Jump to ContentJump to Main Navigation
Show Summary Details
More options …

Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco


IMPACT FACTOR 2018: 6.636

CiteScore 2018: 5.03

SCImago Journal Rank (SJR) 2018: 3.215
Source Normalized Impact per Paper (SNIP) 2018: 3.225

Mathematical Citation Quotient (MCQ) 2017: 1.89

Open Access
Online
ISSN
2191-950X
See all formats and pricing
More options …

Solvability of a product-type system of difference equations with six parameters

Stevo Stević
  • Corresponding author
  • Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, 11000 Beograd, Serbia; and Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 40402, Taiwan, Republic of China; and Department of Healthcare Administration, Asia University, 500 Lioufeng Rd., Wufeng, Taichung 41354, Taiwan, Republic of China
  • Email
  • Other articles by this author:
  • De Gruyter OnlineGoogle Scholar
Published Online: 2016-12-21 | DOI: https://doi.org/10.1515/anona-2016-0145

Abstract

Closed form formulas for well-defined complex-valued solutions to a product-type system of difference equations of interest with six parameters are presented. The form of the solutions is described in detail in terms of the parameters and initial values.

Keywords: System of difference equations; product-type system; closed form formula

MSC 2010: 39A20; 39A45

1 Introduction

Difference equations and systems have been studied a lot recently (see, e.g., [1, 2, 3, 4, 5, 7, 8, 10, 12, 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]). About two decades ago Papaschinopoulos and Schinas started investigating concrete symmetric and related systems of difference equations (see, e.g., [14, 15, 16]). The investigation later motivated numerous authors to conduct research in this direction (see, e.g., [3, 7, 10, 18, 19, 23, 26, 27, 28, 29, 30, 33, 34, 35, 36, 37, 38] and the references therein). On the other hand, the problem of solvability of difference equations and systems has re-attracted some recent attention, especially after the publication of our note [20] which explains a formula in [5] and the publication of [21] in which an open problem from population biology was solved by showing the solvability of the difference equation appearing there (see, e.g., [1, 2, 3, 17, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38]). The ideas in [20] were later developed in [24] and [25] for some related equations, and in [23] and [34] for some related systems. In some of these papers, among others, have appeared some new classes of solvable symmetric or closely related systems of difference equations, as well as solvable cyclic ones [30, 34]. Some max-type systems are solvable too [26]. Classical methods for solving difference equations and systems can be found, e.g., in [9, 11, 13].

Some recent papers have also studied equations and systems whose right-hand sides contain product-type expressions (see, e.g., [22, 33] and the references therein). In the study of the equations and systems in [22] and [33] their methodological connection with the product-type ones was essentially noticed. One of the reasons for this is that some product-type equations and systems are obtained by taking the limits of some parameters in the equations and systems in [22] and [33]. It is well known that the product-type systems of difference equations are solvable in some cases (e.g., if initial values are positive), but, in general, this is not always the case. The problem appears if some initial values are not nonnegative, because power functions are multi-valued in the complex plain in many cases. These facts have motivated us to investigate the solvability of some concrete product-type systems of difference equations with complex initial values in detail. More precisely, the problem is to find systems for which it is possible to find explicit formulas for all their solutions.

Recent papers [29, 30, 32, 37, 38] deal with this problem (this problem appeared in [31] too, as some special cases of the equation treated therein). Papers [29, 32, 37, 38] study some two-dimensional product-type systems of difference equations with different delays, while [30] studies a three-dimensional. For the above-mentioned problem connected to multi-valued functions, all these systems could not be dealt with the transformation methods presented in some of our previous papers such as [3, 20, 21, 23, 24, 25, 27, 34, 35, 36]. Hence, we have been developing some other methods for studying the product-type systems, related to some in [22] and [33], and have some coefficients as the systems in [32] and [38].

Here we continue the study of practical solvability of product-type systems, by investigating the solvability of the following one:

zn+1=αznawnb,wn+1=βwn-1cznd,n0,(1.1)

where a,b,c,d, α,β and z0,w-1,w0, in detail.

If α=0 or β=0, then solutions of (1.1) are trivial or not well-defined. Hence, the case α0β is of some interest and it will be treated in the rest of the paper.

Note that the domain of undefinable solutions [27] to (1.1) is a subset of the set

𝒰={(z0,w-1,w0)3:z0=0 or w-1=0 or w0=0}.

If min{a,b,c}<0 or min{b,c,d}<0, then the domain is 𝒰. Hence, from now on we will also assume that z0,w-1,w0{0}, although many obtained formulas will hold on some larger sets or even on the whole 3 (for example, if min{a,b,c,d}>0).

It is said that the system of the form

zn=f(zn-1,,zn-k,wn-1,,wn-l),wn=g(zn-1,,zn-s,wn-1,,wn-t),n0,

where k,l,s,t, is solvable in closed form if its general solution can be found in terms of initial values z-i, i=1,max{k,s}¯, w-j, j=1,max{l,t}¯, delays k, l, s, t, and index n only. As usual, if m>n we regard that k=mnak is equal to zero.

2 Auxiliary results

In this section we present two auxiliary results which will be used in the proofs of the main results. The following elementary lemma is known (see, e.g., [13]).

Lemma 1.

Let iN0 and

sn(i)(z)=1+2iz+3iz2++nizn-1,n,(2.1)

where zC. Then

sn(0)(z)=1-zn1-z,(2.2)sn(1)(z)=1-(n+1)zn+nzn+1(1-z)2,(2.3)sn(2)(z)=1+z-(n+1)2zn+(2n2+2n-1)zn+1-n2zn+2(1-z)3(2.4)

for every zC{1} and nN.

Formula (2.2) is something which should be known to any mathematician, while (2.3) and (2.4) can be obtained by using the following result, which seems not so well known.

Proposition 2.

Sequences defined in (2.1) satisfy the following recurrent relation:

sn(k)(z)=i=0k-1Cik(-1)k-1-isn(i)(z)-nkzn1-z(2.5)

for every kN and zC{0}.

Proof.

We have

sn(k)(z)-zsn(k)(z)=1+j=2n(jk-(j-1)k)zj-1-nkzn=1+j=2ni=0k-1Cik(-1)k-1-ijizj-1-nkzn=1+i=0k-1Cik(-1)k-1-ij=2njizj-1-nkzn=Ckk+i=0k-1Cik(-1)k-1-i(sn(i)(z)-1)-nkzn=i=0kCik(-1)k-i+i=0k-1Cik(-1)k-1-isn(i)(z)-nkzn=i=0k-1Cik(-1)k-1-isn(i)(z)-nkzk,

where we have used the fact i=0kCik(-1)k-i=0, from which (2.5) follows.∎

Example 3.

If k=2, then from (2.5) and some calculations it follows that

sn(2)(z)=-C02sn(0)(z)+C12sn(1)(z)-n2zn1-z=-1-zn1-z+21-(n+1)zn+nzn+1(1-z)2-n2zn1-z=1+z-(n+1)2zn+(2n2+2n-1)zn+1-n2zn+2(1-z)3.

The following result is a known consequence of the Lagrange interpolation formula, but can be also proved by using some other techniques (see, e.g., [37]).

Lemma 4.

Let

Pk(z)=fkzk+fk-1zk-1++f1z+f0.

If the zeros zj, j=1,,k, of Pk are mutually different, then

j=1kzjlPk(zj)=0

for each l{0,1,,k-2}, and

j=1kzjk-1Pk(zj)=1fk.

3 Main results

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

Theorem 1.

Assume that b,c,dZ, a=0, α,βC{0}, and z0,w-1,w0C{0}. Then system (1.1) is solvable in closed form.

Proof.

Since a=0, system (1.1) is

zn+1=αwnb,wn+1=βwn-1cznd,n0.(3.1)

Using the first equation in (3.1) in the second one, we obtain

wn+1=βαdwn-1c+bd,n.(3.2)

Iterating relation (3.2), we can easily show that

w2n=(βαd)j=0n-1(c+bd)jw0(c+bd)n=αdj=0n-1(c+bd)jβj=0n-1(c+bd)jw0(c+bd)n(3.3)

and

w2n+1=(βαd)j=0n-1(c+bd)jw1(c+bd)n=(βαd)j=0n-1(c+bd)j(βw-1cz0d)(c+bd)n=αdj=0n-1(c+bd)jβj=0n(c+bd)jw-1c(c+bd)nz0d(c+bd)n(3.4)

for every n.

Employing (3.3) and (3.4) into the first equation in (3.1), we get

z2n+1=α1+bdj=0n-1(c+bd)jβbj=0n-1(c+bd)jw0b(c+bd)n(3.5)

and

z2n+2=α1+bdj=0n-1(c+bd)jβbj=0n(c+bd)jw-1bc(c+bd)nz0bd(c+bd)n(3.6)

for every n.

If c+bd1, then from (3.3)–(3.6) and (2.2) we have

z2n+1=α1-c-bd(c+bd)n1-c-bdβb1-(c+bd)n1-c-bdw0b(c+bd)n,(3.7)z2n+2=α1-c-bd(c+bd)n1-c-bdβb1-(c+bd)n+11-c-bdw-1bc(c+bd)nz0bd(c+bd)n,(3.8)w2n=αd1-(c+bd)n1-c-bdβ1-(c+bd)n1-c-bdw0(c+bd)n,(3.9)w2n+1=αd1-(c+bd)n1-c-bdβ1-(c+bd)n+11-c-bdw-1c(c+bd)nz0d(c+bd)n(3.10)

for every n, while if c+bd=1, then we have

z2n+1=α1+bdnβbnw0b,(3.11)z2n+2=α1+bdnβb(n+1)w-1bcz0bd,(3.12)w2n=αdnβnw0,(3.13)w2n+1=αdnβn+1w-1cz0d(3.14)

for every n, finishing the proof of the theorem.∎

Corollary 2.

Consider system (1.1) with b,c,dZ, a=0 and α,βC{0}. Furthermore assume that z0,w-1,w0C{0}. Then the following statements are true:

  • (i)

    If c+bd1 , then the general solution to system ( 1.1 ) is given by ( 3.7 )–( 3.10 ).

  • (ii)

    If c+bd=1 , then the general solution to system ( 1.1 ) is given by ( 3.11 )–( 3.14 ).

Theorem 3.

Assume that a,b,dZ, c=0, α,βC{0}, and z0,w0C{0}. Then system (1.1) is solvable in closed form.

Proof.

In this case system (1.1) becomes

zn+1=αznawnb,wn+1=βznd,n0.(3.15)

The system essentially appeared in [29], but we will give a proof for the sake of completeness. Using the second equation in (3.15) in the first one, we obtain

zn+1=αβbznazn-1bd,n.(3.16)

Now we consider the subcases bd=0 and bd0 separately.

Subcase bd=0. In this case we have

zn=αβbzn-1a,n2,

from which it follows that

zn=(αβb)j=0n-2ajz1an-1=(αβb)j=0n-2aj(αz0aw0b)an-1=αj=0n-1ajβbj=0n-2ajz0anw0ban-1,n2.(3.17)

Using (3.17) in the second equation in (3.15), we get

wn=β(αj=0n-2ajβbj=0n-3ajz0an-1w0ban-2)d=αdj=0n-2ajβz0dan-1,n3.(3.18)

A direct calculation shows that (3.18) holds also for n=2.

From (3.17) and (3.18) we have that

zn=α1-an1-aβb1-an-11-az0anw0ban-1,n2,(3.19)wn=αd1-an-11-aβz0dan-1,n2,(3.20)

in the case a1, while in the case a=1 we have

zn=αnβb(n-1)z0w0b,n2,(3.21)wn=αd(n-1)βz0d,n2.(3.22)

Subcase bd0. Let

γ:=αβb,a1:=a,b1:=bd,x1:=1.

Then (3.16) can be written as

zn+1=γx1zna1zn-1b1,n,(3.23)

from which it follows that

zn+1=γx1(γzn-1a1zn-2b1)a1zn-1b1=γx1+a1zn-1a1a1+b1zn-2b1a1=γx2zn-1a2zn-2b2(3.24)

for n2, where

a2:=a1a1+b1,b2:=b1a1,x2:=x1+a1.(3.25)

Assume that for some k2 we have

zn+1=γxkzn+1-kakzn-kbk,nk,(3.26)

where

ak:=a1ak-1+bk-1,bk:=b1ak-1,xk:=xk-1+ak-1.(3.27)

Then, using (3.23) with nn-k into (3.26), we obtain

zn+1=γxk(γzn-ka1zn-k-1b1)akzn-kbk=γxk+akzn-ka1ak+bkzn-k-1b1ak=γxk+1zn-kak+1zn-k-1bk+1(3.28)

for nk+1, where

ak+1:=a1ak+bk,bk+1:=b1ak,xk+1:=xk+ak.(3.29)

From (3.24), (3.25), (3.28), (3.29), and the induction we see that (3.26) and (3.27) hold for every k and n such that 2kn ((3.26) also holds for k=1 because of (3.23)).

Now note that from the first two equations in (3.27) we get

ak=a1ak-1+b1ak-2,k3.(3.30)

The equalities in (3.27) with k=1 yield

a1=a1a0+b0,b1=b1a0,x1=x0+a0.(3.31)

Since b1=bd0, from the second equation in (3.31) we get a0=1. This, along with x1=1 and the other two relations in (3.31) implies b0=x0=0.

From this and (3.27) with k=0, we obtain

1=a0=a1a-1+b-1,0=b0=b1a-1,0=x0=x-1+a-1.(3.32)

Since b10, from the second equation in (3.32) we get a-1=0. This, along with the other two relations in (3.32), implies b-1=1 and x-1=0.

From this and the second equation in (3.27), we have that (ak)k-1 and (bk)k-1 are solutions to linear equation (3.30) satisfying the initial conditions

a-1=0,a0=1,andb-1=1,b0=0,

respectively, and that (xk)k-1 satisfies the third recurrent relation in (3.27) and

x-1=x0=0,x1=1.

From (3.26) with nn-1 and k=n-1, along with the equality z1=αz0aw0b and the first and third relations in (3.27), we have that

zn=γxn-1z1an-1z0bn-1=(αβb)xn-1(αz0aw0b)an-1z0bn-1=αan-1+xn-1βbxn-1z0aan-1+bn-1w0ban-1=αxnβbxn-1z0anw0ban-1,n0.(3.33)

Using this in the second equation in (3.15), we obtain

wn=αdxn-1β1+bdxn-2z0dan-1w0bdan-2,n.(3.34)

From the third equation in (3.27) and since x1=1, we get

xn=1+j=1n-1aj,n,

which due to the fact a0=1 can be written as

xn=j=0n-1aj,n.(3.35)

Now note that the characteristic equation associated to difference equation (3.30) is λ2-aλ-bd=0, from which it follows that

λ1,2=a±a2+4bd2.

Hence if a2+4bd0, then

an=c1λ1n+c2λ2n.

From this and since a-1=0 and a0=1, we have that

an=λ1n+1-λ2n+1λ1-λ2,(3.36)

which along with the second equation in (3.27) implies

bn=bdλ1n-λ2nλ1-λ2.

Using (3.36) in (3.35) for the case when λ11λ2, which is equivalent to a+bd1, we get

xn=j=0n-1λ1j+1-λ2j+1λ1-λ2=1(λ1-λ2)(λ1λ1n-1λ1-1-λ2λ2n-1λ2-1)=(λ2-1)λ1n+1-(λ1-1)λ2n+1+λ1-λ2(λ1-1)(λ2-1)(λ1-λ2),(3.37)

while if a+bd=1, that is, if one of the characteristic roots is equal to one, say λ2, which implies that λ1=-bd, we get

xn=j=0n-1λ1j+1-1λ1-1=1(λ1-1)(λ1λ1n-1λ1-1-n)=λ1n+1-(n+1)λ1+n(λ1-1)2=(-bd)n+1+(n+1)bd+n(1+bd)2.(3.38)

If a2+4bd=0, then

an=(c^1+c^2n)(a2)n.

From this and since a-1=0 and a0=1, we have that

an=(n+1)(a2)n,(3.39)

which along with the second equation in (3.27) and bd=-a2/4, implies

bn=bdn(a2)n-1=-n(a2)n+1.

Using (3.39) in (3.35) and employing (2.3) for the case a2, we get

xn=j=0n-1(j+1)(a2)j=1-(n+1)(a2)n+n(a2)n+1(1-a2)2,(3.40)

while if a=2, we get

xn=j=0n-1(j+1)=n(n+1)2,(3.41)

as desired.∎

Corollary 4.

Consider system (1.1) with a,b,dZ, c=0 and α,βC{0}. Assume that z0,w0C{0}. Then the following statements are true:

  • (i)

    If bd=0 and a1 , then the general solution to system ( 1.1 ) is given by ( 3.19 ) and ( 3.20 ).

  • (ii)

    If bd=0 and a=1 , then the general solution to system ( 1.1 ) is given by ( 3.21 ) and ( 3.22 ).

  • (iii)

    If bd0, a2+4bd0 and a+bd1 , then the general solution to system ( 1.1 ) is given by formulas ( 3.33 ) and ( 3.34 ), where the sequences (an)n-1 and (xn)n-1 are given by formulas ( 3.36 ) and ( 3.37 ), respectively.

  • (iv)

    If bd0, a2+4bd0 and a+bd=1 , then the general solution to system ( 1.1 ) is given by formulas ( 3.33 ) and ( 3.34 ), where the sequences (an)n-1 and (xn)n-1 are given by formulas ( 3.36 ) and ( 3.38 ), respectively.

  • (v)

    If bd0, a2+4bd=0 and a2 , then the general solution to system ( 1.1 ) is given by formulas ( 3.33 ) and ( 3.34 ), where the sequences (an)n-1 and (xn)n-1 are given by formulas ( 3.39 ) and ( 3.40 ), respectively.

  • (vi)

    If a2+4bd=0 and a=2 , then the general solution to system ( 1.1 ) is given by formulas ( 3.33 ) and ( 3.34 ), where the sequence (an)n-1 is given by formula ( 3.39 ) with a=2 , while (xn)n-1 is given by ( 3.41 ).

Theorem 5.

Assume that a,c,dZ, b=0, α,βC{0}, and z0,w-1,w0C{0}. Then system (1.1) is solvable in closed form.

Proof.

In this case system (1.1) becomes

zn+1=αzna,wn+1=βwn-1cznd,n0.(3.42)

From the first equation in (3.42) we get

zn=αj=0n-1ajz0an,n,(3.43)

which for the case a1 implies that

zn=α1-an1-az0an,n,(3.44)

while for the case a=1 it implies that

zn=αnz0,n.(3.45)

Employing (3.43) in the second equation in (3.42), we obtain

wn+1=βαdj=0n-1ajz0danwn-1c,n.(3.46)

Using (3.46) twice, we obtain

w2n=βαdj=02n-2ajz0da2n-1w2n-2c=βαdj=02n-2ajz0da2n-1(βαdj=02n-4ajz0da2n-3w2n-4c)c=β1+cαdj=02n-2aj+dcj=02n-4ajz0da2n-1+dca2n-3w2n-4c2,(3.47)w2n+1=βαdj=02n-1ajz0da2nw2n-1c=βαdj=02n-1ajz0da2n(βαdj=02n-3ajz0da2n-2w2n-3c)c=β1+cαdj=02n-1aj+dcj=02n-3ajz0da2n+dca2n-2w2n-3c2,n2.(3.48)

Assume that for a natural number k the equations

w2n=βj=0k-1cjαdi=0k-1cij=02n-2i-2ajz0dj=0k-1cja2n-2j-1w2n-2kck,(3.49)w2n+1=βj=0k-1cjαdi=0k-1cij=02n-2i-1ajz0dj=0k-1cja2n-2jw2n-2k+1ck(3.50)

for every nk have been proved.

By using (3.46) with n2n-2k-1 and n2n-2k in (3.49) and (3.50), we get

w2n=βj=0k-1cjαdi=0k-1cij=02n-2i-2ajz0dj=0k-1cja2n-2j-1(βαdj=02n-2k-2ajz0da2n-2k-1w2n-2k-2c)ck=βj=0kcjαdi=0kcij=02n-2i-2ajz0dj=0kcja2n-2j-1w2n-2k-2ck+1,(3.51)w2n+1=βj=0k-1cjαdi=0k-1cij=02n-2i-1ajz0dj=0k-1cja2n-2j(βαdj=02n-2k-1ajz0da2n-2kw2n-2k-1c)ck=βj=0kcjαdi=0kcij=02n-2i-1ajz0dj=0kcja2n-2jw2n-2k-1ck+1(3.52)

for every nk+1.

From (3.47), (3.48), (3.51), (3.52), and induction it follows that (3.49) and (3.50) hold for all k,n such that 1kn. Taking k=n in (3.49) and (3.50), we get

w2n=βj=0n-1cjαdi=0n-1cij=02n-2i-2ajz0dj=0n-1cja2n-2j-1w0cn,w2n+1=βj=0n-1cjαdi=0n-1cij=02n-2i-1ajz0dj=0n-1cja2n-2jw1cn=βj=0n-1cjαdi=0n-1cij=02n-2i-1ajz0dj=0n-1cja2n-2j(βw-1cz0d)cn=βj=0ncjαdi=0n-1cij=02n-2i-1ajz0dj=0ncja2n-2jw-1cn+1

for n.

Subcase a1c, ca2. By multiple use of (2.2) and some calculation, we have

w2n=β1-cn1-cαdi=0n-1ci1-a2n-2i-11-az0daa2n-cna2-cw0cn=β1-cn1-cαd1-a(1-cn1-c-aa2n-cna2-c)z0daa2n-cna2-cw0cn=β1-cn1-cαd(a2-c+(a+c)(1-a)cn-(1-c)a2n+1)(1-a)(1-c)(a2-c)z0daa2n-cna2-cw0cn,n,(3.53)w2n+1=β1-cn+11-cαdi=0n-1ci1-a2n-2i1-az0da2n+2-cn+1a2-cw-1cn+1=β1-cn+11-cαd1-a(1-cn1-c-a2a2n-cna2-c)z0da2n+2-cn+1a2-cw-1cn+1=β1-cn+11-cαd(a2-c+(1-a2)cn+1-(1-c)a2n+2)(1-a)(1-c)(a2-c)z0da2n+2-cn+1a2-cw-1cn+1(3.54)

for every n0.

Subcase a1c, c=a20. In this case we have

w2n=βj=0n-1a2jαdi=0n-1a2ij=02n-2i-2ajz0dj=0n-1a2ja2n-2j-1w0a2n=β1-a2n1-a2αdi=0n-1a2i1-a2n-2i-11-az0dna2n-1w0a2n=β1-a2n1-a2αd1-a(1-a2n1-a2-na2n-1)z0dna2n-1w0a2n=β1-a2n1-a2αd(1-na2n-1-a2n+na2n+1)(a-1)2(a+1)z0dna2n-1w0a2n,n,(3.55)w2n+1=βj=0na2jαdi=0n-1a2ij=02n-2i-1ajz0dj=0na2ja2n-2jw-1a2n+2=β1-a2n+21-a2αdi=0n-1a2i1-a2n-2i1-az0d(n+1)a2nw-1a2n+2=β1-a2n+21-a2αd1-a(1-a2n1-a2-na2n)z0d(n+1)a2nw-1a2n+2=β1-a2n+21-a2αd(1-(n+1)a2n+na2n+2)(a+1)(a-1)2z0d(n+1)a2nw-1a2n+2(3.56)

for every n0.

Subcase a21=c. In this case we have

w2n=βnαdi=0n-11-a2n-2i-11-az0ada2n-1a2-1w0=βnαd1-a(n-aa2n-1a2-1)z0ada2n-1a2-1w0=βnαd(a2n+1+n(1-a2)-a)(a-1)2(a+1)z0ada2n-1a2-1w0,n,(3.57)w2n+1=βj=0n1αdi=0n-1j=02n-2i-1ajz0dj=0na2n-2jw-1=βn+1αdi=0n-11-a2n-2i1-az0da2n+2-1a2-1w-1=βn+1αd1-a(n-a2a2n-1a2-1)z0da2n+2-1a2-1w-1=βn+1αd(a2n+2+n(1-a2)-a2)(a-1)2(a+1)z0da2n+2-1a2-1w-1(3.58)

for every n0.

Subcase a=-1, c=1. In this case we have

w2n=βj=0n-11αdi=0n-1j=02n-2i-2(-1)jz0dj=0n-1(-1)2n-2j-1w0=βnαdi=0n-11-(-1)2n-2i-12z0dj=0n-1(-1)-1w0=βnαndz0-dnw0(3.59)

for every n, and

w2n+1=βj=0n1αdi=0n-1j=02n-2i-1(-1)jz0dj=0n(-1)2n-2jw-1=βn+1αdi=0n-11-(-1)2n-2i2z0dj=0n1w-1=βn+1z0d(n+1)w-1(3.60)

for every n0.

Subcase a=1, c1. In this case, using (2.3), we have

w2n=β1-cn1-cαdi=0n-1(2n-2i-1)ciz0d1-cn1-cw0cn=β1-cn1-cαd((2n-1)1-cn1-c-2c1-ncn-1+(n-1)cn(1-c)2)z0d1-cn1-cw0cn=β1-cn1-cαd(2n-1-(2n+1)c+cn+cn+1)(1-c)2z0d1-cn1-cw0cn(3.61)

for n, and

w2n+1=β1-cn+11-cαdi=0n-1(2n-2i)ciz0d1-cn+11-cw-1cn+1=β1-cn+11-cαd(2n1-cn1-c-2c1-ncn-1+(n-1)cn(1-c)2)z0d1-cn+11-cw-1cn+1=β1-cn+11-cα2d(n-(n+1)c+cn+1)(1-c)2z0d1-cn+11-cw-1cn+1(3.62)

for every n0.

Subcase a=c=1. In this case we have

w2n=βnαdi=0n-1(2n-2i-1)z0dnw0=βnαdn2z0dnw0(3.63)

for n, and

w2n+1=βn+1αdi=0n-1(2n-2i)z0d(n+1)w-1=βn+1αdn(n+1)z0d(n+1)w-1(3.64)

for every n0.∎

Corollary 6.

Consider system (1.1) with a,c,dZ, b=0 and α,βC{0}. Furthermore assume that z0,w-1,w0C{0}. Then the following statements are true:

  • (i)

    If a1c and ca2 , then the general solution to system ( 1.1 ) is given by ( 3.44 ), ( 3.53 ) and ( 3.54 ).

  • (ii)

    If a1c and c=a20 , then the general solution to system ( 1.1 ) is given by ( 3.44 ), ( 3.55 ) and ( 3.56 ).

  • (iii)

    If a21=c , then the general solution to system ( 1.1 ) is given by ( 3.44 ), ( 3.57 ) and ( 3.58 ).

  • (iv)

    If a=-1 and c=1 , then the general solution to system ( 1.1 ) is given by ( 3.44 ), ( 3.59 ) and ( 3.60 ).

  • (v)

    If a=1 and c1 , then the general solution to system ( 1.1 ) is given by ( 3.45 ), ( 3.61 ) and ( 3.62 ).

  • (vi)

    If a=c=1 , then the general solution to system ( 1.1 ) is given by ( 3.45 ), ( 3.63 ) and ( 3.64 ).

Theorem 7.

Assume that a,b,cZ, d=0, α,βC{0}, and z0,w-1,w0C{0}. Then system (1.1) is solvable in closed form.

Proof.

In this case system (1.1) becomes

zn+1=αznawnb,wn+1=βwn-1c,n0.(3.65)

From the second equation in (3.65) it easily follows that

w2n=βj=0n-1cjw0cn,w2n+1=βj=0ncjw-1cn+1,n0,(3.66)

which for the case c1 implies that

w2n=β1-cn1-cw0cn,w2n+1=β1-cn+11-cw-1cn+1,n0,(3.67)

while for the case c=1 we have

w2n=βnw0,w2n+1=βn+1w-1,n0.(3.68)

Employing (3.66) in the first equation in (3.65), we obtain

z2n=αβbj=0n-1cjw-1bcnz2n-1a,(3.69)z2n+1=αβbj=0n-1cjw0bcnz2na,n.(3.70)

Combining (3.69) and (3.70), we have that

z2n=αβbj=0n-1cjw-1bcn(αβbj=0n-2cjw0bcn-1z2n-2a)a=α1+aβbj=0n-1cj+abj=0n-2cj(w-1bcw0ab)cn-1z2n-2a2(3.71)

for n2, and

z2n+1=αβbj=0n-1cjw0bcn(αβbj=0n-1cjw-1bcnz2n-1a)a=α1+aβb(1+a)j=0n-1cj(w-1abw0b)cnz2n-1a2(3.72)

for every n.

Assume that for some natural number k we have proved that

z2n=α(1+a)j=0k-1a2jβbi=0k-1a2i(j=0n-i-1cj+aj=0n-i-2cj)(w-1bcw0ab)j=0k-1a2jcn-j-1z2n-2ka2k(3.73)

and

z2n+1=α(1+a)j=0k-1a2jβb(1+a)i=0k-1a2ij=0n-i-1cj(w-1abw0b)j=0k-1a2jcn-jz2n-2k+1a2k(3.74)

for every nk.

Using (3.71) with nn-k into (3.73), and (3.72) with nn-k into (3.74), we have that

z2n=α(1+a)j=0k-1a2jβbi=0k-1a2i(j=0n-i-1cj+aj=0n-i-2cj)(w-1bcw0ab)j=0k-1a2jcn-j-1×(α1+aβbj=0n-k-1cj+abj=0n-k-2cj(w-1bcw0ab)cn-k-1z2n-2k-2a2)a2k=α(1+a)j=0ka2jβbi=0ka2i(j=0n-i-1cj+aj=0n-i-2cj)(w-1bcw0ab)j=0ka2jcn-j-1z2n-2k-2a2k+2(3.75)

and

z2n+1=α(1+a)j=0k-1a2jβb(1+a)i=0k-1a2ij=0n-i-1cj(w-1abw0b)j=0k-1a2jcn-j(α1+aβb(1+a)j=0n-k-1cj(w-1abw0b)cn-kz2n-2k-1a2)a2k=α(1+a)j=0ka2jβb(1+a)i=0ka2ij=0n-i-1cj(w-1abw0b)j=0ka2jcn-jz2n-2k-1a2k+2(3.76)

for every nk+1.

From (3.71), (3.72), (3.75), (3.76), and induction we have that (3.73) and (3.74) hold for all k,n such that 1kn.

If we choose k=n in (3.73) and (3.74), we get

z2n=α(1+a)j=0n-1a2jβbi=0n-1a2i(j=0n-i-1cj+aj=0n-i-2cj)(w-1bcw0ab)j=0n-1a2jcn-j-1z0a2nz2n+1=α(1+a)j=0n-1a2jβb(1+a)i=0n-1a2ij=0n-i-1cj(w-1abw0b)j=0n-1a2jcn-jz1a2n=α(1+a)j=0n-1a2jβb(1+a)i=0n-1a2ij=0n-i-1cj(w-1abw0b)j=0n-1a2jcn-j(αz0aw0b)a2n=αj=02najβb(1+a)i=0n-1a2ij=0n-i-1cjw-1abj=0n-1a2jcn-jw0bj=0na2jcn-jz0a2n+1

for every n.

Subcase ca21c. In this case we have

z2n=α1-a2n1-aβbi=0n-1a2i(1-cn-i1-c+a1-cn-i-11-c)(w-1bcw0ab)cn-a2nc-a2z0a2n=α1-a2n1-aβb1-c(1-a2n1-a-(c+a)cn-a2nc-a2)(w-1bcw0ab)cn-a2nc-a2z0a2n=α1-a2n1-aβb(c-a2+(a-1)(a+c)cn+(1-c)a2n+1)(1-a)(1-c)(c-a2)(w-1bcw0ab)cn-a2nc-a2z0a2n,(3.77)z2n+1=α1-a2n+11-aβb(1+a)i=0n-1a2i1-cn-i1-cw-1abccn-a2nc-a2w0bcn+1-a2n+2c-a2z0a2n+1=α1-a2n+11-aβb(1+a)1-c(1-a2n1-a2-ccn-a2nc-a2)w-1abccn-a2nc-a2w0bcn+1-a2n+2c-a2z0a2n+1=α1-a2n+11-aβb(1+a)(c-a2+(a2-1)cn+1+(1-c)a2n+2)(1-c)(1-a2)(c-a2)w-1abccn-a2nc-a2w0bcn+1-a2n+2c-a2z0a2n+1(3.78)

for every n.

Subcase a21c, c=a2. In this case we have

z2n=α1-a2n1-aβbi=0n-1a2i(1-a2n-2i1-a2+a1-a2n-2i-21-a2)(w-1ba2w0ab)na2n-2z0a2n=α1-a2n1-aβb1-a2(1-a2n1-a-(a+1)na2n-1)(w-1ba2w0ab)na2n-2z0a2n=α1-a2n1-aβb(1-na2n-1-a2n+na2n+1)(1-a)2(1+a)(w-1ba2w0ab)na2n-2z0a2n,(3.79)z2n+1=α1-a2n+11-aβb(1+a)i=0n-1a2i1-a2n-2i1-a2w-1bna2n+1w0b(n+1)a2nz0a2n+1=α1-a2n+11-aβb(1+a)(1-(n+1)a2n+na2n+2)(1-a2)2w-1bna2n+1w0b(n+1)a2nz0a2n+1(3.80)

for every n.

Subcase a21=c. In this case, using (2.3), we have

z2n=α(1+a)j=0n-1a2jβbi=0n-1a2i(j=0n-i-11+aj=0n-i-21)(w-1bw0ab)j=0n-1a2jz0a2n=α1-a2n1-aβbi=0n-1a2i((1+a)n-a-(1+a)i)(w-1bw0ab)1-a2n1-a2z0a2n=α1-a2n1-aβb(((1+a)n-a)1-a2n1-a2-a2-na2n+(n-1)a2n+2(1-a2)(1-a))(w-1bw0ab)1-a2n1-a2z0a2n=α1-a2n1-aβb(n-a-na2+a2n+1)(1-a2)(1-a)(w-1bw0ab)1-a2n1-a2z0a2n,n,(3.81)z2n+1=αj=02najβb(1+a)i=0n-1a2i(n-i)w-1abj=0n-1a2jw0bj=0na2jz0a2n+1=α1-a2n+11-aβb(1+a)(n1-a2n1-a2-a21-na2n-2+(n-1)a2n(1-a2)2)w-1ab1-a2n1-a2w0b1-a2n+21-a2z0a2n+1=α1-a2n+11-aβb(1+a)(n-(n+1)a2+a2n+2)(1-a2)2w-1ab1-a2n1-a2w0b1-a2n+21-a2z0a2n+1(3.82)

for every n0.

Subcase a=-1, c=1. In this case we have

z2n=βbi=0n-1(-1)2i(j=0n-i-11-j=0n-i-21)(w-1bw0-b)j=0n-1(-1)2jz0(-1)2n=βbn(w-1bw0-b)nz0,(3.83)z2n+1=αj=02n(-1)jw-1-bj=0n-1(-1)2jw0bj=0n(-1)2jz0(-1)2n+1=αw-1-bnw0b(n+1)z-1-1(3.84)

for every n.

Subcase a=1c. In this case we have

z2n=α2nβbi=0n-1(1-cn-i1-c+1-cn-i-11-c)(w-1bcw0b)1-cn1-cz0=α2nβb1-ci=0n-1(2n-(c+1)1-cn1-c)(w-1bcw0b)1-cn1-cz0=α2nβb(2n-1-(2n+1)c+cn+cn+1)(1-c)2(w-1bcw0b)1-cn1-cz0,(3.85)z2n+1=α2n+1β2bi=0n-11-cn-i1-cw-1bc1-cn1-cw0b1-cn+11-cz0=α2n+1β2b(n-(n+1)c+cn+1)(1-c)2w-1bc1-cn1-cw0b1-cn+11-cz0(3.86)

for every n.

Subcase a=c=1. In this case we have

z2n=α2j=0n-11βbi=0n-1(j=0n-i-11+j=0n-i-21)(w-1bw0b)j=0n-11z0=α2nβbi=0n-1(2n-2i-1)(w-1bw0b)nz0=α2nβbn2(w-1w0)bnz0,(3.87)z2n+1=αj=02n1β2bi=0n-1j=0n-i-11w-1bj=0n-11w0bj=0n1z0=α2n+1β2bi=0n-1(n-i)w-1bnw0b(n+1)z0=α2n+1βbn(n+1)w-1bnw0b(n+1)z0(3.88)

for every n, completing the proof.∎

Corollary 8.

Consider system (1.1) with a,b,cZ, d=0 and α,βC{0}. Furthermore assume that z0,w-1,w0C{0}. Then the following statements are true:

  • (i)

    If ca21c , then the general solution to system ( 1.1 ) is given by ( 3.67 ), ( 3.77 ) and ( 3.78 ).

  • (ii)

    If c=a21c , then the general solution to system ( 1.1 ) is given by ( 3.67 ), ( 3.79 ) and ( 3.80 ).

  • (iii)

    If a21=c , then the general solution to system ( 1.1 ) is given by ( 3.68 ), ( 3.81 ) and ( 3.82 ).

  • (iv)

    If a=-1 and c=1 , then the general solution to system ( 1.1 ) is given by ( 3.68 ), ( 3.83 ) and ( 3.84 ).

  • (v)

    If a=1 and c1 , then the general solution to system ( 1.1 ) is given by ( 3.67 ), ( 3.85 ) and ( 3.86 ).

  • (vi)

    If a=c=1 , then the general solution to system ( 1.1 ) is given by ( 3.68 ), ( 3.87 ) and ( 3.88 ).

Theorem 9.

Assume that a,b,c,dZ, abcd0, α,βC{0}, and suppose that z0,w-1,w0C{0}. Then system (1.1) is solvable in closed form.

Proof.

The conditions α,β{0} and z0,w-1,w0{0} along with (1.1) imply zn0wn, n. From the first equation in (1.1), for such a solution, we have

wnb=zn+1αzna,n0,(3.89)

while by taking the second equation in (1.1) to the b-th power it follows that

wn+1b=βbwn-1bcznbd,n0.(3.90)

Putting (3.89) into (3.90), we easily obtain

zn+2=α1-cβbzn+1aznbd+czn-1-ac,n,(3.91)

which is a third order product-type difference equation.

Note also that

z1=αz0aw0bandz2=α(αz0aw0b)a(βw-1cz0d)b=α1+aβbz0a2+bdw0abw-1bc.(3.92)

Let

δ:=α1-cβb,a1=a,b1=bd+c,c1=-ac,y1=1.(3.93)

Then equation (3.91) can be written as

zn+2=δy1zn+1a1znb1zn-1c1,n.(3.94)

Putting (3.94) with nn-1 into itself, we get

zn+2=δy1(δzna1zn-1b1zn-2c1)a1znb1zn-1c1=δy1+a1zna1a1+b1zn-1b1a1+c1zn-2c1a1=δy2zna2zn-1b2zn-2c2(3.95)

for n2, where

a2:=a1a1+b1,b2:=b1a1+c1,c2:=c1a1,y2=y1+a1.(3.96)

Assume that for a k2 we have proved that

zn+2=δykzn+2-kakzn+1-kbkzn-kck(3.97)

for nk, and that

ak=a1ak-1+bk-1,bk=b1ak-1+ck-1,ck=c1ak-1,(3.98)yk=yk-1+ak-1,k2.(3.99)

Then, putting relation (3.94) with nn-k into (3.97), we obtain

zn+2=δyk(δzn+1-ka1zn-kb1zn-k-1c1)akzn+1-kbkzn-kck=δyk+akzn+1-ka1ak+bkzn-kb1ak+ckzn-k-1c1ak=δyk+1zn+1-kak+1zn-kbk+1zn-k-1ck+1(3.100)

for nk+1, where

ak+1:=a1ak+bk,bk+1:=b1ak+ck,ck+1:=c1ak,yk+1=yk+ak.(3.101)

From (3.95), (3.96), (3.100), (3.101), and using induction, we have that (3.97)–(3.99) hold for every k,n such that 2kn. Moreover, due to (3.94), equation (3.97) holds also for k=1.

Choosing k=n in (3.97) and using (3.92), (3.98) and (3.99), we have

zn+2=δynz2anz1bnz0cn=(α1-cβb)yn(α1+aβbz0a2+bdw0abw-1bc)an(αz0aw0b)bnz0cn=α(1-c)yn+(1+a)an+bnβb(yn+an)z0(a2+bd)an+abn+cnw-1bcanw0aban+bbn=αyn+2-cynβbyn+1z0an+2-canw-1bcanw0ban+1(3.102)

for n.

From (3.98) we have that (ak)k4 satisfies the following recurrent relation:

ak=a1ak-1+b1ak-2+c1ak-3.(3.103)

Since bk-1=ak-a1ak-1 and ck=c1ak-1, noting that equation (3.103) is linear, we have that (bk)k and (ck)k also satisfy this equation.

Moreover, ak, bk and ck satisfy (3.103) for k-2. Indeed, from (3.101) with k=0 we get

a1=a1a0+b0,b1=b1a0+c0,c1=c1a0,1=y1=y0+a0.(3.104)

Since c1=-ac0, from the third equation in (3.104) we get a0=1. Using this fact in the other three equalities in (3.104), we get b0=c0=y0=0.

From this and by (3.101) with k=-1 we get

1=a0=a1a-1+b-1,0=b0=b1a-1+c-1,0=c0=c1a-1,0=y0=y-1+a-1.(3.105)

Since c10, from the third equation in (3.105) we get a-1=0. Using this fact in the other three equalities in (3.105), we get b-1=1, c-1=y-1=0.

From this and by (3.101) with k=-2 we get

0=a-1=a1a-2+b-2,1=b-1=b1a-2+c-2,0=c-1=c1a-2,0=y-1=y-2+a-2.(3.106)

Since c10, from the third equation in (3.106) we get a-2=0. Using this fact in the other three equalities in (3.106), we get b-2=y-2=0 and c-2=1

Hence, the sequences (ak)k-2, (bk)k-2 and (ck)k-2 are solutions to linear difference equation (3.103) satisfying the initial conditions

a-2=0,a-1=0,a0=1,b-2=0,b-1=1,b0=0,c-2=1,c-1=0,c0=0,(3.107)

respectively, and (yk)k-2 satisfies recurrent relation (3.99) and

y-2=y-1=y0=0,y1=1.

From (3.99) and since a0=1, we have that

yk=1+j=1k-1aj=j=0k-1aj,k0.(3.108)

Since difference equation (3.103) is solvable, it follows that closed form formulas for (ak)k-2, (bk)k-2 and (ck)k-2 can be found. From this fact, (3.102), (3.108), and since the sum j=0maj, m0, can be calculated by using Lemma 1, we see that equation (3.91) is solvable too.

From the second equation in (1.1) we have that

znd=wn+1βwn-1c,n0,(3.109)

for every well-defined solution, while by taking the first equation in (1.1) to the d-th power, it follows that

zn+1d=αdznadwnbd,n0.(3.110)

Putting (3.109) into (3.110), we easily obtain

wn+2=αdβ1-awn+1awnbd+cwn-1-ac,n0,(3.111)

which is a related difference equation to (3.91) (only with a different coefficient). Note also that (wn)n0 satisfies the following initial condition:

w1=βw-1cz0d.(3.112)

Hence, the above presented procedure for the sequence zn can be repeated and we obtain that for every k such that 1kn we have

wn+2=ηy^kwn+2-kakwn+1-kbkwn-kck,nk-1,(3.113)

where η=αdβ1-a and (ak)k, (bk)k and (ck)k satisfy recurrent relations (3.98) with initial conditions (3.93), and

y^k+1=y^k+ak,k,

with y^1=1, from which it follows that y^k is also given by formula (3.108).

From (3.113) with k=n+1 and by using (3.112), we get

wn+2=ηy^n+1w1an+1w0bn+1w-1cn+1=(αdβ1-a)y^n+1(βw-1cz0d)an+1w0bn+1w-1cn+1=αdy^n+1βy^n+2-ay^n+1z0dan+1w0an+2-aan+1w-1can+1-acan,n0.(3.114)

Note that (ak)k, (bk)k and (ck)k satisfy (3.103), and, as above, since c10, they can be prolonged for k=-2,-1,0, respectively, so that they satisfy (3.107), and (y^k)k can be prolonged also for k=-2,-1,0 and y^-2=y^-1=y^0=0.

The solvability of equation (3.103) shows that closed form formulas for (ak)k-2, (bk)k-2 and (ck)k-2 can be found, from which, along with (3.108) and by Lemma 1, closed form formulas for (y^k)k-2 can be found. These facts along with (3.114) imply that equation (3.111) is solvable too. Direct but time-consuming calculation shows that the sequences zn and wn given by formulas (3.102) and (3.114) are solutions to system (1.1). Hence, system (1.1) is also solvable in this case, finishing the proof of the theorem.∎

Corollary 10.

Consider system (1.1) with a,b,c,dZ, abcd0 and α,βC{0}. Assume further that z0,w-1,w0C{0}. Then the general solution to system (1.1) is given by (3.102) and (3.114), where the sequence (ak)k-2 satisfies the difference equation (3.103) with initial conditions (3.107), and the sequences (yk)k-2 and (y^k)k-2 are both given by (3.108).

Remark 11.

Note that if ac0, then by using the recurrent relations in (3.98) and (3.99) it can be shown, similar as in the proof of Theorem 9, that the sequences ak, bk, ck can be prolonged for every negative index k.

The characteristic polynomial associated to equation (3.103) is

P3(λ)=λ3-a1λ2-b1λ-c1,(3.115)

where a1, b1 and c1 are defined in (3.93). In the case ac0, polynomial (3.115) is of the third degree, so by the Cardano formula (see, e.g., [6]) the zeros of (3.115) are

λ1=a3+B-4A3+B23323+B+4A3+B23323,λ2=a3-(1+i3)B-4A3+B23623-(1-i3)B+4A3+B23623,λ3=a3-(1-i3)B-4A3+B23623-(1+i3)B+4A3+B23623,

where

A:=-(a2+3bd+3c),B:=2a3+9abd-18ac.

Let

Δ:=4A3+B2.

Then if Δ>0, one zero of (3.115) is real and the other two are complex conjugate. If Δ=0, all the zeros are real and at least two of them are equal, while if Δ<0, all the zeros are real and different (see, e.g., [6]).

Case Δ0. Since Δ0, all the zeros λi, i=1,3¯, of polynomial (3.115) are mutually different, and the general solution to (3.103) has the form

un=α1λ1n+α2λ2n+α3λ3n,n,(3.116)

where αi, i=1,3¯, are arbitrary constants. Since for the case c10 the solution can be prolonged for every nonpositive index, we may assume that formula (3.116) holds also for n-3.

From Lemma 4 with P3(t)=j=13(t-λj) we have

j=13λjlP3(λj)=0for l=0,1,

and

j=13λj2P3(λj)=1.

This along with a-2=a-1=0 and a0=1 implies that

an=λ1n+2(λ1-λ2)(λ1-λ3)+λ2n+2(λ2-λ1)(λ2-λ3)+λ3n+2(λ3-λ1)(λ3-λ2)(3.117)

for n-2.

On the other hand, from (3.98) we get

bn=an+1-a1an,(3.118)cn=c1an-1(3.119)

for n-2.

Putting (3.117) and (3.93) into (3.118), we obtain

bn=j=13λj-aP3(λj)λjn+2

for n-2.

Putting (3.117), which also holds for n=-3, and (3.93) into (3.119), we obtain

cn=-j=13acP3(λj)λjn+1

for n-2.

From (3.108) and (3.117) we have

yn=i=0n-1ai=i=0n-1(λ1i+2(λ1-λ2)(λ1-λ3)+λ2i+2(λ2-λ1)(λ2-λ3)+λ3i+2(λ3-λ1)(λ3-λ2))(3.120)

for every n.

Recall also that

y^n=yn,n.(3.121)

Now assume that λi1, i=1,2,3. Then, from formula (3.120), it follows that

yn=y^n=Rn(1)(λ1,λ2,λ3)(3.122)

for n, where

Rn(1)(λ1,λ2,λ3)=λ12(λ1n-1)(λ1-λ2)(λ1-λ3)(λ1-1)+λ22(λ2n-1)(λ2-λ1)(λ2-λ3)(λ2-1)+λ32(λ3n-1)(λ3-λ1)(λ3-λ2)(λ3-1).

It is easy to verify that formula (3.122) holds also for every n-2.

If one of the zeros is equal to one, say λ3, then 1λ1λ21, and we have

yn=y^n=Rn(2)(λ1,λ2)(3.123)

for n, where

Rn(2)(λ1,λ2)=λ12(λ1n-1)(λ1-λ2)(λ1-1)2+λ22(λ2n-1)(λ2-λ1)(λ2-1)2+n(λ1-1)(λ2-1).

It is also easy to verify that formula (3.122) holds also for every n-2.

From the above consideration and Theorem 9 we obtain the following corollary for the case abcd0 and Δ0.

Corollary 12.

Consider system (1.1) with a,b,c,dZ and abcd0. Assume z0,w-1,w0C{0} and Δ0. Then the following statements are true:

  • (i)

    If none of the zeros of characteristic polynomial ( 3.115 ) is equal to one, i.e., if P3(1)0 , then the general solution to ( 1.1 ) is given by formulas ( 3.102 ) and ( 3.114 ), where the sequence (an)n-2 , is given by ( 3.117 ), while (yn)n-2 and (y^n)n-2 are given by ( 3.122 ).

  • (ii)

    If only one of the zeros of characteristic polynomial ( 3.115 ) is equal to one, say λ3 , i.e., if P3(1)=0P3(1) , then the general solution to ( 1.1 ) is given by formulas ( 3.102 ) and ( 3.114 ), where the sequence (an)n-2 is given by ( 3.117 ), while (yn)n-2 and (y^n)n-2 are given by ( 3.123 ).

Remark 13.

Equation (3.115) will have a zero equal to one if

P3(1)=1-a-bd-c+ac=0,

that is, if (a-1)(c-1)=bd so that

P3(λ)=λ3-aλ2-(ac-a+1)λ+ac.

If bd=0, then b=0 or d=0, which implies a=1 or c=1. Assume that a=1 and c1. Then

Δ=-4(1+3c)3+4(1-9c)2=-108c(c-1)2

and

P3(λ)=λ3-λ2-cλ+c=(λ-1)(λ2-c).

Thus, if a=1, b=0, c{0,1}, d, we have that Δ0, so the conditions of Corollary 12 (ii) are satisfied, only one zero of polynomial (3.115) is equal to one and all three zeros are mutually different. Moreover, if c>0 we have that Δ<0 and all three zeros are real and different, while if c<0 we have that Δ>0 and all three zeros are different but two are complex conjugate.

Case Δ=0. If Δ=0 and ac0, then at least two zeros of characteristic polynomial (3.115) are equal, say, λ2 and λ3. It is easy to see that the polynomial would have three equal zeros only if B=0, which along with Δ=0 would also imply A=0. In this case there must be P3(λ1)=P3(λ1)=P3′′(λ1)=0. Hence, from the equality P3′′(λ1)=0, it follows that λj=a/3, j=1,2,3, and consequently

P3(a/3)=-2a3+9abd-18ac27=0andP3(a/3)=-a2+3bd+3c3=0,

which immediately implies 2a3+9abd-18ac=a2+3bd+3c=0. From these two relations it easily follows that bd=8c, and consequently a2=-27c. Thus, we have

P3(λ)=λ3-aλ2+a23λ-a327=(λ-a3)3.

Hence, if a=9a^ for some a^{0}, we have that c=-3a^2{0}, and b,d{0} can be chosen so that bd=8c=-24a^2. Thus, there are polynomials of the form in (3.115) such that all three zeros are equal.

For a=c=1, b=0, d we have that Δ=0 and

P3(λ)=(λ+1)(λ-1)2,

so that there are polynomials with three real zeros such that two of them are equal and different from the third one.

If λ1λ2=λ3, then the general solution of (3.103) has the form

an=α^1λ1n+(α^2+α^3n)λ2n,n,(3.124)

where α^1, α^2 and α^3 are arbitrary constants. Since in our case the conditions a-2=a-1=0 and a0=1 must be satisfied, the solution (an)n-2 of equation (3.103) can be found by letting λ3λ2 in (3.117).

We have

an=limλ3λ2(λ1n+2(λ1-λ2)(λ1-λ3)+λ2n+2(λ2-λ1)(λ2-λ3)+λ3n+2(λ3-λ1)(λ3-λ2))=λ1n+2-(n+2)λ1λ2n+1+(n+1)λ2n+2(λ2-λ1)2

for n-2, that is,

an=λ1n+2+(λ2-2λ1+n(λ2-λ1))λ2n+1(λ2-λ1)2(3.125)

for n-2.

Note that a-2=a-1=0 and a0=1, and that (3.125) is of the form (3.124) with

α^1=λ12(λ2-λ1)2,α^2=λ22-2λ1λ2(λ2-λ1)2,α^3=λ2λ2-λ1.

Using relations (3.125) in (3.118) and (3.119), we get

bn=(λ1-a)λ1n+2(λ2-λ1)2+(λ2(2λ2-3λ1)-a(λ2-2λ1)+n(λ2-λ1)(λ2-a))λ2n+1(λ2-λ1)2,cn=-acλ1n+1+(-λ1+n(λ2-λ1))λ2n(λ2-λ1)2,n-2.

From (3.108) and (3.125) we have

yn=j=0n-1aj=j=0n-1λ1j+2+(λ2-2λ1+j(λ2-λ1))λ2j+1(λ2-λ1)2,n.(3.126)

If we assume λ11λ2=λ3, then (3.126), Lemma 1 and (3.121) imply

yn=y^n=Rn(3)(λ1,λ2)(3.127)

for every n, where

Rn(3)(λ1,λ2)=λ12(λ1n-1)(λ2-λ1)2(λ1-1)+(λ2-2λ1)λ2(λ2n-1)(λ2-λ1)2(λ2-1)+λ22(1-nλ2n-1+(n-1)λ2n)(λ2-λ1)(λ2-1)2.

A direct calculation shows that formulas (3.127) hold also for every n-2.

If we assume that λ11 and λ2=λ3=1, then from (3.126) it follows that

yn=y^n=Rn(4)(λ1)(3.128)

for every n, where

Rn(4)(λ1)=λ12(λ1n-1)(λ1-1)3+(1-2λ1)n(λ1-1)2+(n-1)n2(1-λ1).

A direct calculation shows that formulas (3.128) hold also for every n-2.

If we assume that λ1=1 and λ2=λ31, then from (3.126) it follows that

yn=y^n=dRn(5)(λ2)(3.129)

for every n, where

Rn(5)(λ2)=n(λ2-1)2+(λ2-2)λ2(λ2n-1)(λ2-1)3+λ22(1-nλ2n-1+(n-1)λ2n)(λ2-1)3.

A direct calculation shows that formulas (3.129) hold also for every n-2.

If λ1=λ2=λ3, then the general solution of (3.103) has the form

an=(β^1+β^2n+β^3n2)λ1n,n,(3.130)

where β^1, β^2 and β^3 are arbitrary constants.

To find the solution of equation (3.103) satisfying the conditions a-2=a-1=0 and a0=1, we will let λ2λ1 in formula (3.125). We have

an=limλ2λ1λ1n+2-(n+2)λ1λ2n+1+(n+1)λ2n+2(λ2-λ1)2=limλ2λ1(λ2-λ1)((n+1)λ2n+1-λ1i=0nλ2iλ1n-i)(λ2-λ1)2=limλ2λ1i=0n(λ2n+1-λ2iλ1n+1-i)λ2-λ1=limλ2λ1i=0nλ2i(λ2n+1-i-λ1n+1-i)λ2-λ1=limλ2λ1i=0nλ2ij=0n-iλ2jλ1n-i-j=(n+1)(n+2)2λ1n(3.131)

for n-2.

Note that a-2=a-1=0 and a0=1, and that (3.131) is of the form (3.130) with

β^1=1,β^2=32,β^3=12.

Using relations (3.131), as well as the condition c=-a2/27 in (3.118) and (3.119), we get

bn=n+22(3λ1-a+n(λ1-a))λ1n=-n(n+2)(a3)n+1,cn=-acn(n+1)2λ1n-1=n(n+1)2(a3)n+2

for n-2.

From (3.108) and (3.131) we have

yn=j=0n-1aj=j=0n-1(j+1)(j+2)2λ1j(3.132)

for every n.

If we assume that λ1=λ2=λ31, then from (3.132), Lemma 1 and (3.121) it follows that

yn=y^n=2sn(0)+3λ1sn-1(1)+λ1sn-1(2)2=2-(n+1)(n+2)λ1n+2n(n+2)λ1n+1-n(n+1)λ1n+22(1-λ1)3(3.133)

for every n. A simple calculation shows that (3.133) also holds for n=-2,-1,0.

If we assume λ1=λ2=λ3=1, then from (3.132) and (3.121) it easily follows that

yn=y^n=Rn(6)(1):=j=0n-1(j+1)(j+2)2=n(n+1)(n+2)6(3.134)

for every n.

Note that from (3.134) it immediately follows that Rj(6)(1)=0, j=-2,-1,0.

From the above consideration and Theorem 9 we obtain the following corollary for the case abcd0 and Δ=0.

Corollary 14.

Consider system (1.1) with a,b,c,dZ and abcd0. Assume that z0,w-1,w0C{0}, abcd0 and Δ=0. Then the following statements are true:

  • (i)

    If none of the zeros of characteristic polynomial ( 3.115 ) is equal to one, i.e., if P3(1)0 , and if it has two different zeros, then the general solution to ( 1.1 ) is given by formulas ( 3.102 ) and ( 3.114 ), where the sequence (an)n-2 is given by ( 3.125 ), while (yn)n-2 and (y^n)n-2 are given by ( 3.127 ).

  • (ii)

    If exactly two of the zeros of characteristic polynomial ( 3.115 ) are equal to one, say λ2 and λ3 , i.e., if P3(1)=P3(1)=0P3′′(1) , then the general solution to system ( 1.1 ) is given by formulas ( 3.102 ) and ( 3.114 ), where the sequence (an)n-2 is given by ( 3.125 ), while (yn)n-2 and (y^n)n-2 are given by ( 3.128 ).

  • (iii)

    If all three zeros of characteristic polynomial ( 3.115 ) are equal, then the general solution to system ( 1.1 ) is given by formulas ( 3.102 ) and ( 3.114 ), where the sequence (an)n-2 is given by ( 3.131 ), while (yn)n-2 and (y^n)n-2 are given by ( 3.133 ).

Remark 15.

As we know, polynomial (3.115) has a zero equal to one if (a-1)(c-1)=bd. If λ=1 is a double zero of (3.115), then it must be P3(1)=3-2a-c-bd=0, which is possible only if ac=2-a, that is, if a{±1,±2} and c=(2/a)-1. Note that in this case

P3(λ)=λ3-aλ2+(2a-3)λ+2-a.

If a=1, then c=1 and bd=0, which is not a case treated in Theorem 9 (note that in this case we have P3(λ)=(λ-1)2(λ+1)). If a=2, then c=0, which is also not a case treated in Theorem 9. If a=-1, then c=-3, bd=80, Δ=0, and P3(λ)=(λ-1)2(λ+3). If a=-2, then c=-2, bd=90, Δ=0, and P3(λ)=(λ-1)2(λ+4). Hence, there are two cases in which polynomial (3.115) has exactly two zeros equal to one, that is, the conditions of Corollary 10 (ii) are satisfied.

Remark 16.

For the case of our system it is not possible that all three zeros of (3.115) are equal to one. Namely, in this case there must be λi=a/3=1, i=1,2,3, which implies a=3. On the other hand, c=-a2/27, which in this case implies c=-1/3.

Remark 17.

Assume that only one of the zeros of characteristic polynomial (3.115) is equal to one, say λ1, abcd0 and that Δ=0. Then the other two zeros of (3.115) are equal and different from one, i.e., λ2=λ31 and

λ1=a3+2B3323=1,

which implies that 4B3=3-a. Hence B=2q3 for some q{0} (if q=0, i.e., B=0, the condition Δ=0 would imply A=0, and consequently all three zeros of (3.115) would be equal). Moreover, we have q=(a-3)/2, which also implies that a is an odd number. From this and since 4A3=-B2, we would have that A3=-q6, i.e., A=-q2. Recall that since polynomial (3.115) has a zero equal to one, it must be (a-1)(c-1)=bd. Hence, we have

(a-3)24=q2=a2+3bd+3c=a2+3ac-3a+3,

which is equivalent to (a-1)2=-4ac. This, along with the fact that a=2m+1 for some m, implies that

c=-(a-1)24a=-m22m+1,

so that

k:=m22m+1

must be an integer.

Since m2-2km-k=0, we have that m=k±k2+k, and consequently k2+k must be a perfect square, i.e., k(k+1)=r2, for some r. This would mean that k and k+1 divide r. From this and since k and k+1 are mutually prime numbers if k-1,0, it would follow that r=ck(k+1) for some c, and consequently (ck(k+1))2=k(k+1), which is equivalent to

k(k+1)(c2k(k+1)-1)=0.

The relation c2k(k+1)=1 is not possible since 1 is divided only by two integers ±1. If k=0, then c=0, which would contradict the condition abcd0. If k=-1, then c=1, and consequently bd=0, which would also contradict the condition abcd0. Hence c cannot be an integer, so that polynomial (3.115) cannot have only one zero equal to one under the conditions of Theorem 9.

Remark 18.

The formulas presented in this paper can be used in the investigation of the asymptotic behavior of solutions to system (1.1). We leave the problem to the reader.

References

  • [1]

    M. Aloqeili, Dynamics of a kth order rational difference equation, Appl. Math. Comput. 181 (2006), 1328–1335.  Google Scholar

  • [2]

    A. Andruch-Sobilo and M. Migda, Further properties of the rational recursive sequence xn+1=axn-1/(b+cxnxn-1), Opuscula Math. 26 (2006), no. 3, 387–394.  Google Scholar

  • [3]

    L. Berg and S. Stević, On some systems of difference equations, Appl. Math. Comput. 218 (2011), 1713–1718.  Web of ScienceGoogle Scholar

  • [4]

    L. Berg and S. Stević, On the asymptotics of the difference equation yn(1+yn-1yn-k+1)=yn-k, J. Difference Equ. Appl. 17 (2011), no. 4, 577–586.  Google Scholar

  • [5]

    C. Cinar, On the positive solutions of difference equation, Appl. Math. Comput. 150 (2004), no. 1, 21–24.  Google Scholar

  • [6]

    D. K. Faddeev, Lectures on Algebra (in Russian), Nauka, Moscow, 1984.  Google Scholar

  • [7]

    N. Fotiades and G. Papaschinopoulos, Asymptotic behavior of the positive solutions of a system of k difference equations of exponential form, Dyn. Contin. Discrete Impuls. Syst. Ser. A Math. Anal. 19 (2012), no. 5, 585–597.  Google Scholar

  • [8]

    B. Iričanin and S. Stević, Eventually constant solutions of a rational difference equation, Appl. Math. Comput. 215 (2009), 854–856.  Web of ScienceGoogle Scholar

  • [9]

    C. Jordan, Calculus of Finite Differences, Chelsea Publishing, New York, 1956.  Google Scholar

  • [10]

    G. Karakostas, The dynamics of a cooperative difference system with coefficient a Metzler matrix, J. Difference Equ. Appl. 20 (2014), no. 5–6, 685–693.  Web of ScienceCrossrefGoogle Scholar

  • [11]

    H. Levy and F. Lessman, Finite Difference Equations, Dover Publications, New York, 1992.  Google Scholar

  • [12]

    M. Malin, Multiple solutions for a class of oscillatory discrete problems, Adv. Nonlinear Anal. 4 (2015), no. 3, 221–233.  Web of ScienceGoogle Scholar

  • [13]

    D. S. Mitrinović and J. D. Kečkić, Methods for Calculating Finite Sums (in Serbian), Naučna Knjiga, Beograd, 1984.  Google Scholar

  • [14]

    G. Papaschinopoulos and C. J. Schinas, On a system of two nonlinear difference equations, J. Math. Anal. Appl. 219 (1998), no. 2, 415–426.  CrossrefGoogle Scholar

  • [15]

    G. Papaschinopoulos and C. J. Schinas, On the behavior of the solutions of a system of two nonlinear difference equations, Comm. Appl. Nonlinear Anal. 5 (1998), no. 2, 47–59.  Google Scholar

  • [16]

    G. Papaschinopoulos and C. J. Schinas, Invariants for systems of two nonlinear difference equations, Differ. Equ. Dyn. Syst. 7 (1999), no. 2, 181–196.  Google Scholar

  • [17]

    G. Papaschinopoulos and G. Stefanidou, Asymptotic behavior of the solutions of a class of rational difference equations, Int. J. Differ. Equ. 5 (2010), no. 2, 233–249.  Google Scholar

  • [18]

    G. Stefanidou, G. Papaschinopoulos and C. J. Schinas, On a system of max difference equations, Dyn. Contin. Discrete Impuls. Syst. Ser. A Math. Anal. 14 (2007), no. 6, 885–903.  Google Scholar

  • [19]

    G. Stefanidou, G. Papaschinopoulos and C. J. Schinas, On a system of two exponential type difference equations, Comm. Appl. Nonlinear Anal. 17 (2010), no. 2, 1–13.  Google Scholar

  • [20]

    S. Stević, More on a rational recurrence relation, Appl. Math. E-Notes 4 (2004), 80–85.  Google Scholar

  • [21]

    S. Stević, A short proof of the Cushing–Henson conjecture, Discrete Dyn. Nat. Soc. 2006 (2006), Article ID 37264.  Google Scholar

  • [22]

    S. Stević, On a generalized max-type difference equation from automatic control theory, Nonlinear Anal. 72 (2010), 1841–1849.  Web of ScienceCrossrefGoogle Scholar

  • [23]

    S. Stević, On a system of difference equations, Appl. Math. Comput. 218 (2011), 3372–3378.  Google Scholar

  • [24]

    S. Stević, On the difference equation xn=xn-2/(bn+cnxn-1xn-2), Appl. Math. Comput. 218 (2011), 4507–4513.  Google Scholar

  • [25]

    S. Stević, On the difference equation xn=xn-k/(b+cxn-1xn-k), Appl. Math. Comput. 218 (2012), 6291–6296.  Google Scholar

  • [26]

    S. Stević, Solutions of a max-type system of difference equations, Appl. Math. Comput. 218 (2012), 9825–9830.  Web of ScienceGoogle Scholar

  • [27]

    S. Stević, Domains of undefinable solutions of some equations and systems of difference equations, Appl. Math. Comput. 219 (2013), 11206–11213.  Web of ScienceGoogle Scholar

  • [28]

    S. Stević, Representation of solutions of bilinear difference equations in terms of generalized Fibonacci sequences, Electron. J. Qual. Theory Differ. Equ. 2014 (2014), Article No. 67.  Web of ScienceGoogle Scholar

  • [29]

    S. Stević, First-order product-type systems of difference equations solvable in closed form, Electron. J. Differential Equations 2015 (2015), Article No. 308.  Google Scholar

  • [30]

    S. Stević, Product-type system of difference equations of second-order solvable in closed form, Electron. J. Qual. Theory Differ. Equ. 2015 (2015), Article No. 56.  Web of ScienceGoogle Scholar

  • [31]

    S. Stević, Solvable subclasses of a class of nonlinear second-order difference equations, Adv. Nonlinear Anal. 5 (2016), no. 2, 147–165.  Web of ScienceGoogle Scholar

  • [32]

    S. Stević, M. A. Alghamdi, A. Alotaibi and E. M. Elsayed, Solvable product-type system of difference equations of second order, Electron. J. Differential Equations 2015 (2015), Article No. 169.  Google Scholar

  • [33]

    S. Stević, M. A. Alghamdi, A. Alotaibi and N. Shahzad, Boundedness character of a max-type system of difference equations of second order, Electron. J. Qual. Theory Differ. Equ. 2014 (2014), Article No. 45.  Web of ScienceGoogle Scholar

  • [34]

    S. Stević, J. Diblik, B. Iričanin and Z. Šmarda, On a third-order system of difference equations with variable coefficients, Abstr. Appl. Anal. 2012 (2012), Article ID 508523.  Web of ScienceGoogle Scholar

  • [35]

    S. Stević, J. Diblik, B. Iričanin and Z. Šmarda, On a solvable system of rational difference equations, J. Difference Equ. Appl. 20 (2014), no. 5–6, 811–825.  CrossrefWeb of ScienceGoogle Scholar

  • [36]

    S. Stević, J. Diblik, B. Iričanin and Z. Šmarda, Solvability of nonlinear difference equations of fourth order, Electron. J. Differential Equations 2014 (2014), Article No. 264.  Google Scholar

  • [37]

    S. Stević, B. Iričanin and Z. Šmarda, On a product-type system of difference equations of second order solvable in closed form, J. Inequal. Appl. 2015 (2015), Article No. 327.  Web of ScienceGoogle Scholar

  • [38]

    S. Stević, B. Iričanin and Z. Šmarda, Solvability of a close to symmetric system of difference equations, Electron. J. Differential Equations 2016 (2016), Article No. 159.  Google Scholar

About the article

Received: 2016-06-24

Revised: 2016-08-16

Accepted: 2016-09-13

Published Online: 2016-12-21


Citation Information: Advances in Nonlinear Analysis, Volume 8, Issue 1, Pages 29–51, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2016-0145.

Export Citation

© 2019 Walter de Gruyter GmbH, Berlin/Boston. This work is licensed under the Creative Commons Attribution 4.0 Public License. BY 4.0

Citing Articles

Here you can find all Crossref-listed publications in which this article is cited. If you would like to receive automatic email messages as soon as this article is cited in other publications, simply activate the “Citation Alert” on the top of this page.

[2]
Stevo Stevic
Electronic Journal of Qualitative Theory of Differential Equations, 2017, Number 84, Page 1
[3]
Stevo Stević
Advances in Difference Equations, 2017, Volume 2017, Number 1
[4]
Stevo Stević
Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A. Matemáticas, 2017
[5]
[6]
Stevo Stević
Advances in Difference Equations, 2017, Volume 2017, Number 1

Comments (0)

Please log in or register to comment.
Log in