R E S E A R C H
Open Access
The zeros of difference of meromorphic
solutions for the difference Riccati equation
Chang-Wen Peng
1*and Zong-Xuan Chen
2*Correspondence:
pengcw716@126.com
1School of Mathematics and
Computer Sciences, Guizhou Normal College, Guiyang, Guizhou 550018, China
Full list of author information is available at the end of the article
Abstract
In this paper, we mainly investigate some properties of the transcendental
meromorphic solutionf(z) for the difference Riccati equationf(z+ 1) =p(zf)(fz()+z)+s(qz)(z). We obtain some estimates of the exponents of the convergence of the zeros and poles of
f(z) and the difference
f
(z) =f(z+ 1) –f(z).MSC: 30D35; 39B12
Keywords: difference Riccati equation; Borel exceptional value; admissible meromorphic solution
1 Introduction and main results
Early results for difference equations were largely motivated by the work of Kimura [] on the iteration of analytic functions. Shimomura [] and Yanagihara [] proved the following theorems, respectively.
Theorem A[] For any polynomial P(y),the difference equation
y(z+ ) =Py(z)
has a non-trivial entire solution.
Theorem B[] For any rational function R(y),the difference equation
y(z+ ) =Ry(z)
has a non-trivial meromorphic solution.
Letf be a function transcendental and meromorphic in the plane. The forward differ-ence is defined in the standard way byf(z) =f(z+ ) –f(z). In what follows, we assume the reader is familiar with the basic notions of Nevanlinna’s value distribution theory (see, e.g., [–]). In addition, we use the notationsσ(f) to denote the order of growth of the meromorphic functionf(z), andλ(f) andλ(f) to denote the exponents of convergence of zeros and poles off(z), respectively. Moreover, we say that a meromorphic functiongis small with respect tof ifT(r,g) =S(r,f), whereS(r,f) =o(T(r,f)) outside of a possible ex-ceptional set of finite logarithmic measure. Denote byS(f) the family of all meromorphic
functions which are small compared tof(z). We say that a meromorphic solutionf of a difference equation is admissible if all coefficients of the equation are inS(f).
Recently, a number of papers (including [–]) focused on complex difference equa-tions and difference analogs of Nevanlinna’s theory. As the difference analogs of Nevan-linna’s theory were being investigated, many results on the complex difference equations have been got rapidly. Many papers (including [, , , ]) mainly dealt with the growth of meromorphic solutions of difference equations.
In [], Halburd and Korhonen used value distribution theory to obtain Theorem C.
Theorem C[] Let f(z)be an admissible finite order meromorphic solution of the equa-tion
f(z+ )f(z– ) =c(f(z) –c+)(f(z) –c–) (f(z) –a+)(f(z) –a–)
=:Rz,f(z), (.)
where the coefficients are meromorphic functions,c≡anddegf(R) = .If the order of the
poles of f(z)is bounded,then either f(z)satisfies a difference Riccati equation
f(z+ ) =p(z)f(z) +q(z) f(z) +s(z) ,
where p,q,s∈S(f),or(.)can be transformed by a bilinear change in f(z)to one of the equations
f(z+ )f(z– ) =γf
(z) +δλzf(z) +γ μλz
(f(z) – )(f(z) –γ) ,
f(z+ )f(z– ) =f
(z) +δeiπz/λzf(z) +μλz
f(z) – ,
whereλ∈C,andδ,μ,γ,γ =γ(z– )∈S(f)are arbitrary finite order periodic functions such thatδandγ have periodandμhas period.
From the above, we see that the difference Riccati equations are an important class of difference equations, they will play an important role in research of difference Painlevé equations. Some papers [–, , ] dealt with complex difference Riccati equations.
In this research, we investigate some properties of the difference Riccati equation and prove the following theorems.
Theorem . Let p(z),q(z),s(z)be meromorphic functions of finite order,and let p(z)f(z) + q(z), f(z) +s(z)be relatively prime polynomials in f.Suppose that f(z)is a finite order admissible transcendental meromorphic solution of the difference Riccati equation
f(z+ ) =p(z)f(z) +q(z)
f(z) +s(z) . (.)
Then
(i) λ(
f) =σ(f).Moreover,ifq(z)≡,thenλ(
f) =λ(f) =σ(f);
(ii) λ(f) =σ(f) =σ(f);λ(f f
Theorem . Let p(z),q(z),s(z)be rational functions,and let p(z)f(z) +q(z),f(z) +s(z)be relatively prime polynomials in f.Suppose that f(z)is a finite order transcendental mero-morphic solution of the difference Riccati equation(.).Then:
(i) Ifp(z)≡s(z),and there is a nonconstant rational functionQ(z)satisfying
q(z) =Q(z),then
λ(f) =λ
f f
=σ(f).
(ii) Ifp(z)≡–s(z),s(z)is a nonconstant rational function,and there is a rational functionh(z)satisfyings(z) +q(z) =h(z),then
λ(f) =λ
f f
=σ(f).
(iii) Ifp(z)≡ ±s(z),and there is a nonconstant rational functionm(z)satisfying
(s(z) –p(z))+ q(z) =m(z),then
λ(f) =λ
f f
=σ(f).
(iv) Ifp(z),q(z),s(z)are polynomials anddegp(z),degq(z),degs(z)contain just one maximum,thenf(z)has no nonzero Borel exceptional value.
2 The proof of Theorem 1.1
We need the following lemmas to prove Theorem ..
Lemma .(see [, ]) Let f be a transcendental meromorphic solution of finite orderσ of the difference equation
P(z,f) = ,
where P(z,f)is a difference polynomial in f(z)and its shifts.If P(z,a)≡for a slowly moving target function a,i.e. T(r,a) =S(r,f),then
m
r, f –a
=S(r,f)
outside of a possible exceptional set of finite logarithmic measure.
Lemma .(see []) Let f be a transcendental meromorphic solution of finite orderσ of a difference equation of the form
H(z,f)P(z,f) =Q(z,f),
where H(z,f)is a difference product of total degree n in f(z)and its shifts,and where P(z,f), Q(z,f)are difference polynomials such that the total degreedegQ(z,f)≤n.Then for each ε> ,
mr,P(z,f)=Orσ–+ε+S(r,f)
Lemma .(Valiron-Mohon’ko) (see []) Let f(z)be a meromorphic function.Then for all irreducible rational functions in f,
Rz,f(z)=a(z) +a(z)f(z) +· · ·+am(z)f(z)
m
b(z) +b(z)f(z) +· · ·+bn(z)f(z)n
with meromorphic coefficients ai(z) (i= , , . . . ,m),bj(z) (j= , , . . . ,n),the characteristic
function of R(z,f(z))satisfies
Tr,Rz,f(z)=dT(r,f) +O (r),
where d=degfR=max{m,n}and (r) =maxi,j{T(r,ai),T(r,bj)}.
In the remark of [], p., it is pointed out that Lemma . holds.
Lemma .(see []) Let f be a nonconstant finite order meromorphic function.Then
N(r+ ,f) =N(r,f) +S(r,f), T(r+ ,f) =T(r,f) +S(r,f)
outside of a possible exceptional set of finite logarithmic measure.
Remark . In [], Chiang and Feng proved that iff is a meromorphic function with exponent of convergence of polesλ(f) =λ<∞,η= fixed, then for eachε> ,
Nr,f(z+η)=N(r,f) +Orλ–+ε+O(logr).
Lemma .(see []) Let f(z)be a meromorphic function with orderσ=σ(f) < +∞,and letηbe a fixed non-zero complex number,then for eachε> ,we have
Tr,f(z+η)=T(r,f) +Orσ–+ε+O(logr).
Lemma .(see []) Let g: (, +∞)→R,h: (, +∞)→R be non-decreasing functions. If(i)g(r)≤h(r)outside of an exceptional set of finite linear measure,or(ii) g(r)≤h(r), r∈/ H∪(, ],where H⊂(,∞)is a set of finite logarithmic measure,then for anyα> , there exists r> such that g(r)≤h(αr)for all r>r.
Lemma .(see []) Letη,ηbe two complex numbers such thatη=ηand let f(z)be
a finite order meromorphic function.Letσbe the order of f(z),then for eachε> ,we have
m
r,f(z+η) f(z+η)
=Orσ–+ε.
Proof of Theorem. (i) Suppose thatf(z) is an admissible transcendental meromorphic solution of finite orderσ(f) of (.).
First, we proveλ(
f) =σ(f).
By (.), we have
By Lemma . and (.), we get
mr,f(z+ )=Orσ(f)–+ε+S(r,f) (.)
outside of a possible exceptional set of finite logarithmic measure. From (.) and Lemma ., we have
Tr,f(z+ )=Tr,f(z)+S(r,f). (.)
Hence, by (.) and (.), we conclude that
Nr,f(z+ )=Tr,f(z)+Orσ(f)–+ε+S(r,f) (.)
outside of a possible exceptional set of finite logarithmic measure. By Lemma ., we get
Nr,f(z+ )≤Nr+ ,f(z)=Nr,f(z)+S(r,f). (.)
Hence, by (.) and (.), we conclude that
Nr,f(z)≥Tr,f(z)+Orσ(f)–+ε+S(r,f) (.)
outside of a possible exceptional set of finite logarithmic measure. By Lemma . and (.), we getλ(f) =σ(f).
Second, we proveλ(f) =λ(f) =σ(f) whenq(z)≡. By (.), we have
Pz,f(z):=f(z+ )f(z) +s(z)–p(z)f(z) –q(z) = .
Hence, we get
P(z, ) = –q(z)≡. (.)
Thus, by (.) and Lemma ., we see that
m
r, f
=S(r,f)
outside of a possible exceptional set of finite logarithmic measure. Thus, we have
N
r, f
=T(r,f) +S(r,f) (.)
outside of a possible exceptional set of finite logarithmic measure. By Lemma . and (.), we getλ(f) =σ(f). Soλ(f) =λ(f) =σ(f).
(ii) Suppose thatf(z) is an admissible transcendental meromorphic solution of finite orderσ(f) of (.). By (.), we get
By (.) and Lemma ., we have
m(r,f) =Orσ(f)–+ε+S(r,f) (.)
outside of a possible exceptional set of finite logarithmic measure. From Lemma . and (.), we get
mr,f(z)≤mr,f(z)+m
r,f(z) f(z)
=Orσ(f)–+ε+S(r,f) (.)
outside of a possible exceptional set of finite logarithmic measure. By (.), we get
f=p(z)f(z) +q(z) f(z) +s(z) –f(z) =
p(z)f(z) +q(z) –f(z)(f(z) +s(z))
f(z) +s(z) . (.)
Sincep(z)f(z) +q(z) andf(z) +s(z) are relatively prime polynomials inf, andf(z)(f(z) +s(z)) andf(z) +s(z) have a common factorf(z) +s(z). Thereforep(z)f(z) +q(z) –f(z)(f(z) +s(z)) andf(z) +s(z) are relatively prime polynomials inf. By Lemma . and (.), we get
T(r,f) = T(r,f) +S(r,f). (.)
By (.) and (.), we see that
N(r,f) = T(r,f) +Orσ(f)–+ε+S(r,f) (.)
outside of a possible exceptional set of finite logarithmic measure. Hence, by Lemma . and (.), we get
λ
f
=σ(f) =σ(f).
ByN(r,f) =N(r,ff ·f)≤N(r,ff) +N(r,f) and (.), we get
N
r,f f
≥N(r,f) –N(r,f)≥T(r,f) +Orσ(f)–+ε+S(r,f) (.)
outside of a possible exceptional set of finite logarithmic measure. Hence, by Lemma . and (.), we haveλ(f
f
)≥σ(f). We haveσ(ff)≤σ(f). Thus, we have
λ
f f
=σ
f
f
=σ(f).
3 The proof of Theorem 1.2
Suppose thatf(z) is a transcendental meromorphic solution of finite orderσ(f) of (.). (i) By (.), we get
f = –f
(z) + (s(z) –p(z))f(z) –q(z)
f(z) +s(z) . (.)
Sinces(z)≡p(z) andq(z) =Q(z), by (.), we get
f = –f
(z) –Q(z)
f(z) +s(z) = –
(f(z) –Q(z))(f(z) +Q(z))
f(z) +s(z) . (.)
By (.), we have
Pz,f(z):=f(z+ )f(z) +s(z)f(z+ ) –p(z)f(z) –q(z) = . (.)
By (.), we see that
Pz,Q(z)=Q(z+ )Q(z) +s(z)Q(z+ ) –p(z)Q(z) –q(z) (.)
and
Pz, –Q(z)=Q(z+ )Q(z) –s(z)Q(z+ ) +p(z)Q(z) –q(z). (.)
Ifp(z)≡, thenP(z,Q(z)) =P(z, –Q(z)) =Q(z+ )Q(z) –q(z). IfP(z,Q(z)) =P(z, –Q(z))≡ , thenQ(z+)Q(z) =q(z) =Q(z). Moreover, we getQ(z+)≡Q(z). This is a contradiction
sinceQ(z) is a nonconstant rational function. SoP(z,Q(z)) =P(z, –Q(z))≡.
Suppose thatp(z)≡. IfP(z,Q(z))≡ andP(z, –Q(z))≡, by (.) and (.), we get
p(z)Q(z+ ) –Q(z)≡.
Thus, we know thatQ(z+ )≡Q(z). This is a contradiction sinceQ(z) is a nonconstant rational function. Therefore, we getP(z,Q(z))≡ orP(z, –Q(z))≡. Without loss of gen-erality, we assume thatP(z,Q(z))≡. By Lemma ., we get
m
r, f(z) –Q(z)
=S(r,f)
possibly outside of an exceptional set of finite logarithmic measure. Moreover, we get
N
r, f(z) –Q(z)
=T(r,f) +S(r,f) (.)
possibly outside of an exceptional set of finite logarithmic measure.
Ifzis a common zero off(z) –Q(z) andf(z) +s(z), thenQ(z) +s(z) = . Ifzis a zero
andf(z) +s(z) are relatively prime polynomials inf, we see thatp(z)s(z)≡q(z), that is,
possibly outside of an exceptional set of finite logarithmic measure. Thus, by Lemma ., we see thatλ(f)≥σ(f).
(ii) We divide this proof into the following two cases.
Case Suppose thatq(z)≡. Sinces(z)≡–p(z) andh(z) =s(z) +q(z), by (.), we get
Ifzis a common zero off(z) +s(z) –h(z) andf(z) +s(z), thenh(z) = . Ifzis a zero
possibly outside of an exceptional set of finite logarithmic measure. Thus, by Lemma ., we see thatλ(f)≥σ(f).
By a similar method to above, we have
N
By a similar method to above, we getλ(ff)≥σ(f). Hence,
) –p(z+ ) +m(z+ )≡s(z) –p(z) +m(z), ors(z+ ) –p(z+ ) –m(z+ )≡s(z) –p(z) –m(z).
for allroutside of a possible exceptional set with finite logarithmic measure. Moreover, we get
for allroutside of a possible exceptional set with finite logarithmic measure.
Ifzis a common zero off(z) +s(z)–p(z)–m(z) andf(z) +s(z), then –s(z)+p(z)+m(z)= . If
By (.), we get
Pz,p(z) –s(z)=p(z+ ) –s(z+ )p(z) –s(z) +s(z)–p(z)p(z) –s(z).
That is,
Pz,p(z) –s(z)=p(z)p(z+ ) –s(z+ )–p(z) –s(z).
Byq(z)≡ andm(z) = (p(z) –s(z))+ q(z), we getm(z) = (p(z) –s(z)). Sincem(z) is
a nonconstant rational function, we see thatp(z) –s(z) is a nonconstant rational function too. Sop(z+ ) –s(z+ )≡p(z) –s(z). Therefore, we haveP(z,p(z) –s(z))≡. By Lemma .,
possibly outside of an exceptional set of finite logarithmic measure. Moreover, we get
N
possibly outside of an exceptional set of finite logarithmic measure.
Ifzis a common zero off(z) +s(z) –p(z) andf(z) +s(z), thenp(z) = . Ifzis a zero of
(iv) Suppose thatf(z) is a finite order transcendental meromorphic solution of (.). Without loss of generality, we assume that degp(z) > max{degq(z),degs(z)}. Then
degp(z)≥. Setp(z) =akzk+· · ·+a(ak= ). Leta= . By (.), we have
P(z,a) =a+as(z) –ap(z) +q(z) = –aakzk+· · · ≡ (.)
sinceaak= . By Lemma . and (.), we conclude that
m
r, f –a
=S(r,f)
possibly outside of an exceptional set of finite logarithmic measure. Thus, we get
N
r, f –a
=T(r,f) +S(r,f) (.)
possibly outside of an exceptional set of finite logarithmic measure. Thus, by Lemma . and (.), we get
λ(f –a) =σ(f).
That is,f(z) has no nonzero Borel exceptional value. Theorem . is proved.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
CWP completed the main part of this article, CWP and ZXC corrected the main theorems. All authors read and approved the final manuscript.
Author details
1School of Mathematics and Computer Sciences, Guizhou Normal College, Guiyang, Guizhou 550018, China.2School of
Mathematical Sciences, South China Normal University, Guangzhou, Guangdong 510631, China.
Acknowledgements
The authors thank the referee for his/her valuable suggestions. This work is supported by the State Natural Science Foundation of China (No. 61462016), the Science and Technology Foundation of Guizhou Province (Nos. [2014]2125; [2014]2142), and the Natural Science Research Project of Guizhou Provincial Education Department (No. [2015]422).
Received: 3 July 2015 Accepted: 12 October 2015 References
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