R E S E A R C H
Open Access
Some properties of algebraic difference
equations of first order
Yong Liu
**Correspondence:
[email protected] Department of Mathematics, Shaoxing College of Arts and Sciences, Shaoxing, Zhejiang 312000, P.R. China
Abstract
We prove that ifg(z) is a finite-order transcendental meromorphic solution of
(
cg(z))
2=A(z)g(z)g(z+c) +B(z),whereA(z) andB(z) are polynomials such that degA(z) > 0, then
1≤
ρ
(g) = max
λ
(g),λ
1
g
.
MSC: 30D35; 39B12
Keywords: meromorphic functions; difference equations; value distribution; finite order
1 Introduction
Steinmetz [] and Bank and Kaufman [] proved that the equation
gn=R(z,g)
can be reduced into a list of six simple differential equations by a suitable Möbius trans-formation with polynomial coefficients, which include
g=p(z)g–q(z)(g–ζ)(g–η), (.)
whereζ,ηare constant, andp(z),q(z) are rational functions. Letq(z)∈C. Then equation (.) can be transformed into
g=P(z)g– .
Ishizaki and Korhonen [] investigated meromorphic solutions of
g(z)=P(z)g(z)g(z+ ) –Q(z). (.)
They proved that equation (.) possesses a continuous limit to the equation
g=P(z)g– ,
which extends to solutions in certain cases.
We assume that the reader is familiar with the basic notions of Nevanlinna theory (see, e.g., [, ]). Of late, several scholars [, –] studied the properties of finite-order mero-morphic solutions of algebraic difference equations and obtained many interesting results.
For the special case of (.), Whittaker [] has shown that the equation
g(z+ ) =q(z)g(z),
whereq(z) is a meromorphic function of finite orderρ(q), has a meromorphic solutiong
such thatρ(q)≤ρ(g)≤ρ(q) + . Hereρ(g) denotes the order of growth of the meromor-phic functiong(z).
Chen [] has extended this above result and proved that the Pielou logistic equation
g(z+ ) = R(z)g(z)
P(z) +Q(z)g(z),
whereR(z),P(z), andQ(z) are polynomials withP(z)R(z)Q(z)≡, has a finite-order tran-scendental meromorphic solutiongsuch that ≤ρ(g).
Replacingg(z+ ) withg(z), Ishizaki [] was concerned with the growth and value distributions of transcendental meromorphic solutions of the algebraic difference equa-tion
g(z)=P(z)g(z).
In , Liu [] considered the Nevanlinna growth of an equation related to (.). It is interesting to consider some properties of (.), and our results will be stated in Section .
2 Main results
Theorem . Let c∈C\ {},and let A(z)and B(z)be polynomials such thatdegA(z) > .
If g(z)is a finite-order transcendental meromorphic solution of
cg(z)
=A(z)g(z)g(z+c) +B(z), (.)
then
≤ρ(g) =max
λ(g),λ g
.
Remark It is a curious problem to construct a transcendental meromorphic solution of
(.) for the casedegA> .
Theorem . Let c∈C\ {},and let E(z) = DF((zz))be an irreducible rational function,where D(z)and F(z)are polynomials withdegD(z) =d anddegF(z) =f.If the equation
cg(z)
has a rational solution
g(z) = H(z)
K(z)=
lhzh+· · ·+l
mkzk+· · ·+m
,
where lh(= ), . . . ,l,mk(= ), . . . ,mare constants,degH(z) =h,anddegK(z) =k.
(i) Ifd≥f andd–f is zero or an even number,then
h–k=d–f
.
(ii) Ifd<f,thenh–k=d–f.
Further, Example . shows that there exist rational solutions satisfying Theorem .(i), and Example . shows that there exist rational solutions satisfying Theorem .(ii).
Example . The equation
g(z+c) –g(z)=g(z+c)g(z) +c–z– ( +c)z– c–
has a rational solutiong(z) =z+ , whered= ,f = , andh–k= =d–f.
Example . The equation
g(z+c) –g(z)=g(z+c)g(z) +c
–z(z+c)
z(z+c)
has a rational solutiong(z) =z, whered= ,f= , andh–k= – = d–f.
3 Proof of Theorem 2.1
Lemma .([]) Let w(z)be a transcendental meromorphic solution of finite order of the difference equation
P(z,w) = ,
where P(z,w)is a difference polynomial in w(z)and its shift.If P(z,a)≡for a slowly mov-ing target function a,that is,T(r,a) =S(r,w),then
m r,
w–a
=S(r,w).
The following result obtained by Chiang and Feng [] and Halburd and Korhonen [, ] independently. We state here the form stated in [, Theorem .(b)].
Lemma .([]) Let c,cbe two arbitrary complex numbers,and let w(z)be a
mero-morphic function of finite orderρ.Letε> be given.Then there exists a subset E⊂(,∞)
of finite logarithmic measure such that,for all|z|=r∈/E∪[, ],we have
exp–r(ρ–+ε)≤w(z+c) w(z+c)
Firstly, we prove thatρ(g) =ρ≥. We consider the following two cases separately. Case .. Ifg(z) has infinitely many poles, we can pick a polezofg(z) such thatg(z) =
∞π, whereπ≥, then we deduce by (.) thatg(z
+c) =∞π, whereπ≥m. Substituting
z+cforzinto (.), we have
g(z+ c) –g(z+c)=A(z+c)g(z+ c)g(z+c) +B(z+c). (.)
Then (.) implies thatz+ cis a pole ofgof multiplicityπ≥π≥π.
Sinceg(z) has infinitely many poles, following the previous steps, we pick a polezof
g(z) such that
g(z+nc) =f(ξn) =∞πn,
whereπn≥π for alln∈N. Hence, we can choose a sequence{ξn=z+nc,n∈N}of
poles ofg(z), the multiplicity of which isπn≥π, so we obtainλ(g)≥, and therefore
ρ(g)≥λ(g)≥.
Case .. Ifg(z) is a transcendental meromorphic function with finitely many poles, then we can rewriteg(z) as
g(z) = g(z)
P(z), (.)
whereg(z) is a transcendental entire function, andP(z) is a polynomial. Substituting (.)
into (.), we have
g(z+c)
P(z+c) –
g(z)
P(z)
=A(z)g(z+c)
P(z+c)
g(z)
P(z) +B(z). (.)
By computing (.) we have
P(z)
P(z+c)
g(z+c)
g(z)
+P(z+c)
P(z)
g(z)
g(z+c)
= +A(z) +B(z)P(z)P(z+c)
g(z)g(z+c)
. (.)
We prove thatρ(g) =ρ(g) =ρ≥. Suppose, on the contrary to the assertion, thatρ(g) =
ρ(g) =ρ< . For any givenε( <ε<–ρ(g)), by Lemma . we obtain
g(z+c)
g(z)
≤exprρ(g)–+ε=expo(),
g(z)
g(z+c)
≤exprρ(g)–+ε=expo()
outside a finite logarithmic measure E. Aszk satisfies|g(zk)|=M(rk,g),|zk|=rk∈/ E,
where M is some finite constant, a contradiction, since degA(z) > . Hence we have
By Lemma . we obtain
mm(z(+z)c)≤exprρ(m)–+ε=expo(),
mm(z(+z)c)≤exprρ(m)–+ε=expo()
(.)
outside a finite logarithmic measure. By (.) and (.), as|z| → ∞, we obtain
A(z) + expqc+p=m(z+c)
m(z) exp
qc+p+ m(z)
m(z+c)exp
p
≤m(z+c) m(z) exp
qc+p+ m(z)
m(z+c)exp
p≤M
outside a finite logarithmic measure, whereM is a finite constant. This is impossible,
sincedegA(z) > .
Case . Suppose thatρ(g) > . Then
g(z) =m(z)expl(z), (.)
wherel(z) is a polynomial such thatρ(g) =degl(z) > , andm(z) is a meromorphic function such thatρ(m) <ρ(g). Substituting (.) into (.), we obtain
m(z+c)expl(z+c)–m(z)expl(z)=A(z)m(z+c)expl(z+c)m(z)expl(z). (.)
Let
l(z) =pkzk+pk–zk–+· · ·+pz+p,
wherepk= . Then
l(z+c) =pkzk+ (ckpk+pk–)zk–+Q(z), (.)
l(z+c) –l(z) = (ckpk)zk–+Q(z), (.)
whereQ(z) andQ(z) are polynomials of degree at mostk– . Equalities (.) and (.)
imply that
m(z+c)expckpkzk–+Q(z)+m(z) =A(z) + m(z+c)m(z)expckpkzk–+Q(z),
that is,
expckpkzk–+Q(z)=– m
(z)
m(z+c)+
A(z) + m(z)
m(z+c)exp
ckpkzk–+Q(z)
≤ m(z) m(z+c)
+A(z) + m(z)
m(z+c)exp
By Lemma . we obtain
whereMis a positive constant.
Supposeh=k. Then
a contradiction. Henceh=k, that is,
h–k= =d–f
Ifh=k, then using the similar method as before, we obtain
which is a contradiction, since DF((zz)) → asz→ ∞. Henceh<k. Substitutingg(z) = HK((zz)) into (.), we have
F(z)H(z+c)K(z) – F(z)H(z)H(z+c)K(z)K(z+c) +F(z)H(z)K(z+c)
=D(z)K(z)K(z+c). (.)
Since
degF(z)H(z+c)K(z) – F(z)H(z)H(z+c)K(z)K(z+c) +F(z)H(z)K(z+c)
=f+ h+ k,
degD(z)K(z)K(z+c)=d+ k.
From this and from (.) we have
h–k=d–f
.
Acknowledgements
The author would like to thank the referee for his/her helpful suggestions and comments. The work was supported by the NNSF of China (No. 10771121, 11401387), the NSF of Zhejiang Province, China (No. LQ 14A010007), the NSFC Tianyuan Mathematics Youth Fund (No. 11226094), the NSF of Shandong Province, China (No. ZR2012AQ020 and No. ZR2010AM030), and the Fund of Doctoral Program Research of Shaoxing College of Art and Science (20135018).
Competing interests
The author declares that he has no competing interests.
Author’s contributions
Author read and approved the final manuscript.
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Received: 11 June 2017 Accepted: 10 October 2017
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