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Complex oscillation of meromorphic solutions for difference Riccati equation
Advances in Difference Equations volume 2014, Article number: 247 (2014)
Abstract
In this paper, we investigate zeros and αpoints of meromorphicsolutions f(z) for difference Riccati equations, and we obtain someestimates of exponents of convergence of zeros and αpoints off(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.
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 functionf(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 functionf(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].
In [8], Ishizaki gave some surveys of the basic properties of the difference Riccatiequation
where A(z) is a rational function, which have analogs in thedifferential case [16]. In the proof of the celebrated classification theorem, Halburd and Korhonen [13] were concerned with the difference Riccati equation of the form
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 1be a constant andA(z)=\frac{m(z)}{n(z)}be an irreducible nonconstant rational function,wherem(z)andn(z)are polynomials withdegm(z)=manddegn(z)=n.
If f(z) is a transcendental finite order meromorphic solution of the difference Riccatiequation
then

(i)
if\sigma (f)>0, thenf(z)has at most one Borel exceptional value;

(ii)
\lambda (\frac{1}{f})=\lambda (f)=\sigma (f);

(iii)
ifA(z)\not\equiv {z}^{2}z+1, then the exponent of convergence of fixed points off(z)satisfies\tau (f)=\sigma (f).
In [15], the first author investigated fixed points of meromorphic functionsf(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 αpoints of meromorphic solutionsf(z) for difference Riccati equations (1), and we obtain someestimates of the exponents of convergence of zeros and αpoints off(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 1be a constant andA(z)be a nonconstant rational function.Set\mathrm{\Delta}f(z)=f(z+1)f(z). If there exists a nonconstant rationalfunctions(z)such thatA(z)={s}^{2}(z), then every finite order transcendental meromorphicsolutionf(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 LetA(z)be a nonconstant rational function.If α is a nonzero complex constant, thenevery finite order transcendental meromorphic solutionf(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(z1)(z+1)}{zQ(z)+{z}^{2}(z1)(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 functionQ(z) of order \sigma (Q)=\rho by Ozawa [17]. Thus \sigma (f)=\sigma (Q)=\rho \ge 1.
Also, this solution f(z)=\frac{Q(z)2z(z1)(z+1)}{zQ(z)+{z}^{2}(z1)(z+1)} satisfies
and
Using the same discussion as Lemma 2.1, we easily see that 18{z}^{3}{(z+1)}^{3}{[Q(z)2z(z+1)(2z+1)]}^{2} and [Q(z)+z(z1)(z+1)][Q(z)+z(z+1)(z+2)] (or [Q(z)2z(z1)(z+1)][Q(z)+z(z+1)(z+2)]) have at most finitely many common zeros. Thus,
2 Lemmas for proofs of theorems
Firstly we need the following lemmas for the proof of Theorem 1.1.
Lemma 2.1 LetA(z)be a nonconstant rational function,andf(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})=1f({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
whereP(z,w)is a difference polynomialinw(z). IfP(z,\alpha )\not\equiv 0for a meromorphic function\alpha (z)satisfyingT(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)).
By (1) and the fact that A(z)={s}^{2}(z), we obtain
Since A(z) and s(z) are rational functions, we know thatf(z)s(z) (or f(z)+s(z)) and 1f(z) have the same poles, except possibly finitely many. ByLemma 2.1, we see that A(z)+{f}^{2}(z) and 1f(z) have at most finitely many common zeros. Hence, by (3), weonly need to prove that
Suppose that \lambda (f(z)s(z))<\sigma (f(z)). By \sigma (f(z)s(z))=\sigma (f(z)) and Hadamard factorization theorem,f(z)s(z) can be rewritten in the form
where h(z) is a polynomial with degh(z)\le \sigma (f(z)), {P}_{0}(z) and {Q}_{0}(z) are canonical products ({P}_{0}(z) may be a polynomial) formed by nonzero zeros and poles off(z)s(z), respectively, t is an integer, ift\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
By (5), we obtain
where y(z)=\frac{1}{Q(z)}. Thus, by (6), we have
Substituting (7) into (1), we obtain
By (8) and the fact that A(z)={s}^{2}(z), we have
Since s(z) is a nonconstant rational faction, we see that1s(z)\not\equiv 0 and s(z+1)s(z)\not\equiv 0, so that
Thus, by (6), (9), and Lemma 2.2, we obtain for any given ε(0<\epsilon <\sigma (f(z))\sigma (P(z))),
holds for all r outside of a possible exceptional set with finite logarithmicmeasure.
On the other hand, by y(z)=\frac{1}{Q(z)} and the fact that Q(z) is an entire function, we see that
Thus (10) is a contradiction. Hence, (4) holds, that is, \lambda (\mathrm{\Delta}f(z))=\sigma (f(z)).
Secondly, we prove that \lambda (\frac{\mathrm{\Delta}f(z)}{f(z)})=\sigma (f). By (1), we obtain
Thus, by this and (4), we see that \lambda (\frac{\mathrm{\Delta}f(z)}{f(z)})=\sigma (f).
4 Proof of Theorem 1.2
Suppose that f(z) is a finite order transcendental meromorphic solution of(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 ][1y(z)\alpha ]A(z)y(z)\alpha =0.
Thus
By the condition that A(z) is a nonconstant rational function, we obtain{K}_{0}(z,0)\not\equiv 0. By Lemma 2.2,
holds for all 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)).
Now suppose that n=1. By (2) and \alpha \ne 1, we see that
Using the same discussion as Lemma 2.1, we easily see that f(z)+\frac{A(z)\alpha}{1+\alpha} and 1f(z) have at most finitely many common zeros. Thus, we onlyneed to prove that
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)).
Now in (12), we replace z by z+n1 (n\ge 1), and we obtain
Set g(z)=f(z+n1). Then (14) is transformed as
Since A(z+n1) is a nonconstant rational function too, applying theconclusion for n=1 to (15), 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)
Now we prove
By (2), we have
If \alpha =1, then
Since A(z) is a rational function, A(z)\alpha f(z) and (1f(z)) have the same poles, except possibly finitely many. By(19) and Theorem A, we obtain
If \alpha \ne 1, by (16) and (18), we have
Using the same discussion as Lemma 2.1, we easily see that (1+\alpha ){f}^{2}(z)\alpha f(z)+A(z) and f(z)(1f(z)) have at most finitely many common zeros. Thus, by (20), inorder to prove (17), we only need to prove that
or
Without loss of generality, we prove (21). Suppose that \lambda (f(z)\frac{\alpha}{2(1+\alpha )}n(z))<\sigma (f(z)). Using the same method as in the proof of (4)(11), we seethat f(z)\frac{\alpha}{2(1+\alpha )}n(z) can be rewritten as
where y(z)=\frac{1}{Q(z)}, P(z), Q(z) are nonzero entire functions, such that
Substituting (22) into (2), we obtain
and
By the above equation and (16), we have
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.
Now we prove that
is nonconstant. Since A(z) is a nonconstant rational function and due to (16),n(z) is a nonconstant rational function too. First, ifn(z) is a polynomial with degn(z)=n\ge 1, then
is the maximal degree in R(z) (since \alpha \ne 0,1). Thus R(z) is a polynomial with degR(z)=2n\ge 2. Secondly, if n(z)=\frac{p(z)}{q(z)}, where p(z) and q(z) are polynomials with degp(z)=p<q=degq(z), then R(z)=\frac{s(z)}{t(z)}, where
and
Since p<q,
Thus R(z) is nonconstant. Lastly, if n(z)=\frac{p(z)}{q(z)}, where p(z) and q(z) are polynomials with degp(z)=p\ge q=degp(z), then
where {n}_{1}(z), {p}_{1}(z), and {q}_{1}(z) are polynomials with deg{n}_{1}(z)=pq\ge 0 and deg{p}_{1}(z)<degq(z). By the above discussion, we know thatR(z) is nonconstant. Hence {K}_{1}(z,0)\not\equiv 0, and, by Lemma 2.2, we see that (21) holds.

(iii)
Suppose that there is a rational function m(z) satisfying
A(z)=\frac{{\alpha}^{2}+\alpha}{4}{m}^{2}(z).(23)
In what follows, we prove that
By (2) and (23), we obtain
Using the same discussion as Lemma 2.1, we easily see that {f}^{2}(z)+\alpha f(z)+A(z)\alpha and 1f(z) have at most finitely many common zeros. Thus, by (25), weknow that to prove (24), we only need to prove that
Using the same method as in the proof of (21), we can prove that the above equationholds.
Thus, Theorem 1.2 is proved.
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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).
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YYJ completed the main part of this article, YYJ, ZQM, and MW corrected the maintheorems. All authors read and approved the final manuscript.
An erratum to this article is available at http://dx.doi.org/10.1186/s1366201403399.
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Jiang, YY., Mao, ZQ. & Wen, M. Complex oscillation of meromorphic solutions for difference Riccati equation. Adv Differ Equ 2014, 247 (2014). https://doi.org/10.1186/168718472014247
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DOI: https://doi.org/10.1186/168718472014247
Keywords
 Riccati equation
 meromorphic solution
 difference
 complex oscillation