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The properties of solutions of certain type of difference equations
Advances in Difference Equations volume 2014, Article number: 256 (2014)
Abstract
In this paper, we shall utilize Nevanlinna value distribution theory to study the solvability of the difference equations of the form: $f{(z)}^{n}+p(z){({\mathrm{\Delta}}_{c}f)}^{m}=r(z){e}^{q(z)}$ and $f{(z)}^{n}+p(z){e}^{q(z)}{({\mathrm{\Delta}}_{c}f)}^{m}=r(z)$, and we shall study the growth of their entire solutions. Moreover, we will give a number of examples to show that the results in this paper are the best possible in certain senses. This article extends earlier results by Liu et al. (Czechoslov. Math. J. 61:565576, 2011; Ann. Pol. Math. 102:129142, 2011).
MSC:39A05, 30D35.
1 Introduction and main results
In this paper, we use the basic notions in Nevanlinna theory of meromorphic functions, as found in [1]. In addition, we use $\delta (f)$, $\lambda (f)$, and $\lambda (\frac{1}{f})$ to denote the order, and the exponents of the convergence of zeros and poles of a meromorphic function $f(z)$, respectively. We define difference operators as ${\mathrm{\Delta}}_{c}f=f(z+c)f(z)$, where c is a nonzero constant.
In 2001, Yang [2] started to study the existence and uniqueness of finite order entire solutions of the following type of nonlinear differential equation:
where $L(f)$ is a linear differential polynomial in f with polynomial coefficients, $p(z)$ is a nonvanishing polynomial, $h(z)$ is an entire function, and n is an integer such that $n\ge 4$. Subsequently, several papers have appeared in which the solutions of (1) are studied. The reader is invited to see [3–5].
Recently, Yang and Laine [6] considered the existence of the nonlinear differentialdifference equation of the form
where $L(z,f)$ is a finite sum of product of f, derivatives of f, and their shifts. They obtained the following result.
Theorem A Let $n\ge 4$ be an integer, $L(z,f)$ be a linear differentialdifference polynomial of f, with small meromorphic coefficients, and $h(z)$ be a meromorphic function of finite order. Then (2) possesses at most one admissible transcendental entire solution of finite order, unless $L(z,f)$ vanishes identically. If such a solution $f(z)$ exists, then $f(z)$ is of the same order as $h(z)$.
In particular, it is shown in [6] that the equation
has no transcendental entire solution of finite order, where $P(z)$, $Q(z)$ are polynomials.
Some improvements of Theorem A can be found in [7–10] as well. Replacing $f(z+c)$ with ${\mathrm{\Delta}}_{c}f$, one of the present authors considered the existence of entire solutions of
we get the following result in [11].
Theorem B Let $P(z)$, $Q(z)$ be polynomials, n and m be integers satisfying $n>m\ge 0$. Then (3) has no transcendental entire solution of finite order.
In fact, the special case of (3) with $n=m$, and $p(z)=q(z)=1$, can be viewed as the Fermat type functional equation. It is well known that (3) has no transcendental entire solutions when $n\ge 3$, which can be seen in [12].
It is natural to ask what happens if $P(z)$ or $Q(z)$ is a transcendental entire function in (3). Corresponding to this question, we will investigate the finite order entire (meromorphic) solution of
Theorem 1.1 Consider the nonlinear difference equation of the form
where $p(z)\not\equiv 0$, $q(z)$, $r(z)$ are polynomials, n and m are positive integers. Suppose that $f(z)$ is a transcendental entire function of finite order, not of period c. If $n>m$, then $f(z)$ cannot be a solution of (4).
Remarks (1) In case $n\le m$, Theorem 1.1 is not true. In the special case that $m=n=1$, the function $f(z)={e}^{z}$ solves
Moreover, the entire function $f(z)={e}^{z}+z$ is a solution of the following equation:

(2)
From Theorem 1.1, we get the following result: Let $f(z)$ be a transcendental entire function of finite order, then for $n>m$, $f{(z)}^{n}+p(z){({\mathrm{\Delta}}_{c}f)}^{m}$ assumes zero infinitely often, where $p(z)\not\equiv 0$ is polynomial.
In the following, we will consider the properties of the solutions of the equation
where $p(z)\not\equiv 0$, $q(z)$, $r(z)$ are polynomials. In fact, (5) may have solutions. For example, the function $f(z)=z{e}^{z}$ is a solution of the equation $f{(z)}^{3}\frac{{z}^{3}}{2\pi i}{e}^{2z}{\mathrm{\Delta}}_{2\pi i}f=0$. One solution of the equation $f(z){e}^{z}\frac{1}{2\pi i}{\mathrm{\Delta}}_{2\pi i}f=z$ is the function $f(z)={e}^{z}+z$. In addition, the function $f(z)={e}^{z}+z$ can solve $f(z)+{e}^{z}\frac{1}{4\pi}{({\mathrm{\Delta}}_{2\pi i}f)}^{2}=z$ as well. Hence, here we just study the order of the growth of the solutions. We state our findings as follows.
Theorem 1.2 Let $p(z)\not\equiv 0$, $q(z)$, $r(z)$ be polynomials, n and m be positive integers satisfying $n>m$. Let $f(z)$ be finite order entire solutions of (5), then $\delta (f)=degq(z)$.
Remark We will give the following examples to show that the assumption that $n>m$ in Theorem 1.2 is sharp. Clearly, $f(z)={e}^{z}$ is a solution of the equation $f(z)\frac{1}{2}{\mathrm{\Delta}}_{ln2}f=0$. The function $f(z)=4{e}^{8\pi iz}{e}^{4\pi iz}+z$ solves $f(z){({\mathrm{\Delta}}_{\frac{1}{4}}f)}^{2}=z\frac{1}{16}$. However, $\delta (f)\ne degq(z)$ in the above two examples.
Corollary 1.3 Let $p(z)\not\equiv 0$, $q(z)$, $r(z)$ be polynomials, $n>1$. Let $f(z)$ be finite order entire solutions of the equation
then $\delta (f)=degq(z)$.
According to (6), we will give a further discussion on the existence of meromorphic solutions. We get:
Theorem 1.4 Let $p(z)\not\equiv 0$, $q(z)$, $r(z)$ be polynomials. Assume one of the following assertions holds:

(i)
$n>1$, $f(z)$ is a finite order meromorphic function (not entire),

(ii)
$n>1$, $r(z)\equiv 0$, and $f(z)$ is a finite order entire function with infinitely many zeros,

(iii)
$n\ge 1$, $r(z)\not\equiv 0$, and $f(z)$ is a finite order entire function satisfying $\lambda (f)<\delta (f)$.
Then $f(z)$ cannot be a solution of (6).
Remarks (1) From Theorem 1.4, we know that the solution of the equation $f{(z)}^{n}+p(z){e}^{q(z)}{\mathrm{\Delta}}_{c}f=0$ must be expressed as $f(z)=\alpha (z){e}^{\beta (z)}$, where $\alpha (z)$ and $\beta (z)$ are polynomials.

(2)
The condition that $n>1$ in Theorem 1.4 is sharp. In fact, if $n=1$, then we know the function $f(z)=\frac{{e}^{z}}{z}$ is a solution of the equation $f(z)+(\frac{z}{2\pi i}+1){\mathrm{\Delta}}_{2\pi i}f=0$. Moreover, the function $f(z)=cosz$ can solve $f(z)+\frac{1}{2}{\mathrm{\Delta}}_{\pi}f=0$.

(3)
Some ideas of this paper are from [13].
2 Preliminary lemmas
Lemma 2.1 [[14], Theorem 2.1]
Let $f(z)$ be a meromorphic function of finite order, and let $c\in \mathbb{C}$, then
Remark From Lemma 2.1, we know $T(r,f)=T(r,f(z+c))+S(r,f)$ when $f(z)$ is an entire function of finite order.
Lemma 2.2 [[15], Theorem 2.4.2]
Let $f(z)$ be a transcendental meromorphic solution of
where $A(z,f)$, $B(z,f)$ are differential polynomials in f and its derivatives with small meromorphic coefficients ${a}_{\lambda}$, in the sense of $m(r,{a}_{\lambda})=S(r,f)$ for all $\lambda \in I$. If $d(B(z,f))\le n$, then $m(r,A(z,f))=S(r,f)$.
Lemma 2.3 [[1], Theorem 1.56]
Let ${f}_{j}(z)$ ($j=1,2,3$) be meromorphic functions that satisfy
If ${f}_{1}(z)$ is not a constant, and
where $\lambda <1$ and $T(r)={max}_{1\le j\le 3}\{T(r,{f}_{j})\}$, then either ${f}_{2}(z)\equiv 1$ or ${f}_{3}(z)\equiv 1$.
Lemma 2.4 [[1], Theorem 1.51]
Suppose that ${f}_{j}(z)$ ($j=1,\dots ,n$) ($n\ge 2$) are meromorphic functions and ${g}_{j}(z)$ ($j=1,\dots ,n$) are entire functions satisfying the following conditions.

(1)
${\sum}_{j=1}^{n}{f}_{j}(z){e}^{{g}_{j}(z)}\equiv 0$.

(2)
$1\le j<k\le n$, ${g}_{j}(z){g}_{k}(z)$ are not constants for $1\le j<k\le n$.

(3)
For $1\le j\le n$, $1\le h<k\le n$,
$$T(r,{f}_{j})=o\left\{T(r,{e}^{{g}_{h}{g}_{k}})\right\},\phantom{\rule{1em}{0ex}}r\to \mathrm{\infty},r\notin E,$$
where $E\subset (1,\mathrm{\infty})$ is of finite linear measure.
Then ${f}_{j}(z)\equiv 0$.
3 Proof of Theorem 1.1
If $q(z)$ is a constant or $r(z)\equiv 0$, then the conclusion follows from Theorem B. It remains to consider the case $q(z)$ is a nonconstant polynomial and $r(z)\not\equiv 0$. Assume $f(z)$ is a transcendental entire solution of (4), which is finite order, not of period c. Differentiating (4) and eliminating ${e}^{q(z)}$, we have
If $n{f}^{\prime}(z)({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})f(z)\equiv 0$, then we have $f{(z)}^{n}=Ar(z){e}^{q(z)}$. Writing $f(z)=h(z){e}^{\frac{q(z)}{n}}$, where $h(z)$ satisfies $h{(z)}^{n}=Ar(z)$ and A is a nonzero constant. Substituting $f(z)$ into (4), we get
Clearly, if $A=1$, then $f(z)$ is a period function with period c, which contradicts the assumption. Hence, $A\ne 1$. Let $g={e}^{\frac{q(z)}{n}}$, then ${({\mathrm{\Delta}}_{c}f)}^{m}$ can be expressed as ${\sum}_{i=0}^{m}\left(\genfrac{}{}{0ex}{}{m}{i}\right){(1)}^{i}h{(z)}^{i}h{(z+c)}^{mi}g{(z)}^{i}g{(z+c)}^{mi}$. Furthermore, from Lemma 2.1, we have
From (8) and (9), we get
which contradicts the condition that $n>m$. Therefore, we conclude that $n{f}^{\prime}(z)({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})f(z)\not\equiv 0$. We discuss the following two cases.
Case 1. $n>m+1$. Rewrite (7) in the following forms:
and
Applying Lemma 2.1, Lemma 2.2, and the lemma on the logarithmic derivative to the above two equations, we obtain
and
Hence
which is a contradiction.
Case 2. $n=m+1$. For this case, (4) and (7) now take the following forms:
and
Set $H(z)=(m+1){f}^{\prime}(z)({q}^{\prime}(z)+\frac{{r}^{\prime}(z)}{r(z)})f(z)$. Dividing (11) by $f{(z)}^{m}$, we have
Since $f(z)$ is entire, applying Lemma 2.1 and the lemma on the logarithmic derivative to the above equation, we conclude that
Differentiating $H(z)$, we have
that is
The above equation can be rewritten in the following form:
Suppose ${z}_{0}$ is a zero of $f(z)$ with multiplicity k. If ${z}_{0}$ is a zero of $r(z)$ as well, then the contribution to $N(r,\frac{1}{f})$ is $S(r,f)$. Assuming that ${z}_{0}$ is not a zero of $r(z)$, we will discuss the two subcases:
Subcase 1. Suppose ${z}_{0}$ is a zero of $H(z)$ with multiplicity t. From (13), by simple calculations, we know that $k=1+t\le 2t$, which means that ${z}_{0}$ is a contribution of $S(r,f)$ to $N(r,\frac{1}{f})$ by (12).
Subcase 2. Suppose ${z}_{0}$ is not a zero of $H(z)$. By (13), we get ${k}^{2}k=0$, then such a zero of $f(z)$ must be simple and we find that ${q}^{\prime}+\frac{{r}^{\prime}}{r}+(m+1)\frac{{H}^{\prime}}{H}$ must vanish at ${z}_{0}$. That implies that ${z}_{0}$ makes a contribution of $S(r,f)$ to $N(r,\frac{1}{f})$ by (12).
In a word, $N(r,\frac{1}{f})=S(r,f)$. Therefore, we can assume $f(z)$ to be of the form $f(z)=G(z){e}^{s(z)}$, where $s(z)$ is a nonconstant polynomial, and $G(z)$ is an entire function satisfying $N(r,\frac{1}{G})=T(r,G)=S(r,f)$ by the Hadamard factorization theorem. Substituting this expression into (10), we get
and so
which may be written in the form
From Lemma 2.3, we have either
or
If ${f}_{2}(z)\equiv 1$, then we get
Plugging the above equation into (10), similar to (9), we get $T(r,r{e}^{q}p{({\mathrm{\Delta}}_{c}f)}^{m})\le mT(r,f)+S(r,f)$. Comparing both sides of (9), we get a contradiction. If ${f}_{3}(z)\equiv 1$, then we get $f{(z)}^{m+1}\equiv r(z){e}^{q(z)}$, which means that ${\mathrm{\Delta}}_{c}f\equiv 0$, and this contradicts the assumption.
Therefore, from the discussions above, we find that a transcendental finite order entire function $f(z)$ which is not of period c, cannot be a solution of (4). The proof of Theorem 1.1 is finished completely.
4 Proof of Theorem 1.2
Applying Lemma 2.1 to (5), we get
which means that $(nm)m(r,f)\le m(r,{e}^{q})+S(r,f)$. From the assumption that $n>m$ and the above inequality, we conclude that $\delta (f)\le degq(z)$.
Now we show that $\delta (f)=degq(z)$. Suppose to the contrary that $\delta (f)<degq(z)$. Then $\delta ({f}^{n}+p{e}^{q}{({\mathrm{\Delta}}_{c}f)}^{m})=degq(z)>0$ from Remark below Lemma 2.1, and $\delta (r)=0$. This is a contradiction by (5).
5 Proof of Theorem 1.4
First, we confirm that one period function with period c cannot be a solution of (6). Actually, if $f(z)$ is a period function, then the order of $f(z)$ satisfies $\delta (f)\ge 1$. At the same time, we get $f{(z)}^{n}=r(z)$, which is impossible. It remains to consider the function which is not of period c.

(i)
Suppose that $f(z)$ is a finite order meromorphic solution of (6), and ${z}_{0}$ is a pole of $f(z)$ with multiplicity t. Then we get ${z}_{0}+c$ is a pole of $f(z)$ with multiplicity $\ge nt$. By calculation, we see that ${z}_{0}+kc$ is a pole of $f(z)$ with multiplicity ${n}^{k}t$, where k is a positive integer. Since $n>1$, we conclude that $\lambda (f)=\mathrm{\infty}$, i.e., $\delta (f)=\mathrm{\infty}$. This is a contradiction.

(ii)
Suppose that a finite order entire solution $f(z)$ has infinitely many zeros and $r(z)\equiv 0$. We assume that all zeros of $p(z)$ are in $D=\{z:Rez\le M,Imz\le M\}$, where $M>0$ is some constant. From the assumption, we know that $f(z)$ has infinitely many zeros which are not in D. If ${z}_{0}\notin D$ satisfies $f({z}_{0})=0$, then from (6) we know that ${z}_{0}+kc$ are zeros of $f(z)$. Moreover, we have ${z}_{0}+kc$ are not in D, as $k\to \mathrm{\infty}$. In the same way as (i), we get $\lambda (\frac{1}{f})=\mathrm{\infty}$, i.e., $\delta (f)=\mathrm{\infty}$, which contradicts the assumption that $\delta (f)<\mathrm{\infty}$.

(iii)
Suppose $f(z)$ is an entire solution of finite order satisfying $\lambda (f)<\delta (f)$, then we know $\delta (f)\ge 1$ by the conclusion of Corollary 1.3. Hence, from the Hadamard factorization theorem, $f(z)$ can be written as $f(z)=\alpha (z){e}^{\beta (z)}$, where $\beta (z)$ is a nonconstant polynomial.
Case 1. $n>1$. From the conclusion of Corollary 1.3, we know $\lambda (f)=\delta (\alpha )<\delta (f)=degq(z)=q\ge 1$. So we can rewrite $f(z)$ in the following form:
where ${a}_{q}$ is a nonzero constant, $\delta (\gamma )<q$, and $T(r,\gamma )=S(r,f)$. From the conclusion of Corollary 1.3, we conclude that
where ${b}_{q}$ is a nonzero constant, $\delta (\mu )<q$, and $T(r,\mu )=S(r,f)$. By (6), (15), and (16), we have
where $\nu (z)$ is a polynomial such that $deg\nu (z)\le q1$.
Subcase 1. If $n{a}_{q}={b}_{q}+{a}_{q}$, then (17) can be expressed as
Since $r(z)\not\equiv 0$, we get $\gamma {(z)}^{n}+p(z)\mu (z)(\gamma (z+c){e}^{\nu (z)}\gamma (z))\not\equiv 0$. Comparing the characteristics function of both sides of the above equation, we get a contradiction.
Subcase 2. If $n{a}_{q}\ne {b}_{q}+{a}_{q}$, then we rewrite (17) in the form
Applying Lemma 2.4 to the above equation, we get $r(z)\equiv 0$, which contradicts the assumption.
Case 2. $n=1$. Comparing both sides of (6), we obtain $\delta (f)\ge degq(z)=q$ by Remark below Lemma 2.1. If $\delta (f)=q$, then in the same way as Case 1, the conclusion follows. If $k=\delta (f)>q$, then $f(z)$ can be written as $f(z)=\varphi (z){e}^{{a}_{k}{z}^{k}}$, where $\delta (\varphi )<k$, and $T(r,\varphi )=S(r,f)$. Substituting this expression into (6), we get
where $\delta (\phi )<k$, and $T(r,\phi )=S(r,f)$. Since $r(z)\not\equiv 0$, we get $\varphi (z)+p(z){e}^{q(z)}(\varphi (z+c)\phi (z)\varphi (z))\not\equiv 0$. Comparing the characteristics function of both sides of the above equation, we get a contradiction.
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Acknowledgements
The authors thank the referee for his/her valuable suggestions to improve the present paper. This work was supported by the National Natural Science Foundation of China (No. 11301220 and No. 11371225) and the Tianyuan Fund for Mathematics (No. 11226094), the NSF of Shandong Province, China (No. ZR2012AQ020) and the Fund of Doctoral Program Research of University of Jinan (XBS1211).
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XQ completed the main part of this paper, JD and LY corrected the main theorems. All authors read and approved the final manuscript.
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Qi, X., Dou, J. & Yang, L. The properties of solutions of certain type of difference equations. Adv Differ Equ 2014, 256 (2014). https://doi.org/10.1186/168718472014256
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Keywords
 meromorphic functions
 difference equation
 growth
 finite order