The zeros of difference of meromorphic solutions for the difference Riccati equation
- Chang-Wen Peng^{1}Email author and
- Zong-Xuan Chen^{2}
https://doi.org/10.1186/s13662-015-0668-3
© Peng and Chen 2015
Received: 3 July 2015
Accepted: 12 October 2015
Published: 30 November 2015
Abstract
In this paper, we mainly investigate some properties of the transcendental meromorphic solution \(f(z)\) for the difference Riccati equation \(f(z+1)=\frac{p(z)f(z)+q(z)}{f(z)+s(z)}\). We obtain some estimates of the exponents of the convergence of the zeros and poles of \(f(z)\) and the difference \(\Delta f(z)=f(z+1)-f(z)\).
Keywords
difference Riccati equation Borel exceptional value admissible meromorphic solutionMSC
30D35 39B121 Introduction and main results
Early results for difference equations were largely motivated by the work of Kimura [1] on the iteration of analytic functions. Shimomura [2] and Yanagihara [3] proved the following theorems, respectively.
Theorem A
[2]
Theorem B
[3]
Let f be a function transcendental and meromorphic in the plane. The forward difference is defined in the standard way by \(\Delta f(z)=f(z+1)-f(z)\). In what follows, we assume the reader is familiar with the basic notions of Nevanlinna’s value distribution theory (see, e.g., [4–6]). In addition, we use the notations \(\sigma(f)\) to denote the order of growth of the meromorphic function \(f(z)\), and \(\lambda(f)\) and \(\lambda(\frac{1}{f})\) to denote the exponents of convergence of zeros and poles of \(f(z)\), respectively. Moreover, we say that a meromorphic function g is small with respect to f if \(T(r,g)=S(r,f)\), where \(S(r,f)=o(T(r,f))\) outside of a possible exceptional set of finite logarithmic measure. Denote by \(S(f)\) the family of all meromorphic functions which are small compared to \(f(z)\). We say that a meromorphic solution f of a difference equation is admissible if all coefficients of the equation are in \(S(f)\).
Recently, a number of papers (including [7–21]) focused on complex difference equations and difference analogs of Nevanlinna’s theory. As the difference analogs of Nevanlinna’s theory were being investigated, many results on the complex difference equations have been got rapidly. Many papers (including [7, 9, 12, 17]) mainly dealt with the growth of meromorphic solutions of difference equations.
In [15], Halburd and Korhonen used value distribution theory to obtain Theorem C.
Theorem C
[15]
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 [9–11, 22, 23] dealt with complex difference Riccati equations.
In this research, we investigate some properties of the difference Riccati equation and prove the following theorems.
Theorem 1.1
- (i)
\(\lambda (\frac{1}{f} )=\sigma(f)\). Moreover, if \(q(z)\not\equiv0\), then \(\lambda (\frac{1}{f} )=\lambda (f)=\sigma(f)\);
- (ii)
\(\lambda (\frac{1}{\bigtriangleup f} )=\sigma(\bigtriangleup f)=\sigma(f)\); \(\lambda (\frac{1}{\frac{\bigtriangleup f}{f}} )=\sigma ( \frac{\bigtriangleup f}{f} )=\sigma(f)\).
Theorem 1.2
- (i)If \(p(z)\equiv s(z)\), and there is a nonconstant rational function \(Q(z)\) satisfying \(q(z)=Q^{2}(z)\), then$$\lambda(\Delta f)=\lambda \biggl(\frac{\Delta f}{f} \biggr)=\sigma(f). $$
- (ii)If \(p(z)\equiv-s(z)\), \(s(z)\) is a nonconstant rational function, and there is a rational function \(h(z)\) satisfying \(s^{2}(z)+q(z)=h^{2}(z)\), then$$\lambda(\Delta f)=\lambda \biggl(\frac{\Delta f}{f} \biggr)=\sigma(f). $$
- (iii)If \(p(z)\not\equiv\pm s(z)\), and there is a nonconstant rational function \(m(z)\) satisfying \((s(z)-p(z))^{2}+4q(z)=m^{2}(z)\), then$$\lambda(\Delta f)=\lambda \biggl(\frac{\Delta f}{f} \biggr)=\sigma(f). $$
- (iv)
If \(p(z)\), \(q(z)\), \(s(z)\) are polynomials and \(\deg p(z)\), \(\deg q(z)\), \(\deg s(z)\) contain just one maximum, then \(f(z)\) has no nonzero Borel exceptional value.
2 The proof of Theorem 1.1
We need the following lemmas to prove Theorem 1.1.
Lemma 2.1
Lemma 2.2
(see [18])
Lemma 2.3
(Valiron-Mohon’ko) (see [5])
In the remark of [15], p.15, it is pointed out that Lemma 2.4 holds.
Lemma 2.4
(see [9])
Remark 2.1
Lemma 2.5
(see [12])
Lemma 2.6
(see [24])
Let \(g:(0,+\infty)\rightarrow R\), \(h:(0,+\infty)\rightarrow R\) be non-decreasing functions. If (i) \(g(r)\leq h(r)\) outside of an exceptional set of finite linear measure, or (ii) \(g(r)\leq h(r)\), \(r \notin{H\cup(0,1]}\), where \(H\subset(1,\infty)\) is a set of finite logarithmic measure, then for any \(\alpha>1\), there exists \(r_{0}>0\) such that \(g(r)\leq h(\alpha r)\) for all \(r>r_{0}\).
Lemma 2.7
(see [12])
Proof of Theorem 1.1
(i) Suppose that \(f(z)\) is an admissible transcendental meromorphic solution of finite order \(\sigma (f)\) of (1.2).
First, we prove \(\lambda (\frac{1}{f} )=\sigma(f)\).
Second, we prove \(\lambda (\frac{1}{f} )=\lambda(f)=\sigma(f)\) when \(q(z)\not\equiv0\).
3 The proof of Theorem 1.2
Suppose that \(f(z)\) is a transcendental meromorphic solution of finite order \(\sigma(f)\) of (1.2).
If \(p(z)\equiv0\), then \(P(z,Q(z))=P(z,-Q(z))=Q(z+1)Q(z)-q(z)\). If \(P(z,Q(z))=P(z,-Q(z))\equiv0\), then \(Q(z+1)Q(z)=q(z)=Q^{2}(z)\). Moreover, we get \(Q(z+1)\equiv Q(z)\). This is a contradiction since \(Q(z)\) is a nonconstant rational function. So \(P(z,Q(z))=P(z,-Q(z))\not\equiv0\).
(ii) We divide this proof into the following two cases.
We affirm \(h(z)\not\equiv0\). In fact, if \(h(z)\equiv0\), then \(q(z)+s^{2}(z)=0\), that is, \(q(z)=-s^{2}(z)\). Therefore, we get \(p(z)f(z)+q(z)=-s(z)(f(z)+s(z))\), and this is a contradiction since \(p(z)f(z)+q(z)\) and \((f(z)+s(z))\) are relatively prime polynomials in f.
If \(P(z,-(s(z)+h(z)))\equiv0\) and \(P(z,h(z)-s(z))\equiv0\), then \(s(z+1)h(z)+p(z)h(z)\equiv0\). Moreover, \(s(z+1)=-p(z)=s(z)\) since \(-p(z)=s(z)\). This is a contradiction since \(s(z)\) is a nonconstant rational function.
(iii) First, we prove \(\lambda(\Delta f)=\sigma(f)\). We divide this proof into the following two cases.
We affirm that \(m(z)-s(z)-p(z)\not\equiv0\) and \(m(z)+s(z)+p(z)\not\equiv0\). In fact, if \(m(z)-s(z)-p(z)\equiv0\) or \(m(z)+s(z)+p(z)\equiv0\), then \(m(z)=\pm(s(z)+p(z))\). Substituting \(m(z)=\pm(s(z)+p(z))\) into \((s(z)-p(z))^{2}+4q(z)=m^{2}(z)\), we get \(q(z)=s(z)p(z)\). This is a contradiction since \(p(z)f(z)+q(z)\) and \(f(z)+s(z)\) are relatively prime polynomials in f.
We affirm that \(s(z)-p(z)+m(z)\) or \(s(z)-p(z)-m(z)\) is nonconstant rational function. In fact, if there are two constants \(c_{1}\) and \(c_{2}\), such that \(s(z)-p(z)+m(z)=c_{1}\) and \(s(z)-p(z)-m(z)=c_{2}\), then we get \(s(z)-p(z)=\frac{c_{1}+c_{2}}{2}\). Furthermore, we have \(m(z)=\frac{c_{1}-c_{2}}{2}\), this is a contradiction since \(m(z)\) is a nonconstant rational function. Hence, we conclude that \(s(z)-p(z)+m(z)\) or \(s(z)-p(z)-m(z)\) is a nonconstant rational function. Thus, we get \(s(z+1)-p(z+1)+m(z+1)\not\equiv s(z)-p(z)+m(z)\), or \(s(z+1)-p(z+1)-m(z+1)\not\equiv s(z)-p(z)-m(z)\). So, we get \(P_{1}=P (z,\frac{m(z)}{2}-\frac{s(z)-p(z)}{2} )\not\equiv0\), or \(P_{2}=P (z,-\frac{m(z)}{2}-\frac{s(z)-p(z)}{2} )\not\equiv 0\).
(iv) Suppose that \(f(z)\) is a finite order transcendental meromorphic solution of (1.2).
Theorem 1.2 is proved.
Declarations
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).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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