R E S E A R C H
Open Access
Solvability of a class of product-type
systems of difference equations
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
A solvable class of product-type systems of difference equations with two dependent variables on the complex domain is presented. The main results complement some recent ones in the literature, while their proofs contain some refined methodological details. We provide closed form formulas for general solutions to the system or give procedures for how to get them.
MSC: 39A20; 39A45
Keywords: system of difference equations; product-type system; solvable system; solutions in closed form
1 Introduction
Nonlinear difference equations and systems have been studied a lot in the last few decades (see,e.g., [–]). Two of the topics of recent interest are symmetric and closely related systems (see, for example, [, –, , , , , , , –]), whose investigation was considerably influenced by some papers by Papaschinopoulos and Schinas (see, for example, [–]), and the solvable difference equations and systems (see, for example, [, , –, –] and the references therein). For some classical methods for solving the equations and systems see, for example, [, –]. It has been shown recently that many nonlinear equations and systems can be solved by transforming them to linear ones (see, for example, [, , , , , ] and the related references therein).
Some of the equations and systems that we have studied recently, such as the ones in [] and [] (see also []), are obtained by adding constants to the right-hand sides of some product-type equations/systems or by taking the maximum of some constants and the right-hand sides of the equations/systems. This means that they are related to the product-type ones, which are usually some kind of boundary cases. The case of positive initial values and multipliers is simple, since in that case the equations/systems can easily be treated by one of the simplest transformation methods. The case of complex initial values is more complex due to the fact that complex functions need not be single valued. Hence, our methods in [, , ] and other related papers cannot be applied. This has motivated us to start investigating product-type systems on the complex domain. In a series of papers, see [, , , –], we have obtained some results in the area (during the study of the equation in [] we came across a product-type equation). In our first papers on the topic (see [, , ]) the systems have not had coefficients different from
one. However, not long after that we have introduced two coefficients and also got solvable systems (this was done for the first time in [], and somewhat later in [–]). We have also observed the fact that there are only a few solvable product-type systems of difference equations with two dependent variables. Hence, our aim is to describe all such product-type systems and present formulas for the general solutions for each of them.
This paper complements our previous results on the solvability of product-type systems of difference equations with two dependent variables, by studying the following one:
zn=αzan–wbn–, wn=βwcn–zdn–, n∈N, ()
where a,b,c,d∈Z, α,β∈Candz–,w–,w–∈C. To do this we will modify ideas and
methods from our previous papers, for example, the ones in [, –]. If the initial values belong to the following set:
(z–,w–,w–)∈C:z–= orw–= orw–=
,
and if some of the exponentsa,b,c,dare negative, then such solutions are not defined. Hence, this set of initial values will not be taken into consideration. Besides, ifα= or β = , we getzn= andwn= for everyn∈N, respectively, which are quite simple
cases, or also obtain solutions which are not well defined, so, we will also assume that α= =β. We use the conventionil=kai= , whenl<k, throughout the paper.
2 Main results
Our main results are presented here. The following three cases will be considered sepa-rately:
(i) c= ,ac=bd; (ii) c= ,ac=bd; (iii) ac=bd.
Clearly, in case (i) fromc= andac=bdit immediately follows thatbd= , but we have chosen to writeac=bdat this point, to point out that the whole analysis essentially depends on the values of the quantitiescandac–bd, that is, if they are equal to zero or not.
First, we will consider case (i), then case (iii) and at the end case (ii), for the presentational reasons.
Theorem Assume that a,b,d∈Z,c= ,bd= ,α,β∈C\{}and z–,w–,w–∈C\{}.
Then the following statements hold.
(a) Ifa= ,then the general solution to system()is given by the following formulas:
zn=α
–an+
–a βb–an
– –a zan+
– wba
n
– wba
n–
– , n≥, ()
and
wn=αd
–an
–aβzdan
(b) Ifa= ,then the general solution to system()is given by the following formulas:
zn=αn+βb(n–)z–wb–wb–, n≥, ()
and
wn=αdnβzd–, n≥. ()
Proof Sincec= , we have
zn=αzan–wbn–, wn=βzdn–, n∈N. ()
From () andbd= is obtained
zn=αβbzan–, n≥. ()
From () we easily get
zn=
αβb
n–
j=ajzan–
, n≥,
which, along with
z=α+aza
–wab–wb–, ()
yields
zn=α
n
j=ajβb
n–
j=ajzan+
– wba
n
– wba
n–
– , n≥. ()
By using equation () along with the formula for the sum of the geometric progression we see that equation () holds whena= , while equation () is directly obtained fora= .
From (), the equalitywn=βzdn–, and the conditionbd= , we obtain
wn=αd n–
j=ajβ+bd
n–
j=ajzdan
– wbda
n–
– wbda
n–
–
=αd
n–
j=ajβzdan
– , n≥. ()
By using equation () along with the formula for the sum of the geometric progression we see that equation () holds whena= , while equation () is directly obtained fora= ,
completing the proof of the theorem.
Theorem Assume that a,b,c,d∈Z,ac=bd,α,β∈C\ {}and z–,w–,w–∈C\ {}.
Then system()is solvable in closed form.
Proof Sinceα,β∈C\ {}andz–,w–,w–∈C\ {}, using () it easily follows thatzn= forn≥– andwn= forn≥–. Thus, from () we have
wbn–= zn αza n–
Equalities (), (), (), (), (), and the induction show that (), (), and (), hold for everyk,n∈Nsuch that ≤k≤n+ . Note that () holds also fork= .
Fork=n+ , () becomes
zn+=μyn+zan+z
bn+
z
cn+
–
=α–cβbyn+α+aza–wab–wb–an+
αz–a wb–bn+ zcn+
–
=α(–c)yn++(+a)an++bn+βbyn+zaan++abn++cn+
–
×waban++bbn+
– w
ban+
–
=αyn+–cyn+βbyn+zan+–can+
– w
ban+
– w
ban+
– , n∈N, ()
where we have also used the factz=αza–wb–, (), (), and ().
Further note that () implies that (ak)k≥is a solution to the equation
ak=aak–+bak–+cak–. ()
Sincec= , from (), we get
ak–=
ak–aak––bak–
c
, ()
from which it follows that we can calculatealfor every non-positivel, that is, fork≤. A direct calculation shows that
a= , a–=a–= . ()
Moreover, it is shown that (ak)k≥–, (bk)k≥–, (ck)k≥–are solutions to () such that
a–= , a–= , a= ;
b–= , b–= , b= ;
c–= , c–= , c= ;
respectively, whereas (yk)k≥–satisfies () and
y–=y–=y= , y= . ()
From () and sincea= , we get
yk= k–
j=
aj. ()
The solvability of () shows that for (ak)k≥– can be found a closed form formula.
Therefore, using equation () and known formulas for the following sums:
s(mj)(ζ) = m
k=
wherej= , (see,e.g., [, ]), the closed form formula for (yk)k∈N is easily obtained. This along with () shows the solvability of ().
From (), we also have
znd–= wn
βwcn–, n∈N, ()
and
znd=αdzadn–wbdn–, n∈N. ()
Equalities () and () yield
wn+=αdβ–awanwnc–wbdn––ac, n∈N. ()
We have
w=βwc–zd–. ()
Similarly as above we get
wn+=ηykwank+–kw bk
n–kw ck
n–k–, n≥k– , ()
whereη=αdβ–a, (ak)k
∈N, (bk)k∈Nand (ck)k∈Nare defined by () and (), whereas (yk)k∈N satisfies () and (), so it is given by ().
From () withk=n+ and () is obtained
wn+=ηyn+wan+w
bn+
– w
cn+
–
=αdβ–ayn+βwc–zd–an+wbn+
– w
cn+
–
=αdyn+β(–a)yn++an+zdan+
– w
can++cn+
– w
bn+
–
=αdyn+βyn+–ayn+zdan+
– w
an+–aan+
– w
an+–aan+
– , ()
forn∈N.
The solvability of () along with () shows that for (ak)k≥–we can find a closed form
formula, from which along with () the formulas forykcan also be obtained, as described above. These facts along with () imply the solvability of equation (). It is not difficult to show that formulas () and () really represent solutions to system (). Thus, system
() is also solvable in this case, as claimed.
Corollary Consider system () with a,b,c,d ∈ Z, ac = bd, α,β ∈ C \ {}, and z–,w–,w–∈C\ {}.Then the general solution to system()is given by()and(),
where(ak)k∈Nsatisfies equation()with initial conditions(),while(yk)k∈Nis given by ()and can be found by using formulas for the sums in().
Theorem Assume that a,b,c,d∈Z, c= ,ac=bd,α,β∈C\ {}and z–,w–,w–∈
where
λ=
a
. ()
(d) Ifa= –canda+c= ,then the general solution to system()is given by the
following formulas:
zn=α
(–c)n+(c+)n+
βb(n–) nza(n+)
– w
b(n+)
– wbn– ()
and
wn=αd
n(n+) β
(n+)((–a)n+) zd(n+)
– w
c(n+)
– wcn–. ()
Proof Similar to the proof of Theorem we havezn= forn≥– andwn= forn≥–, and for every such solution () holds, from which, along with the conditionac=bd, it follows that
zn+=α–cβbzan+zcn, () forn∈N.
Letν=α–cβb,
ˆ
a=a, bˆ=c, yˆ= . ()
Then
zn+=νˆyzanˆ+z
ˆ
b
n, ()
forn∈N.
From () is obtained
zn+=νˆy
νzaˆ
nz
ˆ
b
n–
aˆ zbˆ
n
=νˆy+aˆzaˆaˆ+bˆ
n z
ˆ
baˆ
n–
=νˆyzaˆ
n z
ˆ
b
n–, ()
forn∈N, where
ˆ
a:=aˆaˆ+bˆ, bˆ:=bˆaˆ, yˆ:=yˆ+aˆ. ()
Suppose that
zn+=νˆykznaˆk+–kz
ˆ
bk
n+–k, ()
holds for somek∈Nand for everyn≥k– , and that
ˆ
Then from (), withn→n–kand (), we get
zn+=νˆyk
νzaˆ
n+–kz
ˆ
b
n–k
aˆk
zbˆk
n+–k =νˆyk+aˆkzaˆaˆk+bˆk
n+–k z
ˆ
baˆk
n–k =νˆyk+zaˆk+
n+–kz
ˆ
bk+
n–k, ()
forn≥k, where
ˆ
ak+:=aˆaˆk+bˆk, bˆk+:=bˆaˆk, ˆyk+:=ˆyk+aˆk. () Equalities (), (), (), (), and the induction show that () and (), hold for every
k,n∈Nsuch that ≤k≤n+ . Note that the equality in () also holds fork= (see ()). Fork=n+ , () becomes
zn+=νˆyn+zaˆn+z
ˆ
bn+
=α–cβbyˆn+α+aza–wab–wb–aˆn+
αz–a wb–bˆn+
=α(–c)yˆn++(+a)aˆn++bˆn+βbˆyn+zaaˆn++abˆn+
–
×wabaˆn++bbˆn+
– w
baˆn+
–
=αˆyn+–cyˆn+βbyˆn+zaaˆn+
– w
baˆn+
– w
baˆn+
– , n∈N, ()
where we have used the factz=αza–wb–, (), and ().
Now note that (aˆk)k≥is a solution to the equation
ˆ
ak=aˆaˆk–+bˆaˆk–. ()
As in the proof of Theorem , it is shown that (aˆk)k≥– and (bˆk)k≥– are solutions to
equation (), satisfying the initial conditions
ˆ
a–= , aˆ= ;
ˆ
b–= , bˆ= ,
()
respectively, whereas (yˆk)k≥–satisfies the third equation in () and
ˆ
y–=yˆ= , yˆ= . ()
This andaˆ= imply
ˆ yk=
k–
j=
ˆ
aj. ()
Since () also holds, using the conditionac=bd, we get
Similar to the above is obtained
wn+=ηˆykwanˆk+–kw
ˆ
bk
n–k, ()
fork,n∈N, ≤k≤n+ , whereη=αdβ–a, sequences (aˆk)k∈N, and (bˆk)k∈N satisfy () and (), whereas (yˆk)k∈N satisfies the third equation in () and (), so, formula () holds.
From () withk=n+ and () it follows that
wn+=ηˆyn+waˆn+w
ˆ
bn+
–
=αdβ–ayˆn+βwc–zd–aˆn+wbˆn+
–
=αdyˆn+βyˆn+–ayˆn+zdaˆn+
– w
caˆn+
– w
caˆn
–, n∈N. ()
The characteristic polynomial associated to the linear difference equation () is the following:
P(λ) =λ–aλ–c,
from which it follows that the corresponding characteristic roots are given by the formulas in ().
Sincea–= anda= , then, ifa= –canda+c= , we easily get
an=
λn+–λn+ λ–λ
()
and consequently
ˆ yn=
n–
j=
λj+–λj+ λ–λ
=(λ– )λ n+
– (λ– )λn++λ–λ
(λ– )(λ– )(λ–λ)
. ()
Ifa= –canda+c= , then one of the characteristic roots, sayλ
, is equal to one, and
we have
an=
λn+– λ–
()
and
ˆ yn=
n–
j=
λj+– λ–
=n– (n+ )λ+λ n+
( –λ)
. ()
Ifa= –canda+c= , then we have
and
ˆ yn=
n–
j=
(j+ )λj= – (n+ )λ n
+nλn+
( –λ)
. ()
Ifa= –canda+c= , then we have
an=n+ , n∈N, ()
and
ˆ yn=
n–
j=
(j+ ) =n(n+ )
. ()
Using ()-() into () and () and by some calculations equations (), (), (), (), (), (), (), and () are obtained. By some standard, but time-consuming cal-culations, it is shown that these formulas really represent solutions to system () in each
if these four cases.
Remark Note that ifa= –canda+c= , then (a– )= , from which it follows that
a= and consequentlyc= –. Hence, equations () and () can also be written in the following, more concrete, forms:
zn=αn
+n+
βb(n–) nz(n+)
– w
b(n+) – wbn–
and
wn=αd
n(n+) β–
(n+)(n–) zd(n+)
– w –(n+) – w––n.
Competing interests
The author declares that he has no competing interests.
Received: 5 October 2016 Accepted: 17 November 2016
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