )$, then the sequences $(x_{2m})_{m\ge-1}$ and $(x_{2m+1})_{m\ge-1}$ are constant.

• (c) If $v_{-1}=v_{0}=(\alpha b+\beta )/(1-a\alpha )$, then the sequences $(y_{2m})_{m\ge-1}$ and $(y_{2m+1})_{m\ge-1}$ are constant.

• (d) If $\vert a\alpha \vert >1$ and $u_{-1}=(a\beta +b)/(1-a\alpha )\ne u_{0}$, then $x_{2m}\to0$ and $\vert x_{2m+1}\vert \to\infty$, as $m\to\infty$.

• (e) If $\vert a\alpha \vert >1$ and $u_{-1}\ne(a\beta +b)/(1-a\alpha )=u_{0}$, then $x_{2m+1}\to0$ and $\vert x_{2m}\vert \to\infty$, as $m\to\infty$.

• (f) If $\vert a\alpha \vert >1$ and $v_{-1}=(a\beta +b)/(1-a\alpha )\ne v_{0}$, then $y_{2m}\to0$ and $\vert y_{2m+1}\vert \to\infty$, as $m\to\infty$.

• (g) If $\vert a\alpha \vert >1$ and $v_{-1}\ne(a\beta +b)/(1-a\alpha )=v_{0}$, then $y_{2m+1}\to0$ and $\vert y_{2m}\vert \to\infty$, as $m\to\infty$.

• (h) If $\vert a\alpha \vert >1$, $u_{-1}\ne(a\beta +b)/(1-a\alpha )\ne u_{0}$ and $\vert L_{1}\vert <1$, then $x_{2m}\to0$, as $m\to\infty$.

• (i) If $\vert a\alpha \vert >1$, $u_{-1}\ne(a\beta +b)/(1-a\alpha )\ne u_{0}$ and $\vert L_{1}\vert >1$, then $\vert x_{2m}\vert \to\infty$, as $m\to\infty$.

• (j) If $\vert a\alpha \vert >1$, $u_{-1}\ne(a\beta +b)/(1-a\alpha )\ne u_{0}$ and $L_{1}=1$, then $(x_{2m})_{m\ge-1}$ is constant.

• (k) If $\vert a\alpha \vert >1$, $u_{-1}\ne(a\beta +b)/(1-a\alpha )\ne u_{0}$ and $L_{1}=-1$, then $(x_{4m})_{m\ge-1}$ and $(x_{4m+2})_{m\ge-1}$ are convergent.

• (l) If $\vert a\alpha \vert >1$, $u_{-1}\ne(a\beta +b)/(1-a\alpha )\ne u_{0}$ and $\vert L_{2}\vert <1$, then $x_{2m+1}\to0$, as $m\to\infty$.

• (m) If $\vert a\alpha \vert >1$, $u_{-1}\ne(a\beta +b)/(1-a\alpha )\ne u_{0}$ and $\vert L_{2}\vert >1$, then $\vert x_{2m+1}\vert \to\infty$, as $m\to\infty$.

• (n) If $\vert a\alpha \vert >1$, $u_{-1}\ne(a\beta +b)/(1-a\alpha )\ne u_{0}$ and $L_{2}=1$, then $(x_{2m+1})_{m\ge-1}$ is constant.

• (o) If $\vert a\alpha \vert >1$, $u_{-1}\ne(a\beta +b)/(1-a\alpha )\ne u_{0}$ and $L_{2}=-1$, then $(x_{4m+1})_{m\ge-1}$ and $(x_{4m+3})_{m\ge-1}$ are convergent.

• (p) If $\vert a\alpha \vert >1$, $v_{-1}\ne(\alpha b+\beta )/(1-a\alpha )\ne v_{0}$ and $\vert L_{3}\vert <1$, then $y_{2m}\to0$, as $m\to\infty$.

• (q) If $\vert a\alpha \vert >1$, $v_{-1}\ne(\alpha b+\beta )/(1-a\alpha )\ne v_{0}$ and $\vert L_{3}\vert >1$, then $\vert y_{2m}\vert \to\infty$, as $m\to\infty$.

• (r) If $\vert a\alpha \vert >1$, $v_{-1}\ne(\alpha b+\beta )/(1-a\alpha )\ne v_{0}$ and $L_{3}=1$, then $(y_{2m})_{m\ge-1}$ is constant.

• (s) If $\vert a\alpha \vert >1$, $v_{-1}\ne(\alpha b+\beta )/(1-a\alpha )\ne v_{0}$ and $L_{3}=-1$, then $(y_{4m})_{m\ge-1}$ and $(y_{4m+2})_{m\ge-1}$ are convergent.

• (t) If $\vert a\alpha \vert >1$, $v_{-1}\ne(\alpha b+\beta )/(1-a\alpha )\ne v_{0}$ and $\vert L_{4}\vert <1$, then $y_{2m+1}\to0$, as $m\to\infty$.

• (u) If $\vert a\alpha \vert >1$, $v_{-1}\ne(\alpha b+\beta )/(1-a\alpha )\ne v_{0}$ and $\vert L_{4}\vert >1$, then $\vert y_{2m+1}\vert \to\infty$, as $m\to\infty$.

• (v) If $\vert a\alpha \vert >1$, $v_{-1}\ne(\alpha b+\beta )/(1-a\alpha )\ne v_{0}$ and $L_{4}=1$, then $(y_{2m+1})_{m\ge-1}$ is constant.

• (w) If $\vert a\alpha \vert >1$, $v_{-1}\ne(\alpha b+\beta )/(1-a\alpha )\ne v_{0}$ and $L_{4}=-1$, then $(y_{4m+1})_{m\ge-1}$ and $(y_{4m+3})_{m\ge-1}$ are convergent.

Proof

Let

$$\begin{aligned} &p_{m}=\frac{a\beta +b+(a\alpha )^{m}(u_{-1}(1-a\alpha )-a\beta -b)}{a\beta +b+(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)}, \\ &\hat{p}_{m}=\frac{a\beta +b+(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)}{a\beta +b+(a\alpha )^{m+1}(u_{-1}(1-a\alpha )-a\beta -b)}, \\ &q_{m}=\frac{\alpha b+\beta +(a\alpha )^{m}(v_{-1}(1-a\alpha )-\alpha b-\beta )}{\alpha b+\beta +(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )}, \\ &\hat{q}_{m}=\frac{\alpha b+\beta +(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )}{\alpha b+\beta +(a\alpha )^{m+1}(v_{-1}(1-a\alpha )-\alpha b-\beta )} \end{aligned}$$

for $m\in \mathbb {N}_{0}$.

(a) By using (30) we have

$$\begin{aligned}& \begin{aligned}[b] p_{m}&=\frac{1+(a\alpha )^{m}(u_{-1}(1-a\alpha )-a\beta -b)(a\beta +b)^{-1}}{1+(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)(a\beta +b)^{-1}} \\ &=1+(u_{-1}-u_{0}) (1-a\alpha ) (a\beta +b)^{-1}(a \alpha )^{m}+o\bigl((a\alpha )^{m}\bigr), \end{aligned} \end{aligned}$$

(34)

$$\begin{aligned}& \begin{aligned}[b] \hat{p}_{m}&=\frac{1+(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)(a\beta +b)^{-1}}{1+(a\alpha )^{m+1}(u_{-1}(1-a\alpha )-a\beta -b)(a\beta +b)^{-1}} \\ &=1+\frac{(1-a\alpha )(u_{0}-a\alpha u_{-1}-a\beta -b)}{a\beta +b}(a\alpha )^{m}+o\bigl((a\alpha )^{m} \bigr), \end{aligned} \end{aligned}$$

(35)

$$\begin{aligned}& \begin{aligned}[b] q_{m}&=\frac{1+(a\alpha )^{m}(v_{-1}(1-a\alpha )-\alpha b-\beta )(\alpha b+\beta )^{-1}}{1+(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )(\alpha b+\beta )^{-1}} \\ &=1+(v_{-1}-v_{0}) (1-a\alpha ) (\alpha b+\beta )^{-1}(a \alpha )^{m}+o\bigl((a\alpha )^{m}\bigr), \end{aligned} \end{aligned}$$

(36)

$$\begin{aligned}& \begin{aligned}[b] \hat{q}_{m}&=\frac{1+(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )(\alpha b+\beta )^{-1}}{1+(a\alpha )^{m+1}(v_{-1}(1-a\alpha )-\alpha b-\beta )(\alpha b+\beta )^{-1}} \\ &=1+\frac{(1-a\alpha )(v_{0}-a\alpha v_{-1}-\alpha b-\beta )}{\alpha b+\beta }(a\alpha )^{m}+o\bigl((a\alpha )^{m} \bigr) \end{aligned} \end{aligned}$$

(37)

for sufficiently large m.

From (34)-(37), by using the condition $\vert a\alpha \vert <1$ and a well-known criterion for the convergence of products, the statement easily follows.

(b) By using the condition $u_{-1}=u_{0}=(a\beta +b)/(1-a\alpha )$ in (22) and (23), the statement immediately follows.

(c) By using the condition $v_{-1}=v_{0}=(\alpha b+\beta )/(1-a\alpha )$ in (24) and (25), the statement immediately follows.

(d) By using the condition $u_{-1}=(a\beta +b)/(1-a\alpha )\ne u_{0}$, we get

$$\begin{aligned} &p_{m}=\frac{a\beta +b}{a\beta +b+(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)}, \end{aligned}$$

(38)

$$\begin{aligned} &\hat{p}_{m}=\frac{a\beta +b+(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)}{a\beta +b}. \end{aligned}$$

(39)

Letting $m\to\infty$ in (38) and (39) and using the condition $\vert a\alpha \vert >1$, we have $p_{m}\to0$ and $\vert \hat{p}_{m}\vert \to\infty$, from which along with (22) and (23) the statement easily follows.

(e) By using the condition $u_{-1}\ne(a\beta +b)/(1-a\alpha )=u_{0}$, we get

$$\begin{aligned} &p_{m}=\frac{a\beta +b+(a\alpha )^{m}(u_{-1}(1-a\alpha )-a\beta -b)}{a\beta +b}, \end{aligned}$$

(40)

$$\begin{aligned} &\hat{p}_{m}=\frac{a\beta +b}{a\beta +b+(a\alpha )^{m+1}(u_{-1}(1-a\alpha )-a\beta -b)}. \end{aligned}$$

(41)

Letting $m\to\infty$ in (40) and (41) and using the condition $\vert a\alpha \vert >1$, we have $\vert p_{m}\vert \to\infty$ and $\hat{p}_{m}\to0$, from which along with (22) and (23) the statement easily follows.

(f) By using the condition $v_{-1}=(a\beta +b)/(1-a\alpha )\ne v_{0}$, we get

$$\begin{aligned} &q_{m}=\frac{\alpha b+\beta }{\alpha b+\beta +(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )}, \end{aligned}$$

(42)

$$\begin{aligned} &\hat{q}_{m}=\frac{\alpha b+\beta +(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )}{\alpha b+\beta }. \end{aligned}$$

(43)

Letting $m\to\infty$ in (42) and (43) and using the condition $\vert a\alpha \vert >1$, we have $q_{m}\to0$ and $\vert \hat{q}_{m}\vert \to\infty$, from which along with (24) and (25) the statement easily follows.

(g) By using the condition $v_{-1}\ne(a\beta +b)/(1-a\alpha )= v_{0}$, we get

$$\begin{aligned} &q_{m}=\frac{\alpha b+\beta +(a\alpha )^{m}(v_{-1}(1-a\alpha )-\alpha b-\beta )}{\alpha b+\beta }, \end{aligned}$$

(44)

$$\begin{aligned} &\hat{q}_{m}=\frac{\alpha b+\beta }{\alpha b+\beta +(a\alpha )^{m+1}(v_{-1}(1-a\alpha )-\alpha b-\beta )}. \end{aligned}$$

(45)

Letting $m\to\infty$ in (44) and (45) and using the condition $\vert a\alpha \vert >1$, we have $\vert q_{m}\vert \to\infty$ and $\hat{q}_{m}\to0$, from which along with (24) and (25) the statement easily follows.

(h), (i) Note that $\lim_{m\to\infty}p_{m}=L_{1}$. Hence, from the assumptions $\vert L_{1}\vert <1$, that is, $\vert L_{1}\vert >1$ along with (22), the statements easily follow.

(j) The statement immediately follows by using the condition $L_{1}=1$ in (22).

(k) Since $L_{1}=-1$ and by using (30), we have that

$$\begin{aligned} p_{m}&=\frac{a\beta +b+(a\alpha )^{m}(u_{-1}(1-a\alpha )-a\beta -b)}{a\beta +b-(a\alpha )^{m}(u_{-1}(1-a\alpha )-a\beta -b)} \\ &=-\frac{1+\frac{a\beta +b}{(a\alpha )^{m}(u_{-1}(1-a\alpha )-a\beta -b)}}{1-\frac {a\beta +b}{(a\alpha )^{m}(u_{-1}(1-a\alpha )-a\beta -b)}} \\ &=- \biggl(1+\frac{2(a\beta +b)}{(a\alpha )^{m}(u_{-1}(1-a\alpha )-a\beta -b)}+o \biggl(\frac{1}{(a\alpha )^{m}} \biggr) \biggr). \end{aligned}$$

(46)

From (46), by using the condition $\vert a\alpha \vert >1$ and a well-known criterion for the convergence of products, the statement easily follows.

(l), (m) Note that $\lim_{m\to\infty}\hat{p}_{m}=L_{2}$. Hence, from the assumptions $\vert L_{2}\vert <1$, that is, $\vert L_{2}\vert >1$ along with (23), the statements easily follow.

(n) The statement immediately follows by using the condition $L_{2}=1$ in (23).

(o) Since $L_{2}=-1$ and by using (30), we have that

$$\begin{aligned} \hat{p}_{m}&=\frac{a\beta +b+(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)}{a\beta +b-(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)} \\ &=-\frac{1+\frac{a\beta +b}{(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)}}{1-\frac{a\beta +b}{(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)}} \\ &=- \biggl(1+\frac{2(a\beta +b)}{(a\alpha )^{m}(u_{0}(1-a\alpha )-a\beta -b)}+o \biggl(\frac{1}{(a\alpha )^{m}} \biggr) \biggr). \end{aligned}$$

(47)

From (47), by using the condition $\vert a\alpha \vert >1$ and a well-known criterion for the convergence of products, the statement easily follows.

(p), (q) Note that $\lim_{m\to\infty}q_{m}=L_{3}$. Hence, from the assumptions $\vert L_{3}\vert <1$, that is, $\vert L_{3}\vert >1$ along with (24), the statements easily follow.

(r) The statement immediately follows by using the condition $L_{3}=1$ in (24).

(s) Since $L_{3}=-1$ and by using (30), we have that

$$\begin{aligned} q_{m}&=\frac{\alpha b+\beta +(a\alpha )^{m}(v_{-1}(1-a\alpha )-\alpha b-\beta )}{\alpha b+\beta -(a\alpha )^{m}(v_{-1}(1-a\alpha )-\alpha b-\beta )} \\ &=-\frac{1+\frac{\alpha b+\beta }{(a\alpha )^{m}(v_{-1}(1-a\alpha )-\alpha b-\beta )}}{1-\frac{\alpha b+\beta }{(a\alpha )^{m}(v_{-1}(1-a\alpha )-\alpha b-\beta )}} \\ &=- \biggl(1+\frac{2(\alpha b+\beta )}{(a\alpha )^{m}(v_{-1}(1-a\alpha )-\alpha b-\beta )}+o \biggl(\frac{1}{(a\alpha )^{m}} \biggr) \biggr). \end{aligned}$$

(48)

From (48), by using the condition $\vert a\alpha \vert >1$ and a well-known criterion for the convergence of products, the statement easily follows.

(t), (u) Note that $\lim_{m\to\infty}\hat{q}_{m}=L_{4}$. Hence, from the assumptions $\vert L_{4}\vert <1$, that is, $\vert L_{4}\vert >1$ along with (25), the statements easily follow.

(v) The statement immediately follows by using the condition $L_{4}=1$ in (25).

(w) Since $L_{4}=-1$ and by using (30), we have that

$$\begin{aligned} \hat{q}_{m}&=\frac{\alpha b+\beta +(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )}{\alpha b+\beta -(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )} \\ &=-\frac{1+\frac{\alpha b+\beta }{(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )}}{1-\frac {\alpha b+\beta }{(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )}} \\ &=- \biggl(1+\frac{2(\alpha b+\beta )}{(a\alpha )^{m}(v_{0}(1-a\alpha )-\alpha b-\beta )}+o \biggl(\frac{1}{(a\alpha )^{m}} \biggr) \biggr). \end{aligned}$$

(49)

From (49), by using the condition $\vert a\alpha \vert >1$ and a well-known criterion for the convergence of products, the statement easily follows. □

Let

$$M_{1}:=\frac{u_{-1}(u_{-1}-b-a\beta )}{u_{0}(u_{0}-b-a\beta )},\qquad M_{2}:=\frac {v_{-1}(v_{-1}-\beta -\alpha b)}{v_{0}(v_{0}-\beta -\alpha b)}. $$

Theorem 2

Assume that $a\alpha =-1$ and $(x_{n}, y_{n})_{n\ge-2}$ is a well-defined solution to system (14). Then the following statements are true.

• (a) If $\vert M_{1}\vert <1$, then $x_{2m}\to0$ and $\vert x_{2m+1}\vert \to\infty$, as $m\to\infty$.

• (b) If $\vert M_{1}\vert >1$, then $x_{2m+1}\to0$ and $\vert x_{2m}\vert \to\infty$, as $m\to\infty$.

• (c) If $M_{1}=1$, then $(x_{n})_{n\ge-2}$ is four-periodic.

• (d) If $M_{1}=-1$, then $(x_{n})_{n\ge-2}$ is eight-periodic.

• (e) If $\vert M_{2}\vert <1$, then $y_{2m}\to0$ and $\vert y_{2m+1}\vert \to\infty$, as $m\to\infty$.

• (f) If $\vert M_{2}\vert >1$, then $y_{2m+1}\to0$ and $\vert y_{2m}\vert \to\infty$, as $m\to\infty$.

• (g) If $M_{2}=1$, then $(y_{n})_{n\ge-2}$ is four-periodic.

• (h) If $M_{2}=-1$, then $(y_{n})_{n\ge-2}$ is eight-periodic.

Proof

First, note that since $a\alpha =-1$, from (22)-(25) we have

$$\begin{aligned}& x_{4m}=x_{0}M_{1}^{m},\qquad x_{4m+2}=x_{-2}M_{1}^{m+1},\qquad x_{4m+1}=\frac{x_{1}}{M_{1}^{m}},\qquad x_{4m+3}=\frac{x_{-1}}{M_{1}^{m+1}}, \end{aligned}$$

(50)

$$\begin{aligned}& y_{4m}=y_{0}M_{2}^{m},\qquad y_{4m+2}=y_{-2}M_{2}^{m+1},\qquad y_{4m+1}=\frac{y_{1}}{M_{2}^{m}},\qquad y_{4m+3}=\frac{y_{-1}}{M_{2}^{m+1}}, \end{aligned}$$

(51)

for $m\in \mathbb {N}_{0}$. From (50) and (51) all the statements easily follow. □

Let

$$N_{1}:=\frac{u_{-1}}{u_{0}},\qquad N_{2}:=\frac{v_{-1}}{v_{0}}. $$

Theorem 3

Assume that $a\alpha =1$ and $(x_{n}, y_{n})_{n\ge-2}$ is a well-defined solution to system (14). Then the following statements hold true.

• (a) If $a\beta +b=0$ and $\vert N_{1}\vert <1$, then $x_{2m}\to0$ and $\vert x_{2m+1}\vert \to\infty$, as $m\to\infty$;

• (b) If $a\beta +b=0$ and $\vert N_{1}\vert >1$, then $\vert x_{2m}\vert \to\infty$ and $x_{2m+1}\to0$, as $m\to\infty$;

• (c) If $a\beta +b=0$ and $N_{1}=1$, then $(x_{2m})_{m\ge-1}$ and $(x_{2m+1})_{m\ge-1}$ are constant;

• (d) If $a\beta +b=0$ and $N_{1}=-1$, then $(x_{4m+i})_{m\ge-1}$, $i=\overline {0,3}$, are constant.

• (e) If $a\beta +b\ne0$ and $(u_{-1}-u_{0})/(a\beta +b)>0$, then $\vert x_{2m}\vert \to\infty$, as $m\to\infty$;

• (f) If $a\beta +b\ne0$ and $(u_{-1}-u_{0})/(a\beta +b)<0$, then $x_{2m}\to0$, as $m\to\infty$;

• (g) If $a\beta +b\ne0$ and $u_{-1}=u_{0}$, then $(x_{2m})_{m\ge-1}$ is constant;

• (h) If $a\beta +b\ne0$ and $(u_{0}-u_{-1})/(a\beta +b)>1$, then $\vert x_{2m+1}\vert \to\infty$, as $m\to\infty$;

• (i) If $a\beta +b\ne0$ and $(u_{0}-u_{-1})/(a\beta +b)<1$, then $x_{2m+1}\to0$, as $m\to\infty$;

• (j) If $a\beta +b\ne0$ and $u_{-1}-u_{0}=a\beta +b$, then $(x_{2m+1})_{m\ge-1}$ is constant;

• (k) If $\alpha b+\beta =0$ and $\vert N_{2}\vert <1$, then $y_{2m}\to0$ and $\vert y_{2m+1}\vert \to\infty$, as $m\to\infty$;

• (l) If $\alpha b+\beta =0$ and $\vert N_{2}\vert >1$, then $\vert y_{2m}\vert \to\infty$ and $y_{2m+1}\to0$, as $m\to\infty$;

• (m) If $\alpha b+\beta =0$ and $N_{2}=1$, then $(y_{2m})_{m\ge-1}$ and $(y_{2m+1})_{m\ge-1}$ are constant;

• (n) If $\alpha b+\beta =0$ and $N_{2}=-1$, then $(y_{4m+i})_{m\ge-1}$, $i=\overline {0,3}$, are constant.

• (o) If $\alpha b+\beta \ne0$ and $(v_{-1}-v_{0})/(\alpha b+\beta )>0$, then $\vert y_{2m}\vert \to\infty$, as $m\to\infty$;

• (p) If $\alpha b+\beta \ne0$ and $(v_{-1}-v_{0})/(\alpha b+\beta )<0$, then $y_{2m}\to0$, as $m\to\infty$;

• (q) If $\alpha b+\beta \ne0$ and $v_{-1}=v_{0}$, then $(y_{2m})_{m\ge -1}$ is constant.

• (r) If $\alpha b+\beta \ne0$ and $(v_{0}-v_{-1})/(\alpha b+\beta )<1$, then $y_{2m+1}\to0$, as $m\to\infty$;

• (s) If $\alpha b+\beta \ne0$ and $(v_{0}-v_{-1})/(\alpha b+\beta )>1$, then $\vert y_{2m+1}\vert \to\infty$, as $m\to\infty$;

• (t) If $\alpha b+\beta \ne0$ and $v_{-1}-v_{0}=\alpha b+\beta$, then $(y_{2m+1})_{m\ge-1}$ is constant.

Proof

Let

$$\begin{aligned}& r_{m} =\frac{u_{-1}+(a\beta +b)m}{u_{0}+(a\beta +b)m},\qquad \hat{r}_{m}= \frac{u_{0}+(a\beta +b)m}{u_{-1}+(a\beta +b)(m+1)}, \\& s_{m} =\frac{v_{-1}+(\alpha b+\beta )m}{v_{0}+(\alpha b+\beta )m}, \qquad \hat{s}_{m}= \frac{v_{0}+(\alpha b+\beta )m}{v_{-1}+(\alpha b+\beta )(m+1)},\quad m\in \mathbb {N}_{0}. \end{aligned}$$

(a)-(d) Since in this case we have

$$x_{2m}=x_{-2} \biggl(\frac{u_{-1}}{u_{0}} \biggr)^{m+1},\qquad x_{2m+1}=x_{-1} \biggl( \frac{u_{0}}{u_{-1}} \biggr)^{m+1},\quad m\in \mathbb {N}_{0}, $$

these statements easily follow.

(e), (f) By using (30) we have

$$\begin{aligned} r_{m}&=\frac{u_{-1}+(a\beta +b)m}{u_{0}+(a\beta +b)m}= \biggl(1+\frac {u_{-1}}{(a\beta +b)m} \biggr) \biggl(1+\frac{u_{0}}{(a\beta +b)m} \biggr)^{-1} \\ &= \biggl(1+\frac{u_{-1}}{(a\beta +b)m}+O \biggl(\frac{1}{m^{2}} \biggr) \biggr) \biggl(1-\frac {u_{0}}{(a\beta +b)m}+O \biggl(\frac{1}{m^{2}} \biggr) \biggr) \\ &=1+\frac{u_{-1}-u_{0}}{(a\beta +b)m}+O \biggl(\frac{1}{m^{2}} \biggr) \end{aligned}$$

(52)

for sufficiently large m.

From (52), by using the fact that for every $k\in \mathbb {N}$

$$\begin{aligned} \sum_{j=k}^{m}\frac{1}{j}\to\infty, \quad \mbox{as } m\to\infty, \end{aligned}$$

(53)

and a known criterion for convergence of products, the statements easily follow.

(g) Using the condition $u_{-1}=u_{0}$ in (26), the statement immediately follows.

(h), (i) By using (30) we have

$$\begin{aligned} \hat{r}_{m}&=\frac{u_{0}+(a\beta +b)m}{u_{-1}+(a\beta +b)(m+1)}= \biggl(1+\frac {u_{0}}{(a\beta +b)m} \biggr) \biggl(1+\frac{u_{-1}+a\beta +b}{(a\beta +b)m} \biggr)^{-1} \\ &= \biggl(1+\frac{u_{0}}{(a\beta +b)m} \biggr) \biggl(1-\frac{u_{-1}+a\beta +b}{(a\beta +b)m}+O \biggl( \frac{1}{m^{2}} \biggr) \biggr) \\ &=1+\frac{u_{0}-u_{-1}-a\beta -b}{(a\beta +b)m}+O \biggl(\frac{1}{m^{2}} \biggr) \end{aligned}$$

(54)

for sufficiently large m.

From (54), (53), (27) and a known criterion for convergence of products, the statements easily follow.

(j) Using the condition $u_{0}=u_{-1}+a\beta +b$ in (27), the statement immediately follows.

(k)-(n) Since in this case we have

$$y_{2m}=y_{-2} \biggl(\frac{v_{-1}}{v_{0}} \biggr)^{m+1},\qquad y_{2m+1}=y_{-1} \biggl( \frac{v_{0}}{v_{-1}} \biggr)^{m+1},\quad m\in \mathbb {N}_{0}, $$

these statements easily follow.

(o), (p) By using (30) we have

$$\begin{aligned} s_{m}&=\frac{v_{-1}+(\alpha b+\beta )m}{v_{0}+(\alpha b+\beta )m}= \biggl(1+\frac {v_{-1}}{(\alpha b+\beta )m} \biggr) \biggl(1+\frac{v_{0}}{(\alpha b+\beta )m} \biggr)^{-1} \\ &= \biggl(1+\frac{v_{-1}}{(\alpha b+\beta )m} \biggr) \biggl(1-\frac{v_{0}}{(\alpha b+\beta )m}+O \biggl( \frac{1}{m^{2}} \biggr) \biggr) \\ &=1+\frac{v_{-1}-v_{0}}{(\alpha b+\beta )m}+O \biggl(\frac{1}{m^{2}} \biggr) \end{aligned}$$

(55)

for sufficiently large m.

From (55), (53), (28) and a known criterion for convergence of products, the statements easily follow.

(q) Using the condition $v_{0}=v_{-1}$ in (28), the statement immediately follows.

(r), (s) By using (30) we have

$$\begin{aligned} \hat{s}_{m}&=\frac{v_{0}+(\alpha b+\beta )m}{v_{-1}+(\alpha b+\beta )(m+1)}= \biggl(1+\frac {v_{0}}{(\alpha b+\beta )m} \biggr) \biggl(1+\frac{v_{-1}+\alpha b+\beta }{(\alpha b+\beta )m} \biggr)^{-1} \\ &= \biggl(1+\frac{v_{0}}{(\alpha b+\beta )m} \biggr) \biggl(1-\frac{v_{-1}+\alpha b+\beta }{(\alpha b+\beta )m}+O \biggl( \frac{1}{m^{2}} \biggr) \biggr) \\ &=1+\frac{v_{0}-v_{-1}-\alpha b-\beta }{(\alpha b+\beta )m}+O \biggl(\frac{1}{m^{2}} \biggr) \end{aligned}$$

(56)

for sufficiently large m.

From (56), (53), (29) and a known criterion for convergence of products, the statements easily follow.

(t) Using the condition $v_{0}=v_{-1}+\alpha b+\beta$ in (29), the statement immediately follows. □

5 Domain of undefinable solutions to system (2)

In Section 2 we proved that solutions to system (2), for which $x_{-j}=0$ or $y_{-j}=0$ for some $j\in\{1,2\}$, are not defined. The set of all such initial values is characterized here.

Definition 1

Consider the system of difference equations

$$\begin{aligned} \begin{aligned} &x_{n}=f(x_{n-1}, \ldots,x_{n-s},y_{n-1},\ldots,y_{n-s},n), \\ &y_{n}=g(x_{n-1},\ldots,x_{n-s},y_{n-1}, \ldots,y_{n-s},n),\quad n\in \mathbb {N}_{0}, \end{aligned} \end{aligned}$$

(57)

where $s\in \mathbb {N}$, and $x_{-i},y_{-i}\in \mathbb {R}$, $i=\overline {1,s}$. The string of vectors

$$(x_{-s},y_{-s}),\ldots,(x_{-1},y_{-1}),(x_{0},y_{0}), \ldots,(x_{n_{0}},y_{n_{0}}), $$

where $n_{0}\ge-1$, is called an undefined solution of system (57) if

$$x_{j}=f(x_{j-1},\ldots,x_{j-s},y_{j-1}, \ldots,y_{j-s},j) $$

and

$$y_{j}=g(x_{j-1},\ldots,x_{j-s},y_{j-1}, \ldots,y_{j-s},j) $$

for $0\le j< n_{0}+1$, and $x_{n_{0}+1}$ or $y_{n_{0}+1}$ is not a defined number, that is, the quantity

$$f(x_{n_{0}},\ldots,x_{n_{0}-s+1},y_{n_{0}}, \ldots,y_{n_{0}-s+1},n_{0}+1) $$

or

$$g(x_{n_{0}},\ldots,x_{n_{0}-s+1},y_{n_{0}}, \ldots,y_{n_{0}-s+1},n_{0}+1) $$

is not defined.

The set of all initial values $(x_{-s},y_{-s}),\ldots,(x_{-1},y_{-1})$ which generate undefined solutions to system (57) is called domain of undefinable solutions of the system.

The next result characterizes the domain of undefinable solutions to system (2) when $a_{n}b_{n}\alpha _{n}\beta _{n}\ne0$, $n\in \mathbb {N}_{0}$.

Theorem 4

Assume that $a_{n}b_{n}\alpha _{n}\beta _{n}\ne0$, $n\in \mathbb {N}_{0}$. Then the domain of undefinable solutions to system (2) is the following set:

$$\begin{aligned} {\mathcal{U}}={}&\bigcup_{m\in \mathbb {N}_{0}} \biggl\{ (x_{-2},x_{-1},y_{-2},y_{-1})\in \mathbb {R}^{4}: \\ &{}\frac{1}{x_{-1}x_{-2}}=g_{0}^{-1}\circ f_{1}^{-1} \circ\cdots\circ g_{2m-2}^{-1}\circ f_{2m-1}^{-1} \circ g_{2m}^{-1}\circ f_{2m+1}^{-1}(0) \\ &{} \textit{or } \frac{1}{x_{-1}x_{-2}}=g_{0}^{-1}\circ f_{1}^{-1}\circ\cdots\circ g_{2m-2}^{-1} \circ f_{2m-1}^{-1} \circ g_{2m}^{-1}(0) \\ &{} \textit{or } \frac{1}{y_{-1}y_{-2}}=f_{0}^{-1}\circ g_{1}^{-1}\circ\cdots\circ f_{2m-2}^{-1} \circ g_{2m-1}^{-1} \circ f_{2m}^{-1}(0) \\ &{} \textit{or } \frac{1}{y_{-1}y_{-2}}=f_{0}^{-1}\circ g_{1}^{-1}\circ\cdots\circ g_{2m-1}^{-1} \circ f_{2m}^{-1} \circ g_{2m+1}^{-1}(0) \biggr\} \\ &{} \cup \bigl\{ (x_{-2},x_{-1},y_{-2},y_{-1}) \in \mathbb {R}^{4}: \\ &{}x_{-2}=0\textit{ or }x_{-1}=0\textit{ or }y_{-2}=0\textit{ or }y_{-1}=0 \bigr\} , \end{aligned}$$

(58)

where

$$f_{n}(t)=a_{n}t+b_{n}, \qquad g_{n}(t)= \alpha _{n}t+\beta _{n},\quad n\in \mathbb {N}_{0}. $$

Proof

We have already proved that the set

$$\bigl\{ (x_{-2},x_{-1},y_{-2},y_{-1})\in \mathbb {R}^{4}: x_{-2}=0\mbox{ or }x_{-1}=0\mbox{ or }y_{-2}=0\mbox{ or }y_{-1}=0 \bigr\} $$

belongs to the domain of undefinable solutions to system (2).

If $x_{-j}\ne0\ne y_{-j}$, $j=\overline {1,2}$ (i.e., $x_{n}\ne0\ne y_{n}$ for every $n\ge-2$), then such a solution $(x_{n},y_{n})_{n\ge-2}$ is not defined if and only if

$$\begin{aligned} a_{n}+b_{n}y_{n-1}y_{n-2}=0\quad \mbox{or} \quad \alpha _{n}+\beta _{n}x_{n-1}x_{n-2}=0 \end{aligned}$$

(59)

for some $n\in \mathbb {N}_{0}$, which is equivalent to

$$\begin{aligned} v_{n-1}=-b_{n}/a_{n}\quad \mbox{or}\quad u_{n-1}=-\beta _{n}/\alpha _{n} \end{aligned}$$

(60)

for some $n\in \mathbb {N}_{0}$.

Note that

$$\begin{aligned} f_{n}^{-1}(0)=-b_{n}/a_{n}\quad \mbox{and} \quad g_{n}^{-1}(0)=-\beta _{n}/\alpha _{n},\quad n \in \mathbb {N}_{0}. \end{aligned}$$

(61)

We have

$$\begin{aligned}& v_{2m-1} =(g_{2m-1} \circ f_{2m-2}\circ\cdots\circ f_{2}\circ g_{1}\circ f_{0}) (v_{-1}), \end{aligned}$$

(62)

$$\begin{aligned}& v_{2m} =(g_{2m}\circ f_{2m-1}\circ\cdots\circ g_{2}\circ f_{1}\circ g_{0}) (u_{-1}), \end{aligned}$$

(63)

$$\begin{aligned}& u_{2m-1} =(f_{2m-1} \circ g_{2m-2}\circ\cdots\circ g_{2}\circ f_{1}\circ g_{0}) (u_{-1}), \end{aligned}$$

(64)

$$\begin{aligned}& u_{2m} =(f_{2m}\circ g_{2m-1}\circ\cdots\circ f_{2}\circ g_{1}\circ f_{0}) (v_{-1}) \end{aligned}$$

(65)

for $m\in \mathbb {N}_{0}$.

From (61) and (62) we have that

$$-\frac{b_{2m}}{a_{2m}}=v_{2m-1}=(g_{2m-1} \circ f_{2m-2} \circ\cdots \circ f_{2}\circ g_{1}\circ f_{0}) (v_{-1}) $$

for some $m\in \mathbb {N}_{0}$ if and only if

$$\begin{aligned} \frac{1}{y_{-1}y_{-2}}= f_{0}^{-1}\circ g_{1}^{-1} \circ \cdots \circ f_{2m-2}^{-1}\circ g_{2m-1}^{-1} \circ f_{2m}^{-1}(0). \end{aligned}$$

(66)

From (61) and (63) we have that

$$-\frac{b_{2m+1}}{a_{2m+1}}=v_{2m}=(g_{2m}\circ f_{2m-1} \circ\cdots \circ g_{2}\circ f_{1}\circ g_{0}) (u_{-1}) $$

for some $m\in \mathbb {N}_{0}$ if and only if

$$\begin{aligned} \frac{1}{x_{-1}x_{-2}}= g_{0}^{-1}\circ f_{1}^{-1} \circ \cdots\circ g_{2m-2}^{-1}\circ f_{2m-1}^{-1} \circ g_{2m}^{-1}\circ f_{2m+1}^{-1}(0). \end{aligned}$$

(67)

From (61) and (64) we have that

$$-\frac{\beta _{2m}}{\alpha _{2m}}=u_{2m-1}=(f_{2m-1} \circ g_{2m-2} \circ \cdots \circ g_{2}\circ f_{1}\circ g_{0}) (u_{-1}) $$

for some $m\in \mathbb {N}_{0}$ if and only if

$$\begin{aligned} \frac{1}{x_{-1}x_{-2}}=g_{0}^{-1}\circ f_{1}^{-1} \circ \cdots\circ g_{2m-2}^{-1}\circ f_{2m-1}^{-1} \circ g_{2m}^{-1}(0). \end{aligned}$$

(68)

From (61) and (65) we have that

$$-\frac{\beta _{2m+1}}{\alpha _{2m+1}}=u_{2m}=(f_{2m}\circ g_{2m-1} \circ \cdots \circ f_{2}\circ g_{1}\circ f_{0}) (v_{-1}) $$

for some $m\in \mathbb {N}_{0}$ if and only if

$$\begin{aligned} \frac{1}{y_{-1}y_{-2}}= f_{0}^{-1}\circ g_{1}^{-1} \circ \cdots\circ g_{2m-1}^{-1}\circ f_{2m}^{-1} \circ g_{2m+1}^{-1}(0). \end{aligned}$$

(69)

From (66)-(69) we see that the first union in (58) also belongs to the domain of undefinable solutions, finishing the proof of the theorem. □

Remark 1

Quantities

$$\begin{aligned} &g_{0}^{-1}\circ f_{1}^{-1}\circ\cdots \circ g_{2m-2}^{-1}\circ f_{2m-1}^{-1}\circ g_{2m}^{-1}\circ f_{2m+1}^{-1}(0), \end{aligned}$$

(70)

$$\begin{aligned} &g_{0}^{-1}\circ f_{1}^{-1}\circ\cdots \circ g_{2m-2}^{-1}\circ f_{2m-1}^{-1} \circ g_{2m}^{-1}(0), \end{aligned}$$

(71)

$$\begin{aligned} &f_{0}^{-1}\circ g_{1}^{-1}\circ\cdots \circ f_{2m-2}^{-1}\circ g_{2m-1}^{-1} \circ f_{2m}^{-1}(0), \end{aligned}$$

(72)

$$\begin{aligned} &f_{0}^{-1}\circ g_{1}^{-1}\circ\cdots \circ g_{2m-1}^{-1}\circ f_{2m}^{-1} \circ g_{2m+1}^{-1}(0) \end{aligned}$$

(73)

can be calculated for every $m\in \mathbb {N}_{0}$.

Indeed, note that

$$\begin{aligned} &g_{0}^{-1}\circ f_{1}^{-1}\circ\cdots \circ g_{2m-2}^{-1}\circ f_{2m-1}^{-1}\circ g_{2m}^{-1}\circ f_{2m+1}^{-1}(0)= \Biggl( \prod_{j=0}^{m} \bigl(g_{2j}^{-1} \circ f_{2j+1}^{-1}\bigr) \Biggr) (t) \Big|_{t=0}, \end{aligned}$$

(74)

$$\begin{aligned} &g_{0}^{-1}\circ f_{1}^{-1}\circ\cdots \circ g_{2m-2}^{-1}\circ f_{2m-1}^{-1} \circ g_{2m}^{-1}(0)= \Biggl(\prod_{j=0}^{m-1} \bigl(g_{2j}^{-1}\circ f_{2j+1}^{-1}\bigr) \Biggr) (t) \Big|_{t=g_{2m}^{-1}(0)}, \end{aligned}$$

(75)

$$\begin{aligned} &f_{0}^{-1}\circ g_{1}^{-1}\circ\cdots \circ f_{2m-2}^{-1}\circ g_{2m-1}^{-1} \circ f_{2m}^{-1}(0)= \Biggl(\prod_{j=0}^{m-1} \bigl(f_{2j}^{-1}\circ g_{2j+1}^{-1}\bigr) \Biggr) (t)\Big|_{t=f_{2m}^{-1}(0)}, \end{aligned}$$

(76)

$$\begin{aligned} &f_{0}^{-1}\circ g_{1}^{-1}\circ\cdots \circ g_{2m-1}^{-1}\circ f_{2m}^{-1} \circ g_{2m+1}^{-1}(0)= \Biggl(\prod_{j=0}^{m} \bigl(f_{2j}^{-1}\circ g_{2j+1}^{-1}\bigr) \Biggr) (t) \Big|_{t=0}, \end{aligned}$$

(77)

and also that

$$\begin{aligned} \bigl(g_{2j}^{-1}\circ f_{2j+1}^{-1}\bigr) (t)=\frac{t}{\alpha _{2j}a_{2j+1}}-\frac{b_{2j+1}}{\alpha _{2j}a_{2j+1}}-\frac{\beta _{2j}}{\alpha _{2j}},\quad j\in \mathbb {N}_{0}, \end{aligned}$$

(78)

and

$$\begin{aligned} \bigl(f_{2j}^{-1}\circ g_{2j+1}^{-1}\bigr) (t)=\frac{t}{a_{2j}\alpha _{2j+1}}-\frac{\beta _{2j+1}}{a_{2j}\alpha _{2j+1}}-\frac{b_{2j}}{a_{2j}},\quad j\in \mathbb {N}_{0}. \end{aligned}$$

(79)

On the other hand, if

$$h_{j}(t)=c_{j}t+d_{j},\quad j\in \mathbb {N}_{0}, $$

it is easy to see that

$$\begin{aligned} (h_{0}\circ h_{1}\circ\cdots\circ h_{n}) (t)= \Biggl(\prod_{j=0}^{n} c_{j} \Biggr)t+\sum_{i=0}^{n}d_{j}\prod _{j=0}^{i-1}c_{j},\quad n\in \mathbb {N}_{0}. \end{aligned}$$

(80)

From (74)-(80) explicit formulas for the quantities in (70)-(73) are easily obtained.