R E S E A R C H
Open Access
Product-type system of difference
equations with a complex structure of
solutions
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
The solvability of the following system of difference equations
zn+1=
α
zanwnb, wn+1=β
wcn–2zn–2d , n∈N0,wherea,b,c,d∈Z,
α,β
,w–2,w–1,w0,z–2,z–1,z0∈C\ {0}, is studied in detail by usingseveral methods. The system has the most complex structure of solutions of all the related systems studied so far, and some of the forms of solutions appear for the first time.
MSC: 39A20; 39A45
Keywords: system of difference equations; product-type system; solvable in closed form
1 Introduction
Various types of difference equations and systems have been investigated in the last twenty years [–]. Many of the systems studied there, such as the ones in [, , –, –, – , –, –], 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., [–] for widely known methods). For some recent results, see, e.g., [, , , –, –]. 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 [] 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, [, , , , , ] and the references therein).
During the study of some equations and systems (for example, those in [] and []), we have noticed the importance of some classes of solvable ones in their investigation. Elim-inating the constant addends in the equations and systems in these two papers, product-type ones were obtained. It is known that product-product-type equations and systems are solvable if initial values are positive. But the standard method for solving them by using the loga-rithm is not suitable if some of the values are not positive. The corresponding product-type
system in [] is a special case of the following one:
xn=xan–kybn–l, yn=xdn–sycn–m, n∈N,
where k,l,m,s∈N. This observation has motivated us to investigate the solvability of product-type systems with non-positive initial values. However, there are several obsta-cles in the investigation. An obvious one is that the functionsfa(z) =za, whenais not an
integer, are multi-valued. The obstacle naturally suggests the choicea∈Zin dealing with the systems. Our first paper on product-type systems [] investigated the casek=m= , l=s= , and our original intention was to study, as usual, the long-term behavior of their solutions. Not so long after that we realized that a related system with three dependent variables was also solvable [], and that the casek=m= ,l=s= , was solvable ei-ther []. Bearing in mind that calculations in these papers were quite technical, we put aside the problem of investigating the long-term behavior of their solutions, and concen-trated our efforts on studying the solvability of related systems. In the next phase of the investigation we realized that adding the constant multipliers kept the solvability of such a system [], which was also verified for an extension of the system in [], in our paper []. Our further investigation showed that the solutions of solvable product-type systems can have several different forms for different values of parametersa,b,c,d, and that the forms can be obtained by a detailed analysis, which we did for the first time in [] and []. Later we investigated another system in [] where such analysis was unnecessary. A technical difficulty in studying the solvability problem led us to finding another method for getting solutions in []. Paper [] was the first paper which successfully dealt with an associated polynomial to a product-type system of the fourth order in detail. We would also like to say that the max-type system in [] was also solved by using product-type systems, as well as some equations in [].
An important fact connected to the investigation is that there are only several solvable product-type systems with two dependent variables due to the finite number of combina-tions of the sums of two delays which cannot exceed four. Our general task is to present all solvable product-type systems with two dependent variables.
Here we show that the following system
zn+=αzanwbn, wn+=βwcn–zdn–, n∈N, ()
wherea,b,c,d∈Z,α,β∈Candw–,w–,w,z–,z–,z∈C, is solvable, complementing
our previous results in [, , –, –]. In fact, we will consider the case when
α,β,w–,w–,w,z–,z–,z∈C\ {}, since otherwise trivial or not well-defined solutions
to system () are obtained. The system has the most complex structure of solutions to all the related systems studied so far, and some of the forms of solutions appear for the first time. The complexity forced us to use and develop almost all methods and tricks that we have used so far. We have to also mention that we regard thatki=–kai= ,k∈Z.
2 Auxiliary results
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/.
Two different proofs of the following known lemma can be found, for example, in [] and [].
Lemma Ifλj,j= ,k,are mutually different zeros of the polynomial
P(t) =aktk+ak–tk–+· · ·+at+a,
with aka= ,then
k
j=
λlj
P(λj)
=
for l= ,k– ,and
k
j=
λkj–
P(λj)
= ak
.
Lemma Let i∈Nand
s(ni)(z) = + iz+ iz+· · ·+nizn–, n∈N, ()
where z∈C. 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.
3 Main results
This section formulates and proves our main results. The first two results deal with the case whenc= and one of the parametersbanddis also zero.
Theorem Assume that a,d∈Z,b=c= ,α,β,w–,w–,w,z–,z–,z∈C\ {}.Then
the following statements are true.
(a) Ifa= ,then the solution to system()is given by
zn=α –an
–azan
, n∈N, ()
wn=βαd –an–
–a zdan–
, n≥. ()
(b) Ifa= ,then the solution to system()is given by
zn=αnz, n∈N, ()
wn=βαd(n–)zd, n≥. ()
Proof Sinceb=c= , we have
zn+=αzan, wn+=βzdn–, n∈N, ()
from which the following is obtained:
zn=α n–
j=ajzan
, n∈N. ()
Employing () in the second equation in (), we get
wn=βαd n–
j=ajzdan–
, n≥. ()
Theorem Assume that a,b∈Z,c=d= ,α,β,z,w∈C\ {}.Then the following
state-ments are true.
(a) Ifa= ,then the solution to system()is given by
zn=α –an
–aβb–an
– –a zan
wba
n–
, n≥, ()
wn=β, n∈N. ()
(b) Ifa= ,then the solution to system()is given by()and
zn=αnβb(n–)zwb, n≥. ()
Proof Sincec=d= , we have
zn+=αzanwbn, wn+=β, n∈N. ()
Employing the second equation in () in the first one, we get
zn+=αβbzna, n∈N,
from which, along with the factz=αzawb, it follows that
zn=
αβb
n–
j=ajzan– =α
n–
j=ajβb n–
j=ajzan wba
n–
, n≥. ()
From (), formulas () and () easily follow.
Theorem Assume that a,b,c,d∈Z,ac=bd,α,β,w–,w–,w,z–,z–,z∈C\{}.Then
system()is solvable in closed form.
Proof The assumptionα,β,w–,w–,w,z–,z–,z∈C\ {}along with () impliesznwn=
,n≥–. Hence, from () we have
wbn=zn+
αza n
, n∈N, ()
and
wbn+=βbwbcn–zbdn–, n∈N, ()
which together imply
zn+=α–cβbzan+zcn–zbdn––ac, n≥. ()
We also have
z=αzawb, z=α+aβbzbd–za wbc–wab,
z=α+a+a
βb(+a)zabd– zbd–zawabc–wbc–wab.
Now we follow our method presented, for example, in [], p. , and [], Theorem ..
Further, from () we have
Settingk=n– in () and employing (), we get
zn+=δyn–zan–z bn– z
cn– z
dn–
=α–cβbyn–α+a+aβb(+a)zabd– zbd–zawabc–wbc–waban–
×α+aβbzbd–zawbc–wabbn–αzawbcn–zdn–
=α(–c)yn–+(+a+a)an–+(+a)bn–+cn–βbyn–+b(+a)an–+bbn–
×zabdan–+bdbn–
– z
bdan– – z
aa
n–+abn–+acn–+dn–
×wabcan–+bcbn–
– w
bcan– – w
aban–+abbn–+bcn–
=αyn+–cyn–βbyn+zbdan
– z bdan– – z
an+–can–
w
bcan
– w bcan– – w
ban+
()
forn≥.
The equalities in () show that fork≥
ak=aak–+bak–+cak–+dak–. ()
The same equation is also satisfied by sequencesbk,ckanddk,k∈N, due to the relations
bk=ak+–aak,ck=bk+–bak,dk=dak–.
The assumptiond=bd–ac= , along with () and (), shows that it must be
a–=a–=a–= , a= , ()
and
y–=y–=y–=y= , y= , ()
(see, for example, the corresponding calculations in []). From () and sincey=a, we obtain
yk= k–
j=
aj, k∈N. ()
It is clear that closed-form formulas for solutions to problem ()-() can be easily found, from which along with () and Lemma closed-form formulas foryncan also be
found, from which along with () the solvability of system () follows. Further, we have
znd–= wn+
βwc n–
, n∈N ()
and
which together imply
wn+=αdβ–awan+wcn+wbdn–ac, n∈N. ()
We also have
w=βwc–zd–, w=βw–c zd– and w=βwczd. ()
Following the lines of the above method, it is proved that
wn+=ηykwank+–kwnbk+–kwcnk+–kwdnk+–k, ()
fork,n∈N,n≥k– , whereη=αdβ–a, (ak)k∈N, (bk)k∈N, (ck)k∈N and (dk)k∈N satisfy ()
and (), and where (yk)k∈Nsatisfies () and ().
From () withk=n+ and by using (), we get
wn+=ηyn+wan+w bn+ w
cn+ w
dn+
=αdβ–ayn+βwczdan+βwc–zd–bn+βwc–zd–cn+wdn+
=αdyn+β(–a)yn++an++bn++cn+wccn+ – w
cbn+ – w
can++dn+
×zdcn+ – z
dbn+ – z
dan+
=αdyn+βyn+–ayn+wc(an+–aan+)
– w
c(an+–aan+)
– w
an+–aan+
×zd(an+–aan+)
– z
d(an+–aan+)
– z
dan+
()
forn∈N.
From this and since the closed-form formulas forakandykcan be found as above, the
solvability of () follows. It is easily checked that () and () present a solution to
sys-tem ().
Corollary Assume that a,b,c,d∈Z,ac=bd,α,β,w–,w–,w,z–,z–,z∈C\{}.Then
the general solution to system()is given by()and(),where(ak)k≥–satisfies()and
(),while(yk)k≥–is given by()and().
Now we conduct a detailed analysis of the form of sequencesakandykappearing in the
proof of Theorem . The reason why equation () is solvable whenac=bdis based on the fact that its characteristic polynomial
p(λ) =λ–aλ–cλ+ac–bd ()
is of the forth degree, so, solvable by radicals. The equationp(λ) = is equivalent to
λ–a λ+
s
–
a +s
λ–
as –c
λ+s
+bd–ac
Now choose parametersso that (as– c)= (a+ s)(s+ bd– ac) (see, for example,
Recall that solutions to () are found in the following form:s=u+v, by posing the conditionuv= –p/. Sinceu+v= –qanduv= –p/, we have thatu andv are
where
= (ac– bd), ()
=
ac–abd+c, ()
Q= –a– c. ()
According to Lemma , we see that the nature ofλi’s,i= , , depends on the signs of
=
–, ()
P= –a ()
and
D= ac– bd– a. ()
All zeros of pare different and none of them is equal to .By Lemma we see that such
a situation appears if= andp()= . From () we see thatis certainly negative if
< , that is, if ac< bd. For example, ifa= ,c= andbd= , then< ,p() = –=
and
p(λ) =λ–λ– λ– .
Since due to (),Pcannot be positive,pcannot have two pairs of complex-conjugate
zeros.
It is well known that in the case the general solution to () has the form
an=γλn+γλn+γλn+γλn, n∈N, ()
whereγi,i= , , are constants.
By Lemma we also have
j=
λlj
p(λj)
= forl= , and
j=
λj
p(λj)
= . ()
From this and (), it is obtained
an=
λn+
(λ–λ)(λ–λ)(λ–λ)
+ λ
n+
(λ–λ)(λ–λ)(λ–λ)
+ λ
n+
(λ–λ)(λ–λ)(λ–λ)
+ λ
n+
(λ–λ)(λ–λ)(λ–λ)
()
forn≥–.
Using () into () and applying () four times, we get
yn= n–
j=
i=
λji+
p(λi)
=
i=
λi(λni – ) p(λi)(λi– )
, n∈N, ()
All zeros of pare different and one of them is equal to .Polynomialphas as a zero if
p() = –a–c+ac–bd= , that is, if
(a– )(c– ) =bd. ()
Hence
p(λ) =λ–aλ–cλ+a+c–
= (λ– )λ+ ( –a)λ+ ( –a)λ+ –a–c. ()
We may assume thatλ= . To calculate the other three zeros ofp, the following equation
should be solved:
λ+ ( –a)λ+ ( –a)λ+ –a–c= . ()
By using the change of variablesλ=s+a– , equation () is transformed in () with
p=( –a)(a+ )
and q= –a–c– (a– )
–
(a– )
. ()
Hence
λj=
a–
+sj, j= , , ()
where
sj=
–q –
q
+ p
ε
j–+
–q +
q
+ p
ε
j–, j= , , ()
εis a complex zero of the equationt= , andpandqare given in ().
For example, if a= and c= , thenbd= and = , so by Lemma ,p has four
different zeros such that exactly one of them is equal to , and
p(λ) =λ– λ– λ+ = (λ– )
λ– λ– λ– . ()
Formula () holds withλ= . Further, we have
yn= n–
j=
p()+
n–
j=
i=
λji+
p(λi)
= n – a–c+
i=
λi(λni – ) p(λi)(λi– )
()
forn∈N. In fact, using () it is shown that () also holds forn= –j,j= , . From this and by Corollary , we get the following result.
(a) If(a– )(c– )=bd,then the general solution to system()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–c= ,then the general solution to system()is given by()and(),where(an)n≥–is given by()withλ= ,(yn)n≥–is given by(),
λ= ,whileλj,j= , ,are given by()and().
phas exactly one double zero which is different from .Ifa= ,c= andbd= –, then
p(λ) =λ– λ+ = (λ– )
λ+ λ+ ,
is a polynomial with exactly one double zeroλ,= = and two complex conjugate zeros
λ,= –±i
√ .
In such cases, that is, whenλ=λ,λi= λj, ≤i,j≤, we have
an= (γ+γn)λn+γλn+γλn, n∈N, ()
whereγi,i= , , are constants, and the solution satisfying () can be obtained, for
exam-ple, by lettingλ→λin (), which yields (see [])
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)
. ()
Combining () and () and using Lemma , we have
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– )
. ()
phas exactly one double zero equal to .Polynomialp has as a double zero if ()
holds and
p() = – a–c= , ()
that is,c= – a, which implies that
p(λ) =λ–aλ+ (a– )λ+ – a= (λ– )
It will be exactly a double zero ifp() = – a= , that is,a= . C\ {}.Then the following statements are true.
or
Hence, for everya= , polynomialphas two pairs of equal zeros. It will have integer
coefficients only ifa= aˆ, for someaˆ∈Z\ {}. Note that sincea(±√)/= , for every a∈Z, cannot be a double zero ofp.
In such cases the general solution to () has the following form:
an= (γ+γn)λn+ (γ+γn)λn, n∈N, ()
from which along with () and by Lemma , it follows that
=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,ac=bd,=D= ,andα,β,z–,z–,z,w–,w–,
w∈C\ {}.Then the following statements are true.
(a) If(a– )(c– )=bd,then the general solution to system()is given by()and(),
where(an)n≥–is given by(),(yn)n≥–is given by(),whileλ,are given by(). (b) Characteristic polynomial()cannot have two pairs of double zeros such that one
of them is equal to.
phas a triple zero.In this case it must be== which is equivalent to== ,
that is, ifbd= ac/ and
= c
a+ c/.
If c= , thenbd= and consequentlyac=bd, which is impossible. Ifc= –a/, then bd= –a/. Letλ=at, then
p(λ) =λ–aλ+
a
λ– a
=a
t–t+ t –
=a
t–t+
t–
=a
t–
t+
=
λ–a
λ+a
.
Hence, for everya= ,a/ is a triple zero ofp, andpcannot have a zero of the fourth
order. Fora= ˆa,aˆ∈Z\ {}, polynomialphas integer coefficients, and fora= it has
a triple zero equal to , which could be also obtained by further analyzing the polynomial in ().
Letλj,j= , , be the zeros of polynomialp. In the case, we may assume thatλ=λ=λ
andλ=λ. Then, in the case, the general solution to () has the following form:
an=
c+cn+cn
λn+cλn, n∈N, ()
Since () must hold, the following system also holds:
By solving system (), we get
c= –
from which along with () it follows that
an=
Combining () and (), using Lemma and by some calculation, we get
yn=
Combining () and (), using Lemma and by some calculation, we get
Corollary Assume that a,b,c,d∈Z,ac=bd== ,D= ,andα,β,z–,z–,z,
w–,w–,w∈C\ {}.Then the following statements are true.
(a) If the triple zero of polynomial()is different from,then the general solution to system()is given by()and(),where(an)n≥–is given by(),(yn)n≥–is given
by(),λj=a/,j= , andλ= –a/.
(b) If the triple zero of polynomial()is equal to,sayλ=λ=λ= ,which is
equivalent toa= ,c= –andbd= –,then the general solution to system()is given by()and(),where(an)n≥–is given by(),(yn)n≥–is given by(),while
λ= –.
Now we deal with the caseac–bd= ,c= .
Theorem Assume that a,b,c,d∈Z,ac=bd,c= andα,β,z–,z–,z,w–,w–,w∈ C\ {}.Then system()is solvable in closed form.
Proof As above, the conditionα,β,w–,w–,w,z–,z–,z∈C\ {}along with () implies
znwn= ,n≥–. Hence, ()-() hold, which along with the conditionac=bdyields
zn+=α–cβbzna+zcn–, n≥. ()
Letδ=α–cβb,
a=a, b= , c=c, y= . ()
Then equation () can be written as
zn+=δyzna+zbnz c
n–, n≥. ()
Further, from (), we get
zn+=δy
δza nz
b n–z
c n–
a
zb nz
c n–
=δy+azaa+b
n z
ba+c n– z
ca n–
=δyza nz
b n–z
c
n–, ()
forn≥, where
a:=aa+b, b:=ba+c, c:=ca, y:=y+a. ()
Suppose
zn+=δykzank+–kz bk
n+–kz ck
n–k ()
fork∈N\ {}and everyn≥k+ , and
yk=yk–+ak–. ()
Then, by using relation () withn→n–kinto (), we obtain
zn+=δyk
δza n+–kz
b n–kz
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–kz
ck+
n–k–, ()
forn≥k+ , where
ak+:=aak+bk, bk+:=bak+ck, ck+:=cak, ()
yk+:=yk+ak. ()
From (), (), ()-() and the induction is obtained that ()-() hold fork,n∈
Nsuch that ≤k≤n– .
Settingk=n– in () and employing (), we obtain
zn+=δyn–zan–z bn– z
cn–
=α–cβbyn–α+a+aβb(+a)zabd– zbd–zawabc–wbc–waban–
×α+aβbz–bdzawbc–wabbn–αzawbcn–
=α(–c)yn–+(+a+a)an–+(+a)bn–+cn–βbyn–+b(+a)an–+bbn–
×zabdan–+bdbn–
– z
bdan– – z
aa
n–+abn–+acn–
×wabcan–+bcbn–
– w
bcan– – w
aban–+abbn–+bcn–
=αyn+–cyn–βbyn+zbdan
– z bdan– – z
aan+ w
bcan
– w bcan– – w
ban+
()
forn≥.
From (), we get
ak=aak–+bak–+cak–, k≥, ()
which along withbk=ak+–aakandck=cak–implies thatbk andck also satisfy the
equation.
Using the conditionc=c= along with () and (), we easily get
a–= , a–= , a= ()
and
From (), () and sincey=a, we obtain
yk= k–
j=
aj, k∈N. ()
It is clear that closed-form formulas for solutions to () can be easily found, from which along with () and Lemma closed-form formulas foryncan be also found, from
which along with () the solvability of () follows.
On the other hand, we also have that ()-() hold, from which along with the condition ac=bdit follows that
wn+=αdβ–awan+wcn+ ()
forn∈N.
As above, it can be proved that for allk,n∈Nsuch that ≤k≤n
wn+=ηykwank+–kwnbk+–kwcnk+–k, n≥k– , ()
whereη=αdβ–a, (ak)k∈N, (bk)k∈Nand (ck)k∈Nsatisfy () and (), while (yk)k∈Nsatisfies
() and ().
From () withk=n+ and by using (), we get
wn+=ηyn+wan+w bn+ w
cn+
=αdβ–ayn+βwczdan+βwc–zd–bn+βwc–zd–cn+
=αdyn+β(–a)yn++an++bn++cn+wccn+ – w
cbn+ – w
can+
×zdcn+ – z
dbn+ – z
dan+
=αdyn+βyn+–ayn+wc(an+–aan+)
– w
c(an+–aan+)
– w
can+
×zd(an+–aan+)
– z
d(an+–aan+)
– z
dan+
()
forn∈N.
From this and since the closed-form formulas for (ak)k≥–and (yk)k≥–can be found as
above, the solvability of () follows. It is easily checked that () and () present a solution to system () in this case.
Now we conduct a detailed analysis of the form of sequencesakandykappearing in the
proof of Theorem . The reason why equation () is solvable whenc= is based on the fact that its characteristic polynomial
p(λ) =λ–aλ–c ()
is of the third degree, so, solvable by radicals.
By using the change of variablesλ=s+a, the equationp(λ) = is transformed into the
following one:
s–a
s–
a+ c
By using formula (), we get () with
All zeros of pare different and one of them is equal to .Polynomialpwill have a zero
equal to ifa+c= , so that
The general solution to () in this case has the following form:
an=αλn+αλn+αλn, n∈N, ()
This along with () implies that
Ifλi= ,i= , , then from formula () it follows that
yn=
λ(λn– ) (λ–λ)(λ–λ)(λ– )
+ λ
(λn– )
(λ–λ)(λ–λ)(λ– )
+ λ
(λn– )
(λ–λ)(λ–λ)(λ– )
()
forn∈N(in fact, () holds for everyn≥–).
If one of the zeros is equal to one, sayλ, then =λ=λ= , and we have
yn=
λ (λn – )
(λ–λ)(λ– )
+ λ
(λn– )
(λ–λ)(λ– )
+ n
(λ– )(λ– )
()
forn∈N(in fact, () holds for everyn≥–).
Corollary Assume that a,b,c,d∈Z,ac=bd,c= ,= andα,β,z–,z–,z,w–,
w–,w∈C\ {}.Then the following statements are true.
(a) Ifa+c= ,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 one of the zeros of characteristic polynomial()is equal to,sayλ,i.e.,if
a+c= anda= ,then the general solution to()is given by formulas()and
(),where(an)n≥–is given by()withλ= ,(yn)n≥–is given by(),andλ,
are given by().
One of the zeros is double.In this case it must be
= , that is,c(a+ c) = .
Since the casec= is excluded, it must bec= –a/, from which it follows that
p(λ) =λ–aλ+
a
.
Since in this case it must be alsop(λ) = , it follows thatλ,= a/ is a double zero and
p(λ) =
λ–a
λ+a
.
In order thatc∈Z, it is clear that it must bea= aˆ for someaˆ∈Z. Since a/= when a∈Z, the polynomial cannot have as a double zero, and consequently cannot have as a triple zero.
Ifλ=λ=λ, then the general solution to () has the following form:
an=αˆλn+ (αˆ+αˆn)λn, n∈N, ()
whereαˆi∈R,i= , . Since, in our case, condition () must be satisfied, the solution
(an)n≥–to () can be found by lettingλ→λin (), so that
an=
λn++ (λ– λ+n(λ–λ))λn+
(λ–λ)
()
From () and () and the fact thata= , we have
yn= n–
j=
aj= n–
j=
λj++ (λ– λ+j(λ–λ))λj+
(λ–λ)
()
for everyn∈N.
From () and Lemma , it follows that
yn=
λ(λn – ) (λ–λ)(λ– )
+(λ– λ)λ(λ
n – )
(λ–λ)(λ– )
+λ
( –nλn–+ (n– )λn)
(λ–λ)(λ– )
()
forn∈N(in fact, () holds also for everyn≥–).
If we assume thatλ= andλ=λ= , which is possible ifa= –, then from () it
follows that
yn=
n (λ– )
+(λ– )λ(λ
n – )
(λ– )
+λ
( –nλn–+ (n– )λn)
(λ– )
()
for everyn∈N(in fact, () holds also for everyn≥–).
Corollary Assume that a,b,c,d∈Z,ac=bd,c= ,
= ,andα,β,z–,z–,z,w–,
w–,w∈C\ {}.Then the following statements are true.
(a) Ifa+c= ,then the general solution to()is given by()and(),where(an)n≥–
is given by(),(yn)n≥–is given by(),λ= –a/andλ,= a/.
(b) If only one of the zeros of polynomial()is equal to one,sayλ,that is,ifa= –
andc= ,then the general solution to system()is given by()and(),where (an)n≥–is given by()withλ= ,(yn)n≥–is given by(),λ= andλ,= –. (c) It is not possible that two zeros of polynomial()are equal to one.
Triple zero case.In this case it must bep(λ) =p(λ) =p(λ) = . Fromp(λ) = it is
obtained thatλ=a/. Sincep(λ) = λ– aλ, we see thata/ is its root only ifa= , which would imply thatp(λ) =λ–c, but this polynomial has a triple zero if and only
ifc= , which contradicts the assumptionc= . Hence, polynomial () cannot have a triple zero.
Competing interests
The author declares that he has no competing interests.
Author’s contributions
The author has contributed solely to the writing of this paper. He read and approved the manuscript.
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Received: 28 February 2017 Accepted: 24 April 2017 References
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