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 [1â€“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 ${P}_{0}\left(z\right),â€¦,{P}_{n}\left(z\right)$ be polynomials such that there exists an integer l, $0â‰¤lâ‰¤n$, such that

(1.1)

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

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

Then we have $\mathrm{Ïƒ}\left(f\right)â‰¥1$.

In this paper, we use the basic notions of Nevanlinnaâ€™s theory (see [8, 9]). In addition, we use the notation $\mathrm{Ïƒ}\left(f\right)$ to denote the order of growth of the meromorphic function $f\left(z\right)$, and $\mathrm{Î»}\left(f\right)$ to denote the exponent of convergence of zeros of $f\left(z\right)$.

Chen [1] weakened the condition (1.1) of Theorem A and proved the following results.

Theorem B ([1])

Let ${P}_{n}\left(z\right),â€¦,{P}_{0}\left(z\right)$ be polynomials such that ${P}_{n}{P}_{0}â‰¢0$ and

$deg\left({P}_{n}+â‹¯+{P}_{0}\right)=max\left\{deg{P}_{j}:j=0,â€¦,n\right\}â‰¥1.$
(1.3)

Then every finite order meromorphic solution $f\left(z\right)$ (â‰¢0) of equation (1.2) satisfies $\mathrm{Ïƒ}\left(f\right)â‰¥1$, and $f\left(z\right)$ assumes every nonzero value $aâˆˆ\mathbb{C}$ infinitely often and $\mathrm{Î»}\left(fâˆ’a\right)=\mathrm{Ïƒ}\left(f\right)$.

Theorem C ([1])

Let $F\left(z\right)$, ${P}_{n}\left(z\right),â€¦,{P}_{0}\left(z\right)$ be polynomials such that $F{P}_{n}{P}_{0}â‰¢0$ and (1.3). Then every finite order transcendental meromorphic solution $f\left(z\right)$ of the equation

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

satisfies $\mathrm{Ïƒ}\left(f\right)â‰¥1$ and $\mathrm{Î»}\left(f\right)=\mathrm{Ïƒ}\left(f\right)$.

Theorem D ([1])

Let $F\left(z\right)$, ${P}_{n}\left(z\right),â€¦,{P}_{0}\left(z\right)$ be polynomials such that $F{P}_{n}{P}_{0}â‰¢0$. Suppose that $f\left(z\right)$ is a meromorphic solution with infinitely many poles of (1.2) (or (1.4)). Then $\mathrm{Ïƒ}\left(f\right)â‰¥1$.

For the linear difference equation with transcendental coefficients

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

Chiang and Feng proved the following result.

Theorem E ([2])

Let ${A}_{0}\left(z\right),â€¦,{A}_{n}\left(z\right)$ be entire functions such that there exists an integer l, $0â‰¤lâ‰¤n$, such that

(1.6)

If $f\left(z\right)$ is a meromorphic solution of (1.5), then we have $\mathrm{Ïƒ}\left(f\right)â‰¥\mathrm{Ïƒ}\left({A}_{l}\right)+1$.

Laine and Yang proved the following theorem.

Theorem F ([5])

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

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

we have $\mathrm{Ïƒ}\left(f\right)â‰¥\mathrm{Ïƒ}+1$.

Remark 1.1 If ${A}_{0},â€¦,{A}_{n}$ are meromorphic functions satisfying (1.6), then Theorem E does not hold. For example, the equation

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

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

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 $\mathrm{Ïƒ}\left(f\right)â‰¥\mathrm{Ïƒ}\left({A}_{l}\right)+1$.

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 $\mathrm{Ïƒ}\left(f\right)â‰¥\mathrm{Ïƒ}\left({A}_{l}\right)+1$.

In this note, we consider this question and prove the following results.

Theorem 1.1 Let ${c}_{1}$, ${c}_{2}$ (), a be nonzero constants, ${h}_{1}\left(z\right)$ be a nonzero meromorphic function with $\mathrm{Ïƒ}\left({h}_{1}\right)<1$, $B\left(z\right)$ be a nonzero meromorphic function.

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

1. (i)

$\mathrm{Ïƒ}\left(B\right)>1$ and $\mathrm{Î´}\left(\mathrm{âˆž},B\right)>0$;

2. (ii)

$\mathrm{Ïƒ}\left(B\right)<1$;

3. (iii)

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

then every meromorphic solution f (â‰¢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 $\mathrm{Ïƒ}\left(f\right)â‰¥max\left\{\mathrm{Ïƒ}\left(B\right),1\right\}+1$.

Further, if $\mathrm{Ï†}\left(z\right)$ (â‰¢0) is a meromorphic function with

$\mathrm{Ïƒ}\left(\mathrm{Ï†}\right)

then

$\mathrm{Î»}\left(fâˆ’\mathrm{Ï†}\right)=\mathrm{Ïƒ}\left(f\right)â‰¥max\left\{\mathrm{Ïƒ}\left(B\right),1\right\}+1.$

Corollary Under conditions of Theorem  1.1, every finite order solution $f\left(z\right)$ (â‰¢0) of (1.8) has infinitely many fixed points, satisfies $\mathrm{Ï„}\left(f\right)=\mathrm{Ïƒ}\left(f\right)$, and for any nonzero constant c,

$\mathrm{Î»}\left(f\left(z\right)âˆ’c\right)=\mathrm{Ïƒ}\left(f\right)â‰¥max\left\{\mathrm{Ïƒ}\left(B\right),1\right\}+1.$

Example 1.1

The equation

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

satisfies conditions of Theorem 1.1 and has a solution $f\left(z\right)={e}^{{z}^{2}}$ satisfying $\mathrm{Î»}\left(f\right)=0$ and $\mathrm{Ï„}\left(f\right)=\mathrm{Ïƒ}\left(f\right)=2$. 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 $F\left(z\right)$ (â‰¢0) be a meromorphic function with $\mathrm{Ïƒ}\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

$\mathrm{Î»}\left(f\right)=\mathrm{Ïƒ}\left(f\right)â‰¥max\left\{\mathrm{Ïƒ}\left(B\right),1\right\}+1$

with at most one possible exceptional solution with $\mathrm{Ïƒ}\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

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

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 $\mathrm{Ïƒ}\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 ${\mathrm{Î·}}_{1}$, ${\mathrm{Î·}}_{2}$, let f be a meromorphic function of finite order Ïƒ. Then, for each $\mathrm{Îµ}>0$, we have

$m\left(r,\frac{f\left(z+{\mathrm{Î·}}_{1}\right)}{f\left(z+{\mathrm{Î·}}_{2}\right)}\right)=O\left({r}^{\mathrm{Ïƒ}âˆ’1+\mathrm{Îµ}}\right).$

Lemma 2.2 (see [11])

Suppose that $P\left(z\right)=\left(\mathrm{Î±}+i\mathrm{Î²}\right){z}^{n}+â‹¯$ (Î±, Î² are real numbers, ) is a polynomial with degree $nâ‰¥1$, that $A\left(z\right)$ (â‰¢0) is an entire function with $\mathrm{Ïƒ}\left(A\right). Set $g\left(z\right)=A\left(z\right){e}^{P\left(z\right)}$, $z=r{e}^{i\mathrm{Î¸}}$, $\mathrm{Î´}\left(P,\mathrm{Î¸}\right)=\mathrm{Î±}cosn\mathrm{Î¸}âˆ’\mathrm{Î²}sinn\mathrm{Î¸}$. Then, for any given $\mathrm{Îµ}>0$, there exists a set ${H}_{1}âŠ‚\left[0,2\mathrm{Ï€}\right)$ that has the linear measure zero such that for any $\mathrm{Î¸}âˆˆ\left[0,2\mathrm{Ï€}\right)\mathrm{âˆ–}\left({H}_{1}âˆª{H}_{2}\right)$, there is $R>0$ such that for $|z|=r>R$, we have that

1. (i)

if $\mathrm{Î´}\left(P,\mathrm{Î¸}\right)>0$, then

$exp\left\{\left(1âˆ’\mathrm{Îµ}\right)\mathrm{Î´}\left(P,\mathrm{Î¸}\right){r}^{n}\right\}<|g\left(r{e}^{i\mathrm{Î¸}}\right)|
(2.1)
2. (ii)

if $\mathrm{Î´}\left(P,\mathrm{Î¸}\right)<0$, then

$exp\left\{\left(1+\mathrm{Îµ}\right)\mathrm{Î´}\left(P,\mathrm{Î¸}\right){r}^{n}\right\}<|g\left(r{e}^{i\mathrm{Î¸}}\right)|
(2.2)

where ${H}_{2}=\left\{\mathrm{Î¸}âˆˆ\left[0,2\mathrm{Ï€}\right);\mathrm{Î´}\left(P,\mathrm{Î¸}\right)=0\right\}$ is a finite set.

Lemma 2.3 Let ${c}_{1}$, ${c}_{2}$ (), 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 $\mathrm{Ïƒ}\left(f\right)>max\left\{\mathrm{Ïƒ}\left(F\right),\mathrm{Ïƒ}\left({A}_{j}\right)\phantom{\rule{0.25em}{0ex}}\left(j=0,1,2\right)\right\}$, then $\mathrm{Î»}\left(f\right)=\mathrm{Ïƒ}\left(f\right)$.

Proof Suppose that $\mathrm{Ïƒ}\left(f\right)=\mathrm{Ïƒ}$, $max\left\{\mathrm{Ïƒ}\left(F\right),\mathrm{Ïƒ}\left({A}_{j}\right)\phantom{\rule{0.25em}{0ex}}\left(j=0,1,2\right)\right\}=\mathrm{Î±}$. Then $\mathrm{Ïƒ}>\mathrm{Î±}$. 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

$\begin{array}{rcl}T\left(r,f\right)& =& T\left(r,\frac{1}{f}\right)+O\left(1\right)\\ =& m\left(r,\frac{1}{f}\right)+N\left(r,\frac{1}{f}\right)+O\left(1\right)\\ â‰¤& N\left(r,\frac{1}{f}\right)+m\left(r,\frac{1}{F}\right)+\underset{j=0}{\overset{2}{âˆ‘}}m\left(r,{A}_{j}\right)\\ +m\left(r,\frac{f\left(z+{c}_{2}\right)}{f\left(z\right)}\right)+m\left(r,\frac{f\left(z+{c}_{1}\right)}{f\left(z\right)}\right)+O\left(1\right).\end{array}$
(2.5)

For any given Îµ ($0<\mathrm{Îµ}), and for sufficiently large r, we have that

$m\left(r,\frac{1}{F}\right)â‰¤T\left(r,F\right)â‰¤{r}^{\mathrm{Î±}+\mathrm{Îµ}},\phantom{\rule{2em}{0ex}}m\left(r,{A}_{j}\right)â‰¤{r}^{\mathrm{Î±}+\mathrm{Îµ}}\phantom{\rule{1em}{0ex}}\left(j=0,1,2\right).$
(2.6)

By Lemma 2.1, we obtain

$m\left(r,\frac{f\left(z+{c}_{2}\right)}{f\left(z\right)}\right)â‰¤M{r}^{\mathrm{Ïƒ}âˆ’1+\mathrm{Îµ}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}m\left(r,\frac{f\left(z+{c}_{1}\right)}{f\left(z\right)}\right)â‰¤M{r}^{\mathrm{Ïƒ}âˆ’1+\mathrm{Îµ}},$
(2.7)

where M (>0) is some constant.

By $\mathrm{Ïƒ}\left(f\right)=\mathrm{Ïƒ}$, there exists a sequence $\left\{{r}_{n}\right\}$ satisfying ${r}_{1}<{r}_{2}<â‹¯$â€‰, ${r}_{n}â†’\mathrm{âˆž}$ such that

$\underset{nâ†’\mathrm{âˆž}}{lim}\frac{logT\left({r}_{n},f\right)}{log{r}_{n}}=\mathrm{Ïƒ}.$
(2.8)

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

$T\left({r}_{n},f\right)â‰¥{r}_{n}^{\mathrm{Ïƒ}âˆ’\mathrm{Îµ}}.$
(2.9)

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

${r}_{n}^{\mathrm{Ïƒ}âˆ’\mathrm{Îµ}}â‰¤T\left({r}_{n},f\right)â‰¤N\left({r}_{n},\frac{1}{f}\right)+4{r}_{n}^{\mathrm{Î±}+\mathrm{Îµ}}+2M{r}_{n}^{\mathrm{Ïƒ}âˆ’1+\mathrm{Îµ}}.$
(2.10)

Since $\mathrm{Îµ} and Îµ is arbitrary, by (2.10), we obtain

$\underset{nâ†’\mathrm{âˆž}}{\stackrel{Â¯}{lim}}\frac{logN\left({r}_{n},\frac{1}{f}\right)}{log{r}_{n}}=\mathrm{Ïƒ}.$

Hence, $\mathrm{Î»}\left(f\right)=\mathrm{Ïƒ}\left(f\right)=\mathrm{Ïƒ}$.â€ƒâ–¡

Proof of Theorem 1.1 Suppose that $f\left(z\right)$ (â‰¢0) is a meromorphic solution of equation (1.8) with $\mathrm{Ïƒ}\left(f\right)<\mathrm{âˆž}$.

1. (1)

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

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

Clearly, $\mathrm{Ïƒ}\left(f\right)â‰¥\mathrm{Ïƒ}\left(B\right)$ by (1.8). By Lemma 2.1, we see that for any given Îµ ($0<\mathrm{Îµ}<\frac{\mathrm{Ïƒ}\left(B\right)âˆ’1}{3}$),

$m\left(r,\frac{f\left(z+{c}_{j}\right)}{f\left(z\right)}\right)=O\left({r}^{\mathrm{Ïƒ}\left(f\right)âˆ’1+\mathrm{Îµ}}\right)\phantom{\rule{1em}{0ex}}\left(j=1,2\right),$
(2.12)

and

$m\left(r,{h}_{1}\left(z\right){e}^{az}\right)â‰¤T\left(r,{h}_{1}\left(z\right){e}^{az}\right)â‰¤{r}^{1+\mathrm{Îµ}}.$
(2.13)

By (1.8), we have that

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

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

$\begin{array}{rcl}\frac{\mathrm{Î´}}{2}T\left(r,B\right)& â‰¤& m\left(r,B\right)\\ â‰¤& m\left(r,{h}_{1}\left(z\right){e}^{az}\right)+m\left(r,\frac{f\left(z+{c}_{2}\right)}{f\left(z\right)}\right)+m\left(r,\frac{f\left(z+{c}_{1}\right)}{f\left(z\right)}\right)\\ â‰¤& {r}^{1+\mathrm{Îµ}}+O\left({r}^{\mathrm{Ïƒ}\left(f\right)âˆ’1+\mathrm{Îµ}}\right).\end{array}$
(2.15)

By $\mathrm{Ïƒ}\left(B\right)=\mathrm{Ïƒ}$, there is a sequence ${r}_{j}$ ($1<{r}_{1}<{r}_{2}<â‹¯$â€‰, ${r}_{j}â†’\mathrm{âˆž}$) satisfying

$T\left({r}_{j},B\right)>{r}_{j}^{\mathrm{Ïƒ}\left(B\right)âˆ’\mathrm{Îµ}}.$
(2.16)

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

$\frac{\mathrm{Î´}}{2}{r}_{j}^{\mathrm{Ïƒ}\left(B\right)âˆ’\mathrm{Îµ}}â‰¤{r}_{j}^{1+\mathrm{Îµ}}+M{r}_{j}^{\mathrm{Ïƒ}\left(f\right)âˆ’1+\mathrm{Îµ}},$
(2.17)

where M (>0) is some constant. Combining (2.17) and $\mathrm{Îµ}<\frac{\mathrm{Ïƒ}\left(B\right)âˆ’1}{3}$, it follows that

$\frac{\mathrm{Î´}}{2}{r}_{j}^{\mathrm{Ïƒ}\left(B\right)âˆ’\mathrm{Îµ}}\left(1+o\left(1\right)\right)â‰¤M{r}_{j}^{\mathrm{Ïƒ}\left(f\right)âˆ’1+\mathrm{Îµ}}.$

So that, it follows that $\mathrm{Ïƒ}\left(f\right)â‰¥\mathrm{Ïƒ}\left(B\right)+1=max\left\{\mathrm{Ïƒ}\left(B\right),1\right\}+1$.

1. (2)

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

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, ${h}_{0}\left(z\right)$ (â‰¢0) is a meromorphic function with $\mathrm{Ïƒ}\left({h}_{0}\right)<1$.

Now we need to prove $\mathrm{Ïƒ}\left(f\right)â‰¥2$. Contrary to the assertion, suppose that $\mathrm{Ïƒ}\left(f\right)=\mathrm{Î±}<2$. We will deduce a contradiction. Set $z=r{e}^{i\mathrm{Î¸}}$. Then

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

In what follows, we divide this proof into three subcases: (a) ; (b) $arga=argb$ and ; (c) $a=b$.

Subcase (a). Since and (2.18), it is easy to see that there exists a ray $argz={\mathrm{Î¸}}_{0}$ such that

$\left\{\begin{array}{c}\mathbf{Re}\left\{az\right\}=\mathrm{Î´}\left(az,{\mathrm{Î¸}}_{0}\right)|a|r=|a|rcos\left(arga+{\mathrm{Î¸}}_{0}\right)<0,\hfill \\ \mathbf{Re}\left\{bz\right\}=\mathrm{Î´}\left(bz,{\mathrm{Î¸}}_{0}\right)|b|r=|b|rcos\left(argb+{\mathrm{Î¸}}_{0}\right)>0.\hfill \end{array}$
(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 ${\mathrm{Îµ}}_{1}$ ($0<{\mathrm{Îµ}}_{1}) and for sufficiently large r,

$|{h}_{0}\left(r{e}^{i{\mathrm{Î¸}}_{0}}\right){e}^{br{e}^{i{\mathrm{Î¸}}_{0}}}|â‰¥exp\left\{\left(1âˆ’{\mathrm{Îµ}}_{1}\right)|b|\mathrm{Î´}\left(bz,{\mathrm{Î¸}}_{0}\right)r\right\},$
(2.20)

and

$|{h}_{1}\left(r{e}^{i{\mathrm{Î¸}}_{0}}\right){e}^{ar{e}^{i{\mathrm{Î¸}}_{0}}}|â‰¤exp\left\{\left(1âˆ’{\mathrm{Îµ}}_{1}\right)|a|\mathrm{Î´}\left(az,{\mathrm{Î¸}}_{0}\right)r\right\}<1.$
(2.21)

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

$\begin{array}{rcl}exp\left\{\left(1âˆ’{\mathrm{Îµ}}_{1}\right)|b|\mathrm{Î´}\left(bz,{\mathrm{Î¸}}_{0}\right)r\right\}& â‰¤& |{h}_{0}\left(r{e}^{i{\mathrm{Î¸}}_{0}}\right){e}^{br{e}^{i{\mathrm{Î¸}}_{0}}}|\\ â‰¤& |\frac{f\left(r{e}^{i{\mathrm{Î¸}}_{0}}+{c}_{2}\right)}{f\left(r{e}^{i{\mathrm{Î¸}}_{0}}\right)}|+|{h}_{1}\left(r{e}^{i{\mathrm{Î¸}}_{0}}\right){e}^{ar{e}^{i{\mathrm{Î¸}}_{0}}}||\frac{f\left(r{e}^{i{\mathrm{Î¸}}_{0}}+{c}_{1}\right)}{f\left(r{e}^{i{\mathrm{Î¸}}_{0}}\right)}|\\ â‰¤& 2exp\left\{{r}^{\mathrm{Ïƒ}\left(f\right)âˆ’1+{\mathrm{Îµ}}_{1}}\right\}.\end{array}$
(2.22)

By $\mathrm{Î´}\left(bz,{\mathrm{Î¸}}_{0}\right)=cos\left(argb+{\mathrm{Î¸}}_{0}\right)>0$, $\mathrm{Ïƒ}\left(f\right)=\mathrm{Î±}<2$ and ${\mathrm{Îµ}}_{1}<\frac{2âˆ’\mathrm{Î±}}{2}$, it is easy to see that (2.22) is a contradiction. Hence, $\mathrm{Ïƒ}\left(f\right)â‰¥2$.

Subcase (b). By $arga=argb$ and , 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 ${\mathrm{Î¸}}_{1}=âˆ’arga$, then $\mathrm{Î´}\left(az,{\mathrm{Î¸}}_{1}\right)=\mathrm{Î´}\left(bz,{\mathrm{Î¸}}_{1}\right)=1$ and

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

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

$|{h}_{0}\left(r{e}^{i{\mathrm{Î¸}}_{1}}\right){e}^{br{e}^{i{\mathrm{Î¸}}_{1}}}|â‰¥exp\left\{\left(1âˆ’{\mathrm{Îµ}}_{2}\right)|b|r\right\},$
(2.24)

and

$|{h}_{1}\left(r{e}^{i{\mathrm{Î¸}}_{1}}\right){e}^{ar{e}^{i{\mathrm{Î¸}}_{1}}}|â‰¤exp\left\{\left(1+{\mathrm{Îµ}}_{2}\right)|a|r\right\}.$
(2.25)

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

$\begin{array}{rcl}exp\left\{\left(1âˆ’{\mathrm{Îµ}}_{2}\right)|b|r\right\}& â‰¤& |{h}_{0}\left(r{e}^{i{\mathrm{Î¸}}_{1}}\right){e}^{br{e}^{i{\mathrm{Î¸}}_{1}}}|\\ â‰¤& |\frac{f\left(r{e}^{i{\mathrm{Î¸}}_{1}}+{c}_{2}\right)}{f\left(r{e}^{i{\mathrm{Î¸}}_{1}}\right)}|+|{h}_{1}\left(r{e}^{i{\mathrm{Î¸}}_{1}}\right){e}^{ar{e}^{i{\mathrm{Î¸}}_{1}}}||\frac{f\left(r{e}^{i{\mathrm{Î¸}}_{1}}+{c}_{1}\right)}{f\left(r{e}^{i{\mathrm{Î¸}}_{1}}\right)}|\\ â‰¤& exp\left\{{r}^{\mathrm{Ïƒ}\left(f\right)âˆ’1+{\mathrm{Îµ}}_{2}}\right\}+exp\left\{\left(1+{\mathrm{Îµ}}_{2}\right)|a|r\right\}exp\left\{{r}^{\mathrm{Ïƒ}\left(f\right)âˆ’1+{\mathrm{Îµ}}_{2}}\right\}.\end{array}$
(2.26)

Since ${\mathrm{Îµ}}_{2}<2âˆ’\mathrm{Î±}$, we have that $\mathrm{Ïƒ}\left(f\right)âˆ’1+{\mathrm{Îµ}}_{2}=\mathrm{Î±}âˆ’1+{\mathrm{Îµ}}_{2}<1$. Combining this and (2.26), we obtain

$exp\left\{\left(1âˆ’{\mathrm{Îµ}}_{2}\right)|b|r\right\}
(2.27)

By ${\mathrm{Îµ}}_{2}<\frac{|b|âˆ’|a|}{2\left(|b|+|a|\right)}$, 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, $\mathrm{Ïƒ}\left(f\right)â‰¥2$ 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 ${e}^{az}\left[{h}_{1}\left(z\right)f\left(z+{c}_{1}\right)+{h}_{0}\left(z\right)f\left(z\right)\right]=âˆ’f\left(z+{c}_{2}\right)$ is a rational function. So that ${h}_{1}\left(z\right)f\left(z+{c}_{1}\right)+{h}_{0}\left(z\right)f\left(z\right)â‰¡0$, that is, $f\left(z+{c}_{2}\right)â‰¡0$, a contradiction.

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

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

Using the same method as in the proof of (1), we can obtain $\mathrm{Ïƒ}\left(f\right)â‰¥2$.

1. (4)

Suppose that $\mathrm{Ï†}\left(z\right)$ (â‰¢0) is a meromorphic function with $\mathrm{Ïƒ}\left(\mathrm{Ï†}\right). Set $g\left(z\right)=f\left(z\right)âˆ’\mathrm{Ï†}\left(z\right)$. Substituting $f\left(z\right)=g\left(z\right)+\mathrm{Ï†}\left(z\right)$ into (1.8), we obtain

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

If $\mathrm{Ï†}\left(z+{c}_{2}\right)+{h}_{1}\left(z\right){e}^{az}\mathrm{Ï†}\left(z+{c}_{1}\right)+B\left(z\right)\mathrm{Ï†}\left(z\right)â‰¡0$, then $\mathrm{Ï†}\left(z\right)$ is a nonzero meromorphic solution of (1.8). Thus, by the proof above, we have that $\mathrm{Ïƒ}\left(\mathrm{Ï†}\right)â‰¥max\left\{\mathrm{Ïƒ}\left(B\right),1\right\}+1$. This contradicts our condition that $\mathrm{Ïƒ}\left(\mathrm{Ï†}\right). Hence, $\mathrm{Ï†}\left(z+2\right)+{h}_{1}\left(z\right){e}^{az}\mathrm{Ï†}\left(z+1\right)+B\left(z\right)\mathrm{Ï†}\left(z\right)â‰¢0$, and

$\mathrm{Ïƒ}\left(\mathrm{Ï†}\left(z+{c}_{2}\right)+{h}_{1}\left(z\right){e}^{az}\mathrm{Ï†}\left(z+{c}_{1}\right)+B\left(z\right)\mathrm{Ï†}\left(z\right)\right)

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

$\mathrm{Î»}\left(fâˆ’\mathrm{Ï†}\right)=\mathrm{Î»}\left(g\right)=\mathrm{Ïƒ}\left(g\right)â‰¥max\left\{\mathrm{Ïƒ}\left(B\right),1\right\}+1.$

Thus, Theorem 1.1 is proved.â€ƒâ–¡

## 3 Proof of Theorem 1.2

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

$\mathrm{Ïƒ}\left({f}_{0}\right)

If ${f}^{âˆ—}\left(z\right)$ ($â‰¢{f}_{0}\left(z\right)$) is another meromorphic solution of (1.9) satisfying $\mathrm{Ïƒ}\left({f}^{âˆ—}\right), then

$\mathrm{Ïƒ}\left({f}^{âˆ—}âˆ’{f}_{0}\right)

But ${f}^{âˆ—}âˆ’{f}_{0}$ is a solution of the corresponding homogeneous equation (1.8) of (1.9). By Theorem 1.1, we have $\mathrm{Ïƒ}\left({f}^{âˆ—}âˆ’{f}_{0}\right)â‰¥max\left\{\mathrm{Ïƒ}\left(B\right),1\right\}+1$, a contradiction. Hence equation (1.9) possesses at most one exceptional solution ${f}_{0}$ with $\mathrm{Ïƒ}\left({f}_{0}\right).

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

$max\left\{\mathrm{Ïƒ}\left(B\right),1\right\}+1â‰¤\mathrm{Ïƒ}\left(f\right)<\mathrm{âˆž}.$

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

$\mathrm{Î»}\left(f\right)=\mathrm{Ïƒ}\left(f\right).$

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 $f\left(z+\mathrm{Î·}\right)$ 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 ${f}^{â€³}+{e}^{âˆ’z}{f}^{â€²}+Q\left(z\right)f=0$ 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).

## Author information

Authors

### Corresponding author

Correspondence to Yanxia Liu.

### Competing interests

The author declares that they have no competing interests.

## Rights and permissions

Reprints and permissions

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