R E S E A R C H
Open Access
Solvable product-type system of
difference equations with two dependent
variables
Stevo Stevi´c
**Correspondence: [email protected]
Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, Beograd, 11000, Serbia
Operator Theory and Applications Research Group, Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia
Abstract
It has been recently noticed that there is a finite number of two-dimensional classes of product-type systems of difference equations solvable in closed form. We present a new class of this type. A detailed analysis of the form of its solutions is given. Our results complement the previous ones on such systems and present one of the final steps in describing the forms of their solutions.
MSC: 39A20; 39A45
Keywords: system of difference equations; product-type system; solvable in closed form
1 Introduction
Many types of difference equations and systems have been studied so far. A part of the studies can be found in [–]. Some types of the systems essentially obtained by sym-metrization of scalar ones were studied in [–], which was a motivation for further in-vestigations in the field [, , , , –]. Historically, perhaps the first main problem of interest in the whole area was finding formulas for their solutions. For known methods for finding the formulas the reader can consult, for example, [–]. A note of ours from has influenced some investigation in this direction since that time (see, for example, [, –] and the references therein).
In the study of some classes of equations and systems, product-type ones appear as boundary cases. Finding formulas for positive solutions to the equations and systems in the boundary cases is a routine problem, so not of theoretical interest nowadays. It can be of practical interest only if another system or equation is reduced to such one. However, if all solutions are not positive, the problem is very complicated. The boundary cases of equations and systems have motivated us to study them for the case of non-positive ini-tial values. In fact, the equations and systems on the complex domain have attracted our special attention. Our study started in [], where a system with two dependent variables was investigated. The form of the system in [] strikingly suggested the study of the solv-ability of the other systems of related forms (see, e.g., [, ]). Since the system, as well as a couple of other ones later studied (see, e.g., []), was of the form
zn=zan–mw
a
n–m, wn=w
a
n–mz
a
n–m, n∈N, ()
it naturally suggested the study of the solvability of this, as well as of some related sys-tems. This motivated us to include some coefficients in () and study the solvability of such systems, which was for the first time done in [], where we showed the solvability theoretically and gave some hints on how to deal with more concrete cases, that is, for some special values of parametersa,b,candd. Later we realized that complete pictures of the form of the solutions of this type of systems could be given by studying all the quan-tities appearing there in detail. References [] and [] were the first ones which gave the complete pictures of the forms of the solutions to the systems studied therein. Later in [] we devised another method which deals with the solvability problem, although technically somewhat complex. For some quite recent results on product-type systems see [], [] and [].
To finish the project of studying the solvability of product-type systems with two depen-dent variables (see [, –] and the related references therein), we have to study a few more. Here we study the system
zn+=αzanwbn, wn+=βwcn–zdn–, n∈N, ()
wherea,b,c,d∈Z,α,β,z–,z,w–,w–,w∈C. In fact, we assume thatα,β,z–,z,w–,
w–,w∈C\{}, to avoid dealing with non-defined or trivial solutions. We will give a
com-plete picture of the forms of the solutions to system () for all the values of the parameters and initial values.
2 Auxiliary results
Some classical auxiliary results that are employed in the section that follows are quoted in this one.
Lemma (see,e.g., [, ]) Let
Rk(s) =bk k
j=
(s–sj),
sj=st,j=t,and bk= .Then k
j=
sm j Rk(sj)
= ,
for each m∈ {, , . . . ,k– },and k
j=
sk–
j Rk(sj)=
bk
.
Four more or less widely known formulas are listed in the following lemma (see, e.g., [, ]). A recurrent relation connecting this type of sums is given in [].
Lemma Let
s(nm)(z) = n
j=
jmzj–, n∈N,
Then
s()n (z) = –z n
–z,
s()n (z) = – (n+ )z
n+nzn+
( –z) ,
s()n (z) = +z– (n+ )
zn+ (n+ n– )zn+–nzn+
( –z) ,
s()n (z) = n
zn(z– )– nzn(z– )+ nzn(z– ) – (zn– )(z+ z+ )
( –z) ,
for every z∈C\ {}and n∈N.
The following lemma describes the nature of the zeros of a polynomial of the fourth order in detail (see []).
Lemma Let
P(t) =t+bt+ct+dt+e,
=c– bd+ e, = c– bcd+ be+ d– ce,
=
–, P= c– b,
Q=b+ d– bc, D= e– c+ bc– bd– b.
(a) If< ,then two zeros ofPare real and different,and two are complex conjugate. (b) If> ,then all the zeros ofPare real or none is.More precisely,
◦ ifP< andD< ,then all four zeros ofPare real and different;
◦ ifP> orD> ,then there are two pairs of complex conjugate zeros ofP.
(c) If= ,then and only thenPhas a multiple zero.The following cases can occur:
◦ ifP< ,D< and= ,then two zeros ofPare real and equal and two are real and simple;
◦ ifD> or(P> and(D= orQ= )),then two zeros ofPare real and equal and two are complex conjugate;
◦ if= andD= ,there is a triple zero ofPand one simple,all real;
◦ ifD= ,then
.◦ ifP< there are two double real zeros ofP;
.◦ ifP> andQ= there are two double complex conjugate zeros ofP;
.◦ if= ,then all four zeros ofPare real and equal to–b/.
3 Main results
The main results in this paper are proved in this section.
Using ()-() in the first equality in (), we get
zm=α+bd m–
j=(c+bd)jβb
m–
j=(c+bd)jzbd(c+bd)m–
w
bc(c+bd)m–
– , ()
zm+=α+bd
m–
j=(c+bd)jβb
m–
j=(c+bd)jwb(c+bd)m
, ()
zm+=α+bd
m–
j=(c+bd)jβb
m
j=(c+bd)jzbd(c+bd)m
– w
bc(c+bd)m
– . ()
From ()-() and some calculations, we easily get ()-(), as desired.
Theorem Assume that a,c,d∈Z,b= ,α,β,z–,z,w–,w–,w∈C\ {}.Then system
()is solvable in closed form.
Proof Sinceb= system () becomes
zn+=αzan, wn+=βwcn–zdn–, n∈N, ()
which is system (.) in []. Hence, ifc= the theorem follows from Theorem . in [], while the casec= follows from equations (.) and (.) in [], as well as the
second equation in ().
The cased= has been recently studied in [], where, among others, the following theorem was proved.
Theorem Assume that a,b,c∈Z,d= ,α,β,z,w–,w–,w∈C\ {}.Then system()
is solvable in closed form.
Theorem Assume that a,b,c,d∈Z,abcd= ,α,β,z–,z,w–,w–,w∈C\ {}.Then
system()is solvable in closed form.
Proof Fromα,β,z–,z,w–,w–,w∈C\ {}and () we getznwn= forn∈N. Hence, wbn=zn+
αza n
, n∈N, ()
wbn+=βbwbcn–zbdn–, n∈N, ()
and consequently
zn+=α–cβbzan+zbdn–+cz–n–ac, ()
forn≥. Note also that
z=αzawb, z=α
αzawbaβwc–zd–b=α+aβbzbd–zawbc–wab ,
z=αzawb=α+a+a
βb(+a)zabd– za+bdwabc–wbc–wab.
()
Letδ=α–cβb,
then
=α(–c)yn–+(+a+a)an–+(+a)bn–+cn–βbyn–+b(+a)an–+bbn–
×zabdan–+bdbn–
– z
(a+bd)an–+abn–+acn–+dn–
w
abcan–+bcbn–
– w
bcan– –
×waban–+abbn–+bcn–
=αyn+–cyn–βbyn+zbdan
– z
an+–can–
w
bcan
– w
bcan– – w
ban+
, ()
forn≥.
From () one sees thatak,bk,ckanddkare solutions to
ˆ
xk+=axˆk++bxˆk++cxˆk++dxˆk, k∈N, ()
and, along with () (fork= , , –, –), we also obtain
a–= , a–= , a–= , a= , ()
y–=y–=y–=y= , y= , ()
and
yk= k–
j=
aj. ()
The solvability of () is well known, from which, along with (), a formula forakis obtained. Using it in (), a formula forykis obtained by Lemma . Hence, () is solvable.
We have
znd–= wn+
βwc n–
, n∈N, ()
znd+=αdznadwbdn , n∈N, ()
so that
wn+=αdβ–awan+wnbd+cw–n–ac, n∈N. ()
We also have
w=βwc–zd– and w=βwc–zd. ()
As above we get
wn+=ηykwnak+–kwbnk+–kwcnk+–kwndk–k, n≥k– , ()
whereη=αdβ–a,a
ksatisfies () and (), andykis given by (). From () withk=n+ and by using () we get
wn+=ηyn+wan+w
bn+ w
cn+ w
dn+ –
=αdβ–ayn+βwc–zdan+βwc–zd–bn+wcn+ w
=αdyn+β(–a)yn++an++bn+zdbn+
As we have already seen, formulas forakandykcan be found. Using them in () we show the solvability of (). Some calculations show that () and () present a solution
to (), from which the result follows.
Corollary Assume that a,b,c,d∈Z,abcd= ,α,β,z–,z,w–,w–,w∈C\ {}.Then
the general solution to()is given by()and(),where aksatisfies()and(),and yk is given by()and().
Theorem gives a general form of solutions to system () whenabcd= , but does not present explicit formulas for sequencesanandyninvolved in the solutions. Now we give some explicit formulas for them in more concrete cases, following some arguments related to the system in []. Sinceac= , we can find the zeros of the characteristic polynomial associated to ()
p(λ) =λ–aλ– (bd+c)λ+ac. ()
To do this, we consider the following equivalent equation with a parameter []:
or
Forsgiven in () we solve equations () and (). So, the zeros of polynomial () are
λ=
not a zero of the polynomial. In fact, there are many polynomials of the form in () such that< . For example, they are those for which holds ac<abd, that is,< .
Sinceλj=λi,i=j,
an=γλn+γλn+γλn+γλn, n∈N, ()
Equalities (), along with Lemma applied to polynomial (), yield
By using the change of variablesλ=t+a– and some simple calculations, we get
t+pt˜ +q˜= ,
Using the standard arguments, as those in getting (), we obtain
λj=
Equality () holds with, say,λ= . Further, we have
yn= n–
j=
p()+
n–
j=
i=
λji+ p(λi)
= n
– a–ac+
i=
λi(λni – ) p(λi)(λi– )
, ()
forn∈N. It is easily shown that () also holds forn= –j,j= , . This analysis, along with Corollary , implies the following result.
Corollary Assume that a,b,c,d∈Z,abcd= ,α,β,z–,z,w–,w–,w∈C\ {}and
= .Then the following statements are true:
(a) If(a– )(c– )=bd,then the general solution to()is given by()and(),where
(an)n≥–is given by(),(yn)n≥–is given by(),whileλj,j= , ,are given by
()-().
(b) If(a– )(c– ) =bdand – a=ac,then the general solution to()is given by() and(),where(an)n≥–is given by()withλ= ,(yn)n≥–is given by(),λ= , whileλj,j= , ,are given by().
is the only double zero of p. Polynomialphas a double zero equal to if () holds
and
p() = – a–ac= , ()
that is, if and only if
c=
a– . ()
Then we have
p(λ) =λ–aλ+ (a– )λ+ – a= (λ– )
λ+ ( –a)λ+ – a,
and consequently
λ,= , λ,=
a– ±√a+ a–
. ()
From () we must havea= andc= –, ora= andc= , ora= – andc= –, or a= – andc= –.
Ifa=c= , then
p(λ) =λ–λ–λ+ = (λ– )
λ+λ+ ,
and consequently
λ,= , λ,=
–±i√
.
Ifa= ,c= –, then
p(λ) =λ– λ+ λ– = (λ– )
λ–λ– ,
and consequently
λ,= , λ,=
±√
. ()
Ifa=c= –, then
p(λ) =λ+ λ– λ+ = (λ– )
λ+ λ+ ,
and consequently
λ,= , λ,=
–±i√
. ()
Ifa= – andc= –, then
p(λ) =λ+λ– λ+ = (λ– )
λ+ λ+ ,
and consequently
λ,= , λ,=
–±i√
. ()
In these four cases, we have []
an=
n( –λ)( –λ) + λλ– λ– λ+
( –λ)( –λ)
+ λ
n+
(λ– )(λ–λ)
+ λ
n+
(λ– )(λ–λ)
()
and
yn= n–
j=
j( –λ)( –λ) + λλ– λ– λ+
( –λ)( –λ)
+ λ
j+
(λ– )(λ–λ)
+ λ
j+
(λ– )(λ–λ)
= (n– )n
( –λ)( –λ)
+n(λλ– λ– λ+ ) ( –λ)( –λ)
+ λ
(λn– )
(λ– )(λ–λ)
+ λ
(λn– )
(λ– )(λ–λ)
. ()
phas only one double zero different from. Assume thatλ=m∈ {/ , }is a double zero
ofp. Then we have
m–am– (bd+c)m+ac= and m– am–bd–c= . ()
From (), we get
p(λ) =λ–aλ+
am– mλ+ m– am
= (λ–m)λ+ (m–a)λ+m(m– a), ()
and consequently
λ,=m, λ,=
a– m±√–m+ am+a
. ()
Hence, if we additionally assume that a= m, am– m∈Z, m– am∈Z, we
get a family of polynomials of the form in () which have double zeros different from . For example, ifa=m∈Z\ {, }, then from () it follows that
p(λ) = (λ–a)
λ+aλ+a.
Since, in the caseλ=λ,λi=λj, ≤i,j≤, we have
an= (γ+γn)λn+γλn+γλn, n∈N, ()
whereγiandi= , are constants, and the solution satisfying () is
an=
λn+
((n+ )(λ–λ)(λ–λ) –λ(λ–λ–λ))
(λ–λ)(λ–λ)
+ λ
n+
(λ–λ)(λ–λ)
+ λ
n+
(λ–λ)(λ–λ)
. ()
From () and () and by Lemma , we get
yn= n–
j=
λj+((j+ )(λ–λ)(λ–λ) –λ(λ–λ–λ))
(λ–λ)(λ–λ)
+ λ
j+
(λ–λ)(λ–λ)
+ λ
j+
(λ–λ)(λ–λ)
= λ
–nλn++ (n– )λn+
(λ–λ)(λ–λ)( –λ)
+(λ
– λλ– λλ+ λλλ)(λn– )
(λ–λ)(λ–λ)(λ– )
+ λ
(λn– )
(λ–λ)(λ–λ)(λ– )
+ λ
(λn– )
(λ–λ)(λ–λ)(λ– )
. ()
Corollary Assume that a,b,c,d∈Z,abcd= andα,β,z–,z,w–,w–,w∈C\ {}.
Then the following statements are true:
(a) If only one of the zeros ofpis double and different from,sayλandλ,then the general solution to()is given by()and(),where(an)n≥–is given by(),
(yn)n≥–is given by(),whileλj,j= , ,are given by(),wherem∈ {/ , }, m=a= mandam– m, m– am∈Z.
(),whileλj,j= , ,are given by()whena= ,c= –,by()whena=c= –,
and by()whena= –,c= –.
phas two pairs of different double zeros. In this case it must beD= , which implies
thata= or bd= c– a. The casea= is impossible due to the conditionabcd= .
In the other case we have= if and only if
a
c+a
– c
=
ac+
c–a
,
that is,
a=±a– ac+ c. ()
From () we have
a– ac+ c= ,
from which it follows thata/c=
(±i), which is impossible due to the rationality of
a/c, or is
a– ac+ c=a– c= ,
which impliesc=a/.
Assumec=a/. Then
p(λ) =λ–aλ+
a
λ+ a
=
λ–aλ
–
a
(for more details see [], p.). Hence
λ,=
a ( +
√
), λ,=
a ( –
√
), ()
are two double zeros ofp, for eacha= .
Since
a(±√)/= ,
for everya∈Z,pcannot have two pairs of double zeros such that one of them is equal
to .
The general solution to () in this case is of the following form:
an= (γ+γn)λn+ (γ+γn)λn, n∈N, ()
for some constantsγi,i= , . The solution with initial conditions () is
an=
λn+
(n(λ–λ)+λ– λλ+ λ)
(λ–λ)
+λ n+
(n(λ–λ)+λ– λλ+ λ)
(λ–λ)
From (), () and Lemma , we get
yn= n–
j=
λj+(j(λ–λ)+λ– λλ+ λ)
(λ–λ)
+λ j+
(j(λ–λ)+λ– λλ+ λ)
(λ–λ)
=λ
–nλn++ (n– )λn+
(λ–λ)( –λ)
+(λ
– λλ+ λλ)(λn– )
(λ–λ)(λ– )
+λ
–nλn++ (n– )λn+
(λ–λ)( –λ)
+(λ
– λλ+ λλ)(λn– )
(λ–λ)(λ– )
. ()
Corollary Assume that a,b,c,d∈Z,abcd= andα,β,z–,z,w–,w–,w∈C\ {}.
Then the following statements are true:
(a) If polynomialphas two pairs of double zeros both different from,then the general solution to()is given by()and(),where(an)n≥–is given by(),(yn)n≥–is
given by(),whileλj,j= , ,are given by().
(b) The characteristic polynomial()cannot have two pairs of double zeros such that one of them is equal to.
Triple zero case. In this case we must have== or equivalently== , that
is,
a= or bd= c.
Sinceabcd= , the casea= is not possible. Ifc=bd/, then
= c
a+ c.
Since the casec= is excluded, we must havec= –a/.
We have
p(λ) =λ–aλ+
a
λ– a
=
λ–a
λ+a
(see, for example, [], p.), and consequently
λj= a
, j= , , λ= – a
. ()
Thus, for everya= ,a/ is a triple zero ofp, andpcannot have a zero of the fourth
order. Hence
an=
γ+γn+γn
λn+γλn, n∈N, ()
Further, by using the initial conditions in (), we obtain
Then the following statements are true:
(a) If polynomial()has a triple zero different from,then the general solution to system()is given by()and(),where(an)n≥–is given by(),(yn)n≥–is given
()is solvable in closed form.
Proof We modify our method in [, ]. We have
and consequently
zn+=αβbznazbdn–, n∈N. ()
Letδ=αβb,
a=a, b= , c=bd, y= . ()
Then clearly
zn+=δyzanz b
n–z
c
n–, n∈N. ()
Hence,
zn+=δy
δza
n–z
b
n–z
c
n–
a
zb
n–z
c
n–
=δy+azaa+b
n– z
ba+c
n– z
ca
n–
=δyza
n–z
b
n–z
c
n–,
forn≥, where
a:=aa+b, b:=ba+c, c:=ca, y:=y+a.
Assume
zn+=δykzank+–kznbk–kzcnk–k–, ()
for ak≥ and everyn≥k, and
ak=aak–+bk–, bk=bak–+ck–, ck=cak–, ()
yk=yk–+ak–. ()
Further, by (), it follows that
zn+=δyk
δza
n–kz b
n–k–z
c
n–k–
ak
zbk n–kz
ck n–k–
=δyk+akzaak+bk n–k z
bak+ck n–k– z
cak n–k–
=δyk+zak+
n–kz bk+
n–k–z
ck+
n–k–,
forn≥k+ , where
ak+:=aak+bk, bk+:=bak+ck,
ck+:=cak, yk+:=yk+ak.
Settingk=nin (), and employing (), we get
zn+=δynzanzbnzc–n
=αβbynαzawbanzbn
z
cn
–
=αyn+anβbynzcn
–z
aan+bn
w
ban
=αyn+βbynzbdan– – z
an+ w
ban
, ()
forn≥.
From () we see thatak,bkandckare solutions to
˜
xk+=ax˜k++bx˜k++cx˜k, k∈N, ()
and that along with () and () (fork= , –, –), we obtain
a–= , a–= , a= , ()
y–=y–=y= , y= , ()
and
yk= k–
j=
aj. ()
The solvability of () is well known, from which along with () is obtained a formula forak, which along with () and Lemma yields a formula foryk. Hence, () is solvable.
Using (), in the second equation in (), is obtained:
wn=αdyn–βbdyn–+zbd
a
n– – z
dan–
w
bdan–
, n≥. ()
It is shown that equations () and () are solutions to system (), so it is solvable, as
claimed.
Theorem gives a general form of solutions to system () in the casec= ,abd= , but it does not present explicit formulas foranandyninvolved in the solutions. We give some explicit formulas for them, following also some arguments in []. Sincebd= , we find the zeros of the characteristic polynomial associated to ()
p(λ) =λ–aλ–bd. ()
Forλ=s+a
, the equationp(λ) = becomes
s–a
s–
a+ bd
= . ()
We know that
sj= √
εj
–
– +εj
+
–
j= , , where
It is well known that the general solution to () in this case is
If one of the zeros ofpis , say,λ, then
yn=
λ (λn – )
(λ–λ)(λ– )
+ λ
(λn– )
(λ–λ)(λ– )
+ n
(λ– )(λ– )
, ()
forn∈N. Moreover, equations () and () hold forn≥–.
Corollary Assume that a,b,c,d∈Z,abd= ,c= ,α,β,z–,z,w∈C\ {}and=
.Then the following statements are true:
(a) Ifa+bd= ,then the general solution to()is given by()and(),where
(an)n≥–is given by(),(yn)n≥–is given by(),whileλj,j= , ,are given by().
(b) Ifa+bd= ,thenphas a unique zero equal to,sayλ,and the general solution to ()is given by formulas()and(),where(an)n≥–is given by()withλ= ,
(yn)n≥–is given by(),whileλj,j= , ,are given by().
phas a double zero. Since it must be= , we havebd= –a/, so that
p(λ) =λ–aλ+
a
.
The following condition must also be satisfied:p(λ) = . Hence,λ= –a/,λ,= a/,
and consequently
p(λ) =
λ–a
λ+a
.
Due tobd∈Z, we geta= aˆ, for someaˆ∈Z. Now note that a/= , for everya∈Z, so that cannot be a double zero ofp.
Hence
an=αˆλn+ (αˆ+αˆn)λn, n∈N, ()
whereαˆi,i= , , are constants. Using initial conditions () we obtain
an=
λn++ (λ– λ+n(λ–λ))λn+
(λ–λ)
, ()
forn≥–.
From () and (), it follows that
yn= n–
j=
λj++ (λ– λ+j(λ–λ))λj+
(λ–λ)
, ()
forn∈N.
Equation () along with Lemma yields
yn=
λ(λn – ) (λ–λ)(λ– )
+(λ– λ)λ(λ n
– )
(λ–λ)(λ– )
+λ
( –nλn–+ (n– )λn)
(λ–λ)(λ– )
, ()
On the other hand, ifλ= =λ=λ(a= –), then we get
yn= n (λ– )
+(λ– )λ(λ n
– )
(λ– )
+λ
( –nλn–+ (n– )λn)
(λ– )
, ()
forn∈N. Moreover, () and () hold forn≥–.
Corollary Assume that a,b,c,d∈Z,abd= ,c= ,α,β,z–,z,w∈C\ {}and=
.Then the following statements are true:
(a) Ifa+bd= ,then the general solution to()is given by()and(),where
(an)n≥–is given by(),(yn)n≥–is given by(),whereλ= –a/andλ,= a/. (b) If only one of the zeros of the polynomial()is equal to,say,λ,then the general
solution to system()is given by()and(),where(an)n≥–is given by()
withλ= ,while(yn)n≥–is given by().
(c) It is not possible that two zeros of polynomial()are equal to one.
Case when all the zeros of pare equal. We havep(λ) =p(λ) =p(λ) = . So,p(λ) =
would implyλ=a/. Fromp(λ) = λ– aλ, we see thata/ is a unique zero ofpifa= ,
which contradicts the assumptionabd= . Hence, the case is not possible.
Competing interests
The author declares that he has no competing interests.
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Received: 11 May 2017 Accepted: 3 August 2017 References
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