Open Access

Two-dimensional product-type system of difference equations solvable in closed form

Advances in Difference Equations20162016:253

https://doi.org/10.1186/s13662-016-0980-6

Received: 27 August 2016

Accepted: 21 September 2016

Published: 3 October 2016

Abstract

A solvable two-dimensional product-type system of difference equations of interest is presented. Closed form formulas for its general solution are given.

Keywords

system of difference equationsproduct-type systemsolvable in closed form

MSC

39A1039A20

1 Introduction

Concrete nonlinear difference equations and systems have become of some interest recently. Experts have proposed various classes of the equations and systems hoping that their studies will lead to some new general results or will bring about some new methods in the theory (see, e.g., [122]). Many of the papers study or are motivated by the study of symmetric systems (see, e.g., [47, 9, 10, 14, 1722]). It turned out that some of the equations and systems can be solved, which motivated some experts to work on the topic (see, e.g., [1, 8, 11, 1317, 1922]; for some old results see, e.g., [2325]). One of the motivations for the renewed interest in the area has been Stević’s method/idea for transforming some nonlinear equations into solvable linear ones (see, for example, [11, 13, 19, 20] and numerous related references therein). It also turned out that many classes of nonlinear difference equations and systems can be transformed to solvable ones by using some tricks and suitable changes of variables (see, e.g., [8, 13, 16, 19] and the related references therein).

Numerous recent equations and systems are closely related to product-type ones, which are solvable for the case of positive initial values (see, e.g., the equation in [12], which is a kind of perturbation of some product-type and the system in [18]; see also the related references therein). If the initial values are not positive, then there appear several problems. Thus, it is of some interest to describe the product-type systems with complex initial values which are solvable. A detailed study of the problem has been started recently by Stević et al. in [14, 15, 17, 21, 22] (some subclasses of the class of difference equations studied in [16] are also product-type ones). During the investigation we realized that the solvability of some product-type systems is preserved if some coefficients/multipliers are added. The first system of this type was studied in [14]. Based on this idea, quite recently in [22] it has been shown that the solvability of the system studied in [17] is preserved if two coefficients/multipliers are added. On the other hand, it can be seen that there are only several classes of product-type systems of difference equations which can be practically solved in closed form, due to the well-known fact that roots of the polynomials of degree \(d\ge5\) cannot be solved by radicals. Hence, it is of interest to find all the classes of practically solvable product-type systems of difference equations and present formulas for their solutions in terms of the initial values and parameters.

Here we present a new class of product-type systems of difference equations which are solvable under some natural assumptions. Namely, we investigate the solvability of the system
$$\begin{aligned} z_{n+1}=\alpha z_{n}^{a}w_{n-1}^{b}, \qquad w_{n+1}=\beta w_{n-1}^{c}z_{n-1}^{d}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(1)
where \(a,b,c,d\in {\mathbb {Z}}\), \(\alpha ,\beta \in {\mathbb {C}}\) and \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\). It is interesting that none of the subclasses of the class in (1) has been previously treated in our papers on product-type systems, so that all the formulas presented here should be new. The formulas are obtained by further developing the methods in our previous papers, especially the ones in [14] and [22].
A solution to system (1) need not be defined if its initial values belong to the set
$${\mathcal {U}}=\bigl\{ (z_{-1}, z_{0}, w_{-1}, w_{0})\in {\mathbb {C}}^{4}: z_{-1}=0 \mbox{ or } z_{0}=0 \mbox{ or } w_{-1}=0 \mbox{ or } w_{0}=0\bigr\} . $$
Thus, from now on we will assume that \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\). Since the cases \(\alpha =0\) and \(\beta =0\) are trivial or produce solutions which are not well defined we will also assume that \(\alpha \beta \ne0\).

Let us also note that we will use the convention \(\sum_{i=k}^{l}a_{i}=0\), when \(l< k\), throughout the paper.

2 Main results

The main results in this paper are proved in this section.

Theorem 1

Assume that \(b,c,d\in {\mathbb {Z}}\), \(a=0\), \(\alpha ,\beta \in {\mathbb {C}}\setminus\{0\}\), and \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\). Then system (1) is solvable in closed form.

Proof

Since \(a=0\) system (1) is
$$\begin{aligned} z_{n+1}=\alpha w_{n-1}^{b},\qquad w_{n+1}= \beta w_{n-1}^{c}z_{n-1}^{d},\quad n\in {\mathbb {N}}_{0}. \end{aligned}$$
(2)
Using the first equation in (2) in the second one, we obtain
$$\begin{aligned} w_{n+1}=\beta \alpha ^{d}w_{n-1}^{c}w_{n-3}^{bd}, \quad n\ge2, \end{aligned}$$
(3)
from which it follows that
$$\begin{aligned} w_{2n+1}=\beta \alpha ^{d}w_{2n-1}^{c}w_{2n-3}^{bd}, \quad n\in {\mathbb {N}}, \end{aligned}$$
(4)
and
$$\begin{aligned} w_{2n+2}=\beta \alpha ^{d}w_{2n}^{c}w_{2n-2}^{bd}, \quad n\in {\mathbb {N}}. \end{aligned}$$
(5)
Case \(bd=0\) . In this case equations (4) and (5) become
$$\begin{aligned} w_{2n+1}=\beta \alpha ^{d}w_{2n-1}^{c},\quad n \in {\mathbb {N}}, \end{aligned}$$
(6)
and
$$\begin{aligned} w_{2n+2}=\beta \alpha ^{d}w_{2n}^{c},\quad n \in {\mathbb {N}}, \end{aligned}$$
(7)
from which it follows that
$$\begin{aligned} w_{2n+1}&=\bigl(\beta \alpha ^{d} \bigr)^{\sum_{j=0}^{n-1}c^{j}}w_{1}^{c^{n}}=\bigl(\beta \alpha ^{d}\bigr)^{\sum _{j=0}^{n-1}c^{j}}\bigl(\beta w_{-1}^{c}z_{-1}^{d} \bigr)^{c^{n}} \\ &=\beta ^{\sum_{j=0}^{n}c^{j}}\alpha ^{d\sum _{j=0}^{n-1}c^{j}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}}, \quad n\in {\mathbb {N}}, \end{aligned}$$
(8)
$$\begin{aligned} w_{2n}&=\bigl(\beta \alpha ^{d} \bigr)^{\sum_{j=0}^{n-2}c^{j}}w_{2}^{c^{n-1}}=\bigl(\beta \alpha ^{d} \bigr)^{\sum _{j=0}^{n-2}c^{j}}\bigl(\beta w_{0}^{c}z_{0}^{d} \bigr)^{c^{n-1}} \\ &=\beta ^{\sum_{j=0}^{n-1}c^{j}}\alpha ^{d\sum _{j=0}^{n-2}c^{j}}w_{0}^{c^{n}}z_{0}^{dc^{n-1}}, \quad n\ge2. \end{aligned}$$
(9)
Hence
$$\begin{aligned} &w_{2n+1}=\beta ^{\frac{1-c^{n+1}}{1-c}}\alpha ^{d\frac {1-c^{n}}{1-c}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}}, \quad n\in {\mathbb {N}}, \end{aligned}$$
(10)
$$\begin{aligned} &w_{2n}=\beta ^{\frac{1-c^{n}}{1-c}}\alpha ^{d\frac {1-c^{n-1}}{1-c}}w_{0}^{c^{n}}z_{0}^{dc^{n-1}}, \quad n\ge2, \end{aligned}$$
(11)
when \(c\ne1\), and
$$\begin{aligned} &w_{2n+1}=\beta ^{n+1}\alpha ^{dn}w_{-1}z_{-1}^{d}, \quad n\in {\mathbb {N}}, \end{aligned}$$
(12)
$$\begin{aligned} &w_{2n}=\beta ^{n}\alpha ^{d(n-1)}w_{0}z_{0}^{d}, \quad n\ge2, \end{aligned}$$
(13)
when \(c=1\).
By using (8) and (9) in the first equation in (2) with \(n\to2n\) and \(n\to2n-1\), respectively, we get
$$\begin{aligned} &z_{2n+1}=\alpha w_{2n-1}^{b}=\alpha \beta ^{b\sum _{j=0}^{n-1}c^{j}}w_{-1}^{bc^{n}},\quad n\ge2, \end{aligned}$$
(14)
$$\begin{aligned} &z_{2n}=\alpha w_{2n-2}^{b}=\alpha \beta ^{b\sum _{j=0}^{n-2}c^{j}}w_{0}^{bc^{n-1}}, \quad n\ge3. \end{aligned}$$
(15)
Hence, from (14) and (15) we have
$$\begin{aligned} &z_{2n+1}=\alpha \beta ^{b\frac{1-c^{n}}{1-c}}w_{-1}^{bc^{n}}, \quad n\ge2, \end{aligned}$$
(16)
$$\begin{aligned} &z_{2n}=\alpha \beta ^{b\frac{1-c^{n-1}}{1-c}}w_{0}^{bc^{n-1}}, \quad n\ge3, \end{aligned}$$
(17)
when \(c\ne1\), and
$$\begin{aligned} &z_{2n+1}=\alpha \beta ^{bn}w_{-1}^{b}, \quad n\ge2, \end{aligned}$$
(18)
$$\begin{aligned} &z_{2n}=\alpha \beta ^{b(n-1)}w_{0}^{b}, \quad n\ge3, \end{aligned}$$
(19)
when \(c=1\).
Case \(bd\ne0\) . Let \(\gamma :=\beta \alpha ^{d}\)
$$\begin{aligned} a_{1}:=c,\qquad b_{1}=bd,\qquad x_{1}:=1. \end{aligned}$$
(20)
Then (4) and (5) can be written as
$$\begin{aligned} w_{2n+1}=\gamma ^{x_{1}}w_{2n-1}^{a_{1}}w_{2n-3}^{b_{1}}, \quad n\in {\mathbb {N}}, \end{aligned}$$
(21)
and
$$\begin{aligned} w_{2n+2}=\gamma ^{x_{1}}w_{2n}^{a_{1}}w_{2n-2}^{b_{1}}, \quad n\in {\mathbb {N}}. \end{aligned}$$
(22)
By using (21) with \(n\to n-1\) into (21), we get
$$\begin{aligned} w_{2n+1}&=\gamma ^{x_{1}}\bigl(\gamma w_{2n-3}^{a_{1}}w_{2n-5}^{b_{1}} \bigr)^{a_{1}}w_{2n-3}^{b_{1}} \\ &=\gamma ^{x_{1}+a_{1}}w_{2n-3}^{a_{1}a_{1}+b_{1}}w_{2n-5}^{b_{1}a_{1}} \\ &=\gamma ^{x_{2}}w_{2(n-2)+1}^{a_{2}}w_{2(n-3)+1}^{b_{2}}, \end{aligned}$$
(23)
for \(n\ge2\), where
$$\begin{aligned} a_{2}:=a_{1}a_{1}+b_{1}, \qquad b_{2}:=b_{1}a_{1},\qquad x_{2}:=x_{1}+a_{1}. \end{aligned}$$
(24)
Assume that
$$\begin{aligned} w_{2n+1}&=\gamma ^{x_{k}}w_{2(n-k)+1}^{a_{k}}w_{2(n-k-1)+1}^{b_{k}}, \end{aligned}$$
(25)
for some \(k\ge2\) and every \(n\ge k\), where
$$\begin{aligned} a_{k}:=a_{1}a_{k-1}+b_{k-1}, \qquad b_{k}:=b_{1}a_{k-1},\qquad x_{k}:=x_{k-1}+a_{k-1}. \end{aligned}$$
(26)
Using (21) with \(n\to n-k\) into (25) we get
$$\begin{aligned} w_{2n+1}&=\gamma ^{x_{k}}\bigl(\gamma w_{2(n-k-1)+1}^{a_{1}}w_{2(n-k-2)+1}^{b_{1}} \bigr)^{a_{k}}w_{2(n-k-1)+1}^{b_{k}} \\ &=\gamma ^{x_{k}+a_{k}}w_{2(n-k-1)+1}^{a_{1}a_{k}+b_{k}}w_{2(n-k-2)+1}^{b_{1}a_{k}} \\ &=\gamma ^{x_{k+1}}w_{2(n-k-1)+1}^{a_{k+1}}w_{2(n-k-2)+1}^{b_{k+1}}, \end{aligned}$$
(27)
for every \(n\ge k+1\), where
$$\begin{aligned} a_{k+1}:=a_{1}a_{k}+b_{k}, \qquad b_{k+1}:=b_{1}a_{k},\qquad x_{k+1}:=x_{k}+a_{k}. \end{aligned}$$
(28)

Equalities (23), (24), (27), (28), along with the induction show that (25) and (26) hold for all natural numbers k and n such that \(2\le k\le n\). Moreover, because of (21), equality (25) holds for \(1\le k\le n\).

For \(n=k\), (25) becomes
$$\begin{aligned} w_{2n+1}=\gamma ^{x_{n}}w_{1}^{a_{n}}w_{-1}^{b_{n}}, \quad n\in {\mathbb {N}}. \end{aligned}$$
(29)
Using the equalities \(w_{1}=\beta w_{-1}^{c}z_{-1}^{d}\), \(a_{n+1}=ca_{n}+b_{n}\), and \(x_{n+1}=x_{n}+a_{n}\) in (29) it follows that
$$\begin{aligned} w_{2n+1}&=\bigl(\beta \alpha ^{d} \bigr)^{x_{n}}\bigl(\beta w_{-1}^{c}z_{-1}^{d} \bigr)^{a_{n}}w_{-1}^{b_{n}} \\ &=\alpha ^{dx_{n}}\beta ^{x_{n+1}}w_{-1}^{ca_{n}+b_{n}}z_{-1}^{da_{n}} \\ &=\alpha ^{dx_{n}}\beta ^{x_{n+1}}w_{-1}^{a_{n+1}}z_{-1}^{da_{n}}, \quad n\in {\mathbb {N}}. \end{aligned}$$
(30)
Using (30) in the first equation in (2), we get
$$\begin{aligned} z_{2n+1}=\alpha ^{1+bdx_{n-1}}\beta ^{bx_{n}}w_{-1}^{ba_{n}}z_{-1}^{bda_{n-1}}, \quad n\ge2. \end{aligned}$$
(31)
By using the same procedure it is proved that
$$\begin{aligned} w_{2n+2}=\gamma ^{x_{k}}w_{2(n-k+1)}^{a_{k}}w_{2(n-k)}^{b_{k}}, \end{aligned}$$
(32)
for all natural numbers k and n such that \(1\le k\le n\), where \((a_{k})_{k\in {\mathbb {N}}}\), \((b_{k})_{k\in {\mathbb {N}}}\), \((x_{k})_{k\in {\mathbb {N}}}\) satisfy (20) and (26).
For \(n=k\), (32) becomes
$$\begin{aligned} w_{2n+2}=\gamma ^{x_{n}}w_{2}^{a_{n}}w_{0}^{b_{n}}, \quad n\in {\mathbb {N}}. \end{aligned}$$
(33)
Since \(w_{2}=\beta w_{0}^{c}z_{0}^{d}\), \(x_{n+1}=x_{n}+a_{n}\), and \(a_{n+1}=ca_{n}+b_{n}\), from (33) we have
$$\begin{aligned} w_{2n+2}&=\bigl(\beta \alpha ^{d} \bigr)^{x_{n}}\bigl(\beta w_{0}^{c}z_{0}^{d} \bigr)^{a_{n}}w_{0}^{b_{n}} \\ &=\alpha ^{dx_{n}}\beta ^{x_{n+1}}w_{0}^{a_{n+1}}z_{0}^{da_{n}}, \quad n\in {\mathbb {N}}. \end{aligned}$$
(34)
Using (34) in the first equation in (2), we get
$$\begin{aligned} z_{2n+2}=\alpha ^{1+bdx_{n-1}}\beta ^{bx_{n}}w_{0}^{ba_{n}}z_{0}^{bda_{n-1}}, \quad n\ge2. \end{aligned}$$
(35)
From the first two equations in (26) we have
$$\begin{aligned} a_{k}=a_{1}a_{k-1}+b_{1}a_{k-2}, \quad k\ge3. \end{aligned}$$
(36)
From (36) and since \(b_{k}=b_{1}a_{k-1}\), we see that \((b_{k})_{k\in {\mathbb {N}}}\) is also a solution of (36).
From (26) with \(k=1\) one obtains
$$\begin{aligned} a_{1}&=a_{1}a_{0}+b_{0},\qquad b_{1}=b_{1}a_{0},\qquad x_{1}=x_{0}+a_{0}. \end{aligned}$$
(37)
From this and since \(b_{1}=bd\ne0\), from the second equation in (37) we get \(a_{0}=1\), which along with the fact \(x_{1}=1\) and the other two relations in (37) implies \(b_{0}=x_{0}=0\).
This and (26) with \(k=0\) imply
$$\begin{aligned} 1&=a_{0}=a_{1}a_{-1}+b_{-1},\qquad 0=b_{0}=b_{1}a_{-1},\qquad 0=x_{0}=x_{-1}+a_{-1}, \end{aligned}$$
(38)
which along with \(b_{1}\ne0\) and the second equation in (38) implies \(a_{-1}=0\). This along with the other two relations in (38) implies that we must have \(b_{-1}=1\) and \(x_{-1}=0\).
Hence \((a_{k})_{k\ge-1}\) and \((b_{k})_{k\ge-1}\) are solutions to (36) satisfying the (shifted) initial conditions
$$\begin{aligned} a_{-1}=0, \qquad a_{0}=1;\qquad b_{-1}=1, \qquad b_{0}=0, \end{aligned}$$
(39)
while \((x_{k})_{k\ge-1}\) satisfies the third equation in (26) and
$$\begin{aligned} x_{-1}=x_{0}=0,\qquad x_{1}=1. \end{aligned}$$
(40)
From the third equation in (26) along with \(x_{1}=1\) and \(a_{0}=1\), we have
$$\begin{aligned} x_{k}=1+\sum_{j=1}^{k-1}a_{j}= \sum_{j=0}^{k-1}a_{j}. \end{aligned}$$
(41)
The characteristic equation associated to (36) is \(\lambda ^{2}-c\lambda -bd=0\), from which it follows that
$$\lambda _{1,2}=\frac{c\pm\sqrt{c^{2}+4bd}}{2}, $$
are the corresponding characteristic roots.
If \(c^{2}+4bd\ne0\), then
$$a_{n}=c_{1}\lambda _{1}^{n}+c_{2} \lambda _{2}^{n}, $$
which along with \(a_{-1}=0\) and \(a_{0}=1\) yields
$$\begin{aligned} a_{n}=\frac{\lambda _{1}^{n+1}-\lambda _{2}^{n+1}}{\lambda _{1}-\lambda _{2}}. \end{aligned}$$
(42)
From this and since \(b_{n}=b_{1}a_{n-1}\), we have
$$\begin{aligned} b_{n}=bd\frac{\lambda _{1}^{n}-\lambda _{2}^{n}}{\lambda _{1}-\lambda _{2}}. \end{aligned}$$
(43)
If \(c+bd\ne1\), which is equivalent to \(\lambda _{1}\ne1\ne \lambda _{2}\), from (41) and (42), it follows that
$$\begin{aligned} x_{n}=&\sum_{j=0}^{n-1} \frac{\lambda _{1}^{j+1}-\lambda _{2}^{j+1}}{\lambda _{1}-\lambda _{2}}=\frac{(\lambda _{2}-1)\lambda _{1}^{n+1}-(\lambda _{1}-1)\lambda _{2}^{n+1}+\lambda _{1}-\lambda _{2}}{(\lambda _{1}-1)(\lambda _{2}-1)(\lambda _{1}-\lambda _{2})}. \end{aligned}$$
(44)
If \(c+bd=1\), that is, if one of the characteristic roots is one, say \(\lambda _{2}\), then \(\lambda _{1}=-bd\), so that
$$\begin{aligned} x_{n}=\sum_{j=0}^{n-1} \frac{\lambda _{1}^{j+1}-1}{\lambda _{1}-1}=\frac{1}{(\lambda _{1}-1)} \biggl(\lambda _{1}\frac{\lambda _{1}^{n}-1}{\lambda _{1}-1}-n \biggr) =\frac{(-bd)^{n+1}+(n+1)bd+n}{(1+bd)^{2}}. \end{aligned}$$
(45)
If \(c^{2}+4bd=0\), then
$$a_{n}=(\hat{c}_{1}+\hat{c}_{2} n) \biggl( \frac{c}{2} \biggr)^{n}. $$
This along with \(a_{-1}=0\) and \(a_{0}=1\) yields
$$\begin{aligned} a_{n}=(n+1) \biggl(\frac{c}{2} \biggr)^{n}. \end{aligned}$$
(46)
Using the relation \(b_{n}=b_{1}a_{n-1}\) along with the fact \(bd=-c^{2}/4\), we get
$$\begin{aligned} b_{n}=bdn \biggl(\frac{c}{2} \biggr)^{n-1}=-n \biggl( \frac{c}{2} \biggr)^{n+1}. \end{aligned}$$
(47)
From (41) and (46), we have
$$\begin{aligned} x_{n}=&\sum_{j=0}^{n-1}(j+1) \biggl(\frac{c}{2} \biggr)^{j}=\frac{1-(n+1)(\frac{c}{2})^{n}+n(\frac{c}{2})^{n+1}}{ (1-\frac{c}{2} )^{2}}, \end{aligned}$$
(48)
if \(c\ne2\). If \(c=2\), we obtain
$$\begin{aligned} x_{n}=&\sum_{j=0}^{n-1}(j+1)= \frac{n(n+1)}{2}, \end{aligned}$$
(49)
completing the proof of the result. □

Corollary 1

Consider system (1) with \(b,c,d\in {\mathbb {Z}}\), \(a=0\), and \(\alpha ,\beta \in {\mathbb {C}}\setminus\{0\}\). Assume that \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\). Then the following statements are true.
  1. (a)

    If \(bd=0\) and \(c\ne1\), then the general solution to system (1) is given by (10), (11), (16), and (17).

     
  2. (b)

    If \(bd=0\) and \(c=1\), then the general solution to system (1) is given by (12), (13), (18), and (19).

     
  3. (c)

    If \(bd\ne0\), \(c^{2}+4bd\ne0\), and \(c+bd\ne1\), then the general solution to system (1) is given by (30), (31), (34), and (35), where the sequence \((a_{n})_{n\ge-1}\) is given by formula (42), while \((x_{n})_{n\ge-1}\) is given by (44).

     
  4. (d)

    If \(bd\ne0\), \(c^{2}+4bd\ne0\), and \(c+bd=1\), then the general solution to system (1) is given by (30), (31), (34), and (35), where the sequence \((a_{n})_{n\ge-1}\) is given by formula (42), while \((x_{n})_{n\ge-1}\) is given by (45).

     
  5. (e)

    If \(bd\ne0\), \(c^{2}+4bd=0\), and \(c\ne2\), then the general solution to system (1) is given by (30), (31), (34), and (35), where the sequence \((a_{n})_{n\ge-1}\) is given by formula (46), while \((x_{n})_{n\ge-1}\) is given by (48).

     
  6. (f)

    If \(bd\ne0\), \(c^{2}+4bd=0\), and \(c=2\), then the general solution to system (1) is given by (30), (31), (34), and (35), where the sequence \((a_{n})_{n\ge-1}\) is given by formula (46) with \(c=2\), while \((x_{n})_{n\ge-1}\) is given by (49).

     

Theorem 2

Assume that \(a,c,d\in {\mathbb {Z}}\), \(b=0\), \(\alpha ,\beta \in {\mathbb {C}}\setminus\{0\}\), and \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\). Then system (1) is solvable in closed form.

Proof

Since \(b=0\), we have
$$\begin{aligned} z_{n+1}=\alpha z_{n}^{a}, \qquad w_{n+1}= \beta w_{n-1}^{c}z_{n-1}^{d}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$
(50)
From the first equation in (50) we get
$$\begin{aligned} z_{n}=\alpha ^{\sum_{j=0}^{n-1}a^{j}}z_{0}^{a^{n}}, \quad n \in {\mathbb {N}}. \end{aligned}$$
(51)
Hence, if \(a\ne1\), we have
$$\begin{aligned} z_{n}=\alpha ^{\frac{1-a^{n}}{1-a}}z_{0}^{a^{n}},\quad n \in {\mathbb {N}}, \end{aligned}$$
(52)
while if \(a=1\),
$$\begin{aligned} z_{n}=\alpha ^{n}z_{0},\quad n\in {\mathbb {N}}. \end{aligned}$$
(53)
Using (51) in the second equation in (50), it follows that
$$\begin{aligned} w_{n+1}=\beta \alpha ^{d\sum _{j=0}^{n-2}a^{j}}z_{0}^{da^{n-1}}w_{n-1}^{c}, \quad n\ge2. \end{aligned}$$
(54)
Using (54) twice, we get
$$\begin{aligned} w_{2n}&=\beta \alpha ^{d\sum _{j=0}^{2n-3}a^{j}}z_{0}^{da^{2n-2}}w_{2n-2}^{c} \\ &=\beta \alpha ^{d\sum_{j=0}^{2n-3}a^{j}}z_{0}^{da^{2n-2}}\bigl(\beta \alpha ^{d\sum _{j=0}^{2n-5}a^{j}}z_{0}^{da^{2n-4}}w_{2n-4}^{c} \bigr)^{c} \\ &=\beta ^{1+c}\alpha ^{d\sum_{j=0}^{2n-3}a^{j}+dc\sum _{j=0}^{2n-5}a^{j}}z_{0}^{da^{2n-2}+dca^{2n-4}}w_{2n-4}^{c^{2}}, \end{aligned}$$
(55)
for every \(n\ge3\), and
$$\begin{aligned} w_{2n+1}&=\beta \alpha ^{d\sum _{j=0}^{2n-2}a^{j}}z_{0}^{da^{2n-1}}w_{2n-1}^{c} \\ &=\beta \alpha ^{d\sum_{j=0}^{2n-2}a^{j}}z_{0}^{da^{2n-1}}\bigl(\beta \alpha ^{d\sum _{j=0}^{2n-4}a^{j}}z_{0}^{da^{2n-3}}w_{2n-3}^{c} \bigr)^{c} \\ &=\beta ^{1+c}\alpha ^{d\sum_{j=0}^{2n-2}a^{j}+dc\sum _{j=0}^{2n-4}a^{j}}z_{0}^{da^{2n-1}+dca^{2n-3}}w_{2n-3}^{c^{2}}, \quad n\ge 2. \end{aligned}$$
(56)
Assume that, for a natural number k, it has been proved that
$$\begin{aligned} w_{2n}=\beta ^{\sum_{j=0}^{k-1}c^{j}}\alpha ^{d\sum _{i=0}^{k-1}c^{i}\sum _{j=0}^{2n-2i-3}a^{j}}z_{0}^{d\sum _{j=0}^{k-1}c^{j}a^{2n-2j-2}}w_{2n-2k}^{c^{k}}, \end{aligned}$$
(57)
for \(n\ge k+1\), and
$$\begin{aligned} w_{2n+1}=\beta ^{\sum_{j=0}^{k-1}c^{j}}\alpha ^{d\sum _{i=0}^{k-1}c^{i}\sum _{j=0}^{2n-2i-2}a^{j}}z_{0}^{d\sum _{j=0}^{k-1}c^{j}a^{2n-2j-1}}w_{2n-2k+1}^{c^{k}}, \end{aligned}$$
(58)
for every \(n\ge k\).
Using (54) with \(n\to2n-2k-1\) and \(n\to2n-2k\), in (57) and (58), we obtain
$$ \begin{aligned}[b] w_{2n}={}&\beta ^{\sum_{j=0}^{k-1}c^{j}}\alpha ^{d\sum _{i=0}^{k-1}c^{i}\sum_{j=0}^{2n-2i-3}a^{j}}z_{0}^{d\sum _{j=0}^{k-1}c^{j}a^{2n-2j-2}} \\ &{}\times\bigl(\beta \alpha ^{d\sum _{j=0}^{2n-2k-3}a^{j}}z_{0}^{da^{2n-2k-2}}w_{2n-2k-2}^{c} \bigr)^{c^{k}} \\ ={}&\beta ^{\sum_{j=0}^{k}c^{j}}\alpha ^{d\sum_{i=0}^{k}c^{i}\sum _{j=0}^{2n-2i-3}a^{j}}z_{0}^{d\sum _{j=0}^{k}c^{j}a^{2n-2j-2}}w_{2n-2k-2}^{c^{k+1}}, \end{aligned} $$
(59)
for \(n\ge k+2\), and
$$\begin{aligned} w_{2n+1}={}&\beta ^{\sum_{j=0}^{k-1}c^{j}}\alpha ^{d\sum _{i=0}^{k-1}c^{i}\sum_{j=0}^{2n-2i-2}a^{j}}z_{0}^{d\sum _{j=0}^{k-1}c^{j}a^{2n-2j-1}} \\ &{}\times\bigl(\beta \alpha ^{d\sum _{j=0}^{2n-2k-2}a^{j}}z_{0}^{da^{2n-2k-1}}w^{c}_{2n-2k-1} \bigr)^{c^{k}} \\ ={}&\beta ^{\sum_{j=0}^{k}c^{j}}\alpha ^{d\sum_{i=0}^{k}c^{i}\sum _{j=0}^{2n-2i-2}a^{j}}z_{0}^{d\sum _{j=0}^{k}c^{j}a^{2n-2j-1}}w_{2n-2k-1}^{c^{k+1}}, \end{aligned}$$
(60)
for every \(n\ge k+1\).

From (55), (56), (59), (60), and the induction it follows that (57) holds for all natural numbers k and n such that \(1\le k\le n-1\), while (58) holds for all k and n such that \(1\le k\le n\).

By taking \(k=n-1\) in (57), we get
$$\begin{aligned} w_{2n}=&\beta ^{\sum_{j=0}^{n-2}c^{j}}\alpha ^{d\sum _{i=0}^{n-2}c^{i}\sum_{j=0}^{2n-2i-3}a^{j}}z_{0}^{d\sum _{j=0}^{n-2}c^{j}a^{2n-2j-2}}w_{2}^{c^{n-1}}, \quad n\ge2. \end{aligned}$$
(61)
By using the relation \(w_{2}=\beta w_{0}^{c}z_{0}^{d}\) in (61) we get
$$\begin{aligned} w_{2n}&=\beta ^{\sum_{j=0}^{n-2}c^{j}}\alpha ^{d\sum _{i=0}^{n-2}c^{i}\sum_{j=0}^{2n-2i-3}a^{j}}z_{0}^{d\sum _{j=0}^{n-2}c^{j}a^{2n-2j-2}} \bigl(\beta w_{0}^{c}z_{0}^{d} \bigr)^{c^{n-1}} \\ &=\beta ^{\sum_{j=0}^{n-1}c^{j}}\alpha ^{d\sum_{i=0}^{n-2}c^{i}\sum _{j=0}^{2n-2i-3}a^{j}}z_{0}^{d\sum _{j=0}^{n-1}c^{j}a^{2n-2j-2}}w_{0}^{c^{n}}, \quad n\in {\mathbb {N}}. \end{aligned}$$
(62)
By taking \(k=n\) in (58), and using the relation \(w_{1}=\beta w_{-1}^{c}z_{-1}^{d}\), we get
$$\begin{aligned} w_{2n+1}&=\beta ^{\sum_{j=0}^{n-1}c^{j}}\alpha ^{d\sum _{i=0}^{n-1}c^{i}\sum_{j=0}^{2n-2i-2}a^{j}}z_{0}^{d\sum _{j=0}^{n-1}c^{j}a^{2n-2j-1}}w_{1}^{c^{n}} \\ &=\beta ^{\sum_{j=0}^{n-1}c^{j}}\alpha ^{d\sum_{i=0}^{n-1}c^{i}\sum _{j=0}^{2n-2i-2}a^{j}}z_{0}^{d\sum_{j=0}^{n-1}c^{j}a^{2n-2j-1}} \bigl(\beta w_{-1}^{c}z_{-1}^{d} \bigr)^{c^{n}} \\ &=\beta ^{\sum_{j=0}^{n}c^{j}}\alpha ^{d\sum_{i=0}^{n-1}c^{i}\sum _{j=0}^{2n-2i-2}a^{j}}z_{0}^{d\sum _{j=0}^{n-1}c^{j}a^{2n-2j-1}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}}, \end{aligned}$$
(63)
for \(n\in {\mathbb {N}}\). It is also easy to check that (63) holds also for \(n=0\) when \(c\ne0\).
Subcase \(a\ne1\ne c\) , \(c\ne a^{2}\) . In this case we have
$$\begin{aligned} w_{2n} &= \beta ^{\frac{1-c^{n}}{1-c}}\alpha ^{d\sum_{i=0}^{n-2}c^{i}\frac {1-a^{2n-2i-2}}{1-a}}z_{0}^{d\frac {a^{2n}-c^{n}}{a^{2}-c}}w_{0}^{c^{n}}, \\ &=\beta ^{\frac{1-c^{n}}{1-c}}\alpha ^{\frac{d}{1-a} (\frac {1-c^{n-1}}{1-c}-a^{2}\frac{a^{2n-2}-c^{n-1}}{a^{2}-c} )}z_{0}^{d\frac {a^{2n}-c^{n}}{a^{2}-c}}w_{0}^{c^{n}}, \\ &=\beta ^{\frac{1-c^{n}}{1-c}}\alpha ^{\frac {d(a^{2}-c+(1-a^{2})c^{n}+(c-1)a^{2n})}{(1-a)(1-c)(a^{2}-c)}}z_{0}^{d\frac {a^{2n}-c^{n}}{a^{2}-c}}w_{0}^{c^{n}}, \end{aligned}$$
(64)
for \(n\in {\mathbb {N}}\), and
$$\begin{aligned} w_{2n+1}&= \beta ^{\frac{1-c^{n+1}}{1-c}}\alpha ^{d\sum_{i=0}^{n-1}c^{i}\frac {1-a^{2n-2i-1}}{1-a}}z_{0}^{ad\frac {a^{2n}-c^{n}}{a^{2}-c}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}} \\ &=\beta ^{\frac{1-c^{n+1}}{1-c}}\alpha ^{\frac{d}{1-a} (\frac {1-c^{n}}{1-c}-a\frac{a^{2n}-c^{n}}{a^{2}-c} )}z_{0}^{ad\frac {a^{2n}-c^{n}}{a^{2}-c}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}} \\ &=\beta ^{\frac{1-c^{n+1}}{1-c}}\alpha ^{\frac {d(a^{2}-c+(a+c)(1-a)c^{n}-(1-c)a^{2n+1})}{(1-a)(1-c)(a^{2}-c)}}z_{0}^{ad\frac {a^{2n}-c^{n}}{a^{2}-c}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}}, \end{aligned}$$
(65)
for every \(n\in {\mathbb {N}}\).
Subcase \(a\ne1\ne c\) , \(c=a^{2}\) . In this case we have
$$\begin{aligned} w_{2n}&=\beta ^{\sum_{j=0}^{n-1}a^{2j}}\alpha ^{d\sum _{i=0}^{n-2}a^{2i}\sum_{j=0}^{2n-2i-3}a^{j}}z_{0}^{d\sum _{j=0}^{n-1}a^{2j}a^{2n-2j-2}}w_{0}^{a^{2n}}, \\ &=\beta ^{\frac{1-a^{2n}}{1-a^{2}}}\alpha ^{d\sum_{i=0}^{n-2}a^{2i}\frac {1-a^{2n-2i-2}}{1-a}}z_{0}^{dna^{2n-2}}w_{0}^{a^{2n}}, \\ &=\beta ^{\frac{1-a^{2n}}{1-a^{2}}}\alpha ^{\frac{d}{1-a} (\frac {1-a^{2n-2}}{1-a^{2}}-(n-1)a^{2n-2} )}z_{0}^{dna^{2n-2}}w_{0}^{a^{2n}}, \\ &=\beta ^{\frac{1-a^{2n}}{1-a^{2}}}\alpha ^{\frac {d(1-na^{2n-2}+(n-1)a^{2n})}{(a-1)^{2}(a+1)}}z_{0}^{dna^{2n-2}}w_{0}^{a^{2n}}, \quad n\ge2, \end{aligned}$$
(66)
$$\begin{aligned} w_{2n+1}&=\beta ^{\sum_{j=0}^{n}a^{2j}}\alpha ^{d\sum _{i=0}^{n-1}a^{2i}\sum_{j=0}^{2n-2i-2}a^{j}}z_{0}^{d\sum _{j=0}^{n-1}a^{2j}a^{2n-2j-1}}w_{-1}^{a^{2n+2}}z_{-1}^{da^{2n}} \\ &=\beta ^{\frac{1-a^{2n+2}}{1-a^{2}}}\alpha ^{d\sum _{i=0}^{n-1}a^{2i}\frac {1-a^{2n-2i-1}}{1-a}}z_{0}^{dna^{2n-1}}w_{-1}^{a^{2n+2}}z_{-1}^{da^{2n}} \\ &=\beta ^{\frac{1-a^{2n+2}}{1-a^{2}}}\alpha ^{\frac{d}{1-a} (\frac {1-a^{2n}}{1-a^{2}}-na^{2n-1} )}z_{0}^{dna^{2n-1}}w_{-1}^{a^{2n+2}}z_{-1}^{da^{2n}} \\ &=\beta ^{\frac{1-a^{2n+2}}{1-a^{2}}}\alpha ^{\frac {d(1-na^{2n-1}-a^{2n}+na^{2n+1})}{(a+1)(a-1)^{2}}} z_{0}^{dna^{2n-1}}w_{-1}^{a^{2n+2}}z_{-1}^{da^{2n}}, \end{aligned}$$
(67)
for every \(n\in {\mathbb {N}}\).
Subcase \(a^{2}\ne1=c\) . In this case we have
$$\begin{aligned} w_{2n}&=\beta ^{\sum_{j=0}^{n-1}1}\alpha ^{d\sum _{i=0}^{n-2}\sum _{j=0}^{2n-2i-3}a^{j}}z_{0}^{d\sum_{j=0}^{n-1}a^{2n-2j-2}}w_{0}, \\ &=\beta ^{n}\alpha ^{d\sum_{i=0}^{n-2}\frac {1-a^{2n-2i-2}}{1-a}}z_{0}^{d\frac {a^{2n}-1}{a^{2}-1}}w_{0}, \\ &=\beta ^{n}\alpha ^{\frac{d}{1-a} (n-1-a^{2}\frac {a^{2n-2}-1}{a^{2}-1} )}z_{0}^{d\frac{a^{2n}-1}{a^{2}-1}}w_{0}, \\ &=\beta ^{n}\alpha ^{\frac{d(a^{2n}-na^{2}+n-1)}{(a-1)^{2}(a+1)}}z_{0}^{d\frac {a^{2n}-1}{a^{2}-1}}w_{0}, \quad n\in {\mathbb {N}}, \end{aligned}$$
(68)
$$\begin{aligned} w_{2n+1}&=\beta ^{\sum_{j=0}^{n}1}\alpha ^{d\sum _{i=0}^{n-1}\sum _{j=0}^{2n-2i-2}a^{j}}z_{0}^{d\sum _{j=0}^{n-1}a^{2n-2j-1}}w_{-1}z_{-1}^{d} \\ &=\beta ^{n+1}\alpha ^{d\sum_{i=0}^{n-1}\frac {1-a^{2n-2i-1}}{1-a}}z_{0}^{ad\frac {a^{2n}-1}{a^{2}-1}}w_{-1}z_{-1}^{d} \\ &=\beta ^{n+1}\alpha ^{\frac{d}{1-a} (n-a\frac{a^{2n}-1}{a^{2}-1} )}z_{0}^{ad\frac{a^{2n}-1}{a^{2}-1}}w_{-1}z_{-1}^{d} \\ &=\beta ^{n+1}\alpha ^{\frac {d(a^{2n+1}+n(1-a^{2})-a)}{(a-1)^{2}(a+1)}}z_{0}^{ad\frac {a^{2n}-1}{a^{2}-1}}w_{-1}z_{-1}^{d}, \end{aligned}$$
(69)
for every \(n\in {\mathbb {N}}\).
Subcase \(a=-1\) , \(c=1\) . In this case we have
$$\begin{aligned} w_{2n}&=\beta ^{\sum_{j=0}^{n-1}1}\alpha ^{d\sum _{i=0}^{n-2}\sum _{j=0}^{2n-2i-3}(-1)^{j}}z_{0}^{d\sum _{j=0}^{n-1}(-1)^{2n-2j-2}}w_{0}, \\ &=\beta ^{n}z_{0}^{dn}w_{0}, \quad n\in {\mathbb {N}}, \end{aligned}$$
(70)
$$\begin{aligned} w_{2n+1}&=\beta ^{\sum_{j=0}^{n}1}\alpha ^{d\sum_{i=0}^{n-1}\sum _{j=0}^{2n-2i-2}(-1)^{j}}z_{0}^{d\sum _{j=0}^{n-1}(-1)^{2n-2j-1}}w_{-1}z_{-1}^{d} \\ &=\beta ^{n+1}\alpha ^{d\sum_{i=0}^{n-1}\frac {1-(-1)^{2n-2i-1}}{2}}z_{0}^{d\sum _{j=0}^{n-1}(-1)}w_{-1}z_{-1}^{d} \\ &=\beta ^{n+1}\alpha ^{dn}z_{0}^{-dn}w_{-1}z_{-1}^{d}, \end{aligned}$$
(71)
for every \(n\in {\mathbb {N}}_{0}\).
Subcase \(a=1\) , \(c\ne1\) . In this case we have
$$\begin{aligned} w_{2n}&=\beta ^{\sum_{j=0}^{n-1}c^{j}}\alpha ^{d\sum _{i=0}^{n-2}c^{i}\sum_{j=0}^{2n-2i-3}1}z_{0}^{d\sum _{j=0}^{n-1}c^{j}}w_{0}^{c^{n}} \\ &=\beta ^{\frac{1-c^{n}}{1-c}}\alpha ^{d\sum _{i=0}^{n-2}(2n-2i-2)c^{i}}z_{0}^{d\frac {1-c^{n}}{1-c}}w_{0}^{c^{n}} \\ &=\beta ^{\frac{1-c^{n}}{1-c}}\alpha ^{d ((2n-2)\frac {1-c^{n-1}}{1-c}-2c\frac {1-(n-1)c^{n-2}+(n-2)c^{n-1}}{(1-c)^{2}} )}z_{0}^{d\frac {1-c^{n}}{1-c}}w_{0}^{c^{n}} \\ &=\beta ^{\frac{1-c^{n}}{1-c}}\alpha ^{\frac {2d(n-1-nc+c^{n})}{(1-c)^{2}}}z_{0}^{d\frac {1-c^{n}}{1-c}}w_{0}^{c^{n}}, \quad n\in {\mathbb {N}}, \end{aligned}$$
(72)
$$\begin{aligned} w_{2n+1}&=\beta ^{\sum_{j=0}^{n}c^{j}}\alpha ^{d\sum_{i=0}^{n-1}c^{i}\sum _{j=0}^{2n-2i-2}1}z_{0}^{d\sum _{j=0}^{n-1}c^{j}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}} \\ &=\beta ^{\frac{1-c^{n+1}}{1-c}}\alpha ^{d\sum _{i=0}^{n-1}(2n-2i-1)c^{i}}z_{0}^{d\frac {1-c^{n}}{1-c}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}} \\ &=\beta ^{\frac{1-c^{n+1}}{1-c}}\alpha ^{d ((2n-1)\frac {1-c^{n}}{1-c}-2c\sum _{i=1}^{n-1}ic^{i-1} )}z_{0}^{d\frac {1-c^{n}}{1-c}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}} \\ &=\beta ^{\frac{1-c^{n+1}}{1-c}}\alpha ^{d ((2n-1)\frac {1-c^{n}}{1-c}-2c\frac {1-nc^{n-1}+(n-1)c^{n}}{(1-c)^{2}} )}z_{0}^{d\frac {1-c^{n}}{1-c}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}} \\ &=\beta ^{\frac{1-c^{n+1}}{1-c}}\alpha ^{\frac {d(2n-1-(2n+1)c+c^{n}+c^{n+1})}{(1-c)^{2}}}z_{0}^{d\frac {1-c^{n}}{1-c}}w_{-1}^{c^{n+1}}z_{-1}^{dc^{n}}, \end{aligned}$$
(73)
for every \(n\in {\mathbb {N}}\).
Subcase \(a=c=1\) . In this case we have
$$\begin{aligned} w_{2n}&=\beta ^{\sum_{j=0}^{n-1}1}\alpha ^{d\sum _{i=0}^{n-2}\sum _{j=0}^{2n-2i-3}1}z_{0}^{d\sum_{j=0}^{n-1}1}w_{0} \\ &=\beta ^{n}\alpha ^{d\sum_{i=0}^{n-2}(2n-2i-2)}z_{0}^{dn}w_{0} \\ &=\beta ^{n}\alpha ^{d(n-1)n}z_{0}^{dn}w_{0}, \quad n\in {\mathbb {N}}, \end{aligned}$$
(74)
$$\begin{aligned} w_{2n+1}&=\beta ^{\sum_{j=0}^{n}1}\alpha ^{d\sum_{i=0}^{n-1}\sum _{j=0}^{2n-2i-2}1}z_{0}^{d\sum_{j=0}^{n-1}1}w_{-1}z_{-1}^{d} \\ &=\beta ^{n+1}\alpha ^{d\sum _{i=0}^{n-1}(2n-2i-1)}z_{0}^{dn}w_{-1}z_{-1}^{d} \\ &=\beta ^{n+1}\alpha ^{dn^{2}}z_{0}^{dn}w_{-1}z_{-1}^{d}, \end{aligned}$$
(75)
for every \(n\in {\mathbb {N}}\). □

Corollary 2

Consider system (1) with \(a,c,d\in {\mathbb {Z}}\), \(b=0\), and \(\alpha ,\beta \in {\mathbb {C}}\setminus\{0\}\). Assume that \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\). Then the following statements are true.
  1. (a)

    If \(a\ne1\ne c\) and \(c\ne a^{2}\), then the general solution to system (1) is given by (52), (64), and (65).

     
  2. (b)

    If \(a\ne1\ne c\) and \(c=a^{2}\ne0\), then the general solution to system (1) is given by (52), (66), and (67).

     
  3. (c)

    If \(a^{2}\ne1=c\), then the general solution to system (1) is given by (52), (68), and (69).

     
  4. (d)

    If \(a=-1\) and \(c=1\), then the general solution to system (1) is given by (52), (70), and (71).

     
  5. (e)

    If \(a=1\) and \(c\ne1\), then the general solution to system (1) is given by (53), (72), and (73).

     
  6. (f)

    If \(a=c=1\), then the general solution to system (1) is given by (53), (74), and (75).

     

Theorem 3

Assume that \(a,b,c\in {\mathbb {Z}}\), \(d=0\), \(\alpha ,\beta \in {\mathbb {C}}\setminus\{0\}\), and \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\). Then system (1) is solvable in closed form.

Proof

In this case system (1) becomes
$$\begin{aligned} z_{n+1}=\alpha z_{n}^{a}w_{n-1}^{b}, \qquad w_{n+1}=\beta w_{n-1}^{c}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$
(76)
From the second equation in (76) it easily follows that
$$\begin{aligned} w_{2n}=\beta ^{\sum_{j=0}^{n-1}c^{j}}w_{0}^{c^{n}},\quad n\in {\mathbb {N}}\quad\mbox{and}\quad w_{2n+1}=\beta ^{\sum_{j=0}^{n}c^{j}}w_{-1}^{c^{n+1}}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(77)
which, for the case \(c\ne1\), implies that
$$\begin{aligned} w_{2n}=\beta ^{\frac{1-c^{n}}{1-c}}w_{0}^{c^{n}},\quad n\in {\mathbb {N}}, \end{aligned}$$
(78)
and
$$\begin{aligned} w_{2n+1}=\beta ^{\frac {1-c^{n+1}}{1-c}}w_{-1}^{c^{n+1}}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(79)
while, for the case \(c=1\), we have
$$\begin{aligned} w_{2n}=\beta ^{n}w_{0},\quad n\in {\mathbb {N}}, \end{aligned}$$
(80)
and
$$\begin{aligned} w_{2n+1}=\beta ^{n+1}w_{-1},\quad n\in {\mathbb {N}}_{0}. \end{aligned}$$
(81)
Employing (77) in the first equation in (76) we obtain
$$\begin{aligned} &z_{2n}=\alpha \beta ^{b\sum _{j=0}^{n-2}c^{j}}w_{0}^{bc^{n-1}}z_{2n-1}^{a}, \quad n\ge2, \end{aligned}$$
(82)
$$\begin{aligned} &z_{2n+1}=\alpha \beta ^{b\sum _{j=0}^{n-1}c^{j}}w_{-1}^{bc^{n}}z_{2n}^{a}, \quad n\in {\mathbb {N}}. \end{aligned}$$
(83)
Combining (82) and (83) it follows that
$$ \begin{aligned}[b] z_{2n}&=\alpha \beta ^{b\sum _{j=0}^{n-2}c^{j}}w_{0}^{bc^{n-1}} \bigl(\alpha \beta ^{b\sum_{j=0}^{n-2}c^{j}}w_{-1}^{bc^{n-1}}z_{2n-2}^{a} \bigr)^{a} \\ &=\alpha ^{1+a}\beta ^{b(1+a)\sum _{j=0}^{n-2}c^{j}}\bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{c^{n-1}}z_{2n-2}^{a^{2}}, \end{aligned} $$
(84)
for \(n\ge2\), and
$$\begin{aligned} z_{2n+1}&=\alpha \beta ^{b\sum _{j=0}^{n-1}c^{j}}w_{-1}^{bc^{n}} \bigl(\alpha \beta ^{b\sum_{j=0}^{n-2}c^{j}}w_{0}^{bc^{n-1}}z_{2n-1}^{a} \bigr)^{a} \\ &=\alpha ^{1+a}\beta ^{b(\sum_{j=0}^{n-1}c^{j}+a\sum _{j=0}^{n-2}c^{j})}\bigl(w_{0}^{ab}w_{-1}^{bc} \bigr)^{c^{n-1}}z_{2n-1}^{a^{2}},\quad n\in {\mathbb {N}}. \end{aligned}$$
(85)
Assume that, for some natural number k we have proved that
$$\begin{aligned} z_{2n} &=\alpha ^{(1+a)\sum_{j=0}^{k-1}a^{2j}}\beta ^{b(1+a)\sum _{i=0}^{k-1}a^{2i}\sum _{j=0}^{n-i-2}c^{j}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\sum _{j=0}^{k-1}a^{2j}c^{n-j-1}}z_{2n-2k}^{a^{2k}}, \end{aligned}$$
(86)
for \(n\ge k+1\) and
$$\begin{aligned} z_{2n+1} ={}&\alpha ^{(1+a)\sum_{j=0}^{k-1}a^{2j}}\beta ^{b\sum _{i=0}^{k-1}a^{2i}(\sum _{j=0}^{n-i-1}c^{j}+a\sum_{j=0}^{n-i-2}c^{j})} \bigl(w_{0}^{ab}w_{-1}^{bc} \bigr)^{\sum _{j=0}^{k-1}a^{2j}c^{n-j-1}}z_{2n-2k+1}^{a^{2k}}, \end{aligned}$$
(87)
for every \(n\ge k\).
By using (84) with \(n\to n-k\) into (86), and (85) with \(n\to n-k\) into (87), it follows that
$$\begin{aligned} z_{2n} ={}&\alpha ^{(1+a)\sum_{j=0}^{k-1}a^{2j}}\beta ^{b(1+a)\sum _{i=0}^{k-1}a^{2i}\sum _{j=0}^{n-i-2}c^{j}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\sum_{j=0}^{k-1}a^{2j}c^{n-j-1}} \\ &{}\times\bigl(\alpha ^{1+a}\beta ^{b(1+a)\sum _{j=0}^{n-k-2}c^{j}}\bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{c^{n-k-1}}z_{2n-2k-2}^{a^{2}}\bigr)^{a^{2k}} \\ ={}&\alpha ^{(1+a)\sum_{j=0}^{k}a^{2j}}\beta ^{b(1+a)\sum_{i=0}^{k}a^{2i}\sum _{j=0}^{n-i-2}c^{j}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\sum _{j=0}^{k}a^{2j}c^{n-j-1}}z_{2n-2k-2}^{a^{2k+2}}, \end{aligned}$$
(88)
for \(n\ge k+2\) and
$$\begin{aligned} z_{2n+1} ={}&\alpha ^{(1+a)\sum_{j=0}^{k-1}a^{2j}}\beta ^{b\sum _{i=0}^{k-1}a^{2i}(\sum _{j=0}^{n-i-1}c^{j}+a\sum_{j=0}^{n-i-2}c^{j})} \bigl(w_{0}^{ab}w_{-1}^{bc} \bigr)^{\sum _{j=0}^{k-1}a^{2j}c^{n-j-1}} \\ &{}\times\bigl(\alpha ^{1+a}\beta ^{b(\sum_{j=0}^{n-k-1}c^{j}+a\sum _{j=0}^{n-k-2}c^{j})}\bigl(w_{0}^{ab}w_{-1}^{bc} \bigr)^{c^{n-k-1}}z_{2n-2k-1}^{a^{2}}\bigr)^{a^{2k}} \\ ={}&\alpha ^{(1+a)\sum_{j=0}^{k}a^{2j}}\beta ^{b\sum_{i=0}^{k}a^{2i}(\sum _{j=0}^{n-i-1}c^{j}+a\sum_{j=0}^{n-i-2}c^{j})} \bigl(w_{0}^{ab}w_{-1}^{bc} \bigr)^{\sum_{j=0}^{k}a^{2j}c^{n-j-1}} z_{2n-2k-1}^{a^{2k+2}}, \end{aligned}$$
(89)
for every \(n\ge k+1\).

From the equalities in (84), (85), (88), (89), and by induction we see that (86) holds for all natural numbers k and n such that \(1\le k\le n-1\), while (87) holds for \(1\le k\le n\).

If we choose \(k=n-1\) in (86) and \(k=n\) in (87) we get
$$\begin{aligned} z_{2n} &=\alpha ^{(1+a)\sum_{j=0}^{n-2}a^{2j}}\beta ^{b(1+a)\sum _{i=0}^{n-2}a^{2i}\sum _{j=0}^{n-i-2}c^{j}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\sum _{j=0}^{n-2}a^{2j}c^{n-j-1}}z_{2}^{a^{2n-2}} \\ &=\alpha ^{(1+a)\sum_{j=0}^{n-2}a^{2j}}\beta ^{b(1+a)\sum _{i=0}^{n-2}a^{2i}\sum _{j=0}^{n-i-2}c^{j}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\sum_{j=0}^{n-2}a^{2j}c^{n-j-1}}\bigl(\alpha ^{1+a}z_{0}^{a^{2}}w_{-1}^{ab}w_{0}^{b} \bigr)^{a^{2n-2}} \\ &=\alpha ^{(1+a)\sum_{j=0}^{n-1}a^{2j}}\beta ^{b(1+a)\sum _{i=0}^{n-2}a^{2i}\sum _{j=0}^{n-i-2}c^{j}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\sum _{j=0}^{n-1}a^{2j}c^{n-j-1}}z_{0}^{a^{2n}}, \end{aligned}$$
(90)
for every \(n\in {\mathbb {N}}\), and
$$ \begin{aligned}[b] z_{2n+1} &=\alpha ^{(1+a)\sum_{j=0}^{n-1}a^{2j}}\beta ^{b\sum _{i=0}^{n-1}a^{2i}(\sum _{j=0}^{n-i-1}c^{j}+a\sum_{j=0}^{n-i-2}c^{j})} \bigl(w_{0}^{ab}w_{-1}^{bc} \bigr)^{\sum _{j=0}^{n-1}a^{2j}c^{n-j-1}} z_{1}^{a^{2n}} \\ &=\alpha ^{(1+a)\sum_{j=0}^{n-1}a^{2j}}\beta ^{b\sum _{i=0}^{n-1}a^{2i}(\sum _{j=0}^{n-i-1}c^{j}+a\sum_{j=0}^{n-i-2}c^{j})}\bigl(w_{0}^{ab}w_{-1}^{bc} \bigr)^{\sum _{j=0}^{n-1}a^{2j}c^{n-j-1}}\bigl(\alpha z_{0}^{a}w_{-1}^{b} \bigr)^{a^{2n}} \\ &=\alpha ^{\sum_{j=0}^{2n} a^{j}}\beta ^{b\sum_{i=0}^{n-1}a^{2i}(\sum _{j=0}^{n-i-1}c^{j}+a\sum_{j=0}^{n-i-2}c^{j})} w_{0}^{ab\sum_{j=0}^{n-1} a^{2j}c^{n-j-1}}w_{-1}^{b\sum_{j=0}^{n} a^{2j}c^{n-j}}z_{0}^{a^{2n+1}}, \end{aligned} $$
(91)
for every \(n\in {\mathbb {N}}\).
Subcase \(c\ne a^{2}\ne1\ne c\) . In this case we have
$$\begin{aligned} z_{2n} &=\alpha ^{\frac{1-a^{2n}}{1-a}}\beta ^{b(1+a)\sum _{i=0}^{n-2}a^{2i}\frac {1-c^{n-i-1}}{1-c}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\frac{a^{2n}-c^{n}}{a^{2}-c}}z_{0}^{a^{2n}} \\ &=\alpha ^{\frac{1-a^{2n}}{1-a}}\beta ^{\frac{b(1+a)}{1-c} (\frac {1-a^{2n-2}}{1-a^{2}}-c\frac{a^{2n-2}-c^{n-1}}{a^{2}-c} )} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\frac{a^{2n}-c^{n}}{a^{2}-c}}z_{0}^{a^{2n}} \\ &=\alpha ^{\frac{1-a^{2n}}{1-a}}\beta ^{\frac {b(1+a)(a^{2}-c+c^{n}-a^{2n}+ca^{2n}-a^{2}c^{n})}{(1-c)(1-a^{2})(a^{2}-c)}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\frac{a^{2n}-c^{n}}{a^{2}-c}}z_{0}^{a^{2n}}, \end{aligned}$$
(92)
$$\begin{aligned} z_{2n+1}&=\alpha ^{\sum_{j=0}^{2n} a^{j}}\beta ^{b\sum _{i=0}^{n-1}a^{2i}(\sum _{j=0}^{n-i-1}c^{j}+a\sum_{j=0}^{n-i-2}c^{j})} w_{0}^{ab\sum_{j=0}^{n-1} a^{2j}c^{n-j-1}}w_{-1}^{b\sum_{j=0}^{n} a^{2j}c^{n-j}}z_{0}^{a^{2n+1}} \\ &=\alpha ^{\frac{1-a^{2n+1}}{1-a}}\beta ^{b\sum_{i=0}^{n-1}a^{2i} (\frac {1-c^{n-i}}{1-c}+a\frac{1-c^{n-i-1}}{1-c} )} w_{0}^{ab\frac{a^{2n}-c^{n}}{a^{2}-c}}w_{-1}^{b\frac {a^{2n+2}-c^{n+1}}{a^{2}-c}}z_{0}^{a^{2n+1}} \\ &=\alpha ^{\frac{1-a^{2n+1}}{1-a}}\beta ^{\frac{b}{1-c} (\frac {1-a^{2n}}{1-a}-(c+a)\frac{a^{2n}-c^{n}}{a^{2}-c} )} w_{0}^{ab\frac{a^{2n}-c^{n}}{a^{2}-c}}w_{-1}^{b\frac {a^{2n+2}-c^{n+1}}{a^{2}-c}}z_{0}^{a^{2n+1}} \\ &=\alpha ^{\frac{1-a^{2n+1}}{1-a}}\beta ^{\frac {b(a^{2}-c-a(1-c)a^{2n}+(c+a)(1-a)c^{n})}{(1-c)(1-a)(a^{2}-c)}} w_{0}^{ab\frac{a^{2n}-c^{n}}{a^{2}-c}}w_{-1}^{b\frac {a^{2n+2}-c^{n+1}}{a^{2}-c}}z_{0}^{a^{2n+1}}, \end{aligned}$$
(93)
for every \(n\in {\mathbb {N}}\).
Subcase \(a^{2}\ne1\ne c\) , \(c=a^{2}\) . In this case we have
$$\begin{aligned} z_{2n} &=\alpha ^{(1+a)\sum_{j=0}^{n-1}a^{2j}}\beta ^{b(1+a)\sum _{i=0}^{n-2}a^{2i}\sum _{j=0}^{n-i-2}a^{2j}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\sum _{j=0}^{n-1}a^{2j}a^{2n-2j-2}}z_{0}^{a^{2n}} \\ &=\alpha ^{\frac{1-a^{2n}}{1-a}}\beta ^{b(1+a)\sum _{i=0}^{n-2}a^{2i}\frac {1-a^{2n-2i-2}}{1-a^{2}}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{na^{2n-2}}z_{0}^{a^{2n}} \\ &=\alpha ^{\frac{1-a^{2n}}{1-a}}\beta ^{\frac{b}{1-a} (\frac {1-a^{2n-2}}{1-a^{2}}-(n-1)a^{2n-2} )} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{na^{2n-2}}z_{0}^{a^{2n}} \\ &=\alpha ^{\frac{1-a^{2n}}{1-a}}\beta ^{\frac {b(1-na^{2n-2}+(n-1)a^{2n})}{(a+1)(a-1)^{2}}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{na^{2n-2}}z_{0}^{a^{2n}}, \end{aligned}$$
(94)
$$\begin{aligned} z_{2n+1}&=\alpha ^{\sum_{j=0}^{2n} a^{j}}\beta ^{b\sum _{i=0}^{n-1}a^{2i}(\sum _{j=0}^{n-i-1}a^{2j}+a\sum_{j=0}^{n-i-2}a^{2j})} w_{0}^{ab\sum_{j=0}^{n-1} a^{2j}a^{2n-2j-2}}w_{-1}^{b\sum_{j=0}^{n} a^{2j}a^{2n-2j}}z_{0}^{a^{2n+1}} \\ &=\alpha ^{\frac{1-a^{2n+1}}{1-a}}\beta ^{b\sum_{i=0}^{n-1}a^{2i}(\frac {1-a^{2n-2i}}{1-a^{2}}+a\frac{1-a^{2n-2i-2}}{1-a^{2}})} w_{0}^{bna^{2n-1}}w_{-1}^{b(n+1)a^{2n}}z_{0}^{a^{2n+1}} \\ &=\alpha ^{\frac{1-a^{2n+1}}{1-a}}\beta ^{\frac{b}{1-a} (\frac {1-a^{2n}}{1-a^{2}}-na^{2n-1} )} w_{0}^{bna^{2n-1}}w_{-1}^{b(n+1)a^{2n}}z_{0}^{a^{2n+1}} \\ &=\alpha ^{\frac{1-a^{2n+1}}{1-a}}\beta ^{\frac {b(1-a^{2n}-na^{2n-1}+na^{2n+1})}{(a+1)(a-1)^{2}}} w_{0}^{bna^{2n-1}}w_{-1}^{b(n+1)a^{2n}}z_{0}^{a^{2n+1}}, \end{aligned}$$
(95)
for every \(n\in {\mathbb {N}}\).
Subcase \(a^{2}\ne1=c\) . In this case we have
$$\begin{aligned} z_{2n} &=\alpha ^{(1+a)\sum_{j=0}^{n-1}a^{2j}}\beta ^{b(1+a)\sum _{i=0}^{n-2}a^{2i}\sum _{j=0}^{n-i-2}1} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\sum_{j=0}^{n-1}a^{2j}}z_{0}^{a^{2n}} \\ &=\alpha ^{\frac{1-a^{2n}}{1-a}}\beta ^{b(1+a)\sum_{i=0}^{n-2}a^{2i}(n-i-1)} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\frac{1-a^{2n}}{1-a}}z_{0}^{a^{2n}} \\ &=\alpha ^{\frac{1-a^{2n}}{1-a}}\beta ^{b(1+a) ((n-1)\frac {1-a^{2n-2}}{1-a^{2}}-a^{2}\frac {1-(n-1)a^{2n-4}+(n-2)a^{2n-2}}{(1-a^{2})^{2}} )} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\frac{1-a^{2n}}{1-a}}z_{0}^{a^{2n}} \\ &=\alpha ^{\frac{1-a^{2n}}{1-a}}\beta ^{\frac{b(n-1-na^{2}+a^{2n})}{(a+1)(a-1)^{2}}} \bigl(w_{0}^{b}w_{-1}^{ab} \bigr)^{\frac{1-a^{2n}}{1-a}}z_{0}^{a^{2n}}, \end{aligned}$$
(96)
$$\begin{aligned} z_{2n+1}&=\alpha ^{\sum_{j=0}^{2n} a^{j}}\beta ^{b\sum _{i=0}^{n-1}a^{2i}(\sum _{j=0}^{n-i-1}1+a\sum_{j=0}^{n-i-2}1)} w_{0}^{ab\sum_{j=0}^{n-1} a^{2j}}w_{-1}^{b\sum_{j=0}^{n} a^{2j}}z_{0}^{a^{2n+1}} \\ &=\alpha ^{\frac{1-a^{2n+1}}{1-a}}\beta ^{b\sum _{i=0}^{n-1}a^{2i}((1+a)n-a-(1+a)i)} w_{0}^{ab\frac{1-a^{2n}}{1-a}}w_{-1}^{b\frac {1-a^{2n+2}}{1-a}}z_{0}^{a^{2n+1}} \\ &=\alpha ^{\frac{1-a^{2n+1}}{1-a}}\beta ^{b (((1+a)n-a)\frac {1-a^{2n}}{1-a^{2}}-a^{2}\frac{1-na^{2n-2}+(n-1)a^{2n}}{(a+1)(1-a)^{2}} )} w_{0}^{ab\frac{1-a^{2n}}{1-a}}w_{-1}^{b\frac {1-a^{2n+2}}{1-a}}z_{0}^{a^{2n+1}} \\ &=\alpha ^{\frac{1-a^{2n+1}}{1-a}}\beta ^{\frac {b(n-a-na^{2}+a^{2n+1})}{(a+1)(1-a)^{2}}} w_{0}^{ab\frac{1-a^{2n}}{1-a}}w_{-1}^{b\frac {1-a^{2n+2}}{1-a}}z_{0}^{a^{2n+1}}, \end{aligned}$$
(97)
for every \(n\ge-1\).
Subcase \(a=-1\) , \(c=1\) . In this case we have
$$\begin{aligned} z_{2n} &=\bigl(w_{0}^{b}w_{-1}^{-b} \bigr)^{\sum _{j=0}^{n-1}(-1)^{2j}}z_{0}^{(-1)^{2n}} \\ &=\bigl(w_{0}^{b}w_{-1}^{-b} \bigr)^{n}z_{0}, \end{aligned}$$
(98)
$$\begin{aligned} z_{2n+1}&=\alpha ^{\sum_{j=0}^{2n}(-1)^{j}}\beta ^{b\sum _{i=0}^{n-1}(-1)^{2i}(\sum _{j=0}^{n-i-1}1-\sum_{j=0}^{n-i-2}1)} w_{0}^{-b\sum_{j=0}^{n-1}(-1)^{2j}}w_{-1}^{b\sum _{j=0}^{n}(-1)^{2j}}z_{0}^{(-1)^{2n+1}} \\ &=\alpha \beta ^{bn}w_{0}^{-bn}w_{-1}^{b(n+1)}z_{0}^{-1}, \end{aligned}$$
(99)
for every \(n\in {\mathbb {N}}\).
Subcase \(a=1\ne c\) . In this case we have
$$\begin{aligned} z_{2n} &=\alpha ^{2\sum_{j=0}^{n-1}1}\beta ^{2b\sum_{i=0}^{n-2}\sum_{j=0}^{n-i-2}c^{j}} \bigl(w_{0}^{b}w_{-1}^{b} \bigr)^{\sum_{j=0}^{n-1}c^{n-j-1}}z_{0} \\ &=\alpha ^{2n}\beta ^{2b\sum_{i=0}^{n-2}\frac{1-c^{n-i-1}}{1-c}} \bigl(w_{0}^{b}w_{-1}^{b} \bigr)^{\frac{1-c^{n}}{1-c}}z_{0} \\ &=\alpha ^{2n}\beta ^{\frac{2b}{1-c} (n-1-c\frac {1-c^{n-1}}{1-c} )} \bigl(w_{0}^{b}w_{-1}^{b} \bigr)^{\frac{1-c^{n}}{1-c}}z_{0} \\ &=\alpha ^{2n}\beta ^{\frac{2b(n-1-nc+c^{n})}{(1-c)^{2}}} \bigl(w_{0}^{b}w_{-1}^{b} \bigr)^{\frac{1-c^{n}}{1-c}}z_{0}, \end{aligned}$$
(100)
$$\begin{aligned} z_{2n+1}&=\alpha ^{\sum_{j=0}^{2n}1}\beta ^{b\sum_{i=0}^{n-1}(\sum _{j=0}^{n-i-1}c^{j}+\sum_{j=0}^{n-i-2}c^{j})} w_{0}^{b\sum_{j=0}^{n-1}c^{n-j-1}}w_{-1}^{b\sum _{j=0}^{n}c^{n-j}}z_{0} \\ &=\alpha ^{2n+1}\beta ^{b\sum_{i=0}^{n-1}(\frac{1-c^{n-i}}{1-c}+\frac {1-c^{n-i-1}}{1-c})} w_{0}^{b\frac{1-c^{n}}{1-c}}w_{-1}^{b\frac{1-c^{n+1}}{1-c}}z_{0} \\ &=\alpha ^{2n+1}\beta ^{\frac{b}{1-c} (2n-c\frac{1-c^{n}}{1-c}-\frac {1-c^{n}}{1-c} )} w_{0}^{b\frac{1-c^{n}}{1-c}}w_{-1}^{b\frac{1-c^{n+1}}{1-c}}z_{0} \\ &=\alpha ^{2n+1}\beta ^{\frac{b(2n-1-(2n+1)c+c^{n}+c^{n+1})}{(1-c)^{2}}} w_{0}^{b\frac{1-c^{n}}{1-c}}w_{-1}^{b\frac{1-c^{n+1}}{1-c}}z_{0}, \end{aligned}$$
(101)
for \(n\in {\mathbb {N}}\).
Subcase \(a=c=1\) . In this case we have
$$\begin{aligned} z_{2n} &=\alpha ^{2\sum_{j=0}^{n-1}1}\beta ^{2b\sum_{i=0}^{n-2}\sum_{j=0}^{n-i-2}1} \bigl(w_{0}^{b}w_{-1}^{b} \bigr)^{\sum_{j=0}^{n-1}1}z_{0} \\ &=\alpha ^{2n}\beta ^{2b\sum_{i=0}^{n-2}(n-i-1)} \bigl(w_{0}^{b}w_{-1}^{b} \bigr)^{n}z_{0} \\ &=\alpha ^{2n}\beta ^{b(n-1)n}\bigl(w_{0}^{b}w_{-1}^{b} \bigr)^{n}z_{0}, \end{aligned}$$
(102)
$$\begin{aligned} z_{2n+1}&=\alpha ^{\sum_{j=0}^{2n}1}\beta ^{b\sum_{i=0}^{n-1}(\sum _{j=0}^{n-i-1}1+\sum_{j=0}^{n-i-2}1)} w_{0}^{b\sum_{j=0}^{n-1}1}w_{-1}^{b\sum_{j=0}^{n}1}z_{0} \\ &=\alpha ^{2n+1}\beta ^{b\sum _{i=0}^{n-1}(2n-2i-1)}w_{0}^{bn}w_{-1}^{b(n+1)}z_{0} \\ &=\alpha ^{2n+1}\beta ^{bn^{2}}w_{0}^{bn}w_{-1}^{b(n+1)}z_{0}, \end{aligned}$$
(103)
for every \(n\in {\mathbb {N}}\), completing the proof. □

Corollary 3

Consider system (1) with \(a,b,c\in {\mathbb {Z}}\), \(d=0\), and \(\alpha ,\beta \in {\mathbb {C}}\setminus\{0\}\). Assume that \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\). Then the following statements are true.
  1. (a)

    If \(c\ne a^{2}\ne1\ne c\), then the general solution to system (1) is given by (78), (79), (92), and (93).

     
  2. (b)

    If \(c=a^{2}\ne1\ne c\), then the general solution to system (1) is given by (78), (79), (94), and (95).

     
  3. (c)

    If \(a^{2}\ne1=c\), then the general solution to system (1) is given by (80), (81), (96), and (97).

     
  4. (d)

    If \(a=-1\) and \(c=1\), then the general solution to system (1) is given by (80), (81), (98), and (99).

     
  5. (e)

    If \(a=1\) and \(c\ne1\), then the general solution to system (1) is given by (78), (79), (100), and (101).

     
  6. (f)

    If \(a=c=1\), then the general solution to system (1) is given by (80), (81), (102), and (103).

     

Theorem 4

Assume that \(a,b,c,d\in {\mathbb {Z}}\), \(bd\ne0\), \(\alpha ,\beta \in {\mathbb {C}}\setminus\{0\}\), and \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\). Then system (1) is solvable in closed form.

Proof

First note that the conditions \(\alpha ,\beta \in {\mathbb {C}}\setminus\{ 0\}\) and \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\) along with the equations in (1) imply \(z_{n}w_{n}\ne0\) for \(n\ge-1\). Hence, for every such a solution the first equation in (1) yields
$$\begin{aligned} w_{n-1}^{b}=\frac{z_{n+1}}{\alpha z_{n}^{a}}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(104)
while from the second one it follows that
$$\begin{aligned} w_{n+1}^{b}=\beta ^{b} w_{n-1}^{bc}z_{n-1}^{bd}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$
(105)
From (104) and (105) one obtains
$$\begin{aligned} z_{n+3}=\alpha ^{1-c}\beta ^{b} z_{n+2}^{a}z_{n+1}^{c}z_{n}^{-ac}z_{n-1}^{bd}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(106)
which is a fourth order product-type difference equation.
Note also that
$$\begin{aligned} z_{1}=\alpha z_{0}^{a}w_{-1}^{b}, \qquad z_{2}=\alpha \bigl(\alpha z_{0}^{a}w_{-1}^{b} \bigr)^{a}w_{0}^{b}=\alpha ^{1+a}z_{0}^{a^{2}}w_{-1}^{ab}w_{0}^{b}. \end{aligned}$$
(107)
Let \(\delta =\alpha ^{1-c}\beta ^{b}\),
$$\begin{aligned} a_{1}=a, \qquad b_{1}=c, \qquad c_{1}=-ac, \qquad d_{1}=bd,\qquad y_{1}=1. \end{aligned}$$
(108)
Then equation (106) can be written as
$$\begin{aligned} z_{n+3}=\delta ^{y_{1}}z_{n+2}^{a_{1}}z_{n+1}^{b_{1}}z_{n}^{c_{1}}z_{n-1}^{d_{1}}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$
(109)
Using (109) with \(n\to n-1\) into (109) we get
$$ \begin{aligned}[b] z_{n+3}&=\delta ^{y_{1}}\bigl(\delta z_{n+1}^{a_{1}}z_{n}^{b_{1}}z_{n-1}^{c_{1}} z_{n-2}^{d_{1}}\bigr)^{a_{1}}z_{n+1}^{b_{1}}z_{n}^{c_{1}}z_{n-1}^{d_{1}}, \\ &=\delta ^{y_{1}+a_{1}}z_{n+1}^{a_{1}a_{1}+b_{1}}z_{n}^{b_{1}a_{1}+c_{1}} z_{n-1}^{c_{1}a_{1}+d_{1}}z_{n-2}^{d_{1}a_{1}} \\ &=\delta ^{y_{2}}z_{n+1}^{a_{2}}z_{n}^{b_{2}}z_{n-1}^{c_{2}}z_{n-2}^{d_{2}}, \end{aligned} $$
(110)
for \(n\in {\mathbb {N}}\), where
$$ \begin{aligned} &a_{2}:=a_{1}a_{1}+b_{1}, \quad b_{2}:=b_{1}a_{1}+c_{1}, \qquad c_{2}:=c_{1}a_{1}+d_{1}, \\ & d_{2}:=d_{1}a_{1},\qquad y_{2}:=y_{1}+a_{1}. \end{aligned} $$
(111)
Assume that, for a k such that \(2\le k \le n+1\), we have proved that
$$\begin{aligned} z_{n+3}=\delta ^{y_{k}}z_{n+3-k}^{a_{k}}z_{n+2-k}^{b_{k}}z_{n+1-k}^{c_{k}}z_{n-k}^{d_{k}}, \end{aligned}$$
(112)
for \(n\ge k-1\), and that
$$\begin{aligned}& \begin{aligned} &a_{k}=a_{1}a_{k-1}+b_{k-1}, \qquad b_{k}=b_{1}a_{k-1}+c_{k-1},\\ &c_{k}=c_{1}a_{k-1}+d_{k-1},\qquad d_{k}=d_{1}a_{k-1}, \end{aligned} \end{aligned}$$
(113)
$$\begin{aligned}& y_{k}:=y_{k-1}+a_{k-1}. \end{aligned}$$
(114)
Using (109) with \(n\to n-k\) into (110) one obtains
$$\begin{aligned} z_{n+3}&=\delta ^{y_{k}}\bigl(\delta z_{n+2-k}^{a_{1}}z_{n+1-k}^{b_{1}}z_{n-k}^{c_{1}}z_{n-k-1}^{d_{1}} \bigr)^{a_{k}}z_{n+2-k}^{b_{k}}z_{n+1-k}^{c_{k}}z_{n-k}^{d_{k}} \\ &=\delta ^{y_{k}+a_{k}}z_{n+2-k}^{a_{1}a_{k}+b_{k}}z_{n+1-k}^{b_{1}a_{k}+c_{k}}z_{n-k}^{c_{1}a_{k}+d_{k}}z_{n-k-1}^{d_{1}a_{k}} \\ &=\delta ^{y_{k+1}}z_{n+2-k}^{a_{k+1}}z_{n+1-k}^{b_{k+1}} z_{n-k}^{c_{k+1}}z_{n-k-1}^{d_{k+1}}, \end{aligned}$$
(115)
for \(n\ge k\), where
$$\begin{aligned}& \begin{aligned} &a_{k+1}:=a_{1}a_{k}+b_{k}, \qquad b_{k+1}:=b_{1}a_{k}+c_{k},\\ &c_{k+1}:=c_{1}a_{k}+d_{k}, \qquad d_{k+1}:=d_{1}a_{k}, \end{aligned} \end{aligned}$$
(116)
$$\begin{aligned}& y_{k+1}:=y_{k}+a_{k}. \end{aligned}$$
(117)
This along with (110), (111), and the method of induction shows that (112), (113), and (114), hold for every k and n such that \(2\le k\le n+1\). In fact (112) holds for \(1\le k\le n+1\) (see (109)).
Hence, choosing \(k=n+1\) in (112), and using (107) we have
$$ \begin{aligned}[b] z_{n+3} ={}&\delta ^{y_{n+1}}z_{2}^{a_{n+1}}z_{1}^{b_{n+1}}z_{0}^{c_{n+1}}z_{-1}^{d_{n+1}} \\ ={}&\bigl(\alpha ^{1-c}\beta ^{b}\bigr)^{y_{n+1}}\bigl( \alpha ^{1+a}z_{0}^{a^{2}}w_{-1}^{ab}w_{0}^{b} \bigr)^{a_{n+1}}\bigl(\alpha z_{0}^{a}w_{-1}^{b} \bigr)^{b_{n+1}}z_{0}^{c_{n+1}}z_{-1}^{d_{n+1}} \\ ={}&\alpha ^{(1-c)y_{n+1}+(1+a)a_{n+1}+b_{n+1}}\beta ^{by_{n+1}}z_{0}^{a^{2}a_{n+1}+ab_{n+1}+c_{n+1}} \\ &{}\times w_{-1}^{aba_{n+1}+bb_{n+1}}w_{0}^{ba_{n+1}}z_{-1}^{d_{n+1}}, \quad n\in {\mathbb {N}}_{0}. \end{aligned} $$
(118)
From (113) we easily see that \((a_{k})_{k\ge4}\) satisfies the difference equation
$$\begin{aligned} a_{k}=a_{1}a_{k-1}+b_{1}a_{k-2}+c_{1}a_{k-3}+d_{1}a_{k-4}. \end{aligned}$$
(119)

Since \(b_{k}=a_{k+1}-a_{1}a_{k}\), \(c_{k}=b_{k+1}-b_{1}a_{k}\), \(d_{k}=d_{1}a_{k-1}\), and from the linearity of equation (119) we see that \((b_{k})_{k\in {\mathbb {N}}}\), \((c_{k})_{k\in {\mathbb {N}}}\), and \((d_{k})_{k\in {\mathbb {N}}}\) are also solutions to the equation.

System (113) with \(k=1\) yields
$$ \begin{aligned} &a_{1}=a_{1}a_{0}+b_{0}, \qquad b_{1}=b_{1}a_{0}+c_{0}, \qquad c_{1}=c_{1}a_{0}+d_{0}, \\ & d_{1}=d_{1}a_{0},\qquad y_{1}=y_{0}+a_{0}. \end{aligned} $$
(120)
The condition \(d_{1}=bd\ne0\) along with the fourth equation in (120) implies \(a_{0}=1\). Using this and \(y_{1}=1\) in the other equalities in (120) we get \(b_{0}=c_{0}=d_{0}=y_{0}=0\). Repeating the procedure for \(k=0,-1, -2\), is easily obtained
$$ \begin{aligned} &a_{-3}=0,\qquad a_{-2}=0,\qquad a_{-1}=0,\qquad a_{0}=1; \\ &b_{-3}=0,\qquad b_{-2}=0,\qquad b_{-1}=1, \qquad b_{0}=0; \\ &c_{-3}=0,\qquad c_{-2}=1,\qquad c_{-1}=0, \qquad c_{0}=0; \\ &d_{-3}=1,\qquad d_{-2}=0,\qquad d_{-1}=0, \qquad d_{0}=0. \end{aligned} $$
(121)
Hence, \((a_{k})_{k\ge-3}\), \((b_{k})_{k\ge-3}\), \((c_{k})_{k\ge-3}\), and \((d_{k})_{k\ge-3}\) are solutions to (119) satisfying initial conditions (121), while \((y_{k})_{k\ge-3}\) satisfies the following conditions:
$$\begin{aligned} y_{-3}=y_{-2}=y_{-1}=y_{0}=0, \qquad y_{1}=1, \end{aligned}$$
(122)
and (114), from which it follows that
$$\begin{aligned} y_{k}=\sum_{j=1}^{k-1}a_{j}. \end{aligned}$$
(123)

Since equation (119) is solvable, it follows that closed form formulas for \((a_{k})_{k\ge-3}\), \((b_{k})_{k\ge-3}\), \((c_{k})_{k\ge-3}\), and \((d_{k})_{k\ge-3}\), can be found. From (123), the form of the solution \(a_{k}\), and by using some known summation formulas it follows that the formula for \((y_{k})_{k\ge-3}\) can also be found. From these facts and (118) we see that equation (106) is solvable too.

From the second equation in (1), we have that for every well-defined solution
$$\begin{aligned} z_{n-1}^{d}=\frac{w_{n+1}}{\beta w_{n-1}^{c}},\quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(124)
while from the first one it follows that
$$\begin{aligned} z_{n+1}^{d}=\alpha ^{d}z_{n}^{ad}w_{n-1}^{bd}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$
(125)
From (124) into (125) one obtains
$$\begin{aligned} w_{n+3}=\alpha ^{d}\beta ^{1-a}w_{n+2}^{a}w_{n+1}^{c}w_{n}^{-ac}w_{n-1}^{bd}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$
(126)
which differs from (106) only by the constant multiplier.
We have
$$\begin{aligned} w_{1}=\beta w_{-1}^{c}z_{-1}^{d} \quad \mbox{and} \quad w_{2}=\beta w_{0}^{c}z_{0}^{d}. \end{aligned}$$
(127)
As above one obtains, for all natural numbers k and n such that \(1\le k \le n+1\),
$$\begin{aligned} w_{n+3}=\eta^{\hat{y}_{k}}w_{n+3-k}^{a_{k}}w_{n+2-k}^{b_{k}}w_{n+1-k}^{c_{k}}w_{n-k}^{d_{k}}, \quad n\ge k-1, \end{aligned}$$
(128)
where \(\eta=\alpha ^{d}\beta ^{1-a}\), \((a_{k})_{k\in {\mathbb {N}}}\), \((b_{k})_{k\in {\mathbb {N}}}\), \((c_{k})_{k\in {\mathbb {N}}}\), and \((d_{k})_{k\in {\mathbb {N}}}\) satisfy (113) with initial conditions (108), while \((\hat{y}_{k})_{k\in {\mathbb {N}}}\) satisfies (114) and (122), so that (123) holds where \(y_{k}\) is replaced by \(\hat{y}_{k}\).
From (128) with \(k=n+1\) and by using (127) we get
$$\begin{aligned} w_{n+3}&=\eta^{\hat{y}_{n+1}}w_{2}^{a_{n+1}}w_{1}^{b_{n+1}}w_{0}^{c_{n+1}}w_{-1}^{d_{n+1}} \\ &=\bigl(\alpha ^{d}\beta ^{1-a}\bigr)^{\hat{y}_{n+1}}\bigl(\beta w_{0}^{c}z_{0}^{d} \bigr)^{a_{n+1}}\bigl(\beta w_{-1}^{c}z_{-1}^{d} \bigr)^{b_{n+1}}w_{0}^{c_{n+1}}w_{-1}^{d_{n+1}} \\ &=\alpha ^{d\hat{y}_{n+1}}\beta ^{(1-a)\hat{y}_{n+1}+a_{n+1}+b_{n+1}}w_{0}^{ca_{n+1}+c_{n+1}}z_{0}^{da_{n+1}}w_{-1}^{cb_{n+1}+d_{n+1}}z_{-1}^{db_{n+1}}, \end{aligned}$$
(129)
for \(n\in {\mathbb {N}}_{0}\).

As above the solvability of (119) shows that formulas for \((a_{k})_{k\ge-3}\), \((b_{k})_{k\ge-3}\), \((c_{k})_{k\ge-3}\), and \((d_{k})_{k\ge-3}\) can be found, and consequently a formula for \((\hat{y}_{k})_{k\ge-3}\). This fact along with (129) implies that equation (126) is solvable too. Hence, system (1) is also solvable in this case, as desired. □

Corollary 4

Consider system (1) with \(a,b,c,d\in {\mathbb {Z}}\), \(bd\ne0\), \(\alpha ,\beta \in {\mathbb {C}}\setminus\{0\}\). Assume that \(z_{-1}, z_{0}, w_{-1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\). Then the general solution to system (1) is given by (118) and (129), where the sequences \((a_{k})_{k\in {\mathbb {N}}}\), \((b_{k})_{k\in {\mathbb {N}}}\), \((c_{k})_{k\in {\mathbb {N}}}\), and \((d_{k})_{k\in {\mathbb {N}}}\) satisfy the difference equation (119) with initial conditions in (121), while \((y_{k})_{k\in {\mathbb {N}}}\) and \((\hat{y}_{k})_{k\in {\mathbb {N}}}\) are given by (123) and satisfy conditions (122).

Declarations

Acknowledgements

The work of Stevo Stević is supported by the Serbian Ministry of Education and Science projects III 41025 and III 44006. The work of Bratislav Iričanin is supported by the Serbian Ministry of Education and Science projects III 41025 and OI 171007. The work of Zdeněk Šmarda is supported by the project FEKT-S-14-2200 of the Brno University of Technology. Some results in the paper are obtained during Bratislav Iričanin’s visit of Faculty of Electrical Engineering and Communication at the 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
(2)
Operator Theory and Applications Research Group, Department of Mathematics, King Abdulaziz University
(3)
Faculty of Electrical Engineering, Belgrade University
(4)
Department of Mathematics, FEEC - Faculty of Electrical Engineering and Communication, Brno University of Technology

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