Producttype system of difference equations with a complex structure of solutions
 Stevo Stević^{1, 2}Email author
https://doi.org/10.1186/s1366201711906
© The Author(s) 2017
Received: 28 February 2017
Accepted: 24 April 2017
Published: 18 May 2017
Abstract
Keywords
system of difference equations producttype system solvable in closed formMSC
39A20 39A451 Introduction
Various types of difference equations and systems have been investigated in the last twenty years [1–31]. Many of the systems studied there, such as the ones in [2, 3, 5–9, 11–13, 17–20, 22–26, 28–31], were essentially obtained as some sorts of symmetrization of scalar ones, and were studied first by Papaschinopoulos and Schinas. One of the basic problems investigated in the field is the solvability (see, e.g., [32–36] for widely known methods). For some recent results, see, e.g., [2, 10, 14, 16–24, 26–31]. Recently several papers presenting formulas for solutions to some difference equations and systems, but without mentioning any theory, have appeared. A theoretical explanation of formulas for such an equation given in our note [14] has attracted some attention. The note shows that the equation is closely related to a known solvable one, and the main idea has been used and developed later in many papers (see, for example, [2, 10, 16, 21, 26, 27] and the references therein).
An important fact connected to the investigation is that there are only several solvable producttype systems with two dependent variables due to the finite number of combinations of the sums of two delays which cannot exceed four. Our general task is to present all solvable producttype systems with two dependent variables.
2 Auxiliary results
Here we quote three lemmas which we employ in the section that follows. The first one contains what was proved in [37], formulated in a compact form.
Lemma 1
 (a)
If \(\Delta <0\), then two zeros of \(P_{4}\) are real and different, and two are complex conjugate;
 (b)If \(\Delta >0\), 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 \(\Delta =0\), then and only then \(P_{4}\) has a multiple zero. The following cases can occur:
 1^{∘} :

if \(P<0\), \(D<0\) and \(\Delta _{0}\ne0\), then two zeros of \(P_{4}\) are real and equal and two are real and simple;
 2^{∘} :

if \(D>0\) or (\(P>0\) and (\(D\ne0\) or \(Q\ne0\))), then two zeros of \(P_{4}\) are real and equal and two are complex conjugate;
 3^{∘} :

if \(\Delta _{0}=0\) and \(D\ne0\), 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 \(\Delta _{0}=0\), then all four zeros of \(P_{4}\) are real and equal to \(b/4\).
Two different proofs of the following known lemma can be found, for example, in [28] and [34].
Lemma 2
The following lemma contains some known summation formulas which we employ in this paper (see, e.g., [34, 36]). For a general method for calculating this type of sums, see [23].
Lemma 3
3 Main results
This section formulates and proves our main results. The first two results deal with the case when \(c=0\) and one of the parameters b and d is also zero.
Theorem 1
 (a)If \(a\ne1\), then the solution to system (1) is given by$$\begin{aligned} &z_{n}=\alpha ^{\frac{1a^{n}}{1a}}z_{0}^{a^{n}},\quad n\in {\mathbb {N}}, \end{aligned}$$(7)$$\begin{aligned} &w_{n}=\beta \alpha ^{d\frac{1a^{n3}}{1a}}z_{0}^{da^{n3}},\quad n \ge4. \end{aligned}$$(8)
 (b)If \(a=1\), then the solution to system (1) is given by$$\begin{aligned} &z_{n}=\alpha ^{n} z_{0}, \quad n\in {\mathbb {N}}, \end{aligned}$$(9)$$\begin{aligned} &w_{n}=\beta \alpha ^{d(n3)}z_{0}^{d},\quad n \ge4. \end{aligned}$$(10)
Proof
Theorem 2
 (a)If \(a\ne1\), then the solution to system (1) is given by$$\begin{aligned} &z_{n}=\alpha ^{\frac{1a^{n}}{1a}}\beta ^{b\frac {1a^{n1}}{1a}}z_{0}^{a^{n}}w_{0}^{ba^{n1}},\quad n\ge2, \end{aligned}$$(14)$$\begin{aligned} &w_{n}=\beta ,\quad n\in {\mathbb {N}}. \end{aligned}$$(15)
 (b)
Proof
Theorem 3
Assume that \(a,b,c,d\in {\mathbb {Z}}\), \(ac\ne bd\), \(\alpha ,\beta , w_{2}, w_{1}, w_{0}, z_{2}\), \(z_{1}, z_{0}\in {\mathbb {C}}\setminus\{0\}\). Then system (1) is solvable in closed form.
Proof
From (25), (26), (30)(32) and the induction, we conclude that (27)(29) hold for \(k,n\in {\mathbb {N}}\) such that \(2\le k\le n1\).
It is clear that closedform formulas for solutions to problem (34)(35) can be easily found, from which along with (37) and Lemma 3 closedform formulas for \(y_{n}\) can also be found, from which along with (33) the solvability of system (21) follows.
From this and since the closedform formulas for \(a_{k}\) and \(y_{k}\) can be found as above, the solvability of (40) follows. It is easily checked that (33) and (43) present a solution to system (1). □
Corollary 1
Assume that \(a,b,c,d\in {\mathbb {Z}}\), \(ac\ne bd\), \(\alpha ,\beta , w_{2}, w_{1}, w_{0}, z_{2}\), \(z_{1}, z_{0}\in {\mathbb {C}}\setminus\{0\}\). Then the general solution to system (1) is given by (33) and (43), where \((a_{k})_{k\ge3}\) satisfies (34) and (35), while \((y_{k})_{k\ge3}\) is given by (36) and (37).
From this and by Corollary 1, we get the following result.
Corollary 2
 (a)
If \((a1)(c1)\ne bd\), then the general solution to system (1) is given by (33) and (43), where \((a_{n})_{n\ge3}\) is given by (64), \((y_{n})_{n\ge3}\) is given by (65), while \(\lambda _{j}\), \(j=\overline {1,4}\), are given by (52)(55).
 (b)
If \((a1)(c1)=bd\) and \(43ac\ne0\), then the general solution to system (1) is given by (33) and (43), where \((a_{n})_{n\ge3}\) is given by (64) with \(\lambda _{1}=1\), \((y_{n})_{n\ge 3}\) is given by (73), \(\lambda _{1}=1\), while \(\lambda _{j}\), \(j=\overline {2,4}\), are given by (70) and (71).
Corollary 3
 (a)
If only one of the zeros of characteristic polynomial (44) is double, say \(\lambda _{1}=\lambda _{2}\) and \((a1)(c1)\ne bd\), then the general solution to system (1) is given by (33) and (43), where \((a_{n})_{n\ge3}\) is given by (75), while \((y_{n})_{n\ge3}\) is given by (76).
 (b)
If only a double zero of characteristic polynomial (44) is equal to 1, say \(\lambda _{1}=\lambda _{2}=1\), then the general solution to system (1) is given by (33) and (43), where \((a_{n})_{n\ge3}\) is given by (80), \((y_{n})_{n\ge3}\) is given by (81), while \(\lambda _{3,4}\) are given by (79).
Hence, for every \(a\ne0\), polynomial \(p_{4}\) has two pairs of equal zeros. It will have integer coefficients only if \(a=4\hat{a}\), for some \(\hat{a}\in {\mathbb {Z}}\setminus\{0\}\). Note that since \(a(1\pm\sqrt {3})/4\ne1\), for every \(a\in {\mathbb {Z}}\), 1 cannot be a double zero of \(p_{4}\).
Corollary 4
 (a)
If \((a1)(c1)\ne bd\), then the general solution to system (1) is given by (33) and (43), where \((a_{n})_{n\ge3}\) is given by (86), \((y_{n})_{n\ge3}\) is given by (87), while \(\lambda _{2,4}\) are given by (84).
 (b)
Characteristic polynomial (44) cannot have two pairs of double zeros such that one of them is equal to 1.
Hence, for every \(a\ne0\), \(a/2\) is a triple zero of \(p_{4}\), and \(p_{4}\) cannot have a zero of the fourth order. For \(a=2\hat{a}\), \(\hat{a}\in {\mathbb {Z}}\setminus\{0\}\), polynomial \(p_{4}\) has integer coefficients, and for \(a=2\) it has a triple zero equal to 1, which could be also obtained by further analyzing the polynomial in (78).
Corollary 5
 (a)
If the triple zero of polynomial (44) is different from 1, then the general solution to system (1) is given by (33) and (43), where \((a_{n})_{n\ge3}\) is given by (91), \((y_{n})_{n\ge3}\) is given by (93), \(\lambda _{j}=a/2\), \(j=\overline {1,3}\) and \(\lambda _{4}=a/2\).
 (b)
If the triple zero of polynomial (44) is equal to 1, say \(\lambda _{1}=\lambda _{2}=\lambda _{3}=1\), which is equivalent to \(a=2\), \(c=2\) and \(bd=3\), then the general solution to system (1) is given by (33) and (43), where \((a_{n})_{n\ge3}\) is given by (90), \((y_{n})_{n\ge3}\) is given by (92), while \(\lambda _{4}=1\).
Now we deal with the case \(acbd=0\), \(c\ne0\).
Theorem 4
Assume that \(a,b,c,d\in {\mathbb {Z}}\), \(ac=bd\), \(c\ne0\) and \(\alpha ,\beta , z_{2}, z_{1}, z_{0}, w_{2}\), \(w_{1}, w_{0}\in {\mathbb {C}}\setminus\{0\}\). Then system (1) is solvable in closed form.
Proof
From (97), (98), (102)(104) and the induction is obtained that (99)(101) hold for \(k, n\in {\mathbb {N}}\) such that \(2\le k\le n1\).
It is clear that closedform formulas for solutions to (106) can be easily found, from which along with (109) and Lemma 3 closedform formulas for \(y_{n}\) can be also found, from which along with (105) the solvability of (94) follows.
From this and since the closedform formulas for \((a_{k})_{k\ge2}\) and \((y_{k})_{k\ge2}\) can be found as above, the solvability of (110) follows. It is easily checked that (105) and (112) present a solution to system (1) in this case. □
All zeros of \(p_{3}\) are different and none of them is equal to 1. In this case it must be \(\Delta\ne0\), that is, \(\Delta_{1}^{2}\ne 4\Delta_{0}^{3}\), from which it follows that \(4a^{6}\ne(2a^{3}+27c)^{2}\), which is equivalent to \(c(4a^{3}+27c)\ne0\). Hence, if \(0\ne c\ne4a^{3}/27\), then all the zeros of \(p_{3}\) are different. If, additionally, \(p_{3}(1)\ne 0\), that is, \(a+c\ne1\), then none of them is equal to 1. For example, such a situation appears if \(a=c=k\in {\mathbb {N}}\).
Corollary 6
 (a)
If \(a+c\ne1\), then the general solution to (1) is given by (105) and (112), where \((a_{n})_{n\ge2}\) is given by (120), \((y_{n})_{n\ge2}\) is given by (122), while \(\lambda _{j}\), \(j=\overline {1,3}\) are given by (116).
 (b)
If one of the zeros of characteristic polynomial (113) is equal to 1, say \(\lambda _{3}\), i.e., if \(a+c=1\) and \(2a\ne3\), then the general solution to (1) is given by formulas (105) and (112), where \((a_{n})_{n\ge2}\) is given by (120) with \(\lambda _{3}=1\), \((y_{n})_{n\ge2}\) is given by (123), and \(\lambda _{1,2}\) are given by (117).
Corollary 7
 (a)
If \(a+c\ne1\), then the general solution to (1) is given by (105) and (112), where \((a_{n})_{n\ge2}\) is given by (125), \((y_{n})_{n\ge2}\) is given by (127), \(\lambda _{1}=a/3\) and \(\lambda _{2,3}=2a/3\).
 (b)
If only one of the zeros of polynomial (113) is equal to one, say \(\lambda _{1}\), that is, if \(a=3\) and \(c=4\), then the general solution to system (1) is given by (105) and (112), where \((a_{n})_{n\ge2}\) is given by (125) with \(\lambda _{1}=1\), \((y_{n})_{n\ge2}\) is given by (128), \(\lambda _{1}=1\) and \(\lambda _{2,3}=2\).
 (c)
It is not possible that two zeros of polynomial (113) are equal to one.
Triple zero case. In this case it must be \(p_{3}(\lambda )=p_{3}'(\lambda )=p_{3}''(\lambda )=0\). From \(p_{3}''(\lambda )=0\) it is obtained that \(\lambda =a/3\). Since \(p_{3}'(\lambda )=3\lambda ^{2}2a\lambda \), we see that \(a/3\) is its root only if \(a=0\), which would imply that \(p_{3}(\lambda )=\lambda ^{3}c\), but this polynomial has a triple zero if and only if \(c=0\), which contradicts the assumption \(c\ne0\). Hence, polynomial (113) cannot have a triple zero.
Declarations
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Authors’ Affiliations
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