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The zeros of difference of meromorphic solutions for the difference Riccati equation
Advances in Difference Equations volumeÂ 2015, ArticleÂ number:Â 365 (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 \(\Delta f(z)=f(z+1)f(z)\).
1 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]
For any polynomial \(P(y)\), the difference equation
has a nontrivial entire solution.
Theorem B
[3]
For any rational function \(R(y)\), the difference equation
has a nontrivial meromorphic solution.
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 Nevanlinnaâ€™s value distribution theory (see, e.g., [4â€“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 [7â€“21]) focused on complex difference equations and difference analogs of Nevanlinnaâ€™s theory. As the difference analogs of Nevanlinnaâ€™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 TheoremÂ C.
Theorem C
[15]
Let \(f(z)\) be an admissible finite order meromorphic solution of the equation
where the coefficients are meromorphic functions, \(c_{2}\not\equiv0\) and \(\deg_{f}(R)=2\). If the order of the poles of \(f(z)\) is bounded, then either \(f(z)\) satisfies a difference Riccati equation
where \(p, q, s\in{S(f)}\), or (1.1) can be transformed by a bilinear change in \(f(z)\) to one of the equations
where \(\lambda\in{\mathbb{C}}\), and \(\delta, \mu, \gamma,\underline{\gamma}=\gamma(z1) \in{S(f)}\) are arbitrary finite order periodic functions such that Î´ and Î³ have period 2 and Î¼ has period 1.
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 PainlevÃ© equations. Some papers [9â€“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
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
Then

(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
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 meromorphic solution of the difference Riccati equation (1.2). Then:

(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 \(p(z)\equivs(z)\), \(s(z)\) 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 TheoremÂ 1.1.
Lemma 2.1
Let f be a transcendental meromorphic solution of finite order Ïƒ of the difference equation
where \(P(z,f)\) is a difference polynomial in \(f(z)\) and its shifts. If \(P(z,a)\not\equiv0\) for a slowly moving target function a, i.e. \(T(r,a)=S(r,f)\), then
outside of a possible exceptional set of finite logarithmic measure.
Lemma 2.2
(see [18])
Let f be a transcendental meromorphic solution of finite order Ïƒ of a difference equation of the form
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 degree \(\deg Q(z,f)\leq n\). Then for each \(\varepsilon>0\),
possibly outside of an exceptional set of finite logarithmic measure.
Lemma 2.3
(ValironMohonâ€™ko) (see [5])
Let \(f(z)\) be a meromorphic function. Then for all irreducible rational functions in f,
with meromorphic coefficients \(a_{i}(z)\) (\(i=0,1,\ldots,m\)), \(b_{j}(z)\) (\(j=0,1,\ldots,n\)), the characteristic function of \(R(z,f(z))\) satisfies
where \(d=\deg_{f}R=\max\{m,n\}\) and \(\Psi(r)=\max_{i,j}\{T(r,a_{i}),T(r,b_{j})\}\).
In the remark of [15], p.15, it is pointed out that LemmaÂ 2.4 holds.
Lemma 2.4
(see [9])
Let f be a nonconstant finite order meromorphic function. Then
outside of a possible exceptional set of finite logarithmic measure.
Remark 2.1
In [12], Chiang and Feng proved that if f is a meromorphic function with exponent of convergence of poles \(\lambda (\frac{1}{f} )=\lambda<\infty\), \(\eta\neq0\) fixed, then for each \(\varepsilon>0\),
Lemma 2.5
(see [12])
Let \(f(z)\) be a meromorphic function with order \(\sigma=\sigma(f)<+\infty\), and let Î· be a fixed nonzero complex number, then for each \(\varepsilon>0\), we have
Lemma 2.6
(see [24])
Let \(g:(0,+\infty)\rightarrow R\), \(h:(0,+\infty)\rightarrow R\) be nondecreasing 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])
Let \(\eta_{1}\), \(\eta_{2}\) be two complex numbers such that \(\eta_{1}\neq\eta_{2}\) and let \(f(z)\) be a finite order meromorphic function. Let Ïƒ be the order of \(f(z)\), then for each \(\varepsilon>0\), we have
Proof of TheoremÂ 1.1
(i) Suppose that \(f(z)\) is an admissible transcendental meromorphic solution of finite order \(\sigma (f)\) of (1.2).
First, we prove \(\lambda (\frac{1}{f} )=\sigma(f)\).
By (1.2), we have
By LemmaÂ 2.2 and (2.1), we get
outside of a possible exceptional set of finite logarithmic measure. From (1.2) and LemmaÂ 2.3, we have
Hence, by (2.2) and (2.3), we conclude that
outside of a possible exceptional set of finite logarithmic measure. By LemmaÂ 2.4, we get
Hence, by (2.4) and (2.5), we conclude that
outside of a possible exceptional set of finite logarithmic measure. By LemmaÂ 2.6 and (2.6), we get \(\lambda (\frac{1}{f} )=\sigma(f)\).
Second, we prove \(\lambda (\frac{1}{f} )=\lambda(f)=\sigma(f)\) when \(q(z)\not\equiv0\).
By (1.2), we have
Hence, we get
Thus, by (2.7) and LemmaÂ 2.1, we see that
outside of a possible exceptional set of finite logarithmic measure. Thus, we have
outside of a possible exceptional set of finite logarithmic measure. By LemmaÂ 2.6 and (2.8), we get \(\lambda(f)=\sigma(f)\). So \(\lambda (\frac{1}{f} )=\lambda(f)=\sigma(f)\).
(ii) Suppose that \(f(z)\) is an admissible transcendental meromorphic solution of finite order \(\sigma(f)\) of (1.2). By (1.2), we get
By (2.9) and LemmaÂ 2.2, we have
outside of a possible exceptional set of finite logarithmic measure. From LemmaÂ 2.7 and (2.10), we get
outside of a possible exceptional set of finite logarithmic measure.
By (1.2), we get
Since \(p(z)f(z)+q(z)\) and \(f(z)+s(z)\) are relatively prime polynomials in f, and \(f(z)(f(z)+s(z))\) and \(f(z)+s(z)\) have a common factor \(f(z)+s(z)\). Therefore \(p(z)f(z)+q(z)f(z)(f(z)+s(z))\) and \(f(z)+s(z)\) are relatively prime polynomials in f. By LemmaÂ 2.3 and (2.12), we get
By (2.11) and (2.13), we see that
outside of a possible exceptional set of finite logarithmic measure. Hence, by LemmaÂ 2.6 and (2.14), we get
By \(N(r,\bigtriangleup f)=N (r,\frac{\bigtriangleup f}{f}\cdot f )\leq N (r,\frac{\bigtriangleup f}{f} )+N(r,f) \) and (2.14), we get
outside of a possible exceptional set of finite logarithmic measure. Hence, by LemmaÂ 2.6 and (2.15), we have \(\lambda (\frac{1}{\frac{\bigtriangleup f}{f}} )\geq\sigma(f)\). We have \(\sigma (\frac{\bigtriangleup f}{f} )\leq\sigma(f)\). Thus, we have
TheoremÂ 1.1 is proved.â€ƒâ–¡
3 The proof of Theorem 1.2
Suppose that \(f(z)\) is a transcendental meromorphic solution of finite order \(\sigma(f)\) of (1.2).
(i) By (1.2), we get
Since \(s(z)\equiv p(z)\) and \(q(z)=Q^{2}(z)\), by (3.1), we get
By (1.2), we have
By (3.3), we see that
and
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 \(P(z,Q(z))=P(z,Q(z))\not\equiv0\).
Suppose that \(p(z)\not\equiv0\). If \(P(z,Q(z))\equiv0\) and \(P(z,Q(z))\equiv0\), by (3.4) and (3.5), we get
Thus, we know that \(Q(z+1)\equiv Q(z)\). This is a contradiction since \(Q(z)\) is a nonconstant rational function. Therefore, we get \(P(z,Q(z))\not\equiv0\) or \(P(z,Q(z))\not\equiv0\). Without loss of generality, we assume that \(P(z,Q(z))\not\equiv0\). By LemmaÂ 2.1, we get
possibly outside of an exceptional set of finite logarithmic measure. Moreover, we get
possibly outside of an exceptional set of finite logarithmic measure.
If \(z_{0}\) is a common zero of \(f(z)Q(z)\) and \(f(z)+s(z)\), then \(Q(z_{0})+s(z_{0})=0\). If \(z_{0}\) is a zero of \(f(z)Q(z)\), and \(z_{0}\) is a pole of \(f(z)+Q(z)\), then \(z_{0}\) is a pole of \(2Q(z)\). Since \(p(z)f(z)+q(z)\) and \(f(z)+s(z)\) are relatively prime polynomials in f, we see that \(p(z)s(z)\not\equiv q(z)\), that is, \(Q^{2}(z)\not\equiv s^{2}(z)\). Hence, we get \(Q(z)+s(z)\not\equiv0\). Thus, we conclude that
Since \(Q(z)\) and \(s(z)\) are rational functions, we get
possibly outside of an exceptional set of finite logarithmic measure. Thus, by LemmaÂ 2.6, we see that \(\lambda(\Delta f)\geq\sigma(f)\).
By (3.1), we get
By a similar method to above, we get \(\lambda (\frac{\Delta f}{f} )\geq\sigma(f)\). Hence,
(ii) We divide this proof into the following two cases.
Case 1 Suppose that \(q(z)\not\equiv0\). Since \(s(z)\equivp(z)\) and \(h^{2}(z)=s^{2}(z)+q(z)\), by (3.1), we get
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.
Therefore, we get \(P(z,(s(z)+h(z)))\not\equiv0\) or \(P(z,h(z)s(z))\not\equiv0\). Without loss of generality, we assume that \(P(z,h(z)s(z))\not\equiv0\). By a similar method to above, we get
possibly outside of an exceptional set of finite logarithmic measure.
If \(z_{0}\) is a common zero of \(f(z)+s(z)h(z)\) and \(f(z)+s(z)\), then \(h(z_{0})=0\). If \(z_{0}\) is a zero of \(f(z)+s(z)h(z)\), and \(z_{0}\) is a pole of \(f(z)+s(z)+h(z)\), then \(z_{0}\) is a pole of \(2h(z)\). Thus, we see that
\(h(z)\) is a rational function, so we get
By (3.9) and (3.10), we see that
possibly outside of an exceptional set of finite logarithmic measure. Thus, by LemmaÂ 2.6, we see that \(\lambda(\Delta f)\geq\sigma(f)\).
By a similar method to above, we get \(\lambda (\frac{\Delta f}{f} )\geq\sigma(f)\). Hence,
Case 2 Suppose that \(q(z)\equiv0\). Since \(s(z)\equivp(z)\), by (3.1), we get
By (3.3), we get
Since \(s(z)\) is a nonconstant rational function, we get \(s(z+1)\not\equiv s(z)\). Therefore, we have \(P(z,2s(z))\not\equiv0\).
By a similar method to above, we have
possibly outside of an exceptional set of finite logarithmic measure. Thus, by LemmaÂ 2.6, we see that \(\lambda(\Delta f)\geq\sigma(f)\).
By (3.11), we get
By a similar method to above, we get \(\lambda (\frac{\Delta f}{f} )\geq\sigma(f)\). Hence,
(iii) First, we prove \(\lambda(\Delta f)=\sigma(f)\). We divide this proof into the following two cases.
Case 1 Suppose that \(q(z)\not\equiv0\). Substituting \((s(z)p(z))^{2}+4q(z)=m^{2}(z)\) into (3.1), we get
From (3.3), we get
and
Since \((s(z)p(z))^{2}+4q(z)=m^{2}(z)\), by (3.13) and (3.14), we see that
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 \(P_{2}=P (z,\frac{m(z)}{2}\frac{s(z)p(z)}{2} )\not\equiv 0\).
Without loss of generality, we assume that \(P_{1}=P (z,\frac{m(z)}{2}\frac{s(z)p(z)}{2} )\not\equiv0\). By LemmaÂ 2.1, we get
for all r outside of a possible exceptional set with finite logarithmic measure. Moreover, we get
for all r outside of a possible exceptional set with finite logarithmic measure.
If \(z_{0}\) is a common zero of \(f(z)+\frac{s(z)p(z)}{2}\frac{m(z)}{2}\) and \(f(z)+s(z)\), then \(\frac{s(z_{0})+p(z_{0})+m(z_{0})}{2} =0\). If \(z_{0}\) is a zero of \(f(z)+\frac{s(z)p(z)}{2}\frac{m(z)}{2}\), and \(z_{0}\) is a pole of \(f(z)+\frac{s(z)p(z)}{2}+\frac{m(z)}{2}\), then \(z_{0}\) is a pole of \(m(z)\). From above, we know that \(\frac{s(z)+p(z)+m(z)}{2} \not\equiv0\). Thus, we conclude that
Since \(s(z)\), \(p(z)\) and \(m(z)\) are rational functions, we have
From (3.15) and (3.16), we get
for all r outside of a possible exceptional set with finite logarithmic measure. Hence, by LemmaÂ 2.6 and (3.17), we get \(\lambda(\Delta f)\geq\sigma (f)\).
By (3.12), we see that
By a similar method to above, we get \(\lambda (\frac{\Delta f}{f} )\geq\sigma(f)\). Thus, we get
Case 2 Suppose that \(q(z)\equiv0\). Since \(s(z)\not\equiv \pm p(z)\), by (3.1), we get
By (3.3), we get
That is,
By \(q(z)\equiv0\) and \(m^{2}(z)=(p(z)s(z))^{2}+4q(z)\), we get \(m^{2}(z)=(p(z)s(z))^{2}\). Since \(m(z)\) is a nonconstant rational function, we see that \(p(z)s(z)\) is a nonconstant rational function too. So \(p(z+1)s(z+1)\not\equiv p(z)s(z)\). Therefore, we have \(P(z,p(z)s(z))\not\equiv0\). By LemmaÂ 2.1, we get
possibly outside of an exceptional set of finite logarithmic measure. Moreover, we get
possibly outside of an exceptional set of finite logarithmic measure.
If \(z_{0}\) is a common zero of \(f(z)+s(z)p(z)\) and \(f(z)+s(z)\), then \(p(z_{0})=0\). If \(z_{0}\) is a zero of \(f(z)+s(z)p(z)\), and \(z_{0}\) is a pole of \(f(z)\), then \(z_{0}\) is a pole of \(s(z)p(z)\). Thus, we see that
Since \(p(z)\) and \(s(z)p(z)\) are rational functions, we get
possibly outside of an exceptional set of finite logarithmic measure. Thus, by LemmaÂ 2.6 and (3.21), we see that \(\lambda(\Delta f)\geq\sigma (f)\).
By (3.18), we get
By a similar method to above, we get \(\lambda (\frac{\Delta f}{f} )\geq\sigma(f)\). Hence,
(iv) Suppose that \(f(z)\) is a finite order transcendental meromorphic solution of (1.2).
Without loss of generality, we assume that \(\deg p(z)>\max\{\deg q(z), \deg s(z)\}\). Then \(\deg p(z)\geq1\). Set \(p(z)=a_{k}z^{k}+\cdots+ a_{0} \) (\(a_{k}\neq0\)). Let \(a\neq0\). By (3.3), we have
since \(aa_{k}\neq0\). By LemmaÂ 2.1 and (3.22), we conclude that
possibly outside of an exceptional set of finite logarithmic measure. Thus, we get
possibly outside of an exceptional set of finite logarithmic measure. Thus, by LemmaÂ 2.6 and (3.23), we get
That is, \(f(z)\) has no nonzero Borel exceptional value.
TheoremÂ 1.2 is proved.
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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).
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CWP completed the main part of this article, CWP and ZXC corrected the main theorems. All authors read and approved the final manuscript.
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Peng, CW., Chen, ZX. The zeros of difference of meromorphic solutions for the difference Riccati equation. Adv Differ Equ 2015, 365 (2015). https://doi.org/10.1186/s1366201506683
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DOI: https://doi.org/10.1186/s1366201506683