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On growth of meromorphic solutions for linear difference equations with meromorphic coefficients
Advances in Difference Equations volume 2013, Article number: 60 (2013)
Abstract
In this paper, we consider the value distribution of meromorphic solutions for linear difference equations with meromorphic coefficients.
MSC:30D35, 39A10.
1 Introduction and preliminaries
Recently, several papers (including [1–7]) have been published regarding value distribution of meromorphic solutions of linear difference equations. We recall the following results. Chiang and Feng proved the following theorem.
Theorem A ([2])
Let ${P}_{0}(z),\dots ,{P}_{n}(z)$ be polynomials such that there exists an integer l, $0\le l\le n$, such that
holds. Suppose $f(z)$ is a meromorphic solution of the difference equation
Then we have $\sigma (f)\ge 1$.
In this paper, we use the basic notions of Nevanlinna’s theory (see [8, 9]). In addition, we use the notation $\sigma (f)$ to denote the order of growth of the meromorphic function $f(z)$, and $\lambda (f)$ to denote the exponent of convergence of zeros of $f(z)$.
Chen [1] weakened the condition (1.1) of Theorem A and proved the following results.
Theorem B ([1])
Let ${P}_{n}(z),\dots ,{P}_{0}(z)$ be polynomials such that ${P}_{n}{P}_{0}\not\equiv 0$ and
Then every finite order meromorphic solution $f(z)$ (≢0) of equation (1.2) satisfies $\sigma (f)\ge 1$, and $f(z)$ assumes every nonzero value $a\in \mathbb{C}$ infinitely often and $\lambda (fa)=\sigma (f)$.
Theorem C ([1])
Let $F(z)$, ${P}_{n}(z),\dots ,{P}_{0}(z)$ be polynomials such that $F{P}_{n}{P}_{0}\not\equiv 0$ and (1.3). Then every finite order transcendental meromorphic solution $f(z)$ of the equation
satisfies $\sigma (f)\ge 1$ and $\lambda (f)=\sigma (f)$.
Theorem D ([1])
Let $F(z)$, ${P}_{n}(z),\dots ,{P}_{0}(z)$ be polynomials such that $F{P}_{n}{P}_{0}\not\equiv 0$. Suppose that $f(z)$ is a meromorphic solution with infinitely many poles of (1.2) (or (1.4)). Then $\sigma (f)\ge 1$.
For the linear difference equation with transcendental coefficients
Chiang and Feng proved the following result.
Theorem E ([2])
Let ${A}_{0}(z),\dots ,{A}_{n}(z)$ be entire functions such that there exists an integer l, $0\le l\le n$, such that
If $f(z)$ is a meromorphic solution of (1.5), then we have $\sigma (f)\ge \sigma ({A}_{l})+1$.
Laine and Yang proved the following theorem.
Theorem F ([5])
Let ${A}_{0},\dots ,{A}_{n}$ be entire functions of finite order so that among those having the maximal order $\sigma :=max\{\sigma ({A}_{k}):0\le k\le n\}$, exactly one has its type strictly greater than the others. Then for any meromorphic solution of
we have $\sigma (f)\ge \sigma +1$.
Remark 1.1 If ${A}_{0},\dots ,{A}_{n}$ are meromorphic functions satisfying (1.6), then Theorem E does not hold. For example, the equation
has a solution $y(z)={e}^{iz}1$, which $\sigma (y)=1<\sigma ({A}_{0})+1$.
This example shows that for the linear difference equation with meromorphic coefficients, the condition (1.6) cannot guarantee that every transcendental meromorphic solution $f(z)$ of (1.7) satisfies $\sigma (f)\ge \sigma ({A}_{l})+1$.
Thus, a natural question to ask is what conditions will guarantee every transcendental meromorphic solution $f(z)$ of (1.7) with meromorphic coefficients satisfies $\sigma (f)\ge \sigma ({A}_{l})+1$.
In this note, we consider this question and prove the following results.
Theorem 1.1 Let ${c}_{1}$, ${c}_{2}$ ($\ne {c}_{1}$), a be nonzero constants, ${h}_{1}(z)$ be a nonzero meromorphic function with $\sigma ({h}_{1})<1$, $B(z)$ be a nonzero meromorphic function.
If $B(z)$ satisfies any one of the following three conditions:

(i)
$\sigma (B)>1$ and $\delta (\mathrm{\infty},B)>0$;

(ii)
$\sigma (B)<1$;

(iii)
$B(z)={h}_{0}(z){e}^{bz}$ where b is a nonzero constant, ${h}_{0}(z)$ (≢0) is a meromorphic function with $\sigma ({h}_{0})<1$,
then every meromorphic solution f (≢0) of the difference equation
satisfies $\sigma (f)\ge max\{\sigma (B),1\}+1$.
Further, if $\phi (z)$ (≢0) is a meromorphic function with
then
Corollary Under conditions of Theorem 1.1, every finite order solution $f(z)$ (≢0) of (1.8) has infinitely many fixed points, satisfies $\tau (f)=\sigma (f)$, and for any nonzero constant c,
Example 1.1
The equation
satisfies conditions of Theorem 1.1 and has a solution $f(z)={e}^{{z}^{2}}$ satisfying $\lambda (f)=0$ and $\tau (f)=\sigma (f)=2$. This example shows that under conditions of Theorem 1.1, a meromorphic solution of (1.8) may have no zero.
Theorem 1.2 Let ${h}_{1}(z)$, ${c}_{1}$, ${c}_{2}$, a, $B(z)$ satisfy conditions of Theorem 1.1, and let $F(z)$ (≢0) be a meromorphic function with $\sigma (F)<max\{\sigma (B),1\}+1$. Then all meromorphic solutions with finite order of the equation
satisfy
with at most one possible exceptional solution with $\sigma (f)<max\{\sigma (B),1\}+1$.
Remark 1.2 Under conditions of Theorem 1.1, equation (1.8) has no rational solution. But equation (1.9) in Theorem 1.2 may have a rational solution. For example, the equation
satisfies conditions of Theorem 1.2 and has a solution $f(z)=z$. This shows that in Theorem 1.2, there exists one possible exceptional solution with $\sigma (f)<max\{\sigma (B),1\}+1$.
2 Proof of Theorem 1.1
We need the following lemmas to prove Theorem 1.1.
Given two distinct complex constants ${\eta}_{1}$, ${\eta}_{2}$, let f be a meromorphic function of finite order σ. Then, for each $\epsilon >0$, we have
Lemma 2.2 (see [11])
Suppose that $P(z)=(\alpha +i\beta ){z}^{n}+\cdots $ (α, β are real numbers, $\alpha +\beta \ne 0$) is a polynomial with degree $n\ge 1$, that $A(z)$ (≢0) is an entire function with $\sigma (A)<n$. Set $g(z)=A(z){e}^{P(z)}$, $z=r{e}^{i\theta}$, $\delta (P,\theta )=\alpha cosn\theta \beta sinn\theta $. Then, for any given $\epsilon >0$, there exists a set ${H}_{1}\subset [0,2\pi )$ that has the linear measure zero such that for any $\theta \in [0,2\pi )\mathrm{\setminus}({H}_{1}\cup {H}_{2})$, there is $R>0$ such that for $z=r>R$, we have that

(i)
if $\delta (P,\theta )>0$, then
$$exp\{(1\epsilon )\delta (P,\theta ){r}^{n}\}<\leftg\left(r{e}^{i\theta}\right)\right<exp\{(1+\epsilon )\delta (P,\theta ){r}^{n}\};$$(2.1) 
(ii)
if $\delta (P,\theta )<0$, then
$$exp\{(1+\epsilon )\delta (P,\theta ){r}^{n}\}<\leftg\left(r{e}^{i\theta}\right)\right<exp\{(1\epsilon )\delta (P,\theta ){r}^{n}\},$$(2.2)
where ${H}_{2}=\{\theta \in [0,2\pi );\delta (P,\theta )=0\}$ is a finite set.
Lemma 2.3 Let ${c}_{1}$, ${c}_{2}$ ($\ne {c}_{1}$), a be nonzero constants, ${A}_{j}(z)$ ($j=0,1,2$), $F(z)$ be nonzero meromorphic functions. Suppose that $f(z)$ is a finite order meromorphic solution of the equation
If $\sigma (f)>max\{\sigma (F),\sigma ({A}_{j})\phantom{\rule{0.25em}{0ex}}(j=0,1,2)\}$, then $\lambda (f)=\sigma (f)$.
Proof Suppose that $\sigma (f)=\sigma $, $max\{\sigma (F),\sigma ({A}_{j})\phantom{\rule{0.25em}{0ex}}(j=0,1,2)\}=\alpha $. Then $\sigma >\alpha $. Equation (2.3) can be rewritten as the form
Thus, by (2.4), we deduce that
For any given ε ($0<\epsilon <min\{\frac{1}{4},\frac{\sigma \alpha}{4}\}$), and for sufficiently large r, we have that
By Lemma 2.1, we obtain
where M (>0) is some constant.
By $\sigma (f)=\sigma $, there exists a sequence $\{{r}_{n}\}$ satisfying ${r}_{1}<{r}_{2}<\cdots $ , ${r}_{n}\to \mathrm{\infty}$ such that
Thus, for sufficiently large ${r}_{n}$, we have that
Substituting (2.6)(2.9) into (2.5), we obtain for sufficiently large ${r}_{n}$
Since $\epsilon <min\{\frac{1}{4},\frac{\sigma \alpha}{4}\}$ and ε is arbitrary, by (2.10), we obtain
Hence, $\lambda (f)=\sigma (f)=\sigma $. □
Proof of Theorem 1.1 Suppose that $f(z)$ (≢0) is a meromorphic solution of equation (1.8) with $\sigma (f)<\mathrm{\infty}$.

(1)
Suppose that $B(z)$ satisfies the condition (i): $\sigma (B)>1$ and $\delta (\mathrm{\infty},B)=\delta >0$. Thus, for sufficiently large r,
$$m(r,B)>\frac{\delta}{2}T(r,B).$$(2.11)
Clearly, $\sigma (f)\ge \sigma (B)$ by (1.8). By Lemma 2.1, we see that for any given ε ($0<\epsilon <\frac{\sigma (B)1}{3}$),
and
By (1.8), we have that
Substituting (2.11)(2.13) into (2.14), we deduce that
By $\sigma (B)=\sigma $, there is a sequence ${r}_{j}$ ($1<{r}_{1}<{r}_{2}<\cdots $ , ${r}_{j}\to \mathrm{\infty}$) satisfying
Thus, by (2.15) and (2.16), we obtain
where M (>0) is some constant. Combining (2.17) and $\epsilon <\frac{\sigma (B)1}{3}$, it follows that
So that, it follows that $\sigma (f)\ge \sigma (B)+1=max\{\sigma (B),1\}+1$.

(2)
Suppose that $B(z)$ satisfies the condition (ii): $\sigma (B)<1$. Using the same method as in (1), we can obtain $\sigma (f)\ge max\{\sigma (B),1\}+1$.

(3)
Suppose that $B(z)$ satisfies the condition (iii): $B(z)={h}_{0}(z){e}^{bz}$, where b is a nonzero constant, ${h}_{0}(z)$ (≢0) is a meromorphic function with $\sigma ({h}_{0})<1$.
Now we need to prove $\sigma (f)\ge 2$. Contrary to the assertion, suppose that $\sigma (f)=\alpha <2$. We will deduce a contradiction. Set $z=r{e}^{i\theta}$. Then
In what follows, we divide this proof into three subcases: (a) $arga\ne argb$; (b) $arga=argb$ and $a\ne b$; (c) $a=b$.
Subcase (a). Since $arga\ne argb$ and (2.18), it is easy to see that there exists a ray $argz={\theta}_{0}$ such that
By (1.8) and (2.19), we see that $f(z)$ cannot be a rational function. By Lemma 2.1, (2.12) holds. By Lemma 2.2 and (2.19), it is easy to see that for any given ${\epsilon}_{1}$ ($0<{\epsilon}_{1}<min\{\frac{1}{2},\frac{2\alpha}{2}\}$) and for sufficiently large r,
and
Thus, by (1.8), (2.12), (2.20) and (2.21), we deduce that
By $\delta (bz,{\theta}_{0})=cos(argb+{\theta}_{0})>0$, $\sigma (f)=\alpha <2$ and ${\epsilon}_{1}<\frac{2\alpha}{2}$, it is easy to see that (2.22) is a contradiction. Hence, $\sigma (f)\ge 2$.
Subcase (b). By $arga=argb$ and $a\ne b$, we see that $f(z)$ cannot be a rational function. By Lemma 2.1, (2.12) holds. By $arga=argb$ and (2.18), we take ${\theta}_{1}=arga$, then $\delta (az,{\theta}_{1})=\delta (bz,{\theta}_{1})=1$ and
Now suppose that $b>a$. By Lemma 2.2, for any given ${\epsilon}_{2}$ ($0<{\epsilon}_{2}<min\{2\alpha ,\frac{ba}{2(b+a)}\}$),
and
Thus, by (1.8), (2.12), (2.24) and (2.25), we deduce that
Since ${\epsilon}_{2}<2\alpha $, we have that $\sigma (f)1+{\epsilon}_{2}=\alpha 1+{\epsilon}_{2}<1$. Combining this and (2.26), we obtain
By ${\epsilon}_{2}<\frac{ba}{2(b+a)}$, we see that (2.27) is a contradiction.
Now suppose that $b<a$. Using the same method as above, we can also deduce a contradiction.
Hence, $\sigma (f)\ge 2$ in Subcase (b).
Subcase (c). We first affirm that $f(z)$ cannot be a nonzero rational function. In fact, if $f(z)$ is a rational function, then ${e}^{az}[{h}_{1}(z)f(z+{c}_{1})+{h}_{0}(z)f(z)]=f(z+{c}_{2})$ is a rational function. So that ${h}_{1}(z)f(z+{c}_{1})+{h}_{0}(z)f(z)\equiv 0$, that is, $f(z+{c}_{2})\equiv 0$, a contradiction.
By Lemma 2.1, (2.12) holds. By $a=b$, equation (1.8) can be rewritten as
Using the same method as in the proof of (1), we can obtain $\sigma (f)\ge 2$.

(4)
Suppose that $\phi (z)$ (≢0) is a meromorphic function with $\sigma (\phi )<max\{\sigma (B),1\}+1$. Set $g(z)=f(z)\phi (z)$. Substituting $f(z)=g(z)+\phi (z)$ into (1.8), we obtain
$$\begin{array}{c}g(z+{c}_{2})+{h}_{1}(z){e}^{az}g(z+{c}_{1})+B(z)g(z)\hfill \\ \phantom{\rule{1em}{0ex}}=[\phi (z+{c}_{2})+{h}_{1}(z){e}^{az}\phi (z+{c}_{1})+B(z)\phi (z)].\hfill \end{array}$$(2.29)
If $\phi (z+{c}_{2})+{h}_{1}(z){e}^{az}\phi (z+{c}_{1})+B(z)\phi (z)\equiv 0$, then $\phi (z)$ is a nonzero meromorphic solution of (1.8). Thus, by the proof above, we have that $\sigma (\phi )\ge max\{\sigma (B),1\}+1$. This contradicts our condition that $\sigma (\phi )<max\{\sigma (B),1\}+1$. Hence, $\phi (z+2)+{h}_{1}(z){e}^{az}\phi (z+1)+B(z)\phi (z)\not\equiv 0$, and
Applying this and Lemma 2.3 to (2.29), we deduce that
Thus, Theorem 1.1 is proved. □
3 Proof of Theorem 1.2
Suppose that ${f}_{0}$ is a meromorphic solution of (1.9) with
If ${f}^{\ast}(z)$ ($\not\equiv {f}_{0}(z)$) is another meromorphic solution of (1.9) satisfying $\sigma ({f}^{\ast})<max\{\sigma (B),1\}+1$, then
But ${f}^{\ast}{f}_{0}$ is a solution of the corresponding homogeneous equation (1.8) of (1.9). By Theorem 1.1, we have $\sigma ({f}^{\ast}{f}_{0})\ge max\{\sigma (B),1\}+1$, a contradiction. Hence equation (1.9) possesses at most one exceptional solution ${f}_{0}$ with $\sigma ({f}_{0})<max\{\sigma (B),1\}+1$.
Now suppose that f is a meromorphic solution of (1.9) with
Since $\sigma (f)>max\{\sigma (B),\sigma (F),\sigma (h(z){e}^{az})\}$, applying Lemma 2.3 to (1.9), we obtain
Thus, Theorem 1.2 is proved.
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Acknowledgements
The author is grateful to the referees for a number of helpful suggestions to improve the paper. This research was partly supported by the National Natural Science Foundation of China (grant no. 11171119).
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Liu, Y. On growth of meromorphic solutions for linear difference equations with meromorphic coefficients. Adv Differ Equ 2013, 60 (2013). https://doi.org/10.1186/16871847201360
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Keywords
 difference equation
 meromorphic coefficient
 growth
 zero