Theory and Modern Applications

# On growth of meromorphic solutions for linear difference equations with meromorphic coefficients

## 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 ) 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 ()

Let ${P}_{0}\left(z\right),\dots ,{P}_{n}\left(z\right)$ be polynomials such that there exists an integer l, $0\le l\le n$, such that

$deg\left({P}_{l}\right)>\underset{0\le j\le n,j\ne l}{max}\left\{deg\left({P}_{j}\right)\right\}$
(1.1)

holds. Suppose $f\left(z\right)$ is a meromorphic solution of the difference equation

${P}_{n}\left(z\right)f\left(z+n\right)+\cdots +{P}_{1}\left(z\right)f\left(z+1\right)+{P}_{0}\left(z\right)f\left(z\right)=0.$
(1.2)

Then we have $\sigma \left(f\right)\ge 1$.

In this paper, we use the basic notions of Nevanlinna’s theory (see [8, 9]). In addition, we use the notation $\sigma \left(f\right)$ to denote the order of growth of the meromorphic function $f\left(z\right)$, and $\lambda \left(f\right)$ to denote the exponent of convergence of zeros of $f\left(z\right)$.

Chen  weakened the condition (1.1) of Theorem A and proved the following results.

Theorem B ()

Let ${P}_{n}\left(z\right),\dots ,{P}_{0}\left(z\right)$ be polynomials such that ${P}_{n}{P}_{0}\not\equiv 0$ and

$deg\left({P}_{n}+\cdots +{P}_{0}\right)=max\left\{deg{P}_{j}:j=0,\dots ,n\right\}\ge 1.$
(1.3)

Then every finite order meromorphic solution $f\left(z\right)$ (0) of equation (1.2) satisfies $\sigma \left(f\right)\ge 1$, and $f\left(z\right)$ assumes every nonzero value $a\in \mathbb{C}$ infinitely often and $\lambda \left(f-a\right)=\sigma \left(f\right)$.

Theorem C ()

Let $F\left(z\right)$, ${P}_{n}\left(z\right),\dots ,{P}_{0}\left(z\right)$ be polynomials such that $F{P}_{n}{P}_{0}\not\equiv 0$ and (1.3). Then every finite order transcendental meromorphic solution $f\left(z\right)$ of the equation

${P}_{n}\left(z\right)f\left(z+n\right)+\cdots +{P}_{1}\left(z\right)f\left(z+1\right)+{P}_{0}\left(z\right)f\left(z\right)=F\left(z\right)$
(1.4)

satisfies $\sigma \left(f\right)\ge 1$ and $\lambda \left(f\right)=\sigma \left(f\right)$.

Theorem D ()

Let $F\left(z\right)$, ${P}_{n}\left(z\right),\dots ,{P}_{0}\left(z\right)$ be polynomials such that $F{P}_{n}{P}_{0}\not\equiv 0$. Suppose that $f\left(z\right)$ is a meromorphic solution with infinitely many poles of (1.2) (or (1.4)). Then $\sigma \left(f\right)\ge 1$.

For the linear difference equation with transcendental coefficients

${A}_{n}\left(z\right)f\left(z+n\right)+\cdots +{A}_{1}\left(z\right)f\left(z+1\right)+{A}_{0}\left(z\right)f\left(z\right)=0,$
(1.5)

Chiang and Feng proved the following result.

Theorem E ()

Let ${A}_{0}\left(z\right),\dots ,{A}_{n}\left(z\right)$ be entire functions such that there exists an integer l, $0\le l\le n$, such that

$\sigma \left({A}_{l}\right)>max\left\{\sigma \left({A}_{j}\right):0\le j\le n,j\ne l\right\}.$
(1.6)

If $f\left(z\right)$ is a meromorphic solution of (1.5), then we have $\sigma \left(f\right)\ge \sigma \left({A}_{l}\right)+1$.

Laine and Yang proved the following theorem.

Theorem F ()

Let ${A}_{0},\dots ,{A}_{n}$ be entire functions of finite order so that among those having the maximal order $\sigma :=max\left\{\sigma \left({A}_{k}\right):0\le k\le n\right\}$, exactly one has its type strictly greater than the others. Then for any meromorphic solution of

${A}_{n}\left(z\right)f\left(z+{C}_{n}\right)+\cdots +{A}_{1}\left(z\right)f\left(z+{C}_{1}\right)+{A}_{0}\left(z\right)f\left(z\right)=0,$
(1.7)

we have $\sigma \left(f\right)\ge \sigma +1$.

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

$y\left(z+1\right)-\left({e}^{i}+\frac{{e}^{i}-1}{{e}^{iz}-1}\right)y\left(z\right)=0$

has a solution $y\left(z\right)={e}^{iz}-1$, which $\sigma \left(y\right)=1<\sigma \left({A}_{0}\right)+1$.

This example shows that for the linear difference equation with meromorphic coefficients, the condition (1.6) cannot guarantee that every transcendental meromorphic solution $f\left(z\right)$ of (1.7) satisfies $\sigma \left(f\right)\ge \sigma \left({A}_{l}\right)+1$.

Thus, a natural question to ask is what conditions will guarantee every transcendental meromorphic solution $f\left(z\right)$ of (1.7) with meromorphic coefficients satisfies $\sigma \left(f\right)\ge \sigma \left({A}_{l}\right)+1$.

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, ${h}_{1}\left(z\right)$ be a nonzero meromorphic function with $\sigma \left({h}_{1}\right)<1$, $B\left(z\right)$ be a nonzero meromorphic function.

If $B\left(z\right)$ satisfies any one of the following three conditions:

1. (i)

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

2. (ii)

$\sigma \left(B\right)<1$;

3. (iii)

$B\left(z\right)={h}_{0}\left(z\right){e}^{bz}$ where b is a nonzero constant, ${h}_{0}\left(z\right)$ (0) is a meromorphic function with $\sigma \left({h}_{0}\right)<1$,

then every meromorphic solution f (0) of the difference equation

$f\left(z+{c}_{2}\right)+{h}_{1}\left(z\right){e}^{az}f\left(z+{c}_{1}\right)+B\left(z\right)f\left(z\right)=0$
(1.8)

satisfies $\sigma \left(f\right)\ge max\left\{\sigma \left(B\right),1\right\}+1$.

Further, if $\phi \left(z\right)$ (0) is a meromorphic function with

$\sigma \left(\phi \right)

then

$\lambda \left(f-\phi \right)=\sigma \left(f\right)\ge max\left\{\sigma \left(B\right),1\right\}+1.$

Corollary Under conditions of Theorem  1.1, every finite order solution $f\left(z\right)$ (0) of (1.8) has infinitely many fixed points, satisfies $\tau \left(f\right)=\sigma \left(f\right)$, and for any nonzero constant c,

$\lambda \left(f\left(z\right)-c\right)=\sigma \left(f\right)\ge max\left\{\sigma \left(B\right),1\right\}+1.$

Example 1.1

The equation

$f\left(z+2\right)-\frac{1}{2}{e}^{2z+3}f\left(z+1\right)-\frac{1}{2}{e}^{4z+4}f\left(z\right)=0$

satisfies conditions of Theorem 1.1 and has a solution $f\left(z\right)={e}^{{z}^{2}}$ satisfying $\lambda \left(f\right)=0$ and $\tau \left(f\right)=\sigma \left(f\right)=2$. This example shows that under conditions of Theorem 1.1, a meromorphic solution of (1.8) may have no zero.

Theorem 1.2 Let ${h}_{1}\left(z\right)$, ${c}_{1}$, ${c}_{2}$, a, $B\left(z\right)$ satisfy conditions of Theorem  1.1, and let $F\left(z\right)$ (0) be a meromorphic function with $\sigma \left(F\right). Then all meromorphic solutions with finite order of the equation

$f\left(z+{c}_{2}\right)+{h}_{1}\left(z\right){e}^{az}f\left(z+{c}_{1}\right)+B\left(z\right)f\left(z\right)=F\left(z\right)$
(1.9)

satisfy

$\lambda \left(f\right)=\sigma \left(f\right)\ge max\left\{\sigma \left(B\right),1\right\}+1$

with at most one possible exceptional solution with $\sigma \left(f\right).

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

$f\left(z+2\right)+{e}^{z}f\left(z+1\right)-{e}^{z}f\left(z\right)=z+2-{e}^{z}$

satisfies conditions of Theorem 1.2 and has a solution $f\left(z\right)=z$. This shows that in Theorem 1.2, there exists one possible exceptional solution with $\sigma \left(f\right).

## 2 Proof of Theorem 1.1

We need the following lemmas to prove Theorem 1.1.

Lemma 2.1 ([2, 10])

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

$m\left(r,\frac{f\left(z+{\eta }_{1}\right)}{f\left(z+{\eta }_{2}\right)}\right)=O\left({r}^{\sigma -1+\epsilon }\right).$

Lemma 2.2 (see )

Suppose that $P\left(z\right)=\left(\alpha +i\beta \right){z}^{n}+\cdots$ (α, β are real numbers, $|\alpha |+|\beta |\ne 0$) is a polynomial with degree $n\ge 1$, that $A\left(z\right)$ (0) is an entire function with $\sigma \left(A\right). Set $g\left(z\right)=A\left(z\right){e}^{P\left(z\right)}$, $z=r{e}^{i\theta }$, $\delta \left(P,\theta \right)=\alpha cosn\theta -\beta sinn\theta$. Then, for any given $\epsilon >0$, there exists a set ${H}_{1}\subset \left[0,2\pi \right)$ that has the linear measure zero such that for any $\theta \in \left[0,2\pi \right)\mathrm{\setminus }\left({H}_{1}\cup {H}_{2}\right)$, there is $R>0$ such that for $|z|=r>R$, we have that

1. (i)

if $\delta \left(P,\theta \right)>0$, then

$exp\left\{\left(1-\epsilon \right)\delta \left(P,\theta \right){r}^{n}\right\}<|g\left(r{e}^{i\theta }\right)|
(2.1)
2. (ii)

if $\delta \left(P,\theta \right)<0$, then

$exp\left\{\left(1+\epsilon \right)\delta \left(P,\theta \right){r}^{n}\right\}<|g\left(r{e}^{i\theta }\right)|
(2.2)

where ${H}_{2}=\left\{\theta \in \left[0,2\pi \right);\delta \left(P,\theta \right)=0\right\}$ is a finite set.

Lemma 2.3 Let ${c}_{1}$, ${c}_{2}$ ($\ne {c}_{1}$), a be nonzero constants, ${A}_{j}\left(z\right)$ ($j=0,1,2$), $F\left(z\right)$ be nonzero meromorphic functions. Suppose that $f\left(z\right)$ is a finite order meromorphic solution of the equation

${A}_{2}\left(z\right)f\left(z+{c}_{2}\right)+{A}_{1}\left(z\right)f\left(z+{c}_{1}\right)+{A}_{0}\left(z\right)f\left(z\right)=F\left(z\right).$
(2.3)

If $\sigma \left(f\right)>max\left\{\sigma \left(F\right),\sigma \left({A}_{j}\right)\phantom{\rule{0.25em}{0ex}}\left(j=0,1,2\right)\right\}$, then $\lambda \left(f\right)=\sigma \left(f\right)$.

Proof Suppose that $\sigma \left(f\right)=\sigma$, $max\left\{\sigma \left(F\right),\sigma \left({A}_{j}\right)\phantom{\rule{0.25em}{0ex}}\left(j=0,1,2\right)\right\}=\alpha$. Then $\sigma >\alpha$. Equation (2.3) can be rewritten as the form

$\frac{1}{f\left(z\right)}=\frac{F\left(z\right)}{f\left(z\right)}\left({A}_{2}\left(z\right)\frac{f\left(z+{c}_{2}\right)}{f\left(z\right)}+{A}_{1}\left(z\right)\frac{f\left(z+{c}_{1}\right)}{f\left(z\right)}+{A}_{0}\left(z\right)\right).$
(2.4)

Thus, by (2.4), we deduce that

$\begin{array}{rcl}T\left(r,f\right)& =& T\left(r,\frac{1}{f}\right)+O\left(1\right)\\ =& m\left(r,\frac{1}{f}\right)+N\left(r,\frac{1}{f}\right)+O\left(1\right)\\ \le & N\left(r,\frac{1}{f}\right)+m\left(r,\frac{1}{F}\right)+\sum _{j=0}^{2}m\left(r,{A}_{j}\right)\\ +m\left(r,\frac{f\left(z+{c}_{2}\right)}{f\left(z\right)}\right)+m\left(r,\frac{f\left(z+{c}_{1}\right)}{f\left(z\right)}\right)+O\left(1\right).\end{array}$
(2.5)

For any given ε ($0<\epsilon ), and for sufficiently large r, we have that

$m\left(r,\frac{1}{F}\right)\le T\left(r,F\right)\le {r}^{\alpha +\epsilon },\phantom{\rule{2em}{0ex}}m\left(r,{A}_{j}\right)\le {r}^{\alpha +\epsilon }\phantom{\rule{1em}{0ex}}\left(j=0,1,2\right).$
(2.6)

By Lemma 2.1, we obtain

$m\left(r,\frac{f\left(z+{c}_{2}\right)}{f\left(z\right)}\right)\le M{r}^{\sigma -1+\epsilon }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}m\left(r,\frac{f\left(z+{c}_{1}\right)}{f\left(z\right)}\right)\le M{r}^{\sigma -1+\epsilon },$
(2.7)

where M (>0) is some constant.

By $\sigma \left(f\right)=\sigma$, there exists a sequence $\left\{{r}_{n}\right\}$ satisfying ${r}_{1}<{r}_{2}<\cdots$ , ${r}_{n}\to \mathrm{\infty }$ such that

$\underset{n\to \mathrm{\infty }}{lim}\frac{logT\left({r}_{n},f\right)}{log{r}_{n}}=\sigma .$
(2.8)

Thus, for sufficiently large ${r}_{n}$, we have that

$T\left({r}_{n},f\right)\ge {r}_{n}^{\sigma -\epsilon }.$
(2.9)

Substituting (2.6)-(2.9) into (2.5), we obtain for sufficiently large ${r}_{n}$

${r}_{n}^{\sigma -\epsilon }\le T\left({r}_{n},f\right)\le N\left({r}_{n},\frac{1}{f}\right)+4{r}_{n}^{\alpha +\epsilon }+2M{r}_{n}^{\sigma -1+\epsilon }.$
(2.10)

Since $\epsilon and ε is arbitrary, by (2.10), we obtain

$\underset{n\to \mathrm{\infty }}{\overline{lim}}\frac{logN\left({r}_{n},\frac{1}{f}\right)}{log{r}_{n}}=\sigma .$

Hence, $\lambda \left(f\right)=\sigma \left(f\right)=\sigma$. □

Proof of Theorem 1.1 Suppose that $f\left(z\right)$ (0) is a meromorphic solution of equation (1.8) with $\sigma \left(f\right)<\mathrm{\infty }$.

1. (1)

Suppose that $B\left(z\right)$ satisfies the condition (i): $\sigma \left(B\right)>1$ and $\delta \left(\mathrm{\infty },B\right)=\delta >0$. Thus, for sufficiently large r,

$m\left(r,B\right)>\frac{\delta }{2}T\left(r,B\right).$
(2.11)

Clearly, $\sigma \left(f\right)\ge \sigma \left(B\right)$ by (1.8). By Lemma 2.1, we see that for any given ε ($0<\epsilon <\frac{\sigma \left(B\right)-1}{3}$),

$m\left(r,\frac{f\left(z+{c}_{j}\right)}{f\left(z\right)}\right)=O\left({r}^{\sigma \left(f\right)-1+\epsilon }\right)\phantom{\rule{1em}{0ex}}\left(j=1,2\right),$
(2.12)

and

$m\left(r,{h}_{1}\left(z\right){e}^{az}\right)\le T\left(r,{h}_{1}\left(z\right){e}^{az}\right)\le {r}^{1+\epsilon }.$
(2.13)

By (1.8), we have that

$-B\left(z\right)=\frac{f\left(z+{c}_{2}\right)}{f\left(z\right)}+{h}_{1}\left(z\right){e}^{az}\frac{f\left(z+{c}_{1}\right)}{f\left(z\right)}.$
(2.14)

Substituting (2.11)-(2.13) into (2.14), we deduce that

$\begin{array}{rcl}\frac{\delta }{2}T\left(r,B\right)& \le & m\left(r,B\right)\\ \le & m\left(r,{h}_{1}\left(z\right){e}^{az}\right)+m\left(r,\frac{f\left(z+{c}_{2}\right)}{f\left(z\right)}\right)+m\left(r,\frac{f\left(z+{c}_{1}\right)}{f\left(z\right)}\right)\\ \le & {r}^{1+\epsilon }+O\left({r}^{\sigma \left(f\right)-1+\epsilon }\right).\end{array}$
(2.15)

By $\sigma \left(B\right)=\sigma$, there is a sequence ${r}_{j}$ ($1<{r}_{1}<{r}_{2}<\cdots$ , ${r}_{j}\to \mathrm{\infty }$) satisfying

$T\left({r}_{j},B\right)>{r}_{j}^{\sigma \left(B\right)-\epsilon }.$
(2.16)

Thus, by (2.15) and (2.16), we obtain

$\frac{\delta }{2}{r}_{j}^{\sigma \left(B\right)-\epsilon }\le {r}_{j}^{1+\epsilon }+M{r}_{j}^{\sigma \left(f\right)-1+\epsilon },$
(2.17)

where M (>0) is some constant. Combining (2.17) and $\epsilon <\frac{\sigma \left(B\right)-1}{3}$, it follows that

$\frac{\delta }{2}{r}_{j}^{\sigma \left(B\right)-\epsilon }\left(1+o\left(1\right)\right)\le M{r}_{j}^{\sigma \left(f\right)-1+\epsilon }.$

So that, it follows that $\sigma \left(f\right)\ge \sigma \left(B\right)+1=max\left\{\sigma \left(B\right),1\right\}+1$.

1. (2)

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

2. (3)

Suppose that $B\left(z\right)$ satisfies the condition (iii): $B\left(z\right)={h}_{0}\left(z\right){e}^{bz}$, where b is a nonzero constant, ${h}_{0}\left(z\right)$ (0) is a meromorphic function with $\sigma \left({h}_{0}\right)<1$.

Now we need to prove $\sigma \left(f\right)\ge 2$. Contrary to the assertion, suppose that $\sigma \left(f\right)=\alpha <2$. We will deduce a contradiction. Set $z=r{e}^{i\theta }$. Then

$\left\{\begin{array}{c}\mathbf{Re}\left\{az\right\}=\delta \left(az,\theta \right)|a|r=|a|rcos\left(arga+\theta \right),\hfill \\ \mathbf{Re}\left\{bz\right\}=\delta \left(bz,\theta \right)|b|r=|b|rcos\left(argb+\theta \right).\hfill \end{array}$
(2.18)

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

$\left\{\begin{array}{c}\mathbf{Re}\left\{az\right\}=\delta \left(az,{\theta }_{0}\right)|a|r=|a|rcos\left(arga+{\theta }_{0}\right)<0,\hfill \\ \mathbf{Re}\left\{bz\right\}=\delta \left(bz,{\theta }_{0}\right)|b|r=|b|rcos\left(argb+{\theta }_{0}\right)>0.\hfill \end{array}$
(2.19)

By (1.8) and (2.19), we see that $f\left(z\right)$ 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}) and for sufficiently large r,

$|{h}_{0}\left(r{e}^{i{\theta }_{0}}\right){e}^{br{e}^{i{\theta }_{0}}}|\ge exp\left\{\left(1-{\epsilon }_{1}\right)|b|\delta \left(bz,{\theta }_{0}\right)r\right\},$
(2.20)

and

$|{h}_{1}\left(r{e}^{i{\theta }_{0}}\right){e}^{ar{e}^{i{\theta }_{0}}}|\le exp\left\{\left(1-{\epsilon }_{1}\right)|a|\delta \left(az,{\theta }_{0}\right)r\right\}<1.$
(2.21)

Thus, by (1.8), (2.12), (2.20) and (2.21), we deduce that

$\begin{array}{rcl}exp\left\{\left(1-{\epsilon }_{1}\right)|b|\delta \left(bz,{\theta }_{0}\right)r\right\}& \le & |{h}_{0}\left(r{e}^{i{\theta }_{0}}\right){e}^{br{e}^{i{\theta }_{0}}}|\\ \le & |\frac{f\left(r{e}^{i{\theta }_{0}}+{c}_{2}\right)}{f\left(r{e}^{i{\theta }_{0}}\right)}|+|{h}_{1}\left(r{e}^{i{\theta }_{0}}\right){e}^{ar{e}^{i{\theta }_{0}}}||\frac{f\left(r{e}^{i{\theta }_{0}}+{c}_{1}\right)}{f\left(r{e}^{i{\theta }_{0}}\right)}|\\ \le & 2exp\left\{{r}^{\sigma \left(f\right)-1+{\epsilon }_{1}}\right\}.\end{array}$
(2.22)

By $\delta \left(bz,{\theta }_{0}\right)=cos\left(argb+{\theta }_{0}\right)>0$, $\sigma \left(f\right)=\alpha <2$ and ${\epsilon }_{1}<\frac{2-\alpha }{2}$, it is easy to see that (2.22) is a contradiction. Hence, $\sigma \left(f\right)\ge 2$.

Subcase (b). By $arga=argb$ and $|a|\ne |b|$, we see that $f\left(z\right)$ cannot be a rational function. By Lemma 2.1, (2.12) holds. By $arga=argb$ and (2.18), we take ${\theta }_{1}=-arga$, then $\delta \left(az,{\theta }_{1}\right)=\delta \left(bz,{\theta }_{1}\right)=1$ and

$\mathbf{Re}\left\{ar{e}^{i{\theta }_{1}}\right\}=|a|r\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathbf{Re}\left\{br{e}^{i{\theta }_{1}}\right\}=|b|r.$
(2.23)

Now suppose that $|b|>|a|$. By Lemma 2.2, for any given ${\epsilon }_{2}$ ($0<{\epsilon }_{2}),

$|{h}_{0}\left(r{e}^{i{\theta }_{1}}\right){e}^{br{e}^{i{\theta }_{1}}}|\ge exp\left\{\left(1-{\epsilon }_{2}\right)|b|r\right\},$
(2.24)

and

$|{h}_{1}\left(r{e}^{i{\theta }_{1}}\right){e}^{ar{e}^{i{\theta }_{1}}}|\le exp\left\{\left(1+{\epsilon }_{2}\right)|a|r\right\}.$
(2.25)

Thus, by (1.8), (2.12), (2.24) and (2.25), we deduce that

$\begin{array}{rcl}exp\left\{\left(1-{\epsilon }_{2}\right)|b|r\right\}& \le & |{h}_{0}\left(r{e}^{i{\theta }_{1}}\right){e}^{br{e}^{i{\theta }_{1}}}|\\ \le & |\frac{f\left(r{e}^{i{\theta }_{1}}+{c}_{2}\right)}{f\left(r{e}^{i{\theta }_{1}}\right)}|+|{h}_{1}\left(r{e}^{i{\theta }_{1}}\right){e}^{ar{e}^{i{\theta }_{1}}}||\frac{f\left(r{e}^{i{\theta }_{1}}+{c}_{1}\right)}{f\left(r{e}^{i{\theta }_{1}}\right)}|\\ \le & exp\left\{{r}^{\sigma \left(f\right)-1+{\epsilon }_{2}}\right\}+exp\left\{\left(1+{\epsilon }_{2}\right)|a|r\right\}exp\left\{{r}^{\sigma \left(f\right)-1+{\epsilon }_{2}}\right\}.\end{array}$
(2.26)

Since ${\epsilon }_{2}<2-\alpha$, we have that $\sigma \left(f\right)-1+{\epsilon }_{2}=\alpha -1+{\epsilon }_{2}<1$. Combining this and (2.26), we obtain

$exp\left\{\left(1-{\epsilon }_{2}\right)|b|r\right\}
(2.27)

By ${\epsilon }_{2}<\frac{|b|-|a|}{2\left(|b|+|a|\right)}$, 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, $\sigma \left(f\right)\ge 2$ in Subcase (b).

Subcase (c). We first affirm that $f\left(z\right)$ cannot be a nonzero rational function. In fact, if $f\left(z\right)$ is a rational function, then ${e}^{az}\left[{h}_{1}\left(z\right)f\left(z+{c}_{1}\right)+{h}_{0}\left(z\right)f\left(z\right)\right]=-f\left(z+{c}_{2}\right)$ is a rational function. So that ${h}_{1}\left(z\right)f\left(z+{c}_{1}\right)+{h}_{0}\left(z\right)f\left(z\right)\equiv 0$, that is, $f\left(z+{c}_{2}\right)\equiv 0$, a contradiction.

By Lemma 2.1, (2.12) holds. By $a=b$, equation (1.8) can be rewritten as

${e}^{-az}f\left(z+{c}_{2}\right)+{h}_{1}\left(z\right)f\left(z+{c}_{1}\right)+{h}_{0}\left(z\right)f\left(z\right)=0.$
(2.28)

Using the same method as in the proof of (1), we can obtain $\sigma \left(f\right)\ge 2$.

1. (4)

Suppose that $\phi \left(z\right)$ (0) is a meromorphic function with $\sigma \left(\phi \right). Set $g\left(z\right)=f\left(z\right)-\phi \left(z\right)$. Substituting $f\left(z\right)=g\left(z\right)+\phi \left(z\right)$ into (1.8), we obtain

$\begin{array}{c}g\left(z+{c}_{2}\right)+{h}_{1}\left(z\right){e}^{az}g\left(z+{c}_{1}\right)+B\left(z\right)g\left(z\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\left[\phi \left(z+{c}_{2}\right)+{h}_{1}\left(z\right){e}^{az}\phi \left(z+{c}_{1}\right)+B\left(z\right)\phi \left(z\right)\right].\hfill \end{array}$
(2.29)

If $\phi \left(z+{c}_{2}\right)+{h}_{1}\left(z\right){e}^{az}\phi \left(z+{c}_{1}\right)+B\left(z\right)\phi \left(z\right)\equiv 0$, then $\phi \left(z\right)$ is a nonzero meromorphic solution of (1.8). Thus, by the proof above, we have that $\sigma \left(\phi \right)\ge max\left\{\sigma \left(B\right),1\right\}+1$. This contradicts our condition that $\sigma \left(\phi \right). Hence, $\phi \left(z+2\right)+{h}_{1}\left(z\right){e}^{az}\phi \left(z+1\right)+B\left(z\right)\phi \left(z\right)\not\equiv 0$, and

$\sigma \left(\phi \left(z+{c}_{2}\right)+{h}_{1}\left(z\right){e}^{az}\phi \left(z+{c}_{1}\right)+B\left(z\right)\phi \left(z\right)\right)

Applying this and Lemma 2.3 to (2.29), we deduce that

$\lambda \left(f-\phi \right)=\lambda \left(g\right)=\sigma \left(g\right)\ge max\left\{\sigma \left(B\right),1\right\}+1.$

Thus, Theorem 1.1 is proved. □

## 3 Proof of Theorem 1.2

Suppose that ${f}_{0}$ is a meromorphic solution of (1.9) with

$\sigma \left({f}_{0}\right)

If ${f}^{\ast }\left(z\right)$ ($\not\equiv {f}_{0}\left(z\right)$) is another meromorphic solution of (1.9) satisfying $\sigma \left({f}^{\ast }\right), then

$\sigma \left({f}^{\ast }-{f}_{0}\right)

But ${f}^{\ast }-{f}_{0}$ is a solution of the corresponding homogeneous equation (1.8) of (1.9). By Theorem 1.1, we have $\sigma \left({f}^{\ast }-{f}_{0}\right)\ge max\left\{\sigma \left(B\right),1\right\}+1$, a contradiction. Hence equation (1.9) possesses at most one exceptional solution ${f}_{0}$ with $\sigma \left({f}_{0}\right).

Now suppose that f is a meromorphic solution of (1.9) with

$max\left\{\sigma \left(B\right),1\right\}+1\le \sigma \left(f\right)<\mathrm{\infty }.$

Since $\sigma \left(f\right)>max\left\{\sigma \left(B\right),\sigma \left(F\right),\sigma \left(h\left(z\right){e}^{az}\right)\right\}$, applying Lemma 2.3 to (1.9), we obtain

$\lambda \left(f\right)=\sigma \left(f\right).$

Thus, Theorem 1.2 is proved.

## References

1. Chen ZX: Growth and zeros of meromorphic solution of some linear difference equations. J. Math. Anal. Appl. 2011, 373: 235-241. 10.1016/j.jmaa.2010.06.049

2. Chiang YM, Feng SJ:On the Nevanlinna characteristic of $f\left(z+\eta \right)$ and difference equations in the complex plane. Ramanujan J. 2008, 16: 105-129. 10.1007/s11139-007-9101-1

3. Ishizaki K: On difference Riccati equations and second order linear difference equations. Aequ. Math. 2011, 81: 185-198. 10.1007/s00010-010-0060-z

4. Ishizaki K, Yanagihara N: Wiman-Valiron method for difference equations. Nagoya Math. J. 2004, 175: 75-102.

5. Laine I: Nevanlinna Theory and Complex Differential Equations. de Gruyter, Berlin; 1993.

6. Li S, Gao ZS: Finite order meromorphic solutions of linear difference equations. Proc. Jpn Acad., Ser. A 2011, 87(5):73-76. 10.3792/pjaa.87.73

7. Zheng XM, Tu J: Growth of meromorphic solutions of linear difference equations. J. Math. Anal. Appl. 2011, 384: 349-356. 10.1016/j.jmaa.2011.05.069

8. Hayman WK: Meromorphic Functions. Clarendon, Oxford; 1964.

9. Laine I, Yang CC: Clunie theorems for difference and q -difference polynomials. J. Lond. Math. Soc. 2007, 76(3):556-566. 10.1112/jlms/jdm073

10. Halburd RG, Korhonen R: Difference analogue of the lemma on the logarithmic derivative with applications to difference equations. J. Math. Anal. Appl. 2006, 314: 477-487. 10.1016/j.jmaa.2005.04.010

11. Chen ZX:The growth of solutions of ${f}^{″}+{e}^{-z}{f}^{\prime }+Q\left(z\right)f=0$ where the order $\left(Q\right)=1$.Sci. China Math. 2002, 45(3):290-300.

## 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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Correspondence to Yanxia Liu.

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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/10.1186/1687-1847-2013-60

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• DOI: https://doi.org/10.1186/1687-1847-2013-60

### Keywords

• difference equation
• meromorphic coefficient
• growth
• zero 