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Article

Solvability of the Class of Two-Dimensional Product-Type Systems of Difference Equations of Delay-Type (1, 3, 1, 1)

1
Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, 11000 Beograd, Serbia
2
Operator Theory and Applications Research Group, Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
Symmetry 2017, 9(10), 200; https://doi.org/10.3390/sym9100200
Submission received: 1 September 2017 / Revised: 21 September 2017 / Accepted: 21 September 2017 / Published: 25 September 2017
(This article belongs to the Special Issue Symmetry: Feature Papers 2017)

Abstract

:
This paper essentially presents the last and important steps in the study of (practical) solvability of two-dimensional product-type systems of difference equations of the following form z n = α z n k a w n l b , w n = β w n m c z n s d , n N 0 , where k , l , m , s N , a , b , c , d Z , and where α , β and the initial values are complex numbers. It is devoted to the most complex case which has not been considered so far (the case k = l = s = 1 and m = 3 ). Closed form formulas for solutions to the system are found in all possible cases. The structure of the solutions to the system is considered in detail. The following five cases: (1) b = 0 ; (2) c = 0 ; (3) d = 0 ; (4) a c 0 ; (5) a = 0 , b c d 0 , are considered separately. Some of the situations appear for the first time in the literature.

1. Introduction

There has been a growing interest in difference equations and systems of difference equations (see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43]). Among several subfields of recent interest we mention here two, on whose intersection is the study in this paper. The first one is the classical subfield of finding solutions to the equations and systems in closed form. Books [6,12,13,14,15,16] contain some old information in the subfield. Some recent ones can be found, for example, in the following papers: [5,27,31,32,33,34,35,36,37,38,39,40,41,42,43] (see also numerous references therein). For some related results and applications of solvable difference equations and systems, see also [1,2,3,4]. The recent interest has been considerably motivated by the fact that some new interesting classes of nonlinear difference equations and systems have been solved by transforming them to known solvable ones. One of our transformations has had some impact in the recent interest. For more information, see, for example, [5,27,34], and the related references therein. Beside the line of investigation, there have been some other several ones which also use some related ideas. The reader can consult the representative paper [34] and find many other related ones in its list of references. Generally speaking, above mentioned lines of investigations use the method of transformation in solving the equations and systems therein. In many of these papers obtained formulas for the solutions to the equations and systems studied therein are used in describing their long-term behavior (for example, in [5,39]).
The second one is the subfield on concrete systems of difference equations. Some of the papers which have had some impact on the growing interest in the subfield are [19,20,21] by Papaschinopoulos and Schinas. One of their main ideas is to consider symmetric systems of difference equations obtained from the following scalar one
x n = f ( x n k , x n l ) , n N 0 ,
where k , l N , k l , for concrete values of function f, that is, to study some concrete systems of difference equations of one of the following forms
x n = f ( x n k , y n l ) , y n = f ( y n k , x n l ) , x n = f ( y n k , x n l ) , y n = f ( x n k , y n l ) , x n = f ( y n k , y n l ) , y n = f ( x n k , x n l ) , n N 0 .
This, among other things, has motivated us to study the solvability of some concrete systems of the form, such as the ones in [5,39,40] (see also the related references therein). The idea naturally evolved into the investigation of more general symmetric systems of difference equations. For example, a few symmetric systems with three variables were studied in [23], while in [22] were studied, among others, the invariants of the following system of nonlinear difference equations
x n + 1 = a + b ( y n + x n 1 + + y n k + 1 + x n k ) y n k 1 y n + 1 = a + b ( x n + y n 1 + + x n k + 1 + y n k ) x n k 1
where k is an odd number and a , b ( 0 , ) , which means that more complex symmetric systems were studied therein. In fact, the study of the invariants of some equations and systems can be regarded as a kind of the study of their solvability, so that paper [22], as well as [21] essentially belong to both subfields.
It is highly expected that the methods and ideas used in the study of symmetric systems of difference equations can produce the same or related results for the systems which are not symmetric, but are close to them. For example, the following max-type system of difference equations
x n + 1 = max { a n y n , b n } y n x n 1 , y n + 1 = max { c n x n , d n } x n y n 1 , n N 0 ,
where a n , b n , c n , d n are sequences of positive numbers and x 1 , x 0 , y 1 , y 0 are positive numbers, is such one, and was studied in [25] (for another max-type system see also [28]). Such system are called close-to-symmetric systems of difference equations and are frequently studied (see, for example, [8,17,24,29,31,33] and the related references therein). In paper [11] was initiated study of cyclic systems of difference equations, which naturally evolved into the study of some close to cyclic systems of difference equations (see, for example, [7,18]).
Studying positive solutions to some classes of equations and systems, such as the ones of the special cases of the following equation
x n = α + x n k p x n l q , n N 0 ,
where α , p , q ( 0 , ) , k , l N , k l (see, for example, [30], as well as [9,10,26] and the related references therein), as well as of the corresponding two-dimensional symmetric systems of difference equations, such as
x n = α + y n k p x n l q , y n = α + x n k p y n l q , n N 0 ,
we came across some product-type equations and systems.
The solvability of the product-type equations and systems in this case is something which should be known to any expert. Namely, if all the initial values are positive a simple inductive argument shows that all other terms are also positive, so that by using the logarithm the equation/system is transformed to a linear one with constant coefficients, which is one of the most known solvable equation/system. If the domain is changed, then several problems occur which prevent using the standard method for solving the equations and systems for the case of positive solutions. Our study of the systems on the complex plane was set off in [39]. It turned out that finding solutions to some related systems is not so simple problem. The next product-type system was solved in [40], but without detailed analysis of the structure of its solutions. The forms of the systems studied in [39,40] are similar, which suggested an investigation of the extensions which include both of them. On the other hand, the occurrence of some multipliers in some cases of the one-dimensional equation in [34] has suggested an investigation of the related systems with additional constant multipliers, which has been done for the first time in [31].
Somewhat later, a detailed analysis has shown that complete lists of formulas can be given for some of concrete systems of difference equations of the following form
z n = α z n k a w n l b , w n = β w n m c z n s d , n N ,
with “small" values of delays k , l , m and s, which means that they are solvable. Since we have studied so far a number of the systems of type (1), to facilitate classification and terminology, from now on we will say that the system is of delay-type ( k , m , l , s ) .
The corresponding lists of formulas for solutions are given first for the systems in [33,43], unlike the systems in above mentioned paper [40] and for the system in [42]. For some systems such as the ones in [35,41] the solutions were obtained more easily, so the analysis was simpler and it was of a different character. Some technical problems in dealing with the systems of the sort in (1) lead us to devising another method for solving them in [32], which has been recently also used in [37]. Recently, we have done, for the first time a detailed analysis of the structure of solutions to a class of product-type systems with an associated polynomial (to the system) of the fourth order in [38], and quite recently in [36].
The main goal of the whole project is to classify solvable product-type systems of difference equations of the form in (1) and present their solutions in closed form in terms of the involved parameters and initial values. Here we continue the project. This paper is a natural continuation of our research in [31,32,33,35,36,37,38,39,40,41,42,43], and essentially presents the last and important steps in the finishing of the project.
Our task here is to show the solvability of system (1) of delay-type ( 1 , 3 , 1 , 1 ) (the case k = l = s = 1 and m = 3 ), that is, of
z n + 1 = α z n a w n b , w n + 1 = β w n 2 c z n d , n N 0 ,
where a , b , c , d Z , α , β , z 0 , w 2 , w 1 , w 0 C \ { 0 } . The case when some of the quantities α , β , z 0 , w 2 , w 1 , w 0 is zero we do not take into consideration because in the case are obtained solutions which are either not defined or trivial, so of not special interest.
The following five cases are considered separately in this paper: (1) b = 0 ; (2) c = 0 ; (3) d = 0 ; (4) a c 0 ; (5) a = 0 , b c d 0 . We would like to point out that the fifth case is not covered with the fourth one, since the condition a = 0 changes the order of an associated polynomial appearing in the study. If k , l Z , then the notation k , l ¯ , denotes the set of all j Z such that k j l , whereas we regard that j = l l 1 c j = 0 for each l Z .

2. Auxiliary Results

The following three lemmas are useful tools in our investigation and have been already used in some of our previous papers devoted to the project on product-type systems. The first one is a consequence of the Langrage formula applied to the functions f s ( t ) = t s , s N (see, for example, [13] or [44], as well as [40] for a proof based on complex analysis).
Lemma 1.
Let
p k ( t ) = c k j = 1 k ( t t j ) ,
c k 0 and t i t j , i j . Then
j = 1 k t j s p k ( t j ) = 0 , 0 s k 2 ,
and
j = 1 k t j k 1 p k ( t j ) = 1 c k .
Further, we need several closed form formulas for some sums which can be found in numerous books (see, for example, [13] or [16]). For a general method for calculating this type of sums consult our recent paper [33] where a recurrent formula for this type of sums is presented, and by using it the sums can be calculated.
Lemma 2.
Let
s n ( i ) ( z ) = j = 1 n j i z j 1 , n N ,
where i N 0 and z C .
Then
s n ( 0 ) ( z ) = 1 z n 1 z ,
s n ( 1 ) ( z ) = 1 ( n + 1 ) z n + n z n + 1 ( 1 z ) 2 ,
s n ( 2 ) ( z ) = 1 + z ( n + 1 ) 2 z n + ( 2 n 2 + 2 n 1 ) z n + 1 n 2 z n + 2 ( 1 z ) 3 ,
s n ( 3 ) ( z ) = n 3 z n ( z 1 ) 3 3 n 2 z n ( z 1 ) 2 + 3 n z n ( z 2 1 ) ( z n 1 ) ( z 2 + 4 z + 1 ) ( 1 z ) 4 ,
for every z C \ { 1 } and n N .
The following lemma describes the nature/type of the zeros of an arbitrary fourth order polynomial equation. The results in the lemma are certainly folklore and were essentially obtained, for example, in [45] (the lemma formulates the results appearing therein in a unified way, although the notation and some quantities are different).
Lemma 3.
Let
P 4 ( t ) = t 4 + b t 3 + c t 2 + d t + e ,
where b , c , d , e are real numbers,
Δ 0 = c 2 3 b d + 12 e , Δ 1 = 2 c 3 9 b c d + 27 b 2 e + 27 d 2 72 c e , Δ = 1 27 ( 4 Δ 0 3 Δ 1 2 ) , P = 8 c 3 b 2 , Q = b 3 + 8 d 4 b c , D = 64 e 16 c 2 + 16 b 2 c 16 b d 3 b 4 .
Then, the following statements hold.
(a) 
If Δ < 0 , then two zeros of P 4 are real and different, and two are complex conjugate.
(b) 
If Δ > 0 , then all the zeros of P 4 are real or none is. More precisely,
1
if P < 0 and D < 0 , then all four zeros of P 4 are real and different;
2
if P > 0 or D > 0 , then there are two pairs of complex conjugate zeros of P 4 .
(c) 
If Δ = 0 , then and only then P 4 has a multiple zero. The following cases can occur:
1
if P < 0 , D < 0 and Δ 0 0 , then two zeros of P 4 are real and equal and two are real and simple;
2
if D > 0 or ( P > 0 and ( D 0 or Q 0 )), then two zeros of P 4 are real and equal and two are complex conjugate;
3
if Δ 0 = 0 and D 0 , there is a triple zero of P 4 and one simple, all real;
4
if D = 0 , then
4.1
if P < 0 there are two double real zeros of P 4 ;
4.2
if P > 0 and Q = 0 there are two double complex conjugate zeros of P 4 ;
4.3
if Δ 0 = 0 , then all four zeros of P 4 are real and equal to b / 4 .

3. Main Results

In this section we state and prove our main results and by using them and some further analysis we get several corollaries. Before this we give a list of first several members of the sequences z n and w n , whose values will be used in the proofs of some of the results.
We have
z 1 = α z 0 a w 0 b , w 1 = β w 2 c z 0 d , z 2 = α ( α z 0 a w 0 b ) a ( β w 2 c z 0 d ) b = α 1 + a β b z 0 a 2 + b d w 2 b c w 0 a b , w 2 = β w 1 c z 1 d = β w 1 c ( α z 0 a w 0 b ) d = α d β z 0 a d w 1 c w 0 b d z 3 = α z 2 a w 2 b = α ( α 1 + a β b z 0 a 2 + b d w 2 b c w 0 a b ) a ( α d β z 0 a d w 1 c w 0 b d ) b = α 1 + a + a 2 + b d β b + a b z 0 a 3 + 2 a b d w 2 a b c w 1 b c w 0 b ( a 2 + b d ) .
As we have already mentioned we will consider the following five cases separately: (1) b = 0 ; (2) c = 0 ; (3) d = 0 ; (4) a c 0 ; (5) a = 0 , b c d 0 . Hence, we will prove five results on the solvability and by further analysis we will get several consequences from them.
Theorem 1.
Assume that a , c , d Z , b = 0 , α , β , z 0 , w 2 , w 1 , w 0 C \ { 0 } . Then system (2) is solvable in closed form.
Proof. 
Since b = 0 , we have
z n + 1 = α z n a , w n + 1 = β w n 2 c z n d , n N 0 ,
from which it follows that
z n = α j = 0 n 1 a j z 0 a n ,
for n N .
From (10) we have that
z n = α 1 a n 1 a z 0 a n , n N ,
when a 1 , and
z n = α n z 0 , n N ,
when a = 1 .
Using (10) in the second equation in (9), is obtained
w n = α d j = 0 n 2 a j β z 0 d a n 1 w n 3 c , n 2 ,
which is equivalent to
w 3 m + i = α d j = 0 3 m + i 2 a j β z 0 d a 3 m + i 1 w 3 ( m 1 ) + i c ,
for m N and i = 1 , 0 , 1 .
From (13) and by induction it is proved that
w 3 m + i = α d l = 0 k 1 c l j = 0 3 ( m l ) + i 2 a j β l = 0 k 1 c l z 0 d l = 0 k 1 c l a 3 ( m l ) + i 1 w 3 ( m k ) + i c k ,
for m k and i = 1 , 0 , 1 , from which for k = m is obtained
w 3 m + i = α d l = 0 m 1 c l j = 0 3 ( m l ) + i 2 a j β l = 0 m 1 c l z 0 d l = 0 m 1 c l a 3 ( m l ) + i 1 w i c m ,
for m N and i = 1 , 0 , 1 .
From (8) and (14), we have
w 3 m 1 = α d l = 0 m 1 c l j = 0 3 ( m l 1 ) a j β l = 0 m 1 c l z 0 d l = 0 m 1 c l a 3 ( m l ) 2 w 1 c m ,
w 3 m = α d l = 0 m 1 c l j = 0 3 ( m l ) 2 a j β l = 0 m 1 c l z 0 d l = 0 m 1 c l a 3 ( m l ) 1 w 0 c m ,
w 3 m + 1 = α d l = 0 m 1 c l j = 0 3 ( m l ) 1 a j β l = 0 m 1 c l z 0 d l = 0 m 1 c l a 3 ( m l ) ( β w 2 c z 0 d ) c m = α d l = 0 m 1 c l j = 0 3 ( m l ) 1 a j β l = 0 m c l z 0 d l = 0 m c l a 3 ( m l ) w 2 c m + 1 ,
for m N .
Now we use the formulas in (15)–(17) in five subcases separately.
Case a 1 c a 3 . From (15)–(17) and by Lemma 2, we obtain
w 3 m 1 = α d l = 0 m 1 c l 1 a 3 ( m l ) 2 1 a β 1 c m 1 c z 0 a d a 3 m c m a 3 c w 1 c m = α d 1 a 1 c m 1 c a a 3 m c m a 3 c β 1 c m 1 c z 0 a d a 3 m c m a 3 c w 1 c m = α d ( a 3 c + ( 1 a ) ( a + a 2 + c ) c m + ( c 1 ) a 3 m + 1 ) ( 1 a ) ( 1 c ) ( a 3 c ) β 1 c m 1 c z 0 a d a 3 m c m a 3 c w 1 c m ,
w 3 m = α d l = 0 m 1 c l 1 a 3 ( m l ) 1 1 a β 1 c m 1 c z 0 d a 2 a 3 m c m a 3 c w 0 c m = α d 1 a 1 c m 1 c a 2 a 3 m c m a 3 c β 1 c m 1 c z 0 d a 2 a 3 m c m a 3 c w 0 c m = α d ( a 3 c + ( 1 a ) ( c + c a + a 2 ) c m + ( c 1 ) a 3 m + 2 ) ( 1 a ) ( 1 c ) ( a 3 c ) β 1 c m 1 c z 0 d a 2 a 3 m c m a 3 c w 0 c m ,
w 3 m + 1 = α d l = 0 m 1 c l 1 a 3 ( m l ) 1 a β 1 c m + 1 1 c z 0 d a 3 m + 3 c m + 1 a 3 c w 2 c m + 1 = α d 1 a 1 c m 1 c a 3 a 3 m c m a 3 c β 1 c m + 1 1 c z 0 d a 3 m + 3 c m + 1 a 3 c w 2 c m + 1 = α d ( a 3 c + ( 1 a 3 ) c m + 1 + ( c 1 ) a 3 m + 3 ) ( 1 a ) ( 1 c ) ( a 3 c ) β 1 c m + 1 1 c z 0 d a 3 m + 3 c m + 1 a 3 c w 2 c m + 1 ,
for m N .
Case a 1 c = a 3 . From (15)–(17) and by Lemma 2, we obtain
w 3 m 1 = α d l = 0 m 1 a 3 l j = 0 3 ( m l 1 ) a j β l = 0 m 1 a 3 l z 0 d l = 0 m 1 a 3 l a 3 ( m l ) 2 w 1 a 3 m = α d l = 0 m 1 a 3 l 1 a 3 ( m l ) 2 1 a β 1 a 3 m 1 a 3 z 0 d m a 3 m 2 w 1 a 3 m = α d 1 a 1 a 3 m 1 a 3 m a 3 m 2 β 1 a 3 m 1 a 3 z 0 d m a 3 m 2 w 1 a 3 m = α d ( m a 3 m + 1 a 3 m m a 3 m 2 + 1 ) ( 1 a ) ( 1 a 3 ) β 1 a 3 m 1 a 3 z 0 d m a 3 m 2 w 1 a 3 m ,
w 3 m = α d l = 0 m 1 a 3 l j = 0 3 ( m l ) 2 a j β l = 0 m 1 a 3 l z 0 d l = 0 m 1 a 3 l a 3 ( m l ) 1 w 0 a 3 m = α d l = 0 m 1 a 3 l 1 a 3 ( m l ) 1 1 a β 1 a 3 m 1 a 3 z 0 d m a 3 m 1 w 0 a 3 m = α d 1 a 1 a 3 m 1 a 3 m a 3 m 1 β 1 a 3 m 1 a 3 z 0 m d a 3 m 1 w 0 a 3 m = α d ( m a 3 m + 2 a 3 m m a 3 m 1 + 1 ) ( 1 a ) ( 1 a 3 ) β 1 a 3 m 1 a 3 z 0 d m a 3 m 1 w 0 a 3 m ,
w 3 m + 1 = α d l = 0 m 1 a 3 l j = 0 3 ( m l ) 1 a j β l = 0 m a 3 l z 0 d l = 0 m a 3 l a 3 ( m l ) w 2 a 3 m + 3 = α d l = 0 m 1 a 3 l 1 a 3 ( m l ) 1 a β 1 a 3 m + 3 1 a 3 z 0 d ( m + 1 ) a 3 m w 2 a 3 m + 3 = α d 1 a 1 a 3 m 1 a 3 m a 3 m β 1 a 3 m + 3 1 a 3 z 0 d ( m + 1 ) a 3 m w 2 a 3 m + 3 = α d ( m a 3 m + 3 ( m + 1 ) a 3 m + 1 ) ( 1 a ) ( 1 a 3 ) β 1 a 3 m + 3 1 a 3 z 0 d ( m + 1 ) a 3 m w 2 a 3 m + 3 ,
for m N .
Case a 1 = c . From (15)–(17) and by Lemma 2, we obtain
w 3 m 1 = α d l = 0 m 1 j = 0 3 ( m l 1 ) a j β l = 0 m 1 1 z 0 d l = 0 m 1 a 3 ( m l ) 2 w 1 = α d l = 0 m 1 1 a 3 ( m l ) 2 1 a β m z 0 a d a 3 m 1 a 3 1 w 1 = α d 1 a m a a 3 m 1 a 3 1 β m z 0 a d a 3 m 1 a 3 1 w 1 = α d ( a 3 m + 1 + m ( 1 a 3 ) a ) ( 1 a ) ( 1 a 3 ) β m z 0 a d a 3 m 1 a 3 1 w 1 ,
w 3 m = α d l = 0 m 1 j = 0 3 ( m l ) 2 a j β l = 0 m 1 1 z 0 d l = 0 m 1 a 3 ( m l ) 1 w 0 = α d l = 0 m 1 1 a 3 ( m l ) 1 1 a β m z 0 d a 2 a 3 m 1 a 3 1 w 0 = α d 1 a m a 2 a 3 m 1 a 3 1 β m z 0 d a 2 a 3 m 1 a 3 1 w 0 = α d ( a 3 m + 2 + m ( 1 a 3 ) a 2 ) ( 1 a ) ( 1 a 3 ) β m z 0 d a 2 a 3 m 1 a 3 1 w 0 ,
w 3 m + 1 = α d l = 0 m 1 j = 0 3 ( m l ) 1 a j β l = 0 m 1 z 0 d l = 0 m a 3 ( m l ) w 2 = α d l = 0 m 1 1 a 3 ( m l ) 1 a β m + 1 z 0 d a 3 m + 3 1 a 3 1 w 2 = α d 1 a m a 3 a 3 m 1 a 3 1 β m + 1 z 0 d a 3 m + 3 1 a 3 1 w 2 = α d ( a 3 m + 3 ( m + 1 ) a 3 + m ) ( 1 a ) ( 1 a 3 ) β m + 1 z 0 d a 3 m + 3 1 a 3 1 w 2 ,
for m N .
Case a = 1 c . From (15)–(17) and by Lemma 2, we obtain
w 3 m 1 = α d l = 0 m 1 c l j = 0 3 ( m l 1 ) 1 β l = 0 m 1 c l z 0 d l = 0 m 1 c l w 1 c m = α d l = 0 m 1 ( 3 ( m l 1 ) + 1 ) c l β 1 c m 1 c z 0 d 1 c m 1 c w 1 c m = α d ( 3 m 2 ) 1 c m 1 c 3 c 1 m c m 1 + ( m 1 ) c m ( 1 c ) 2 β 1 c m 1 c z 0 d 1 c m 1 c w 1 c m = α d ( 3 m 2 ( 3 m + 1 ) c + 2 c m + c m + 1 ) ( 1 c ) 2 β 1 c m 1 c z 0 d 1 c m 1 c w 1 c m ,
w 3 m = α d l = 0 m 1 c l j = 0 3 ( m l ) 2 1 β l = 0 m 1 c l z 0 d l = 0 m 1 c l w 0 c m = α d l = 0 m 1 ( 3 ( m l ) 1 ) c l β 1 c m 1 c z 0 d 1 c m 1 c w 0 c m = α d ( 3 m 1 ) 1 c m 1 c 3 c 1 m c m 1 + ( m 1 ) c m ( 1 c ) 2 β 1 c m 1 c z 0 d 1 c m 1 c w 0 c m = α d ( 3 m 1 ( 3 m + 2 ) c + c m + 2 c m + 1 ) ( 1 c ) 2 β 1 c m 1 c z 0 d 1 c m 1 c w 0 c m ,
w 3 m + 1 = α d l = 0 m 1 c l j = 0 3 ( m l ) 1 1 β l = 0 m c l z 0 d l = 0 m c l w 2 c m + 1 = α d l = 0 m 1 3 ( m l ) c l β 1 c m + 1 1 c z 0 d 1 c m + 1 1 c w 2 c m + 1 = α d 3 m 1 c m 1 c 3 c 1 m c m 1 + ( m 1 ) c m ( 1 c ) 2 β 1 c m + 1 1 c z 0 d 1 c m + 1 1 c w 2 c m + 1 = α 3 d ( m ( m + 1 ) c + c m + 1 ) ( 1 c ) 2 β 1 c m + 1 1 c z 0 d 1 c m + 1 1 c w 2 c m + 1 ,
for m N .
Case a = c = 1 . From (15)–(17), we obtain
w 3 m 1 = α d l = 0 m 1 j = 0 3 ( m l 1 ) 1 β l = 0 m 1 1 z 0 d l = 0 m 1 1 w 1 = α d l = 0 m 1 ( 3 ( m l 1 ) + 1 ) β m z 0 d m w 1 = α d m ( 3 m 1 ) 2 β m z 0 d m w 1 ,
w 3 m = α d l = 0 m 1 j = 0 3 ( m l ) 2 1 β l = 0 m 1 1 z 0 d l = 0 m 1 1 w 0 = α d l = 0 m 1 ( 3 ( m l ) 1 ) β m z 0 d m w 0 = α d m ( 3 m + 1 ) 2 β m z 0 d m w 0 ,
w 3 m + 1 = α d l = 0 m 1 j = 0 3 ( m l ) 1 1 β m + 1 z 0 d ( m + 1 ) w 2 = α 3 d l = 0 m 1 ( m l ) β m + 1 z 0 d ( m + 1 ) w 2 = α 3 d m ( m + 1 ) 2 β m + 1 z 0 d ( m + 1 ) w 2 ,
for m N .
From (11), (12), (18)–(32) the theorem follows. ☐
The following corollary follows from Theorem 1.
Corollary 1.
Assume that a , c , d Z , b = 0 , α , β , z 0 , w 2 , w 1 , w 0 C \ { 0 } . Then the following statements are true.
(a) 
If a 1 c a 3 , then the general solution to system (2) is given by formulas (11), (18)–(20).
(b) 
If a 1 c = a 3 , then the general solution to system (2) is given by formulas (11), (21)–(23).
(c) 
If a 1 = c , then the general solution to system (2) is given by formulas (11), (24)–(26).
(d) 
If a = 1 c , then the general solution to system (2) is given by formulas (12), (27)–(29).
(e) 
If a = c = 1 , then the general solution to system (2) is given by formulas (12), (30)–(32).
Theorem 2.
Assume that a , b , c Z , d = 0 , α , β , z 0 , w 2 , w 1 , w 0 C \ { 0 } . Then system (2) is solvable in closed form.
Proof. 
In this case we have
z n + 1 = α z n a w n b , w n + 1 = β w n 2 c , n N 0 .
Hence
w 3 n + i = β w 3 ( n 1 ) + i c ,
for n N and i = 2 , 0 ¯ , and consequently
w 3 n + i = β j = 0 n 1 c j w i c n ,
for n N and i = 2 , 0 ¯ .
When c 1 relation (35) implies
w 3 n + i = β 1 c n 1 c w i c n ,
for n N and i = 2 , 0 ¯ , and
w 3 n + i = β n w i ,
for n N and i = 2 , 0 ¯ , when c = 1 .
Further, we have
z n = α ( α z n 2 a w n 2 b ) a w n 1 b = α 1 + a z n 2 a 2 w n 2 b a w n 1 b = α 1 + a ( α z n 3 a w n 3 b ) a 2 w n 2 b a w n 1 b = α 1 + a + a 2 w n 3 b a 2 w n 2 b a w n 1 b z n 3 a 3 ,
for n 3 , which can be written as
z 3 n = α 1 + a + a 2 w 3 n 3 b a 2 w 3 n 2 b a w 3 n 1 b z 3 ( n 1 ) a 3
z 3 n + 1 = α 1 + a + a 2 w 3 n 2 b a 2 w 3 n 1 b a w 3 n b z 3 ( n 1 ) + 1 a 3
z 3 n + 2 = α 1 + a + a 2 w 3 n 1 b a 2 w 3 n b a w 3 n + 1 b z 3 ( n 1 ) + 2 a 3 ,
for n N .
Employing the method of induction we get
z 3 n = z 0 a 3 n j = 0 n 1 α 1 + a + a 2 w 3 j b a 2 w 3 j + 1 b a w 3 j + 2 b a 3 ( n j 1 )
z 3 n + 1 = z 1 a 3 n j = 0 n 1 α 1 + a + a 2 w 3 j + 1 b a 2 w 3 j + 2 b a w 3 j + 3 b a 3 ( n j 1 )
z 3 n + 2 = z 2 a 3 n j = 0 n 1 α 1 + a + a 2 w 3 j + 2 b a 2 w 3 j + 3 b a w 3 j + 4 b a 3 ( n j 1 ) ,
for n N 0 .
Using formula (35), along with the first equation in (33) with n = 0 , 1 , into (41)–(43), we get
z 3 n = z 0 a 3 n j = 0 n 1 α 1 + a + a 2 β b a 2 l = 0 j 1 c l w 0 b a 2 c j β b a l = 0 j c l w 2 b a c j + 1 β b l = 0 j c l w 1 b c j + 1 a 3 ( n j 1 ) = α j = 0 3 n 1 a j β b a 2 j = 0 n 1 a 3 ( n j 1 ) l = 0 j 1 c l + b a j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l + b j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l × z 0 a 3 n w 2 b a c j = 0 n 1 c j a 3 ( n j 1 ) w 1 b c j = 0 n 1 c j a 3 ( n j 1 ) w 0 b a 2 j = 0 n 1 c j a 3 ( n j 1 ) ,
z 3 n + 1 = ( α z 0 a w 0 b ) a 3 n j = 0 n 1 α 1 + a + a 2 β b a 2 l = 0 j c l w 2 b a 2 c j + 1 β b a l = 0 j c l w 1 b a c j + 1 β b l = 0 j c l w 0 b c j + 1 a 3 ( n j 1 ) = α j = 0 3 n a j β b a 2 j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l + b a j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l + b j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l × z 0 a 3 n + 1 w 2 b a 2 c j = 0 n 1 c j a 3 ( n j 1 ) w 1 b a c j = 0 n 1 c j a 3 ( n j 1 ) w 0 b j = 0 n c j a 3 ( n j ) ,
z 3 n + 2 = ( α 1 + a β b z 0 a 2 w 2 b c w 0 a b ) a 3 n × j = 0 n 1 α 1 + a + a 2 β b a 2 l = 0 j c l w 1 b a 2 c j + 1 β b a l = 0 j c l w 0 b a c j + 1 β b l = 0 j + 1 c l w 2 b c j + 2 a 3 ( n j 1 ) = α j = 0 3 n + 1 a j β b a 2 j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l + b a j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l + b j = 0 n a 3 ( n j ) l = 0 j c l × z 0 a 3 n + 2 w 2 b c j = 0 n c j a 3 ( n j ) w 1 b a 2 c j = 0 n 1 c j a 3 ( n j 1 ) w 0 a b j = 0 n c j a 3 ( n j ) ,
for n N 0 .
As in the proof of the previous theorem, there are five subcases, depending on the values of parameters a and c, for which we get closed form formulas for solutions to system (2).
Case a 1 c a 3 . From (44)–(46) and by Lemma 2, we have
z 3 n = α j = 0 3 n 1 a j β b a 2 j = 0 n 1 a 3 ( n j 1 ) l = 0 j 1 c l + b a j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l + b j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l × z 0 a 3 n w 2 b a c j = 0 n 1 c j a 3 ( n j 1 ) w 1 b c j = 0 n 1 c j a 3 ( n j 1 ) w 0 b a 2 j = 0 n 1 c j a 3 ( n j 1 ) = α 1 a 3 n 1 a β b a 2 j = 0 n 1 a 3 ( n j 1 ) 1 c j 1 c + b a j = 0 n 1 a 3 ( n j 1 ) 1 c j + 1 1 c + b j = 0 n 1 a 3 ( n j 1 ) 1 c j + 1 1 c × z 0 a 3 n w 2 b a c a 3 n c n a 3 c w 1 b c a 3 n c n a 3 c w 0 b a 2 a 3 n c n a 3 c = α 1 a 3 n 1 a β b ( a 2 + a + 1 ) 1 c 1 a 3 n 1 a 3 b ( a 2 + a c + c ) 1 c a 3 n c n a 3 c z 0 a 3 n w 2 b a c a 3 n c n a 3 c w 1 b c a 3 n c n a 3 c w 0 b a 2 a 3 n c n a 3 c = α 1 a 3 n 1 a β b ( a 3 c + ( c 1 ) a 3 n + 2 ( a 2 + a c + c ) ( a 1 ) c n ) ( 1 c ) ( a 3 c ) ( 1 a ) z 0 a 3 n w 2 b a c a 3 n c n a 3 c w 1 b c a 3 n c n a 3 c w 0 b a 2 a 3 n c n a 3 c ,
z 3 n + 1 = α j = 0 3 n a j β b a 2 j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l + b a j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l + b j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l × z 0 a 3 n + 1 w 2 b a 2 c j = 0 n 1 c j a 3 ( n j 1 ) w 1 b a c j = 0 n 1 c j a 3 ( n j 1 ) w 0 b j = 0 n c j a 3 ( n j ) = α 1 a 3 n + 1 1 a β b a 2 j = 0 n 1 a 3 ( n j 1 ) 1 c j + 1 1 c + b a j = 0 n 1 a 3 ( n j 1 ) 1 c j + 1 1 c + b j = 0 n 1 a 3 ( n j 1 ) 1 c j + 1 1 c × z 0 a 3 n + 1 w 2 b a 2 c a 3 n c n a 3 c w 1 b a c a 3 n c n a 3 c w 0 b a 3 n + 3 c n + 1 a 3 c = α 1 a 3 n + 1 1 a β b ( a 2 + a + 1 ) 1 c 1 a 3 n 1 a 3 b c ( a 2 + a + 1 ) 1 c a 3 n c n a 3 c z 0 a 3 n + 1 w 2 b a 2 c a 3 n c n a 3 c w 1 b a c a 3 n c n a 3 c w 0 b a 3 n + 3 c n + 1 a 3 c = α 1 a 3 n + 1 1 a β b ( a 3 c + ( c 1 ) a 3 n + 3 + ( 1 a 3 ) c n + 1 ) ( 1 c ) ( 1 a ) ( a 3 c ) z 0 a 3 n + 1 w 2 b a 2 c a 3 n c n a 3 c w 1 b a c a 3 n c n a 3 c w 0 b a 3 n + 3 c n + 1 a 3 c ,
z 3 n + 2 = α j = 0 3 n + 1 a j β b a 2 j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l + b a j = 0 n 1 a 3 ( n j 1 ) l = 0 j c l + b j = 0 n a 3 ( n j ) l = 0 j c l × z 0 a 3 n + 2 w 2 b c j = 0 n c j a 3 ( n j ) w 1 b a 2 c j = 0 n 1 c j a 3 ( n j 1 ) w 0 a b j = 0 n c j a 3 ( n j ) = α 1 a 3 n + 2 1 a β b a 2 j = 0 n 1 a 3 ( n j 1 ) 1 c j + 1 1 c + b a j = 0 n 1 a 3 ( n j 1 ) 1 c j + 1 1 c + b j = 0 n a 3 ( n j ) 1 c j + 1 1 c × z 0 a 3 n + 2 w 2 b c a 3 n + 3 c n + 1 a 3 c w 1 b a 2 c a 3 n c n a 3 c w 0 a b a 3 n + 3 c n + 1 a 3 c = α 1 a 3 n + 2 1 a β b ( a 2 + a ) 1 c 1 a 3 n 1 a 3 + b ( 1 a 3 n + 3 ) ( 1 c ) ( 1 a 3 ) b c ( a 2 + a ) 1 c a 3 n c n a 3 c b c 1 c a 3 n + 3 c n + 1 a 3 c × z 0 a 3 n + 2 w 2 b c a 3 n + 3 c n + 1 a 3 c w 1 b a 2 c a 3 n c n a 3 c w 0 a b a 3 n + 3 c n + 1 a 3 c = α 1 a 3 n + 2 1 a β b ( a 3 c + ( c 1 ) a 3 n + 4 + ( 1 a ) ( a 2 + a + c ) c n + 1 ) ( 1 c ) ( 1 a ) ( a 3 c ) z 0 a 3 n + 2 × w 2 b c a 3 n + 3 c n + 1 a 3 c w 1 b a 2 c a 3 n c n a 3 c w 0 a b a 3 n + 3 c n + 1 a 3 c ,
for n N 0 .
Case a 1 c = a 3 . From (44)–(46) and by Lemma 2, we have
z 3 n = α j = 0 3 n 1 a j β b a 2 j = 0 n 1 a 3 ( n j 1 ) l = 0 j 1 a 3 l + b a j = 0 n 1 a 3 ( n j 1 ) l = 0 j a 3 l + b j = 0 n 1 a 3 ( n j 1 ) l = 0 j a 3 l × z 0 a 3 n w 2 b a 4 j = 0 n 1 a 3 j a 3 ( n j 1 ) w 1 b a 3 j = 0 n 1 a 3 j a 3 ( n j 1 ) w 0 b a 2 j = 0 n 1 a 3 j a 3 ( n j 1 ) = α 1 a 3 n 1 a β b a 2 j = 0 n 1 a 3 ( n j 1 ) 1 a 3 j 1 a 3 + b a j = 0 n 1 a 3 ( n j 1 ) 1 a 3 j + 3 1 a 3 + b j = 0 n 1 a 3 ( n j 1 ) 1 a 3 j + 3 1 a 3 × z 0 a 3 n w 2 b n a 3 n + 1 w 1 b n a 3 n w 0 b n a 3 n 1 = α 1 a 3 n 1 a β b a 2 1 a 3 1 a 3 n 1 a 3 n a 3 n 3 + b a 1 a 3 1 a 3 n 1 a 3 n a 3 n + b 1 a 3 1 a 3 n 1 a 3 n a 3 n × z 0 a 3 n w 2 b n a 3 n + 1 w 1 b n a 3 n w 0 b n a 3 n 1 = α 1 a 3 n 1 a β b n a 3 n + 2 a 3 n n a 3 n 1 + 1 ( 1 a ) ( 1 a 3 ) z 0 a 3 n w 2 b n a 3 n + 1 w 1 b n a 3 n w 0 b n a 3 n 1 ,
z 3 n + 1 = α j = 0 3 n a j β b a 2 j = 0 n 1 a 3 ( n j 1 ) l = 0 j a 3 l + b a j = 0 n 1 a 3 ( n j 1 ) l = 0 j a 3 l + b j = 0 n 1 a 3 ( n j 1 ) l = 0 j a 3 l × z 0 a 3 n + 1 w 2 b a 5 j = 0 n 1 a 3 j a 3 ( n j 1 ) w 1 b a 4 j = 0 n 1 a 3 j a 3 ( n j 1 ) w 0 b j = 0 n a 3 j a 3 ( n j ) = α 1 a 3 n + 1 1 a β b a 2 j = 0 n 1 a 3 ( n j 1 ) 1 a 3 j + 3 1 a 3 + b a j = 0 n 1 a 3 ( n j 1 ) 1 a 3 j + 3 1 a 3 + b j = 0 n 1 a 3 ( n j 1 ) 1 a 3 j + 3 1 a 3 × z 0 a 3 n + 1 w 2 b n a 3 n + 2 w 1 b n a 3 n + 1 w 0 b ( n + 1 ) a 3 n = α 1 a 3 n + 1 1 a β b a 2 1 a 3 1 a 3 n 1 a 3 n a 3 n + b a 1 a 3 1 a 3 n 1 a 3 n a 3 n + b 1 a 3 1 a 3 n 1 a 3 n a 3 n × z 0 a 3 n + 1 w 2 b n a 3 n + 2 w 1 b n a 3 n + 1 w 0 b ( n + 1 ) a 3 n = α 1 a 3 n + 1 1 a β b ( n a 3 n + 3 ( n + 1 ) a 3 n + 1 ) ( 1 a ) ( 1 a 3 ) z 0 a 3 n + 1 w 2 b n a 3 n + 2 w 1 b n a 3 n + 1 w 0 b ( n + 1 ) a 3 n ,
z 3 n + 2 = α j = 0 3 n + 1 a j β b a 2 j = 0 n 1 a 3 ( n j 1 ) l = 0 j a 3 l + b a j = 0 n 1 a 3 ( n j 1 ) l = 0 j a 3 l + b j = 0 n a 3 ( n j ) l = 0 j a 3 l × z 0 a 3 n + 2 w 2 b a 3 j = 0 n a 3 j a 3 ( n j ) w 1 b a 5 j = 0 n 1 a 3 j a 3 ( n j 1 ) w 0 a b j = 0 n a 3 j a 3 ( n j ) = α 1 a 3 n + 2 1 a β b a 2 j = 0 n 1 a 3 ( n j 1 ) 1 a 3 j + 3 1 a 3 + b a j = 0 n 1 a 3 ( n j 1 ) 1 a 3 j + 3 1 a 3 + b j = 0 n a 3 ( n j ) 1 a 3 j + 3 1 a 3 × z 0 a 3 n + 2 w 2 b ( n + 1 ) a 3 n + 3 w 1 b n a 3 n + 2 w 0 b ( n + 1 ) a 3 n + 1 = α 1 a 3 n + 2 1 a β b a 2 1 a 3 1 a 3 n 1 a 3 n a 3 n + b a 1 a 3 1 a 3 n 1 a 3 n a 3 n + b 1 a 3 1 a 3 n + 3 1 a 3 ( n + 1 ) a 3 n + 3 × z 0 a 3 n + 2 w 2 b ( n + 1 ) a 3 n + 3 w 1 b n a 3 n + 2 w 0 b ( n + 1 ) a 3 n + 1 = α 1 a 3 n + 2 1 a β b ( ( n + 1 ) a 3 n + 4 a 3 n + 3 ( n + 1 ) a 3 n + 1 + 1 ) ( 1 a ) ( 1 a 3 ) z 0 a 3 n + 2 w 2 b ( n + 1 ) a 3 n + 3 w 1 b n a 3 n + 2 w 0 b ( n + 1 ) a 3 n + 1 ,
for n N 0 .
Case If a 1 = c . From (44)–(46) and by Lemma 2, we have
z 3 n = α j = 0 3 n 1 a j β b a 2 j = 0 n 1 j a 3 ( n j 1 ) + b a j = 0 n 1 ( j + 1 ) a 3 ( n j 1 ) + b j = 0 n 1 ( j + 1 ) a 3 ( n j 1 ) × z 0 a 3 n w 2 b a j = 0 n 1 a 3 ( n j 1 ) w 1 b j = 0 n 1 a 3 ( n j 1 ) w 0 b a 2 j = 0 n 1 a 3 ( n j 1 ) = α 1 a 3 n 1 a β b a 2 a 3 n n a 3 + n 1 ( a 3 1 ) 2 + b a a 3 n + 3 ( n + 1 ) a 3 + n ( a 3 1 ) 2 + b a 3 n + 3 ( n + 1 ) a 3 + n ( a 3 1 ) 2 × z 0 a 3 n w 2 b a 1 a 3 n 1 a 3 w 1 b 1 a 3 n 1 a 3 w 0 b a 2 1 a 3 n 1 a 3 = α 1 a 3 n 1 a β b ( a 2 + a + 1 ) a 3 n + 2 n a 5 ( n + 1 ) a 4 ( n + 1 ) a 3 + ( n 1 ) a 2 + n a + n ( a 3 1 ) 2 × z 0 a 3 n w 2 b a 1 a 3 n 1 a 3 w 1 b 1 a 3 n 1 a 3 w 0 b a 2 1 a 3 n 1 a 3 = α 1 a 3 n 1 a β b a 3 n + 2 n a 3 + n a 2 ( a 1 ) ( a 3 1 ) z 0 a 3 n w 2 b a 1 a 3 n 1 a 3 w 1 b 1 a 3 n 1 a 3 w 0 b a 2 1 a 3 n 1 a 3 ,
z 3 n + 1 = α j = 0 3 n a j β b a 2 j = 0 n 1 ( j + 1 ) a 3 ( n j 1 ) + b a j = 0 n 1 ( j + 1 ) a 3 ( n j 1 ) + b j = 0 n 1 ( j + 1 ) a 3 ( n j 1 ) × z 0 a 3 n + 1 w 2 b a 2 j = 0 n 1 a 3 ( n j 1 ) w 1 b a j = 0 n 1 a 3 ( n j 1 ) w 0 b j = 0 n a 3 ( n j ) = α 1 a 3 n + 1 1 a β b a 3 n + 3 ( n + 1 ) a 3 + n ( a 1 ) ( a 3 1 ) z 0 a 3 n + 1 w 2 b a 2 1 a 3 n 1 a 3 w 1 b a 1 a 3 n 1 a 3 w 0 b 1 a 3 n + 3 1 a 3 ,
z 3 n + 2 = α j = 0 3 n + 1 a j β b a 2 j = 0 n 1 ( j + 1 ) a 3 ( n j 1 ) + b a j = 0 n 1 ( j + 1 ) a 3 ( n j 1 ) + b j = 0 n ( j + 1 ) a 3 ( n j ) × z 0 a 3 n + 2 w 2 b j = 0 n a 3 ( n j ) w 1 b a 2 j = 0 n 1 a 3 ( n j 1 ) w 0 a b j = 0 n a 3 ( n j ) = α 1 a 3 n + 2 1 a β b a 2 a 3 n + 3 ( n + 1 ) a 3 + n ( a 3 1 ) 2 + b a a 3 n + 3 ( n + 1 ) a 3 + n ( a 3 1 ) 2 + b a 3 n + 6 ( n + 2 ) a 3 + n + 1 ( a 3 1 ) 2 × z 0 a 3 n + 2 w 2 b 1 a 3 n + 3 1 a 3 w 1 b a 2 1 a 3 n 1 a 3 w 0 a b 1 a 3 n + 3 1 a 3 = α 1 a 3 n + 2 1 a β b a 3 n + 4 + ( 1 a 3 ) n ( a 3 + a 1 ) ( a 1 ) ( a 3 1 ) z 0 a 3 n + 2 w 2 b 1 a 3 n + 3 1 a 3 w 1 b a 2 1 a 3 n 1 a 3 w 0 a b 1 a 3 n + 3 1 a 3 ,
for n N 0 .
Case If a = 1 c . From (44)–(46) and by Lemma 2, we have
z 3 n = α 3 n β b j = 0 n 1 l = 0 j 1 c l + b j = 0 n 1 l = 0 j c l + b j = 0 n 1 l = 0 j c l z 0 w 2 b c j = 0 n 1 c j w 1 b c j = 0 n 1 c j w 0 b j = 0 n 1 c j = α 3 n β b j = 0 n 1 1 c j 1 c + b j = 0 n 1 1 c j + 1 1 c + b j = 0 n 1 1 c j + 1 1 c z 0 w 2 b c 1 c n 1 c w 1 b c 1 c n 1 c w 0 b 1 c n 1 c = α 3 n β b ( 2 c n + 1 + c n 3 ( c 1 ) n 2 c 1 ) ( 1 c ) 2 z 0 w 2 b c 1 c n 1 c w 1 b c 1 c n 1 c w 0 b 1 c n 1 c ,
z 3 n + 1 = α 3 n + 1 β b j = 0 n 1 l = 0 j c l + b j = 0 n 1 l = 0 j c l + b j = 0 n 1 l = 0 j c l z 0 w 2 b c j = 0 n 1 c j w 1 b c j = 0 n 1 c j w 0 b j = 0 n c j = α 3 n + 1 β 3 b j = 0 n 1 1 c j + 1 1 c z 0 w 2 b c 1 c n 1 c w 1 b c 1 c n 1 c w 0 b 1 c n + 1 1 c = α 3 n + 1 β 3 b c n + 1 ( c 1 ) n c ( 1 c ) 2 z 0 w 2 b c 1 c n 1 c w 1 b c 1 c n 1 c w 0 b 1 c n + 1 1 c ,
z 3 n + 2 = α 3 n + 2 β b j = 0 n 1 l = 0 j c l + b j = 0 n 1 l = 0 j c l + b j = 0 n l = 0 j c l × z 0 w 2 b c j = 0 n c j w 1 b c j = 0 n 1 c j w 0 b j = 0 n c j = α 3 n + 2 β b j = 0 n 1 1 c j + 1 1 c + b j = 0 n 1 1 c j + 1 1 c + b j = 0 n 1 c j + 1 1 c × z 0 w 2 b c 1 c n + 1 1 c w