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# Complex oscillation of meromorphic solutions for difference Riccati equation

- Yang-Yang Jiang
^{1}Email author, - Zhi-Qiang Mao
^{1}and - Min Wen
^{2}

**2014**:247

https://doi.org/10.1186/1687-1847-2014-247

© Jiang et al.; licensee Springer. 2014

**Received:**4 April 2014**Accepted:**5 September 2014**Published:**24 September 2014

The Erratum to this article has been published in Advances in Difference Equations 2015 2015:8

## Abstract

In this paper, we investigate zeros and *α*-points of meromorphicsolutions $f(z)$ for difference Riccati equations, and we obtain someestimates of exponents of convergence of zeros and *α*-points of$f(z)$ and shifts $f(z+n)$, differences $\mathrm{\Delta}f(z)=f(z+1)-f(z)$, and divided differences $\frac{\mathrm{\Delta}f(z)}{f(z)}$.

**MSC:** 30D35, 39B12.

## Keywords

- Riccati equation
- meromorphic solution
- difference
- complex oscillation

## 1 Introduction and main results

In this paper, we assume that the reader is familiar with the standard notations andbasic results of Nevanlinna’s value distribution theory (see [1, 2]). In addition, we use the notions $\sigma (f)$ to denote the order of growth of the meromorphic function$f(z)$, $\lambda (f)$, and $\lambda (\frac{1}{f})$ to denote the exponents of convergence of zeros and polesof $f(z)$, respectively. We say a meromorphic function$f(z)$ is oscillatory if $f(z)$ has infinitely many zeros.

The theory of difference equations, the methods used in their solutions, and their wideapplications have advanced beyond their adolescent stage to occupy a central position inapplicable analysis. The theory of oscillation play an important role in the research ondiscrete equations, and it is systematically introduced in [3]. The complex oscillation is the development and deepening of thecorresponding real oscillation, and it can profoundly reveals the essence of theoscillation problem that the property of oscillation is investigated in complex domain.

Recently, as the difference analogs of Nevanlinna’s theory were being investigated [4–6], many results on the complex difference equations have been got rapidly. Manypapers [4, 7–9] mainly deal with the growth of meromorphic solutions of some differenceequations, and several papers [7, 8, 10–15] deal with analytic properties of meromorphic solutions of some nonlineardifference equations. Especially, there has been an increasing interest in studyingdifference Riccati equations in the complex plane [8, 10, 12, 15].

where *A* is a polynomial, $\delta =\pm 1$. In [10], Chen and Shon investigated the existence and forms of rational solutions,and the Borel exceptional value, zeros, poles, and fixed points of transcendentalsolutions, and they proved the following theorem.

**Theorem A** *Let*$\delta =\pm 1$*be a constant and*$A(z)=\frac{m(z)}{n(z)}$*be an irreducible nonconstant rational function*,*where*$m(z)$*and*$n(z)$*are polynomials with*$degm(z)=m$*and*$degn(z)=n$.

*If*$f(z)$

*is a transcendental finite order meromorphic solution of the difference Riccatiequation*

*then*

- (i)
*if*$\sigma (f)>0$,*then*$f(z)$*has at most one Borel exceptional value*; - (ii)
$\lambda (\frac{1}{f})=\lambda (f)=\sigma (f)$;

- (iii)
*if*$A(z)\not\equiv -{z}^{2}-z+1$,*then the exponent of convergence of fixed points of*$f(z)$*satisfies*$\tau (f)=\sigma (f)$.

In [15], the first author investigated fixed points of meromorphic functions$f(z)$ for difference Riccati equation (1), and obtain someestimates of exponents of convergence of fixed points of $f(z)$ and shifts $f(z+n)$, differences $\mathrm{\Delta}f(z)=f(z+1)-f(z)$, and divided differences $\frac{\mathrm{\Delta}f(z)}{f(z)}$.

In this paper, we investigate zeros and *α*-points of meromorphic solutions$f(z)$ for difference Riccati equations (1), and we obtain someestimates of the exponents of convergence of zeros and *α*-points of$f(z)$ and shifts $f(z+n)$, differences $\mathrm{\Delta}f(z)=f(z+1)-f(z)$, and divided differences $\frac{\mathrm{\Delta}f(z)}{f(z)}$ of meromorphic solutions of (1). We prove the followingtheorem.

**Theorem 1.1**

*Let*$\delta =\pm 1$

*be a constant and*$A(z)$

*be a nonconstant rational function*.

*Set*$\mathrm{\Delta}f(z)=f(z+1)-f(z)$.

*If there exists a nonconstant rationalfunction*$s(z)$

*such that*$A(z)=-{s}^{2}(z)$,

*then every finite order transcendental meromorphicsolution*$f(z)$

*of the difference Riccati equation*(1),

*itsdifference*$\mathrm{\Delta}f(z)$,

*and divided difference*$\frac{\mathrm{\Delta}f(z)}{f(z)}$

*are oscillatory and satisfy*

**Theorem 1.2**

*Let*$A(z)$

*be a nonconstant rational function*.

*If*

*α*

*is a non*-

*zero complex constant*,

*thenevery finite order transcendental meromorphic solution*$f(z)$

*of the difference Riccati equation*

*satisfies*

- (i)
*if*$\alpha \ne -1$,*then*$\lambda (f(z+n)-\alpha )=\sigma (f)$, $n=0,1,2,\dots $ ; - (ii)
*if there is a rational function*$n(z)$*satisfying*$A(z)=\frac{{\alpha}^{2}}{4(1+\alpha )}-(1+\alpha ){n}^{2}(z),$

*then*$\lambda (\frac{\mathrm{\Delta}f(z)}{f(z)}-\alpha )=\sigma (f)$;

- (iii)
*if there is a rational function*$m(z)$*satisfying*$A(z)=\frac{{\alpha}^{2}+\alpha}{4}-{m}^{2}(z),$

*then*$\lambda (\mathrm{\Delta}f(z)-\alpha )=\sigma (f)$.

**Example 1.1**The function $f(z)=\frac{Q(z)-2z(z-1)(z+1)}{zQ(z)+{z}^{2}(z-1)(z+1)}$ satisfies the difference Riccati equation

where $A(z)=-\frac{2}{z(z+1)}$, $Q(z)$ is a periodic function with period 1. Note that for any$\rho \in [1,+\mathrm{\infty})$, there exists a prime periodic entire function$Q(z)$ of order $\sigma (Q)=\rho $ by Ozawa [17]. Thus $\sigma (f)=\sigma (Q)=\rho \ge 1$.

## 2 Lemmas for proofs of theorems

Firstly we need the following lemmas for the proof of Theorem 1.1.

**Lemma 2.1**

*Let*$A(z)$

*be a nonconstant rational function*,

*and*$f(z)$

*be a nonconstant meromorphic function*.

*Then*

*have at most finitely many common zeros*.

*Proof*Suppose that ${z}_{0}$ is a common zero of ${y}_{1}(z)$ and ${y}_{2}(z)$. Then ${y}_{2}({z}_{0})=1-f({z}_{0})=0$. Thus, $f({z}_{0})=1$. Substituting $f({z}_{0})=1$ into ${y}_{1}(z)$, we obtain

Since $A(z)$ is a nonconstant rational function,$A(z)+1$ has only finitely many zeros. Thus,${y}_{1}(z)$ and ${y}_{2}(z)$ have at most finitely many common zeros. □

**Lemma 2.2**

*Let*$w(z)$

*be a nonconstant finite order transcendental meromorphic solution of the differenceequation of*

*where*$P(z,w)$

*is a difference polynomialin*$w(z)$.

*If*$P(z,\alpha )\not\equiv 0$

*for a meromorphic function*$\alpha (z)$

*satisfying*$T(r,\alpha )=S(r,w)$,

*then*

*holds for all* *r* *outside of a possible exceptional set with finitelogarithmic measure*.

## 3 Proof of Theorem 1.1

Suppose that $\delta =1$. We only prove the case $\delta =1$. We can use the same method to prove the case$\delta =-1$.

First, we prove that $\lambda (\mathrm{\Delta}f(z))=\sigma (f(z))$.

*t*is an integer, if$t\ge 0$, then $P(z)={z}^{t}{P}_{0}(z)$, $Q(z)={Q}_{0}(z){e}^{-h(z)}$; if $t<0$, then $P(z)={P}_{0}(z)$, $Q(z)={z}^{-t}{Q}_{0}(z){e}^{-h(z)}$. Combining Theorem A with the property of thecanonical product, we have

*ε*($0<\epsilon <\sigma (f(z))-\sigma (P(z))$),

holds for all *r* outside of a possible exceptional set with finite logarithmicmeasure.

Thus (10) is a contradiction. Hence, (4) holds, that is, $\lambda (\mathrm{\Delta}f(z))=\sigma (f(z))$.

Thus, by this and (4), we see that $\lambda (\frac{\mathrm{\Delta}f(z)}{f(z)})=\sigma (f)$.

## 4 Proof of Theorem 1.2

- (i)First, we prove that the conclusion holds when $n=0$. Set $y(z)=f(z)-\alpha $. Thus, $y(z)$ is transcendental, $T(r,y)=T(r,f)+O(logr)$, and $S(r,y)=S(r,f)$. Substituting $f(z)=y(z)+\alpha $ into (2), we obtain${K}_{0}(z,y)=[y(z+1)+\alpha ][1-y(z)-\alpha ]-A(z)-y(z)-\alpha =0.$

*r*outside of a possible exceptional set with finite logarithmicmeasure. That is,

holds for all *r* outside of a possible exceptional set with finite logarithmicmeasure. Thus, we obtain $\lambda (f(z)-\alpha )=\sigma (f(z))$.

Using the same method as in the proof of (4)-(11), we can prove that (13) holds. Hence$\lambda (f(z+1)-\alpha )=\sigma (f(z))$.

*z*by $z+n-1$ ($n\ge 1$), and we obtain

- (ii)Suppose that there is a rational function $n(z)$ satisfying$A(z)=\frac{{\alpha}^{2}}{4(1+\alpha )}-(1+\alpha ){n}^{2}(z).$(16)

where $R(z)=(1+\alpha ){n}^{2}(z)-n(z)n(z+1)+\frac{2+\alpha}{2(1+\alpha )}n(z+1)-\frac{2+3\alpha}{2(1+\alpha )}n(z)$. Since $\frac{(3+\alpha ){\alpha}^{2}}{4{(1+\alpha )}^{2}}$ is a constant, to prove ${K}_{1}(z,0)\not\equiv 0$, we need to prove that $R(z)$ is nonconstant.

- (iii)Suppose that there is a rational function $m(z)$ satisfying$A(z)=\frac{{\alpha}^{2}+\alpha}{4}-{m}^{2}(z).$(23)

Using the same method as in the proof of (21), we can prove that the above equationholds.

Thus, Theorem 1.2 is proved.

## Notes

## Declarations

### Acknowledgements

The authors thank the referee for his/her valuable suggestions. This work issupported by PhD research startup foundation of Jiangxi Science and Technology NormalUniversity, and it is partly supported by Natural Science Foundation of GuangdongProvince, China (Nos. S2012040006865, S2013040014347) and the Natural ScienceFoundation of Jiangxi, China (No. 20132BAB201008).

## Authors’ Affiliations

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