R E S E A R C H
Open Access
Solution to the solvability problem for a
class of product-type systems of difference
equations
Stevo Stevi´c
**Correspondence: [email protected] Mathematical Institute, 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
Solution to the solvability problem for a class of product-type systems of difference equations is given by presenting explicit formulas for its solutions. In the main/most complicated cases the problem is solved by using two different methods. This is the last system, out of the three non-equivalent ones, for which an associated polynomial, which essentially determines the structure of the solutions, is of the third order, so that the paper finishes the study of this kind of two-dimensional systems.
MSC: 39A20; 39A45
Keywords: solvability problem; system of difference equations; product-type system; solutions in closed form
1 Introduction
Theory of difference equations is a mathematical area of great interest [–]. Concrete systems is one of its subareas which is of some interest nowadays. Papers [–] by Pa-paschinopoulos and Schinas are some of those which have influenced the interest. Later many others have appeared [, –, , , , , –, –]. One of the oldest topics is finding solutions to the equations and systems [, –]. Since the mid of s there has been increasing interest in the topic [, , –, –]. An interesting fact is that many nonlinear equations and systems are related to linear ones although they are of relatively complex forms, which usually do not suggest it on the first site. A line of the investigation related to a solvable nonlinear difference equation can be followed, for example, in [, , , ] (see also the references therein). A more complicated equation and related meth-ods can be found in []. The corresponding and related systems of difference equations were studied, for example, in [, ] and [] (see also the references therein). Another system of interest can be found in []. The equations and systems in these papers are solved by employing suitable changes of variables which transform them to some linear equations and systems which are solvable.
Many equations and systems contain as special cases product-type ones [, ]. Some methods for the study of the long-term behavior of their solutions have used directly or indirectly the equations and systems which can be solved. In [] and [] we studied only positive solutions to the equations and systems appearing therein, so the product-type
ones, which are obtained from them by special choices of their parameters, were solvable. Applying the logarithm is a standard method for solving such equations and systems; it was, among others, also used in solving a system in []. In general, a more complex case is studying non-positive solutions to difference equations and systems, which is also the case for the product-type systems, since the previous method is useless because it produces multi-valued solutions. In our papers [] and [], we have presented a two- and a three-dimensional solvable product-type system in the complex plane, respectively. Many results on the long-term behavior of solutions to the system in [] were presented therein, while in [] we only gave a method for solving the corresponding system. Having published these two papers has motivated us to look for other solvable product-type systems. In [] we have managed to solve another one. The fact that the systems in [] and [] have been special cases of the following system
zn=zan–kwbn–k, wn=wcn–kzdn–k, n∈N, ()
has motivated us to start investigating the solvability of other special cases of the system. Some equations in [] are product-type ones, but they have additional multipliers, which suggested us to study the solvability of the generalizations of system () containing some multipliers. The first step in the study was made in []. Our further investigations sug-gested that closed-form formulas for solutions to product-type systems can be presented for all given values of parameters and multipliers, which for two specific systems of the form in () was done in [] and [] for the first time (in [] and [] general formulas were presented in the main cases, but without detailed analysis and without presenting more concrete formulas for each possible case; this type of problems does not appear for the system in [], where its solutions are presented in all the cases). Product-type sys-tems can be also solved by using the method which we developed in [], but is somewhat technically complicated with respect to the one developed in [, , ]. Another class of product-type systems has been quite recently investigated in detail in [].
This paper is devoted to the study of an extension of system (), with the following delays
k=k= andk=k= , that is, of the system:
zn=αzan–wbn–, wn=βwcn–zn–d , n∈N, ()
wherea,b,c,d∈Z,α,β,z–,z–,w–,w–∈C. It is also a natural continuation of our study
in [, –, –].
This is the last system, out of the three non-equivalent ones, for which an associated polynomial, which essentially determines the structure of the solutions, is of the third order, so that the paper finishes the study of this kind of systems with two dependent variables. The case when some of initial values or parametersαandβare equal to zero is excluded from the consideration since such solutions are either not well defined [] or eventually equal to zero. If a finite sum is of the forml–j=lcj, for somel∈Z, we regard that its value is zero.
2 Auxiliary results
Lemma Let
Q(t) =
k
j= cjtj=ck
k
j=
(t–tj)
such that ti=tj,i=j,and ck= .Then
k
j= ts
j Q(tj)=
for≤s≤k– ,and
k
j= tk–j Q(tj)=
ck.
The second lemma is well known ([, ], see also [] for a more general result).
Lemma Let i∈Nand
s(i)n(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) ()
for every z∈C\ {}and n∈N.
3 Main results
Five cases will be considered separately. Before we formulate and prove our main results, note that
z=αza–wb–, z=α+aβbzbd–za–wbc–wab–,
w=βwc–zd–, w=βwc–z–d.
()
Theorem Assume that a,b,d∈Z,c= ,bd= ,α,β,z–,z–,w–∈C\ {}.Then
(a) ifa= ,the general solution to()is given by
zn=α–an
+
–a βb––anazan+
– wba
n
– , n∈N ()
and
wn=βαd
–an–
–a zdan–
(b) ifa= ,the general solution to()is given by
zn=αn+βbnz–wb–, n∈N ()
and
wn=αd(n–)βz–d, n≥. ()
Proof Sincec= , we have
zn=αzan–wbn–, wn=βzdn–, n∈N. ()
From the equations in () and the assumptionbd= , we get
zn=αβbzan–, n∈N, ()
and consequently
zn=
αβb n–
j=ajzan
, n∈N,
which along with () yields
zn=α n
j=ajβb
n–
j=ajzan+
– wba
n
– , n∈N. ()
From () we easily get () and ().
Using () in the second equality in (), as well as the assumptionbd= , we obtain
wn=αd
n–
j=ajβ+bd
n–
j=ajzdan–
– wbda
n–
–
=αd n–
j=ajβzdan–
– , n≥. ()
From () we easily get () and () forn≥. Forn= , equalities () and () are directly
verified.
Theorem Assume that a,c,d∈Z, c= , b= , α,β,z–,z–,w–,w–∈C\ {}.Then system()is solvable in closed form.
This theorem was essentially proved in [], Theorem , (see also [], Corollary , for the closed form formulas for the solutions in all the cases), since whenb= system () is
zn=αzan–, wn=βwcn–zdn–, n∈N,
which is nothing but the corresponding system in Theorem in [] with indices shifted backward for one. Hence, we omit the proof of the theorem.
Theorem Assume that a,b,c∈Z,c= ,d= ,α,β,z–,w–,w–∈C\ {}.Then system
This theorem was essentially proved in [], since whend= system () is
zn=αzan–wbn–, wn=βwcn–, n∈N,
which is nothing but the corresponding system in Theorem . in [] whose indices are shifted backward for one (see also [], Corollary ., for the closed form formulas for the solutions in all the cases). Hence, the proof of the theorem will be also omitted.
Theorem Assume that a,b,c,d∈Z\ {},ac=bd,α,β,z–,z–,w–,w–∈C\ {}.Then system()is solvable in closed form.
Proof Sinceα,β,z–,z–,w–,w–∈C\ {}, from () we getzn= =wnforn≥–. Hence
wbn–= zn αza n–
, n∈N
and
wbn=βbwbcn–zbdn–, n∈N,
from which it follows that
zn+=α–cβbzanzcn–zbd–acn– , n∈N. ()
Letμ=α–cβb,
a=a, b=c, c=bd–ac, y= . ()
Then
zn+=μyza
nz b
n–z c
n–, n∈N. ()
Equality () implies
zn+=μyμza
n–z b
n–z c
n–
a
zb
n–z c
n–
=μy+azaa+b
n– z ba+c
n– z ca
n–
=μyza
n–z b
n–z c
n–, ()
forn≥, where
a:=aa+b, b:=ba+c, c:=ca, y=y+a. ()
Assume
zn+=μykzan+–kk z bk
n–kz ck
for ak≥ and alln≥k, and
Relations (), (), (), () along with the inductive argument confirm the conjectures in ()-().
The relations in () show that
ak=aak–+bak–+cak–, fork≥. ()
Sincec=bd–ac= , () yields
ak–=
ak–aak––bak–
c . ()
This equality enables us to calculateajforj≤. It is obtained that
a–=a–= , a= , ()
(see detailed corresponding calculations in [] and []) from which along with (), it is obtained
and
yk=
k–
j=
aj, k∈N. ()
Linear equation () can be solved, from which along with () a closed form formula forak is obtained. Using () and Lemma , a closed form formula foryk is obtained. Employing such obtained formulas forakandykin (), we obtain a closed form formula
for the solution to (). From (), we also have
zn–d = wn βwc
n–
, n∈N, ()
znd=αdzadn–wbdn–, n∈N, ()
and consequently
wn+=αdβ–awan+wncwbd–acn– , n∈N. ()
As above, we get
wn+=ηykwan+–kk wn+–kbk wcn–kk , n≥k– , ()
where η=αdβ–a, ak,bk andck are defined by () and (), andyk is defined by ()
and ().
Settingk=n+ in () and using (), we get
wn+=ηyn+wan+w bn+
w cn+
–
=αdβ–ayn+βwc–zd–an+βwc–z–d bn+wcn+
–
=αdyn+β(–a)yn++an++bn+zdbn+
– z dan+
– w cbn+
– w can++cn+
–
=αdyn+βyn+–ayn+zd(an+–aan+)
– z
dan+
– w
c(an+–aan+)
– w
an+–aan+
– ()
forn∈N.
As we have already mentioned, closed form formulas forakandykcan be found, from
which along with () it follows that () is solvable. It is not difficult to see that formulas
() and () present a solution to system (), from which the theorem follows.
Remark Theorem can be also proved by using the method in []. We include the proof also for the benefit of the reader and since in the theorem that follows we follow the lines of the method.
First note that
w=βwczd=β
βwc–zd–cαza–wb–d=αdβ+czcd–z–adwc–wbd–,
w=βwczd=αd(+a)β+c+bdz–bdz–(a+c)dwbcd–wabd+c– .
cn:=acn–+bγn–, γn:=dcn–+cγn–, ()
dn:=adn–+bδn–, δn:=ddn–+cδn–, ()
forn∈N.
Relations ()-(), ()-() define xn,yn,an,bn,cn,dn,un,vn,αn,βn,γn,δn uniquely,
whileznandwnin () are defined uniquely by initial valuesz–,z–,w–,w–. Sinceznand wndefined by () and () are solutions to () such that their values forn= , are the same, it follows that they are closed form formulas for solutions to system (). Hence, we need only to prove that systems ()-() are solvable.
The conditiond= implies
xn–=un–cun–
d , n∈N, ()
which along with the first equality in () yields
un+=aun++cun+ (bd–ac)un–+d, n∈N, ()
and also
yn–=vn–cvn––
d , n∈N, ()
which along with the first equality in () yields
vn+=avn++cvn+ (bd–ac)vn–+ –a, n∈N. ()
We also have
an–=
αn–cαn–
d , ()
bn–=
βn–cβn–
d , ()
cn–=γn–cγn–
d , ()
dn–=δn–cδn–
d , ()
forn∈N, and consequently
αn+=aαn++cαn+ (bd–ac)αn–, ()
βn+=aβn++cβn+ (bd–ac)βn–, ()
γn+=aγn++cγn+ (bd–ac)γn–, ()
δn+=aδn++cδn+ (bd–ac)δn–, ()
forn∈N.
Let
A(tn) =tn+–atn+–ctn+ (ac–bd)tn–, n≥–, ()
where (tn)n≥–is a sequence. ThenAis a linear operator, () can be written asA(un) =d, n∈N, () asA(vn) = –a,n∈N, and ()-() as
A(αn) =A(βn) =A(γn) =A(δn) = , n∈N.
Since the equation
A(tn) = ()
is linear of third order, it can be solved, and specially closed form formulas for αn, βn,
γn,δncan be obtained for the corresponding initial conditions ()-(). Employing the
formulas forαn,βn,γn,δnin ()-(), formulas foran,bn,cn,dnare obtained.
Whenh∈R, the equation
A(tn) =h ()
can be also solved (especially forh=dandh= –a), since a particular solution to the equation can be found in the following form:
tn=g+gn+gn+gn
for some constantsgj,j= , (see, e.g., [, ]). From this and using ()-(), formulas forunandvncan be found, from which along with () and () formulas forxnandyn
are obtained. Using the obtained formulas for these twelve sequences in () and (), we obtain closed form formulas for solutions to system ().
3.1 Forms of solutions to (2) for the caseac=bd
The characteristic polynomial associated to () is
P(λ) =λ–aλ–cλ+ac–bd, ()
which is of the third order due to the assumptionac=bd. By using the change of variables λ=t+a, the equationP(λ) = becomes
t+pt+q= ,
where p= –(a+ c)/ andq= (ac– bd– a)/, which is solved by looking for
solutions in the formt=u+vand posing the condition uv= –p. Thenu andv are solutions of the equationz+qz–p = , from which it follows that
λj= a
+ε
j–
–q –
q
+
p
+ε
j–
–q +
q
+
p
, j= , , ()
If the discriminant
:= p+ q ()
is such that = , then all three zeros are different (if > two zeros are complex con-jugate, if < all three zeros are real), whereas if = , at least two zeros are equal [].
Case = . Since in this case the zeros ofPare different, the general solution to () is
un=αλn+αλn+αλn, n∈N, ()
whereαi,i= , , are constants (in fact, () holds for everyn∈Zdue to the assumption ac=bd; see the relation in ()).
Lemma employed toPshows that
Equalities () and () yield
and
yn= λ
(λn– )
(λ–λ)(λ– )
+ λ
(λn– )
(λ–λ)(λ– )
+ n
(λ– )(λ– )
. ()
Using () it is easily checked that () holds for n≥– (for a detailed explanation, see []).
Corollary Assume that a,b,c,d∈Z\ {},α,β,z–,z–,w–,w–∈C\ {},ac=bd and
= .Then the following statements hold.
(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(),andλj,j= , ,are given by
().
(b) If(a– )(c– ) =bdanda+c= ,then the general solution to system()is given by
()and(),where(an)n≥–is given by()withλ= ,(yn)n≥–is given by(),
λ= ,whileλ,are given by().
Remark As we have already mentioned,Phas a zero equal to if and only if (a– )(c– ) =bd, so that () holds. The condition = will be satisfied if and only if
ac– (a– )(c– ) – a= a+ c.
For example, ifa=c∈Z\ {, }andbd= (a– )= , then = ,
P(λ) =λ–aλ–aλ+ a– = (λ– )
λ– (a– )λ+ – a,
and conditions of Corollary (b) are satisfied. Note thatλ= and
λ,=
a– ±√a+ a–
.
ThusPcan have three different zeros, one of which is equal to .
Case = .In this case,Phas at least two equal zeros. We may assume thatλ=λ. If p= , thenλis not a triple zero ofP, so the general solution to () is
an=gˆλn + (gˆ +gnˆ )λn, n∈N, ()
wheregjˆ,j= , , are constants. Since the solution satisfies (), we can find it by letting λ→λin (), and obtain
an=λ
n+
– (n+ )λλn+ + (n+ )λn+
(λ–λ)
, n≥–, ()
from which along with (), we get
yn= n–
j=
λj+ + (λ– λ+j(λ–λ))λj+
(λ–λ)
If we assumeλ= =λ=λ, then () and Lemma imply
If we assumeλ=λ=λ= , then from () and some calculation, we obtain
yn=
n–
j=
(j+ )(j+ )
=
n(n+ )(n+ )
()
for everyn∈N, which obviously holds forj= –, –, (for more details, see [, ]).
Corollary Assume that a,b,c,d∈Z\ {},α,β,z–,z–,w–,w–∈C\ {},ac=bd and
= .Then the following statements are true.
(a) If(a– )(c– )=bdanda+ c= ,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– ) =bd,a+c= anda= ,then the general solution to system()is given by()and(),where(an)n≥–is given by()withλ= ,(yn)n≥–is given by
(),andλ=a– .
(c) If(a– )(c– ) =bd,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= ,
whileλ,are given by().
(d) If(a– )(c– )=bdanda+ c= ,then the general solution to system()is given by
()and(),where(an)n≥–is given by(),(yn)n≥–is given by(),whileλj=a/, j= , .
(e) If(a– )(c– ) =bdanda+c= anda= ,then the general solution to system()
is given by()and(),where(an)n≥–is given by()withλ= ,while(yn)n≥–is
given by().
Remark Recall that equation () will have a zero equal to one if (a– )(c– ) =bd. If λ=m∈Z\ {}is a double zero ofP, then it must be
P(λ) =λ–aλ–cλ+cm+am–m, ()
P(m) = m– am–c= . ()
From () we have thatc= m– am. Using this in (), we get
P(λ) =λ–aλ+am– mλ+ m–am= (λ–m)(λ–a+ m). ()
From this, and sinceac=bd, it follows thata= m. SinceP(m)= , we also have that it must bea= m. Hence, for everym∈Z\ {}anda∈Zsuch that m=a= m, the polynomials in () are of type (), and they have exactly two equal zerosλ=λ=m=
λ=a– m, and if additionallym= anda= m+ hold, we see that the conditions
of Corollary (a) are satisfied. Form= anda= , it is obtained thatPcan have as a double zero, which means that such polynomials satisfy the conditions of Corollary (b). Fora= m+ , we get a polynomial which has a double zero different from , which is its third zero. This means that such polynomials satisfy the conditions of Corollary (c).
The polynomial has three equal zeros only ifq= , which along with = impliesp= . Hence, we havec= –a/ and ac= bd, which implies thatc=a/, so that
P(λ) =λ–aλ+ a
λ–
a
= λ–
a
Sincecmust be an integer, it follows thata= aˆ for someaˆ ∈Z, so thatc= –aˆ and c=aˆ. Hence, for eachaˆ∈Z\ {}, these relations give polynomials of the form in ()
which have three equal zeros. This means that such polynomials satisfy the conditions of Corollary (d). Whena= , polynomialPhas as a triple zero, which means that there
is a polynomial satisfying the conditions of Corollary (e).
Before we formulate our next result, we want to say that the solution to the difference equation
This simple result is well known and can be essentially found in any book on difference equations and many papers [, , ].
vn+=avn++cvn+ –a, ()
αn+=aαn++cαn, ()
βn+=aβn++cβn, ()
γn+=aγn++cγn, ()
δn+=aδn++cδn ()
forn∈N, while operatorAdefined in () becomes
A(tn) =tn+–atn+–ctn.
Now, equation () is of the second order, so solvable in closed form. Consequently, ()-() are solvable, so that closed form formulas forαn,βn,γn,δncan be found by employing
the corresponding conditions in ()-(). Such obtained formulas along with ()-() yield formulas foran,bn,cn,dn. Further, () is also solvable, since a particular solution to the equation has the form
tn=dˆ +dnˆ +dnˆ
for some constantsdjˆ,j= , . Specially, it is solved forf =dandf = –a, which gives formulas for the sequencesunandvn, and consequently closed form formulas forxnand
yn. Using such obtained formulas in () and (), we get formulas for solutions to system () in this case.
(a)Case a+ c= ,a+c= . In this case, a particular solution to equation () is
upn= d
–a–c, ()
so that
un=cλn+cλn+ d
–a–c, n∈N,
where
λ,=
a±√a+ c
, ()
from which along withu=u= it follows that
c=
d(λ– )
(a+c– )(λ–λ)
and c=
d( –λ)
(a+c– )(λ–λ)
,
and consequently
un= d
–a–c
(λ– )λn– (λ– )λn
λ–λ
+
Using () in (), we get
Particular solution to equation () in this case is
vpn= –a
By using the corresponding conditions in ()-() in formula (), we get
Using (), (), ()-() in () and (), as well as the fact thatλ,are roots of the
polynomialλ–aλ–c, and after some standard calculations, formulas () and () are obtained.
(b)Case a+ c= ,a+c= . In this case, we haveλ
=λ= and a particular solution
to equation () is given by (), so that
un= (c+cn)λn + d
–a–c, n∈N,
whereλis given by (), from which along withu=u= it follows that
c= d
a+c– and c=
d( –λ)
(a+c– )λ
,
and consequently
un= d a+c–
λ+n( –λ)
λn– – . ()
Using () in (), we get
xn=
a+c–
–(n+ )λ+ (n+ )λ+c(n– )λ–cn
λn– +c– . ()
Particular solution to equation () in this case is given by (), so that
vn= (d+dn)λn+ –a
–a–c, n∈N,
whereλis given by (), from which along withv=v= it follows that
d= c
a+c– and d=
c( –λ)
(a+c– )λ
,
and consequently
vn=
a+c–
cλ+ ( –λ)n
λn– +a– . ()
Using () in (), we get
yn= c
d(a+c– )
–(n+ )λ+ (n+ )λ+c(n– )λ–cn
λn– –a. ()
By using the corresponding conditions in ()-() in formula (), we get
αn=d( –n)λn, ()
βn=dnλn– , ()
γn=c( –n)λn, ()
By using ()-() in ()-(), we get
an=c(n– ) – (n+ )λλn, ()
bn=(n+ )λ–cnλn– , ()
cn=c
d
c(n– ) – (n+ )λλn, ()
dn= c d
(n+ )λ–cnλn– . ()
Using the equalities in ()-(), in () and (), formulas () and () are ob-tained.
(c)Case a+ c= ,a+c= . In this case, we have thatλ= =λand that a particular
solution to equation () is given by
up n=
dn
–a, ()
so that
un=cλn+c+ dn
–a, n∈N,
whereλ= –c=a– , from which along withu=u= it follows that
c= –c= d (a– )(λ– )
= d
(a– ),
and consequently
un= d
a– λn– λ–
–n
=d(λ
n
–nλ+n– )
(a– )
=d((a– )
n+n( –a) +n– )
(a– ) . ()
Using () in (), we get
xn=
a(a– )n+–n(a– )– (a– ) +n+
(a– ) . ()
Particular solution to equation () in this case is given by
vpn=n( –a)
–a , ()
so that
vn=dλn+d+n( –a)
–a , n∈N,
from which along withv=v= it follows that
d= –a
( –a) and d= –
–a
and consequently
vn=
–(a– )n++ ( –a)– +a+n( –a)( –a)
( –a) . ()
Using () in (), we get
yn=
–a(a– )n++na( –a)( –a) +a(a– )
d( –a) . ()
By using the corresponding conditions in ()-() in formula (), we get
αn=dλ
λn– – –λ
=d(a– )(a– )
n––
–a , ()
βn=d
λn– λ–
=d(a– ) n–
a– , ()
γn=cλ
λn– – –λ
= (a– )(a– )
n––
a– , ()
δn=c
λn – λ–
= (a– )(a– )
n–
–a . ()
By using ()-() in ()-(), we get
an=a(a– )(a– )
n–
–a , ()
bn=a(a– ) n+–
a– , ()
cn=a(a– )
((a– )n– )
d(a– ) , ()
dn=a(a– )((a– )
n+– )
d( –a) . ()
Using (), (), ()-() in () and (), formulas () and () are obtained. (d)Case a+ c= ,a+c= . In this case, we have thata= andc= –. We also have
thatλ=λ= , and that a particular solution to equation () is given by
upn=dn
, ()
so that
un=c+cn+dn
, n∈N,
from which along withu=u= it follows thatc= andc= –d, and consequently
un=
d(n– )n
Using () in (), we get
xn=n+n+ . ()
Particular solution to equation () in this case is given by
vpn= –n
, ()
so that
vn=d+dn–n
, n∈N,
from which along withv=v= it follows thatd= andd= /, and consequently
vn=
+n–n
. ()
Using () in (), we get
yn= – n(n+ )
d =
bn(n+ )
. ()
By using the corresponding conditions in ()-() in formula (), we get
αn=d( –n), ()
βn=dn, ()
γn=n– , ()
δn= –n. ()
By using ()-() in ()-(), we get
an= –n, ()
bn= n+ , ()
cn=n
d = –bn, ()
dn= –
(n+ )
d =b(n+ ). ()
Using (), (), ()-() in () and (), formulas () and () are obtained. By some calculation it can be proved that formulas ()-() really satisfy system (),
which was checked by the author.
Competing interests
The author declares that he has no competing interests.
Author’s contributions
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Received: 25 April 2017 Accepted: 11 May 2017
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