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Some properties of algebraic difference equations of first order
Advances in Difference Equations volume 2017, Article number: 334 (2017)
Abstract
We prove that if \(g(z)\) is a finiteorder transcendental meromorphic solution of
where \(A(z)\) and \(B(z)\) are polynomials such that \(\deg A(z)>0\), then
Introduction
Steinmetz [1] and Bank and Kaufman [2] proved that the equation
can be reduced into a list of six simple differential equations by a suitable Möbius transformation with polynomial coefficients, which include
where ζ, η are constant, and \(p(z), q(z)\) are rational functions. Let \(q(z)\in\mathbb{C}\). Then equation (1.1) can be transformed into
Ishizaki and Korhonen [3] investigated meromorphic solutions of
They proved that equation (1.2) possesses a continuous limit to the equation
which extends to solutions in certain cases.
We assume that the reader is familiar with the basic notions of Nevanlinna theory (see, e.g., [4, 5]). Of late, several scholars [3, 6–14] studied the properties of finiteorder meromorphic solutions of algebraic difference equations and obtained many interesting results.
For the special case of (1.2), Whittaker [15] has shown that the equation
where \(q(z)\) is a meromorphic function of finite order \(\rho(q)\), has a meromorphic solution g such that \(\rho(q)\leq\rho(g)\leq\rho(q)+1\). Here \(\rho(g)\) denotes the order of growth of the meromorphic function \(g(z)\).
Chen [7] has extended this above result and proved that the Pielou logistic equation
where \(R(z)\), \(P(z)\), and \(Q(z)\) are polynomials with \(P(z)R(z)Q(z)\not\equiv0\), has a finiteorder transcendental meromorphic solution g such that \(1\leq\rho(g)\).
Replacing \(g(z+1)\) with \(\triangle g(z)\), Ishizaki [10] was concerned with the growth and value distributions of transcendental meromorphic solutions of the algebraic difference equation
In 2014, Liu [12] considered the Nevanlinna growth of an equation related to (1.2). It is interesting to consider some properties of (1.2), and our results will be stated in Section 2.
Main results
Theorem 2.1
Let \(c\in\mathbb{C}\setminus\{0\}\), and let \(A(z)\) and \(B(z)\) be polynomials such that \(\deg A(z)>0\). If \(g(z)\) is a finiteorder transcendental meromorphic solution of
then
Remark
It is a curious problem to construct a transcendental meromorphic solution of (2.1) for the case \(\deg A>0\).
Theorem 2.2
Let \(c\in\mathbb{C}\setminus\{0\}\), and let \(E(z)=\frac{D(z)}{F(z)}\) be an irreducible rational function, where \(D(z)\) and \(F(z)\) are polynomials with \(\deg D(z)=d\) and \(\deg F(z)=f\). If the equation
has a rational solution
where \(l_{h}\ (\neq0), \dots, l_{0}, m_{k}\ (\neq0),\dots, m_{0}\) are constants, \(\deg H(z)=h\), and \(\deg K(z)=k\).

(i)
If \(d\geq f\) and \(df\) is zero or an even number, then
$$hk=\frac{df}{2}. $$ 
(ii)
If \(d< f\), then \(hk=\frac{df}{2}\).
Further, Example 2.3 shows that there exist rational solutions satisfying Theorem 2.2(i), and Example 2.4 shows that there exist rational solutions satisfying Theorem 2.2(ii).
Example 2.3
The equation
has a rational solution \(g(z)=z+2\), where \(d=2\), \(f=0\), and \(hk=1=\frac{df}{2}\).
Example 2.4
The equation
has a rational solution \(g(z)=\frac{1}{z}\), where \(d=2\), \(f=4\), and \(hk=1=\frac{df}{2}\).
Proof of Theorem 2.1
Lemma 3.1
([11])
Let \(w(z)\) be a transcendental meromorphic solution of finite order of the difference equation
where \(P(z,w)\) is a difference polynomial in \(w(z)\) and its shift. If \(P(z,a)\not\equiv0\) for a slowly moving target function a, that is, \(T(r, a)=S(r, w)\), then
The following result obtained by Chiang and Feng [16] and Halburd and Korhonen [9, 17] independently. We state here the form stated in [16, Theorem 8.2(b)].
Lemma 3.2
([16])
Let \(c_{1}\), \(c_{2}\) be two arbitrary complex numbers, and let \(w(z)\) be a meromorphic function of finite order ρ. Let \(\varepsilon>0\) be given. Then there exists a subset \(E\subset(1, \infty)\) of finite logarithmic measure such that, for all \(z=r\notin E\cup[0, 1]\), we have
Firstly, we prove that \(\rho(g)=\rho\geq1\). We consider the following two cases separately.
Case 1.1. If \(g(z)\) has infinitely many poles, we can pick a pole \(z_{0}\) of \(g(z)\) such that \(g(z_{0})=\infty^{\pi}\), where \(\pi\geq1\), then we deduce by (2.1) that \(g(z_{0}+c)=\infty^{\pi_{1}}\), where \(\pi _{1}\geq m\). Substituting \(z_{0}+c\) for z into (2.1), we have
Then (3.1) implies that \(z_{0}+2c\) is a pole of g of multiplicity \(\pi _{2}\geq\pi_{1} \geq\pi\).
Since \(g(z)\) has infinitely many poles, following the previous steps, we pick a pole \(z_{0}\) of \(g(z)\) such that
where \(\pi_{n}\geq\pi\) for all \(n \in \mathbb{N}^{0}\). Hence, we can choose a sequence \(\{\xi_{n}=z_{0}+nc, n \in \mathbb{N}^{0}\}\) of poles of \(g(z)\), the multiplicity of which is \(\pi_{n}\geq\pi\), so we obtain \(\lambda(\frac{1}{g})\geq1\), and therefore \(\rho(g)\geq \lambda(\frac{1}{g})\geq1\).
Case 1.2. If \(g(z)\) is a transcendental meromorphic function with finitely many poles, then we can rewrite \(g(z)\) as
where \(g_{1}(z)\) is a transcendental entire function, and \(P(z)\) is a polynomial. Substituting (3.2) into (2.1), we have
By computing (3.3) we have
We prove that \(\rho(g)=\rho(g_{1})=\rho\geq1\). Suppose, on the contrary to the assertion, that \(\rho(g)=\rho(g_{1})=\rho<1\). For any given ε (\(0<\varepsilon<\frac{1\rho(g_{1})}{2}\)), by Lemma 3.2 we obtain
outside a finite logarithmic measure E. As \(z_{k}\) satisfies \(g_{1}(z_{k})=M(r_{k}, g_{1})\), \(z_{k}=r_{k} \notin E\), \(r_{k}\rightarrow\infty\), we deduce by (3.4) and (3.5) that
where M is some finite constant, a contradiction, since \(\deg A(z)>0\). Hence we have \({\rho(g)\geq1}\).
Next, we prove that \(\max\{\lambda(g), \lambda(\frac{1}{g})\}=\rho(g)\). If \(B(z)\not\equiv0\), then we set
Since \(P(z, 0)=B(z)\not\equiv0\), by Lemma 3.1 we deduce that
Hence \(\lambda(g)=\rho(g)\).
If \(B(z)\equiv0\), then (2.1) can be reduced to
Next, we prove that \(\max\{\lambda(g), \lambda(\frac{1}{g})\}=\rho(g)\). Suppose, on the contrary to the assertion, that \(\max\{\lambda(g), \lambda(\frac{1}{g})\}=\alpha<\rho(g)\). We next divide the proof into the following two cases.
Case 1. Suppose that \(\rho(g)=1\). Then we obtain
where \(q\neq0\) and p are constants, and \(m(z)\) is a meromorphic function such that \(\rho(m)=\alpha<1\). Substituting (3.6) into (2.1), we obtain
By computing (3.7) we obtain
that is,
By Lemma 3.2 we obtain
outside a finite logarithmic measure. By (3.9) and (3.10), as \(z\rightarrow\infty\), we obtain
outside a finite logarithmic measure, where \(M_{1}\) is a finite constant. This is impossible, since \(\deg A(z)>0\).
Case 2. Suppose that \(\rho(g)>1\). Then
where \(l(z)\) is a polynomial such that \(\rho(g)=\deg l(z)>1\), and \(m(z)\) is a meromorphic function such that \(\rho(m)<\rho(g)\). Substituting (3.11) into (2.1), we obtain
Let
where \(p_{k}\neq0\). Then
where \(Q(z)\) and \(Q_{1}(z)\) are polynomials of degree at most \(k2\). Equalities (3.12) and (3.14) imply that
that is,
By Lemma 3.2 we obtain
outside a possible set of finite logarithmic measure E. As \(z=r\notin E\cup[0, 1]\), and \(r\rightarrow\infty\), we deduce by (3.15) and (3.16) that
where M is a positive constant.
We can find a sequence \(\{z_{k}\}\) (\(z_{k}\rightarrow\infty\)) such that \(z_{k}=r_{k}\notin E\cup[0, 1]\), and \(cp_{k} z_{k}^{k1}=cp_{k}r_{k}^{k1}\) as \(r_{k}\rightarrow\infty\). We obtain
By (3.17) and (3.18), for any given ε (\(0<\varepsilon<\frac {k\rho(m)}{2}\)), we obtain
which is impossible. Hence we proved that \(\max\{\lambda(g), \lambda (\frac{1}{g})\}=\rho(g)\). Theorem 2.1 is thus proved.
Proof of Theorem 2.2
Suppose that (2.2) has a rational solution \(g(z)\) and has poles \(l_{1}, l_{2},\dots,l_{k}\). Hence \(g(z)\) can be represented as
where \(b_{0},\dots,b_{r}, t_{js_{j}},\dots,t_{j1}\) are constants.
(i) If \(d>f\) and \(df\) is an even number, then (2.2) and (4.1) imply that
Let \(\deg H(z)=h\) and \(\deg K(z)=k\). Suppose \(h< k\). Then
From (4.2) with (4.3) we obtain
This is impossible, since \(\frac{D(z)}{F(z)}\rightarrow\infty\) as \(z\rightarrow\infty\).
Suppose \(h=k\). Then
where \(\beta\in\mathbb{C}\setminus\{0\}\). Relations (4.2) and (4.4) yield that
a contradiction, since \(\frac{D(z)}{F(z)}\rightarrow\infty\) as \(z\rightarrow\infty\). Hence, we obtain that \(h>k\). So \(b_{r}\neq0\) (\(r\geq1\)). As \(z\rightarrow\infty\), we have
where \(\alpha\in\mathbb{C}\setminus\{0\}\). As \(z\rightarrow\infty\), by (4.2) and (4.5) we can deduce
Relation (4.6) implies that
If \(d=f\), then, as \(z\rightarrow\infty\), we obtain
where \(\alpha\in\mathbb{C}\setminus\{0\}\). If \(h< k\), then using the similar method as before, we can obtain a contradiction. If \(h>k\), then \(b_{r}\neq0\) (\(r\geq1\)). By (4.2), as \(z\rightarrow\infty\), we obtain
a contradiction. Hence \(h=k\), that is,
(ii) We next consider the case \(d< f\). Suppose that \(h>k\). Then \(b_{r}\neq0\) (\(r\geq1\)). Using the similar method as before, as \(z\rightarrow\infty\), by (4.2) we obtain that
a contradiction.
If \(h=k\), then using the similar method as before, we obtain
which is a contradiction, since \(\frac{D(z)}{F(z)}\rightarrow0\) as \(z\rightarrow\infty\). Hence \(h< k\). Substituting \(g(z)=\frac{H(z)}{K(z)}\) into (2.2), we have
Since
From this and from (4.8) we have
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Acknowledgements
The author would like to thank the referee for his/her helpful suggestions and comments. The work was supported by the NNSF of China (No. 10771121, 11401387), the NSF of Zhejiang Province, China (No. LQ 14A010007), the NSFC Tianyuan Mathematics Youth Fund (No. 11226094), the NSF of Shandong Province, China (No. ZR2012AQ020 and No. ZR2010AM030), and the Fund of Doctoral Program Research of Shaoxing College of Art and Science (20135018).
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Liu, Y. Some properties of algebraic difference equations of first order. Adv Differ Equ 2017, 334 (2017). https://doi.org/10.1186/s1366201713958
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DOI: https://doi.org/10.1186/s1366201713958
MSC
 30D35
 39B12
Keywords
 meromorphic functions
 difference equations
 value distribution
 finite order