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

Advances in Difference Equations20152015:264

https://doi.org/10.1186/s13662-015-0591-7

  • Received: 11 June 2015
  • Accepted: 3 August 2015
  • Published:

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 \(\beta _{n}\), \(n\in \mathbb {N}_{0}\).

Keywords

  • system of difference equations
  • closed form solution
  • long-term behavior
  • periodic solutions

MSC

  • 39A10
  • 39A20

1 Introduction

Studying concrete nonlinear difference equations and systems is a topic of a great recent interest (see, e.g., [146] 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, 1216, 18, 19, 23, 24, 2629, 3138, 40, 41, 44, 46]). Another topic of interest is solvable difference equations and systems and their applications (see, e.g., [15, 7, 17, 20, 21, 2327, 2937, 3946]). 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, 2327, 2937, 3945].

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, 911, 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.
  1. (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.

     
  2. (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.

     
  3. (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.

     
  4. (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 \(m\to\infty\).

     
  5. (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 \(m\to\infty\).

     
  6. (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 \(m\to\infty\).

     
  7. (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 \(m\to\infty\).

     
  8. (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 \(m\to\infty\).

     
  9. (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 \(m\to\infty\).

     
  10. (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.

     
  11. (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.

     
  12. (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 \(m\to\infty\).

     
  13. (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 \(m\to\infty\).

     
  14. (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.

     
  15. (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.

     
  16. (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 \(m\to\infty\).

     
  17. (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 \(m\to\infty\).

     
  18. (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.

     
  19. (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.

     
  20. (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 \(m\to\infty\).

     
  21. (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 \(m\to\infty\).

     
  22. (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.

     
  23. (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.
  1. (a)

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

     
  2. (b)

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

     
  3. (c)

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

     
  4. (d)

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

     
  5. (e)

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

     
  6. (f)

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

     
  7. (g)

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

     
  8. (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.
  1. (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 \(m\to\infty\);

     
  2. (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 \(m\to\infty\);

     
  3. (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;

     
  4. (d)

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

     
  5. (e)

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

     
  6. (f)

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

     
  7. (g)

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

     
  8. (h)

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

     
  9. (i)

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

     
  10. (j)

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

     
  11. (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 \(m\to\infty\);

     
  12. (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 \(m\to\infty\);

     
  13. (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;

     
  14. (n)

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

     
  15. (o)

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

     
  16. (p)

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

     
  17. (q)

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

     
  18. (r)

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

     
  19. (s)

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

     
  20. (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.

Declarations

Acknowledgements

The work of the first and the second authors was supported by the Serbian Ministry of Education and Science, project III 41025. The work of the first author was also supported by the Serbian Ministry of Education and Science, project III 44006. The work of the second author was also supported by the Serbian Ministry of Education and Science, project OI 171007. The work of the third author was realized in CEITEC - Central European Institute of Technology with research infrastructure supported by project CZ.1.05/1.1.00/02.0068 financed from the European Regional Development Fund. The third author was also supported by the project FEKT-S-14-2200 of Brno University of Technology.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, Beograd, 11000, Serbia
(2)
Operator Theory and Applications Research Group, Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia
(3)
Faculty of Electrical Engineering, Belgrade University, Bulevar Kralja Aleksandra 73, Beograd, 11000, Serbia
(4)
CEITEC - Central European Institute of Technology, Brno University of Technology, Technická 3058/10, Brno, CZ-616 00, Czech Republic
(5)
FEEC - Faculty of Electrical Engineering and Communication, Department of Mathematics, Brno University of Technology, Technická 3058/10, Brno, CZ-616 00, Czech Republic

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