††2010 _Mathematics Subject Classification_ : 39B32, 30D35.†† _Key words and
phrases_ : Exponential polynomial, differential-difference equation, convex
hull, Nevanlinna theory.

# Characterization of exponential polynomial as solution of certain type of
non-linear delay-differential equation

Abhijit Banerjee and Tania Biswas  Department of Mathematics, University of
Kalyani, West Bengal 741235, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, we have characterized the nature and form of solutions of the
following non-linear delay-differential equation:

$f^{n}(z)+\sum_{i=1}^{n-1}b_{i}f^{i}(z)+q(z)e^{Q(z)}L(z,f)=P(z),$

where $b_{i}\in\mathbb{C}$, $L(z,f)$ be a linear delay-differential polynomial
of $f$; $n$ be positive integers; $q$, $Q$ and $P$ respectively be non-zero,
non-constant and any polynomials. Different special cases of our result will
accommodate all the results of ([J. Math. Anal. Appl., 452(2017), 1128-1144.],
[Mediterr. J. Math., 13(2016), 3015-3027], [Open Math., 18(2020), 1292-1301]).
Thus our result can be considered as an improvement of all of them. We have
also illustrated a handful number of examples to show that all the cases as
demonstrated in our theorem actually occurs and consequently the same are
automatically applicable to the previous results.

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##  1\. Introduction, Results and Examples

Throughout the paper, we denote by $f$ a meromorphic function in the complex
plane $\mathbb{C}$ and related to the function, we assume that the readers are
familiar with the basic terms like $T(r,f)$, $N(r,f)$, $m(r,f)$, of Nevanlinna
value distribution theory of meromorphic functions (see [6, 8]). The notation
$S(r,f)$, will be used to define any quantity satisfying $S(r,f)=o(T(r,f))$ as
$r\rightarrow\infty$, possibly outside a set $E$ of $r$ of finite logarithmic
measure. In addition, we will respectively use the symbols $\rho(f)$,
$\lambda(f)$ and $\tau(f)$ to denote the order, exponent of convergent and
type of $f$. The symbol $L(f)$ will be used to represent a linear differential
polynomial in $f$ with polynomial coefficients. Also, throughout this paper,
by $card(S)$, we mean the cardinality of a set $S$, i.e., the number of
elements in $S$.

Considering the non-linear differential equation

$L(f)-p(z)f^{n}(z)=h(z),$

in 2001, Yang [17] investigated about the transcendental finite order entire
solutions $f$ of the equation, where $p(z)$ is a non-vanishing polynomial,
$h(z)$ is entire and $n\geq 4$ is an integer.

In 2010, Yang-Laine [18] showed that the equation

$f(z)^{2}+q(z)f(z+1)=p(z),$

where $p(z)$, $q(z)$ are polynomials, admits no transcendental entire
solutions of finite order.

In the last two decades researchers mainly studied (see [1, 9, 11, 17, 20]
etc.) about the following three distinct features of solutions of shift or
delay-differential or differential equations:
i) existence and non-existence conditions,
ii) order of growth and
iii) different types of forms of solutions.

Next, let us consider the exponential polynomial $f(z)$, defined by the form

$\displaystyle f(z)=P_{1}(z)e^{Q_{1}(z)}+\cdots+P_{k}(z)e^{Q_{k}(z)},$ (1.1)

where $P_{j}$’s and $Q_{j}$’s are polynomials in $z$. Steinmetz [14], showed
that (1.1) can be written in the normalized form

$\displaystyle
f(z)=H_{0}(z)+H_{1}(z)e^{\omega_{1}z^{t}}+\cdots+H_{m}(z)e^{\omega_{m}z^{t}},$
(1.2)

where $H_{j}$ are either exponential polynomials of order $<t$ or ordinary
polynomials in $z$, the leading coefficients $\omega_{j}$ are pairwise
distinct and $m\leq k$.

Let $co(\mathcal{W})$ be the convex hull of a set
$\mathcal{W}\subset\mathbb{C}$ which is the intersection of all convex sets
containing $\mathcal{W}$. If $\mathcal{W}$ contains finitely many elements
then $co(\mathcal{W})$ is obtained as an intersection of finitely many half-
planes, then $co(\mathcal{W})$ is either a compact polygon with a non-empty
interior or a line segment. We denote by $C(co(\mathcal{W}))$, the
circumference of $co(\mathcal{W})$. If $co(\mathcal{W})$ is a line-segment,
then $C(co(\mathcal{W}))$ is equals to twice the length of this line segment.
Throughout the paper, we denote
$W=\\{\bar{\omega}_{1},\bar{\omega}_{2},\ldots,\bar{\omega}_{m}\\}$ and
$W_{0}=W\cup\\{0\\}$.

Now-a-days, to find the form of exponential polynomials as solution of certain
non-linear differential-difference equation has become an interesting topic
among researchers (see [3, 4, 16]). Most probably, in this regard, the first
attempt was made by Wen-Heittokangas-Laine [15]. In 2012, they considered the
equation

$\displaystyle f(z)^{n}+q(z)e^{Q(z)}f(z+c)=P(z),$ (1.3)

where $q(z)$, $Q(z)$, $P(z)$ are polynomials, $n\geq 2$ is an integer,
$c\in\mathbb{C}\backslash\\{0\\}$. Wen-Heittokangas-Laine also pointed out
that for a non-constant polynomial $\alpha(z)$ and $d\in\mathbb{C}$, every
solution $f$ of the form (1.2) reduces to a function which belongs to one of
the following classes:

$\displaystyle\Gamma_{1}$ $\displaystyle=$
$\displaystyle\\{e^{\alpha(z)}+d\\},$ $\displaystyle\Gamma_{0}$
$\displaystyle=$ $\displaystyle\\{e^{\alpha(z)}\\}$

and classified finite order entire(meromorphic) solutions of (1.3) as follows:

######  Theorem A.

[15] Let $n\geq 2$ be an integer, let $c\in\mathbb{C}\backslash\\{0\\}$,
$q(z)$, $Q(z)$, $P(z)$ be polynomials such that $Q(z)$ is not a constant and
$q(z)\not\equiv 0$. Then the finite order entire solutions $f$ of equation
(1.3) satisfies the following conclusions:

  * (a)

Every solution $f$ satisfies $\rho(f)=\deg Q$ and is of mean type.

  * (b)

Every solution $f$ satisfies $\lambda(f)=\rho(f)$ if and only if
$P(z)\not\equiv 0$.

  * (c)

A solution $f$ belongs to $\Gamma_{0}$ if and only if $P(z)\equiv 0$. In
particular, this is the case if $n\geq 3$.

  * (d)

If a solution $f$ belongs to $\Gamma_{0}$ and if $g$ is any other finite order
entire solution of (1.3), then $f=\eta g$, where $\eta^{n-1}=1$.

  * (e)

If $f$ is an exponential polynomial solution of the form (1.2), then
$f\in\Gamma_{1}$. Moreover, if $f\in\Gamma_{1}\backslash\Gamma_{0}$, then
$\rho(f)=1$.

Inspired by Theorem A, in 2016, Liu [12] replaced $f(z+c)$ by $f^{(k)}(z+c)$
in (1.3) and for two polynomials $p_{1}(z)$, $p_{2}(z)$ and a non-constant
polynomial $\alpha(z)$, introduced two new classes of solutions:

$\displaystyle\Gamma_{1}^{\prime}$ $\displaystyle=$
$\displaystyle\\{p_{1}(z)e^{\alpha(z)}+p_{2}(z)\\},$
$\displaystyle\Gamma_{0}^{\prime}$ $\displaystyle=$
$\displaystyle\\{p_{1}(z)e^{\alpha(z)}\\}$

to obtain the following theorem.

######  Theorem B.

[12] Under the same situation as in Theorem A with $k\geq 1$, the finite-order
transcendental entire solution $f$ of

$\displaystyle f(z)^{n}+q(z)e^{Q(z)}f^{(k)}(z+c)=P(z),$ (1.4)

should satisfy the results (a), (b), (d) and

  * (1)

A solution $f$ belongs to $\Gamma_{0}^{\prime}$ if and only if $P(z)\equiv 0$.
In particular, this is the case if $n\geq 3$.

  * (2)

If $f$ is an exponential polynomial solution of (1.4) of the form (1.2), then
$f\in\Gamma_{1}^{\prime}$.

Recently, Liu-Mao-Zheng [13] considered $\Delta_{c}f(z)$ instead of $f(z+c)$
in (1.3) and proved that

######  Theorem C.

[13] Under the same situation as in Theorem A, the finite order entire
solutions $f$ of the equation

$\displaystyle f(z)^{n}+q(z)e^{Q(z)}\Delta_{c}f(z)=P(z),$ (1.5)

satisfies the results (a), (b) and

  * (1)

$\lambda(f)=\rho(f)-1$ if and only if $P(z)\equiv 0$. In particular, this is
the case if $n\geq 3$.

  * (2)

If $n\geq 3$ or $P(z)\equiv 0$, $f$ is of the form $f(z)=A(z)e^{\omega
z^{s}}$, where $s=\deg{Q}$, $\omega$ is a nonzero constant and
$A(z)(\not\equiv 0)$ is an entire function satisfying
$\lambda(A)=\rho(A)=\deg{Q}-1$. In particular, if $\deg{Q}=1$, then $A(z)$
reduces to a polynomial.

  * (3)

If $f$ is an exponential polynomial solution of (1.5) of the form (1.2), then
f is of the form

$f(z)=H_{0}(z)+H_{1}(z)e^{\omega_{1}z},$

where $H_{1}(z)$, $H_{2}(z)$ are non-constant polynomials and $\omega_{1}$ is
a non-zero constant satisfying $e^{\omega_{1}c}=1$.

In 2017, Li-Yang [10] considered the following form of equation

$\displaystyle
f^{n}(z)+a_{n-1}f^{n-1}(z)+\cdots+a_{1}f(z)+q(z)e^{Q(z)}f(z+c)=P(z),$ (1.6)

where $a_{i}\in\mathbb{C}$ and proved the following results.

######  Theorem D.

[10] Under the same situation as in Theorem A, the finite order entire
solutions $f$ of equation (1.6) satisfies the results (a), (d) and

  * (1)

If zero is a Borel exceptional value of $f(z)$, then we have
$a_{n-1}=\cdots=a_{1}=P(z)\equiv 0$.

  * (2)

If $P(z)\equiv 0$, then we have
$z^{n-1}+a_{n-1}z^{n-2}+\cdots+a_{1}=(z+a_{n-1}/n)^{n-1}$. Furthermore, if
there exists $i_{0}\in\\{1,\ldots,n-1\\}$ such that $a_{i_{0}}=0$, then all of
the $a_{j}(j=1,\ldots,n-1)$ must be zero as well and we have
$\lambda(f)<\rho(f)$; otherwise we have $\lambda(f)=\rho(f)$.

  * (3)

A solution $f$ belongs to $\Gamma_{0}$ if and only if $P(z)\equiv 0$ and there
exists an $i_{0}\in\\{1,\ldots,n-1\\}$ such that $a_{i_{0}}=0$.

  * (4)

When $n\geq 3$, if there exists an $i_{0}\in\\{1,\ldots,n-1\\}$ such that
$a_{i_{0}}=0$ and $card\\{z:p(z)=p^{\prime}(z)=p^{\prime\prime}(z)=0\\}\geq 1$
or $card\\{z:p(z)=p^{\prime}(z)=0\\}\geq 2$, where
$p(z)=z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z$, then $f$ belongs to $\Gamma_{0}$
and $a_{n-1}=\cdots=a_{1}=0\equiv P(z)$.

In the same paper, Li-Yang [10] also proved the following result.

######  Theorem E.

[10] If $f$ is an exponential polynomial solution of the form (1.2) of the
equation (1.6) for $n=2$ and $a_{1}\neq 0$, then the following conclusions
hold.

  * (1)

when $m\geq 2$, there exists $i,j\in\\{1,2,\ldots,m\\}$ such that
$\omega_{i}=2\omega_{j}$.

  * (2)

when $m=1$, then $f\in\Gamma_{1}$. Moreover, if
$f\in\Gamma_{1}\backslash\Gamma_{0}$, then $\rho(f)=1$,
$f(z)=Ke^{\frac{1}{c}(2k\pi i-\log\frac{2d+a_{1}}{d}}$,
$Q(z)=\frac{1}{c}(2k\pi i-\log\frac{2d+a_{1}}{d})z$,
$q(z)=-\frac{2d+a_{1}}{d}$ and $d^{2}+a_{1}d=P(z)$, where
$K,d\in\mathbb{C}\backslash\\{0\\}$ and $k\in\mathbb{Z}$.

We now introduce the generalized linear delay-differential operator of $f(z)$,

$\displaystyle L(z,f)=\sum_{i=0}^{k}b_{i}f^{(r_{i})}(z+c_{i})\;(\not\equiv
0),$ (1.7)

where $b_{i},c_{i}\in\mathbb{C}$, $r_{i}$ are non-negative integers,
$c_{0}=0$, $r_{0}=0$. In view of the above theorems it is quiet natural to
characterize the nature of exponential polynomial as solution of certain non-
linear complex equation involving generalized linear delay-differential
operator. In this regard, we consider the following non-linear delay-
differential equation

$\displaystyle
f^{n}(z)+\sum_{i=1}^{n-1}a_{i}f^{i}(z)+q(z)e^{Q(z)}L(z,f)=P(z),$ (1.8)

where $a_{i}\in\mathbb{C}$, $n$ be non-negative integers; $q$, $Q$, $P$
respectively be non-zero, non-constant, any polynomials. We also introduce,
for any polynomials $p_{i}(z)$ and non-constant polynomials $\alpha_{i}(z)$, a
new class of solution as follows:

$\displaystyle\Gamma_{2}^{\prime}$ $\displaystyle=$
$\displaystyle\\{p_{1}(z)e^{\alpha_{1}(z)}+p_{2}(z)e^{\alpha_{2}(z)}+p_{3}(z)\\}$

Now we are at a state to present our main result which improves all the above
mentioned results as follows:

######  Theorem 1.1.

Under the same situation as in Theorem A, the finite order entire solutions
$f$ of equation (1.8) satisfies

  * (i)

Every solution $f$ satisfies $\rho(f)=\deg Q$ and is of mean type.

  * (ii)

If zero is a Borel exceptional value of $f(z)$, then we have
$a_{n-1}=\cdots=a_{1}=P(z)\equiv 0$. Conversely, if $P(z)\equiv 0$ and there
exists an $i_{0}\in\\{1,\ldots,n-1\\}$ such that $a_{i_{0}}=0$, then all of
$a_{j}$’s $(j=1,\ldots,n-1)$ must be zero and we have $\lambda(f)<\rho(f)$;
otherwise we have $\lambda(f)=\rho(f)$.

  * (iii)

If a solution $f$ belongs to $\Gamma_{0}^{\prime}$, then
$a_{n-1}=\cdots=a_{1}=P(z)\equiv 0$. Conversely, let $P(z)\equiv 0$ and there
exists an $i_{0}\in\\{1,\ldots,n-1\\}$ such that $a_{i_{0}}=0$, then either
$\lambda(f)=\rho(f)-1$ for $c_{i}=c_{j}$, $1\leq i,j\leq k$ or $f$ belongs to
$\Gamma_{0}^{\prime}$.

  * (iv)

Let $n\geq 3$. If at least one $a_{i_{0}}=0$ $(i_{0}=1,2,\ldots,n-1)$ and
$p(z)=z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z$ such that
$card\\{z:p(z)=p^{\prime}(z)=p^{\prime\prime}(z)=0\\}\geq 1$ or
$card\\{z:p(z)=p^{\prime}(z)=0\\}\geq 2$, then $P(z)\equiv
0=a_{n-1}=\cdots=a_{1}=0$ and $f\in\Gamma_{0}^{\prime}$. Moreover,
$card\\{z:p(z)=p^{\prime}(z)=0\\}\geq 2$ is not possible.

  * (v)

Let $f$ be given by (1.2), which is a solution of (1.8) for $n=2$ and
$a_{1}\neq 0$. Then the following conclusions hold:

    * (a)

when $m\geq 2$, there exists $i,j\in\\{1,2,\ldots,m\\}$ such that
$\omega_{i}=2\omega_{j}$. In this case, $f\in\Gamma_{2}^{\prime}$.

    * (b)

when $m=1$, then $f$ takes the form
$f(z)=H_{0}(z)+H_{1}(z)e^{\omega_{1}z^{t}}$, i.e., $f\in\Gamma_{1}^{\prime}$.
In this case,

      * (I)

either $t=1$, $\rho(f)=1$ and $H_{0}(z)$, $H_{1}(z)$ are polynomials and
$Q(z)$ is a polynomial of degree $1$

      * (II)

or $H_{0}(z)=-\frac{a_{1}}{2}$, $P(z)=-\frac{a_{1}^{2}}{4}$,
$H_{1}^{2}(z)=\frac{b_{0}a_{1}}{2}q(z)e^{Q_{t-1}(z)}$ and
$L(z,f)=b_{0}H_{0}(z)$

      * (III)

or $H_{0}(z)=-\frac{a_{1}}{2}$, $P(z)=-\frac{a_{1}^{2}}{4}$,
$H_{1}^{2}(z)=-q(z)e^{Q_{t-1}(z)}\mathcal{A}_{1}(z)$ and
$L(z,f)=\mathcal{A}_{1}(z)e^{\omega_{1}z^{t}}$, where
$\mathcal{A}_{1}(z)=\sum_{i=0}^{k}b_{i}\tilde{H}_{1}(z+c_{i})e^{\omega_{1}(z+c_{i})^{t}-\omega_{1}z^{t}}$
such that $\tilde{H}_{1}(z+c_{i})$ are the delay-differential polynomial of
$H_{1}(z)$.

######  Remark 1.1.

Note that Cases (i)-(iv) and (v) of Theorem 1.1 improve Theorems D and E,
respectively. Also, since $L(z,f)$ includes $f^{(k)}(z+c)$ and
$\Delta_{c}f(z)$, Theorem 1.1 improves Theorems B-C as follows:

(I) Cases (i), (ii) and (iii)-(iv) of Theorem 1.1 improve Case (a), (b) and
(1) of each of Theorems B-C, respectively.

(II) Case (v)-(b) of Theorem 1.1 improves, respectively, Case (2) and Case (3)
of Theorem B and Theorem C.

This following three examples clarify Cases (ii)-(iii).

######  Example 1.1.

Take $L(z,f)=f^{\prime\prime}(z+c)$. Then the function $f=e^{2z}$ satisfies
the equation $f^{2}-\frac{1}{4}e^{2z}L(z,f)=0$ such that $e^{2c}=1$. Clearly,
$0=\lambda(f)=\rho(f)-1$. This example clarifies Theorem B as well.

######  Example 1.2.

Let $L(z,f)=\Delta_{c}f(z)$. Then the function $f=e^{\alpha z}$ satisfies the
equation $f^{2}-\frac{1}{2}e^{\alpha z}L(z,f)=0$ such that $e^{\alpha c}=3$.
Clearly, $0=\lambda(f)=\rho(f)-1$. This example also satisfies Theorem C.

######  Example 1.3.

Let $L(z,f)=f(z+1)+f^{\prime}(z+1)-f^{\prime\prime}(z+1)$. Then the function
$f=(z+1)e^{z}$ satisfies the equation $f^{2}-(z+1)e^{z-1}L(z,f)=0$. Note that,
here $c_{1}=c_{2}=c_{3}=1$ and $f\in\Gamma_{0}^{\prime}$.

The next example satisfies Case (iv).

######  Example 1.4.

Let $L(z,f)=f(z+\log 2)+f^{\prime\prime}(z+\pi i)$. Then the function
$f=e^{iz}$ satisfies the equation $f^{3}+qe^{2iz}L(z,f)=0$, where
$q=\frac{1}{e^{-\pi}-2^{i}}$. Note that, here $p(z)=z^{3}$ and
$card\\{z:p(z)=p^{\prime}(z)=p^{\prime\prime}(z)=0\\}=1$. Also,
$a_{2}=a_{1}=0\equiv P(z)$ and $f\in\Gamma_{0}^{\prime}$.

By the following example, it is clear that the Case (v)-(a) occurs
significantly.

######  Example 1.5.

Take $L(z,f)=f^{\prime}(z+\log 4)-4f(z+\log 3)$ and $m=2$. Then the function
$f=e^{2z}-e^{z}+1$ satisfies the equation
$f^{2}-2f+\frac{1}{4}e^{2z}L(z,f)=-1$. Note that here
$f\in\Gamma_{2}^{\prime}$.

The following two examples show that the Case (v)-(b)-(I) actually holds.

######  Example 1.6.

We take $L(z,f)=f(z+c)$. Then the function $f=d+e^{\alpha z}$ satisfies the
equation $f^{2}-df-e^{\alpha z}L(z,f)=0$ such that $e^{\alpha c}=1$. Here,
$P(z)\equiv 0$.
Also, the same function satisfies $f^{2}-3df+e^{\alpha z}L(z,f)=-2d^{2}$ such
that $e^{\alpha c}=-1$. Here, $f\in\Gamma_{1}^{\prime}$. Here, $P(z)\not\equiv
0$. This example is true for Theorem E as well.

######  Example 1.7.

Put $L(z,f)=f(z+\log 2)+f^{\prime}(z+\pi i)+f^{\prime\prime}(z+2\pi i)$ and
$m=1$. Then the function $f=2+3e^{z}$ satisfies the equation
$f^{2}-3f-\frac{3}{2}e^{z}L(z,f)=-2$. Here, $P(z)\not\equiv 0$.
Also, let $L(z,f)=f(z+\log 3)-f^{\prime}(z+\log 4)+f^{\prime\prime}(z+\log 2)$
and $m=1$. Then the function $f=3+e^{z}$ satisfies the equation
$f^{2}-3f-e^{z}L(z,f)=0$. Here, $P(z)\equiv 0$.

Next example shows that the Case (v)-(b)-(II) actually occurs.

######  Example 1.8.

Let $L(z,f)=3f(z)+f^{\prime}(z+\log 2)-3f^{\prime\prime}(z+2\pi i)$ and $m=1$.
Then the function $f=-\frac{a_{1}}{2}+2e^{3z}$ satisfies the equation
$f^{2}+a_{1}f+\frac{8}{3a_{1}}e^{6z}L(z,f)=-\frac{a_{1}^{2}}{4}$. Note that
here $b_{0}=3$, $H_{0}=-\frac{a_{1}}{2}$ and so,
$L(z,f)=-\frac{3a_{1}}{2}=b_{0}H_{0}$.

Next example shows that the Case (v)-(b)-(III) actually occurs.

######  Example 1.9.

Let $L(z,f)=f(z)-f(z+\log 2)+\frac{1}{2}f^{\prime}(z+\log
2)+\frac{2}{9}f^{\prime}(z+\log 3)-\frac{1}{9}f^{\prime\prime}(z+\log 3)$ and
$m=1$. Then the function $f=-\frac{a_{1}}{2}+ze^{2z}$ satisfies the equation
$f^{2}+a_{1}f-ze^{2z}L(z,f)=-\frac{a_{1}^{2}}{4}$. Note that here, $q(z)=-z$,
$Q(z)=2z$, $Q_{t-1}(z)=0$. Also,
$\mathcal{A}_{1}(z)=\sum_{i=0}^{5}\left(b_{i}\tilde{H}_{1}(z+c_{i})e^{2c_{i}}\right)=z$.
So, $H_{1}^{2}(z)=z^{2}=-q(z)e^{Q_{t-1}(z)}\mathcal{A}_{1}(z)$ and
$L(z,f)=\mathcal{A}_{1}(z)e^{2z}=\sum_{i=0}^{5}\left(b_{i}\tilde{H}_{1}(z+c_{i})e^{2c_{i}}\right)e^{2z}=ze^{2z}$.

##  2\. Lemmas

We give the following well-known results which are important to prove our
theorems.

######  Lemma 2.1.

[5] Let $f$ be a non-constant meromorphic function and $c_{1}$, $c_{2}$ be two
complex numbers such that $c_{1}\neq c_{2}$. Let $f(z)$ be a meromorphic
function with finite order $\rho$, then for each $\varepsilon>0$,

$\displaystyle m\left(r,\frac{f(z+c_{1})}{f(z+c_{2})}\right)=S(r,f).$

######  Lemma 2.2.

[8, Corollary 2.3.4] Let $f$ be a transcendental meromorphic function and
$k\geq 1$ be an integer. Then $m\left(r,\frac{f^{(k)}}{f}\right)=S(r,f)$.

Combining Lemmas 2.1-2.2 we have the following lemma:

######  Lemma 2.3.

Let $f(z)$ be a meromorphic function of finite order and let $c\in\mathbb{C}$,
$k\geq 1$ be an integer. Then $m\left(r,\frac{f^{(k)}(z+c)}{f}\right)=S(r,f)$.

###### Proof.

$m\left(r,\frac{f^{(k)}(z+c)}{f}\right)=m\left(r,\frac{f^{(k)}(z+c)}{f(z+c)}.\frac{f(z+c)}{f(z)}\right)=S(r,f).$

∎

######  Lemma 2.4.

[10] Let $f$ be a non-constant meromorphic function of hyper order less than
$1$ and $c\in\mathbb{C}$. Then

$\displaystyle N(r,1/f(z+c))=N(r,0;f(z))+S(r,f).$

######  Lemma 2.5.

[19] Suppose $f_{j}(z)$ $(j=1,2,...,n+1)$ and $g_{k}(z)$ $(k=1,2,...,n)$
$(n\geq 1)$ are entire functions satisfying the following conditions:

  * (i)

$\sum_{j=1}^{n}f_{j}(z)e^{g_{j}(z)}\equiv f_{n+1}(z)$,

  * (ii)

The order of $f_{j}(z)$ is less than the order of $e^{g_{k}(z)}$ for $1\leq
j\leq n+1$ , $1\leq k\leq n$ and furthermore, the order of $f_{j}(z)$ is less
than the order of $e^{g_{h}(z)-g_{k}(z)}$ for $n\geq 2$ and $1\leq j\leq n+1$,
$1\leq h<k\leq n$. Then $f_{j}(z)\equiv 0$, $(j=1,2,...,n+1)$.

######  Lemma 2.6.

[2, 6] Let $f$ be a meromorphic function and suppose that

$R(z)=a_{n}f(z)^{n}+\cdots+a_{0}(z)$

has small meromorphic coefficients $a_{j}(z)$, $a_{n}\neq 0$ in the sense of
$T(r,a_{j})=S(r,f)$. Moreover, assume that

$\overline{N}\left(r,\frac{1}{R}\right)+\overline{N}(r,f)=S(r,f).$

Then

$R(z)=a_{n}\left(f+\frac{a_{n-1}}{na_{n}}\right).$

The following lemma gives the Nevanlinna characteristic and counting functions
of an exponential polynomial.

######  Lemma 2.7.

[14] Let $f(z)$ be given by (1.2). Then

$T(r,f)=C(co(W_{0}))\frac{r^{t}}{2\pi}+o(r^{t}).$

If $H_{0}(z)\not\equiv 0$, then

$m(r,1/f)=o(r^{t}),$

while if $H_{0}(z)\equiv 0$, then

$N(r,1/f)=C(co(W))\frac{r^{t}}{2\pi}+o(r^{t}).$

Next, we proof the following lemmas which are the core parts of our paper.

######  Lemma 2.8.

Let $f$ be given by (1.2) which is a solution of (1.8) for $n=2$ and
$\omega_{i}\neq 2\omega_{j}$. If the points
$0,\omega_{1},\omega_{2},\ldots,\omega_{m}$ are collinear, then $m=1$.

######  Proof.

Assume on the contrary to the assertion that $m\geq 2$. For each
$i\in\\{1,2,\ldots,m\\}$, we may write $\omega_{i}=\xi_{i}\omega$, where the
constants $\xi_{i}\in\mathbb{C}\backslash\\{0\\}$ are distinct, $\xi_{0}=0$
and $\omega\in\mathbb{C}\backslash\\{0\\}$. Moreover, we may suppose that
$\xi_{i}>\xi_{j}$ for $i>j$. Equation (1.8) can be written as

$\displaystyle\sum_{i,j=0}^{m}H_{i}(z)H_{j}(z)e^{(\xi_{i}+\xi_{j})\omega
z^{t}}+a_{1}\sum_{l=0}^{m}H_{l}(z)e^{\xi_{l}\omega z^{t}}$
$\displaystyle\qquad+q(z)e^{Q_{t-1}(z)}\left[{\mathcal{A}_{0}(z)e^{v_{t}z^{t}}}+\sum_{h=1}^{m}\mathcal{A}_{h}(z)e^{(v_{t}+\xi_{h}\omega)z^{t}}\right]=P(z),$
(2.1)

where $Q_{t-1}(z)=Q(z)-v_{t}z^{t}$ with $\deg Q_{t-1}(z)\leq t-1$ and
$\mathcal{A}_{0}(z)=\sum_{i=0}^{k}b_{i}H^{(r_{i})}_{0}(z+c_{i})$,
$\mathcal{A}_{h}(z)=\sum_{i=0}^{k}b_{i}\tilde{H}_{h}(z+c_{i})e^{\omega_{h}(z+c_{i})^{t}-\omega_{h}z^{t}}$,
$h=1,2,\ldots,m$ such that $\tilde{H}_{h}(z+c_{i})$ are the delay-differential
polynomial of $H_{h}(z)$.
Now we consider following two cases to derive contradiction.

Case 1. Let $\xi_{m}>0$. Note that
$\max\\{\xi_{i}+\xi_{j}:i,j=0,1,\ldots,m\\}=2\xi_{m}$. Since $L(z,f)\not\equiv
0$, then at least one of $\mathcal{A}_{h}(z)$, $h=0,1,\ldots,m$ is not
vanishing.
Case 1.1. Let all $\mathcal{A}_{h}(z)=0$, $h=1,2,\ldots,m$. Then
$\mathcal{A}_{0}(z)\not\equiv 0$., i.e., $H_{0}(z)\not\equiv 0$. If
$2\xi_{m}\omega\neq v_{t}$, applying Lemma 2.5 on (2), we obtain
$H_{m}^{2}(z)\equiv 0$, a contradiction. Next, let $2\xi_{m}\omega=v_{t}$.
Since, $\omega_{i}\neq 2\omega_{j}$, applying Lemma 2.5 on (2), we obtain
$H_{1}^{2}(z)\equiv 0$, a contradiction.
Case 1.2. Let at least one of $\mathcal{A}_{h}(z)\not=0$, for
$h=1,2,\ldots,m$.
Case 1.2.1. Let $\mathcal{A}_{0}(z)\neq 0$. Then by Lemma 2.5, from (2), there
exists one $h_{0}\in\\{0,1,\ldots,m\\}$ such that
$2\xi_{m}\omega=v_{t}+\xi_{h_{0}}\omega$. Otherwise, we have
$H_{m}^{2}(z)\equiv 0$, a contradiction.
Case 1.2.1.1. If $h_{0}=m$, then we have $v_{t}=\xi_{m}\omega$. Since
$2\xi_{i}\neq\xi_{j}$, $j=0,1,\ldots,m$ and
$2\xi_{1}\not\in\\{\xi_{i}+\xi_{j}:0\leq i,j\leq m,(i,j)\neq(1,1)\\}$ and
$2\xi_{1}\not\in\\{\xi_{m}+\xi_{i}:i=0,1,\ldots,m\\}$. By Lemma 2.5, we obtain
$H_{1}^{2}(z)\equiv 0$, a contradiction.
Case 1.2.1.2. If $h_{0}\in\\{0,1,\ldots,m-1\\}$, since
$0=\xi_{0}<\xi_{1}<\xi_{2}<\cdots<\xi_{m-1}<\xi_{m}$ and
$2\xi_{i}\neq\xi_{j}$, $i,j=0,1,\ldots,m$, then for $m>h_{0}$,

$\displaystyle
2\xi_{m}-\xi_{h_{0}}+\xi_{m}>\max\\{2\xi_{m}-\xi_{h_{0}}+\xi_{i}:i=0,1,\ldots,m-1\\}.$

Also, $2\xi_{m}=\max\\{\xi_{i}+\xi_{j}:i,j=0,1,\ldots,m\\}$. In view of Lemma
2.5, we obtain $q(z)e^{Q_{t-1}(z)}\mathcal{A}_{m}\equiv 0$, i.e., $q(z)\equiv
0$, a contradiction.
Case 1.2.2. Let $\mathcal{A}_{0}(z)=0$ and $H_{0}(z)\neq 0$. Then (2) becomes

$\displaystyle\sum_{i,j=0}^{m}H_{i}(z)H_{j}(z)e^{(\xi_{i}+\xi_{j})\omega
z^{t}}+a_{1}\sum_{l=0}^{m}H_{l}(z)e^{\xi_{l}\omega
z^{t}}+q(z)e^{Q_{t-1}(z)}\sum_{h=1}^{m}\mathcal{A}_{h}(z)e^{(v_{t}+\xi_{h}\omega)z^{t}}=P(z).$
(2.2)

Then similar as Case 1.2.1, by Lemma 2.5, from (2.2), we have there exists one
$h_{0}\in\\{1,2,\ldots,m\\}$ such that
$2\xi_{m}\omega=v_{t}+\xi_{h_{0}}\omega$ and proceeding similarly as done in
Case 1.2.1, we can get a contradiction.
Case 1.2.3. Let $\mathcal{A}_{0}(z)=0$ and $H_{0}(z)=0$. Then (2) becomes

$\displaystyle\sum_{i,j=1}^{m}H_{i}(z)H_{j}(z)e^{(\xi_{i}+\xi_{j})\omega
z^{t}}+a_{1}\sum_{l=1}^{m}H_{l}(z)e^{\xi_{l}\omega
z^{t}}+q(z)e^{Q_{t-1}(z)}\sum_{h=1}^{m}\mathcal{A}_{h}(z)e^{(v_{t}+\xi_{h}\omega)z^{t}}=P(z).$
(2.3)

Next, similar as Case 1.2.1, by Lemma 2.5, from (2.3), we have there exists
one $h_{0}\in\\{1,2,\ldots,m\\}$ such that
$2\xi_{m}\omega=v_{t}+\xi_{h_{0}}\omega$ and adopting the same method as done
in Case 1.2.1, we get a contradiction.

Case 2. $\xi_{m}<0$. Note that
$\min\\{\xi_{i}+\xi_{j}:i,j=0,1,\ldots,m\\}=2\xi_{1}.$ Similar as Case 1, we
divide the following cases.
Case 2.1. Let all $\mathcal{A}_{h}(z)=0$, $h=1,2,\ldots,m$. Then
$\mathcal{A}_{0}(z)\not\equiv 0$., i.e., $H_{0}(z)\not\equiv 0$. If
$2\xi_{1}\omega\neq v_{t}$, applying Lemma 2.5 on (2), we obtain
$H_{1}^{2}(z)\equiv 0$, a contradiction. Next, let $2\xi_{1}\omega=v_{t}$.
Since, $\omega_{i}\neq 2\omega_{j}$, applying Lemma 2.5 on (2), we obtain
$H_{m}^{2}(z)\equiv 0$, a contradiction.
Case 2.2. Let at least one of $\mathcal{A}_{h}(z)\neq 0$ for $h=1,2,\ldots,m$.
Case 2.2.1. Let $\mathcal{A}_{0}(z)\neq 0$. Then by Lemma 2.5, there exists
one $h_{0}\in\\{0,1,\ldots,m\\}$ such that
$2\xi_{1}\omega=v_{t}+\xi_{h_{0}}\omega$. Otherwise, we have
$H_{1}^{2}(z)\equiv 0$, a contradiction.
Case 2.2.1.1. If $h_{0}=1$, then we have $v_{t}=\xi_{1}\omega$.
Since, $2\xi_{i}\neq\xi_{j}$ and $2\xi_{m}\not\in\\{\xi_{i}+\xi_{j}:0\leq
i,j\leq m,(i,j)\neq(m,m)\\}$ and
$2\xi_{m}\not\in\\{\xi_{1}+\xi_{i}:i=0,2,\ldots,m\\}$. By Lemma 2.5, we obtain
$H_{m}^{2}(z)\equiv 0$, a contradiction.
Case 2.2.1.2. If $h_{0}\in\\{0,2,3,\ldots,m\\}$, since
$0=\xi_{0}<\xi_{1}<\xi_{2}<\cdots<\xi_{m-1}<\xi_{m}$ and
$2\xi_{i}\neq\xi_{j}$, $i,j=0,1,\ldots,m$, then

$\displaystyle
2\xi_{1}-\xi_{h_{0}}+\xi_{1}<\min\\{2\xi_{1}-\xi_{h_{0}}+\xi_{i}:i=0,2,3,\ldots,m\\}.$

Also, $\min\\{\xi_{i}+\xi_{j}:i,j=0,1,\ldots,m\\}=2\xi_{1}$. In view of Lemma
2.5, we obtain $q(z)e^{Q_{t-1}}\mathcal{A}_{1}(z)\equiv 0$, i.e., $q(z)\equiv
0$, a contradiction.
Case 2.2.2. Let $\mathcal{A}_{0}(z)=0$ and $H_{0}(z)\neq 0$. Then in this
case, we get equation (2.2). Similar as Case 2.2.1, by Lemma 2.5, there exists
one $h_{0}\in\\{2,3,\ldots,m\\}$ such that
$2\xi_{1}\omega=v_{t}+\xi_{h_{0}}\omega$ and proceeding similarly as adopted
in Case 2.2.1, we can get a contradiction.
Case 2.2.3. Let $\mathcal{A}_{0}(z)=0$ and $H_{0}(z)=0$. Then, we have
equation (2.3). Similar as Case 2.2.1, by Lemma 2.5, there exists one
$h_{0}\in\\{2,3,\ldots,m\\}$ such that
$2\xi_{1}\omega=v_{t}+\xi_{h_{0}}\omega$. Next, adopting the same method as
executing in Case 2.2.1, we get a contradiction. ∎

######  Lemma 2.9.

If $m\geq 2$ and $\omega_{i}\neq 2\omega_{j}$ for any $i\neq j$, then $f$ of
the form (1.2) is not a solution of (1.8) for $n=2$.

######  Proof.

Suppose on the contrary to the assertion that, $m\geq 2$. Substituting $f$ of
the form (1.2) into (1.8), we get

$\displaystyle F(z)$ $\displaystyle=$ $\displaystyle f^{2}(z)+a_{1}f(z)-P(z)$
(2.6) $\displaystyle=$ $\displaystyle
G(z)+\sum_{\scriptsize{{\begin{array}[]{clcr}i,j=0\\\
\omega_{i}+\omega_{j}\neq
0\end{array}}}}^{m}H_{i}(z)H_{j}(z)e^{(\omega_{i}+\omega_{j})z^{t}}+a_{1}\sum_{l=1}^{m}H_{l}(z)e^{\omega_{l}z^{t}},$

where $G(z)=H_{0}(z)(H_{0}(z)+a_{1})-P(z)$ is either an exponential polynomial
of degree $<t$ or a polynomial in $z$.
Therefore, also

$\displaystyle
F(z)=-q(z)e^{Q(z)}L(z,f)=-q(z)e^{Q(z)}\sum_{h=0}^{m}\mathcal{A}_{h}(z)e^{\omega_{h}z^{t}},\hskip
52.6445pt$ (2.7)

such that $\mathcal{A}_{h}(z)$ is defined as in(2).
Now we set

$\displaystyle X_{1}$ $\displaystyle=$
$\displaystyle\\{\overline{\omega}_{1},\ldots,\overline{\omega}_{m},\overline{\omega}_{i}+\overline{\omega}_{j}:\overline{\omega}_{i}+\overline{\omega}_{j}\neq
0,i,j=1,\ldots,m\\},$ $\displaystyle X_{2}$ $\displaystyle=$
$\displaystyle\\{\overline{\omega}_{1},\ldots,\overline{\omega}_{m},2\overline{\omega}_{1},\ldots,2\overline{\omega}_{m}\\},$
$\displaystyle X_{3}$ $\displaystyle=$
$\displaystyle\\{2\overline{\omega}_{1},\ldots,2\overline{\omega}_{m}\\}.$

Clearly, by the theory of convexity, we have
$\overline{\omega}_{i}+\overline{\omega}_{j}=\frac{1}{2}\cdot
2\overline{\omega}_{i}+(1-\frac{1}{2})\cdot 2\overline{\omega}_{j}$, i.e.,
$co(X_{1})=co(X_{2})$. Since, $X_{3}\subset X_{2}$, we have $co(X_{3})\leq
co(X_{2})$, respectively.
Next, we consider the following cases to show a contradiction.
Case 1. If all $\mathcal{A}_{h}(z)=0$ for $h=1,\ldots,m$, then we have
$\mathcal{A}_{0}(z)\neq 0$, which implies $H_{0}(z)\neq 0$. Then (2.7) becomes
$F(z)=-q(z)e^{Q(z)}\mathcal{A}_{0}(z)$. Then applying Lemma 2.7, we get

$\displaystyle
N\left(r,\frac{1}{F(z)}\right)=N\left(r,\frac{1}{\mathcal{A}_{0}(z)}\right)=o(r^{t}).$
(2.8)

Sub-case 1.1. Let $G(z)\equiv 0$. Applying Lemma 2.7 on (2.6), we have

$\displaystyle
N\left(r,\frac{1}{F(z)}\right)=C(co(X_{1}))\frac{r^{t}}{2\pi}+o(r^{t}).$ (2.9)

Therefore, (2.8) and (2.9) yields a contradiction.
Sub-case 1.2. Let $G(z)\not\equiv 0$. Applying Lemma 2.7 on (2.6), we have
$m\left(r,\frac{1}{F(z)}\right)=o(r^{t})$ and then

$\displaystyle m\left(r,\frac{1}{F(z)}\right)+N\left(r,\frac{1}{F(z)}\right)$
$\displaystyle=$ $\displaystyle T(r,F(z))+O(1)=2T(r,f(z))+S(r,f)$
$\displaystyle=$ $\displaystyle
2\left(C(co(W_{0}))\frac{r^{t}}{2\pi}+o(r^{t})\right)+S(r,f)$
$\displaystyle\implies N\left(r,\frac{1}{F(z)}\right)$ $\displaystyle=$
$\displaystyle 2C(co(W_{0}))\frac{r^{t}}{2\pi}+o(r^{t}).$ (2.10)

Therefore, using (2.8) and (2), we get a contradiction.
Case 2. Let there exists some $h_{0}\in\\{1,2,\ldots,m\\}$ such that
$\mathcal{A}_{h_{0}}(z)\neq 0$. Now, we denote the following set as

$\displaystyle V=\\{\overline{\omega}_{h_{0}}:h_{0}\in\\{1,2,\ldots,m\\}\text{
for which }\mathcal{A}_{h_{0}}(z)\neq 0\\}\text{ and }V_{0}=V\cup\\{0\\}.$

Since, $V\subseteq W$ and $V_{0}\subseteq W_{0}$, then $C(co(V))\leq C(co(W))$
and $C(co(V_{0}))\leq C(co(W_{0}))$, respectively.
Case 2.1. Let $\mathcal{A}_{0}(z)\equiv 0$ and $H_{0}(z)\equiv 0$. Then using
Lemma 2.7 on (2.7), we have

$\displaystyle
N\left(r,\frac{1}{F(z)}\right)=N\left(r,\frac{1}{L(z,f)}\right)+O(\log
r)=C(co(V))\frac{r^{t}}{2\pi}+o(r^{t}).$ (2.11)

Case 2.1.1. If $G(z)\not\equiv 0$. Then similar as Sub-case 1.2, we have
equation (2). From (2) and (2.11), we obtain a contradiction by
$C(co(W_{0}))\geq C(co(V_{0}))\geq C(co(V))=2C(co(W_{0}))$.
Case 2.1.2. If $G(z)\equiv 0$, using Lemma 2.7 on (2.6), we have equation
(2.9). Using (2.9) and (2.11), from $C(co(W))\geq
C(co(V))=C(co(X_{1}))=C(co(X_{2}))\geq C(co(X_{3}))=2C(co(W))$, we get a
contradiction.
Case 2.2. Let $\mathcal{A}_{0}(z)\equiv 0$ and $H_{0}(z)\not\equiv 0$. Then
proceeding similarly as done in Case 2.1, we get a contradiction.
Case 2.3. Let $\mathcal{A}_{0}(z)\neq 0$, which implies $H_{0}(z)\not\equiv
0$. Then using Lemma 2.7 on (2.7), we have
$m\left(r,\frac{1}{L(z,f)}\right)=o(r^{t})$ and then

$\displaystyle N\left(r,\frac{1}{F(z)}\right)$ $\displaystyle=$ $\displaystyle
N\left(r,\frac{1}{L(z,f)}\right)+O(\log r)$ (2.12) $\displaystyle=$
$\displaystyle T(r,L(z,f))+o(r^{t})=C(co(V_{0}))\frac{r^{t}}{2\pi}+o(r^{t}).$

Case 2.3.1. If $G(z)\not\equiv 0$. Then, from (2) and (2.12), we get
$C(co(W_{0}))\geq C(co(V_{0}))=2C(co(W_{0}))$, a contradiction.
Case 2.3.2. If $G(z)\equiv 0$. Using (2.9) and (2.12), we get

$\displaystyle C(co(V_{0}))=C(co(X_{1})).$ (2.13)

Now, since $m\geq 2$, by Lemma 2.8, $co(W_{0})$ can not be a line-segment.
Therefore, $co(W_{0})$ must be a polygon with non-empty interior. If $0$ is
not a boundary point of $co(W_{0})$, then we have $co(W_{0})=co(W)$. Then we
have $C(co(W))=C(co(W_{0}))\geq C(co(V_{0}))=C(co(X_{1}))=C(co(X_{2}))\geq
C(co(X_{3}))=2C(co(W))$, a contradiction. So, $0$ is a boundary point of
$co(W_{0})$. We choose the other non-zero corner points of $co(W_{0})$ among
the points $\overline{\omega}_{1},\ldots,\overline{\omega}_{m}$ are
$u_{1},\ldots,u_{t}$, $t\leq m$ such that
$0\leq\arg(u_{i})\leq\arg(u_{i+1})\leq 2\pi$ for $1\leq i\leq t-1$. Hence,

$\displaystyle
C(co(W_{0}))=|u_{1}|+|u_{2}-u_{1}|+\cdots+|u_{t}-u_{t-1}|+|u_{t}|.$ (2.14)

Let $X_{4}=\\{u_{1},2u_{1},2u_{2},\ldots,2u_{t},u_{t}\\}$. Therefore, the
points $2u_{1},2u_{2},\ldots,2u_{t}$ are the corner points of $co(X_{4})$.
However, since $t\leq m$, $co(X_{4})$ may have more corner points. Then, using
(2.14), we have

$\displaystyle C(co(X_{3}))>C(co(X_{4}))$ $\displaystyle>$
$\displaystyle|2u_{1}-u_{1}|+|2u_{2}-2u_{1}|+\cdots+|2u_{t}-2u_{t-1}|+|u_{t}-2u_{t}|$
$\displaystyle=$
$\displaystyle|u_{1}|+2|u_{2}-u_{1}|+\cdots+2|u_{t}-u_{t-1}|+|u_{t}|$
$\displaystyle>$
$\displaystyle|u_{1}|+|u_{2}-u_{1}|+\cdots+|u_{t}-u_{t-1}|+|u_{t}|$
$\displaystyle=$ $\displaystyle C(co(W_{0}))$

Therefore,

$C(co(X_{1}))=C(co(X_{2}))\geq C(co(X_{3}))>C(co(W_{0}))\geq C(co(V_{0})),$

which contradicts (2.13).
Hence, the proof is completed. ∎

######  Lemma 2.10.

Let $f$ be given by (1.2), which is a solution of (1.8) for $n=2$, then $f$
takes the form,

$f(z)=H_{0}(z)+H_{1}(z)e^{\omega_{1}z^{t}},$

i.e., $f\in\Gamma_{1}^{\prime}$. In this case,

  * (I)

either $t=1$, $\rho(f)=1$ and $H_{0}(z)$, $H_{1}(z)$ are polynomials and
$Q(z)$ is a polynomial of degree $1$

  * (II)

or $H_{0}(z)=-\frac{a_{1}}{2}$, $P(z)=-\frac{a_{1}^{2}}{4}$,
$H_{1}^{2}(z)=\frac{b_{0}a_{1}}{2}q(z)e^{Q_{t-1}(z)}$ and
$L(z,f)=b_{0}H_{0}(z)$

  * (III)

or $H_{0}(z)=-\frac{a_{1}}{2}$, $P(z)=-\frac{a_{1}^{2}}{4}$,
$H_{1}^{2}(z)=-q(z)e^{Q_{t-1}(z)}\mathcal{A}_{1}(z)$ and
$L(z,f)=\mathcal{A}_{1}(z)e^{\omega_{1}z^{t}}$, where
$\mathcal{A}_{1}(z)=\sum_{i=0}^{k}b_{i}\tilde{H}_{1}(z+c_{i})e^{\omega_{1}(z+c_{i})^{t}-\omega_{1}z^{t}}$
such that $\tilde{H}_{1}(z+c_{i})$ are the delay-differential polynomial of
$H_{1}(z)$.

######  Proof.

For $n=2$, (1.8) becomes

$\displaystyle f^{2}(z)+a_{1}f(z)+q(z)e^{Q(z)}L(z,f)=P(z).$ (2.15)

By Lemma 2.9, we have $m=1$, i.e., (1.2) becomes

$\displaystyle f(z)=H_{0}(z)+H_{1}(z)e^{\omega_{1}z^{t}},$ (2.16)

where $H_{0}(z)$, $H_{1}(z)(\not\equiv 0)$ are either exponential polynomials
of order $<t$ or ordinary polynomials in $z$. Substituting (2.16) in (2.15),
we have

$\displaystyle
H_{1}(z)(2H_{0}(z)+a_{1})e^{\omega_{1}z^{t}}+H_{1}^{2}(z)e^{2\omega_{1}z^{t}}+q(z)e^{Q_{t-1}(z)}{\mathcal{A}_{0}(z)e^{v_{t}z^{t}}}$
$\displaystyle\quad+q(z)e^{Q_{t-1}(z)}\mathcal{A}_{1}(z)e^{(v_{t}+\omega_{1})z^{t}}=P(z)-H_{0}(z)(H_{0}(z)+a_{1}),$
(2.17)

where $Q_{t-1}(z)=Q(z)-v_{t}z^{t}$ with $\deg Q_{t-1}(z)\leq q-1$ and
$\mathcal{A}_{0}(z)=\sum_{i=0}^{k}b_{i}H^{(r_{i})}_{0}(z+c_{i})$,
$\mathcal{A}_{1}(z)=\sum_{i=0}^{k}b_{i}\tilde{H}_{1}(z+c_{i})e^{\omega_{1}(z+c_{i})^{t}-\omega_{1}z^{t}}$
such that $\tilde{H}_{h}(z+c_{i})$ are the delay-differential polynomial of
$H_{h}(z)$ for $h=1,2$. Since, $L(z,f)\not\equiv 0$, then at least one of
$\mathcal{A}_{0}(z)$ and $\mathcal{A}_{1}(z)$ is non-vanishing. Next, we
divide the following cases to prove our result.
Case 1. Let $\mathcal{A}_{1}(z)\equiv 0$. Then $\mathcal{A}_{0}(z)\not\equiv
0$, which implies $H_{0}(z)\not\equiv 0$.
If $v_{t}\neq\omega_{1},2\omega_{1}$ or $v_{t}=\omega_{1}$, applying Lemma 2.5
on (2), we have $H_{1}^{2}(z)\equiv 0$, a contradiction.
If $v_{t}=2\omega_{1}$, then, applying Lemma 2.5 on (2), we have

$H_{1}(2H_{0}(z)+a_{1})=0,$
$H_{1}^{2}(z)+q(z)e^{Q_{t-1}(z)}{\mathcal{A}_{0}(z)}=0,$
$P(z)-H_{0}(z)(H_{0}(z)+a_{1})=0.$

Since, $H_{1}(z)\not\equiv 0$. Therefore, solving these three equations, we
have $H_{0}(z)=-\frac{a_{1}}{2}$, $P(z)=-\frac{a_{1}^{2}}{4}$ and
$H_{1}^{2}(z)=\frac{b_{0}a_{1}}{2}q(z)e^{Q_{t-1}}$. Therefore, in this case
$L(z,f)=b_{0}H_{0}(z)$.
Case 2. Let $\mathcal{A}_{0}(z)\equiv 0$. Then $\mathcal{A}_{1}(z)\not\equiv
0$.
Sub-case 2.1. Let $H_{0}(z)\equiv 0$. If $v_{t}=\pm\omega_{1}$, using Lemma
2.5 on (2), we have $H_{1}\equiv 0$, a contradiction.
Sub-case 2.2. Let $H_{0}(z)\not\equiv 0$. If $v_{t}=-\omega_{1}$, in view of
Lemma 2.5, from on (2), we have $H_{1}\equiv 0$, a contradiction. If
$v_{t}=\omega_{1}$, similar as Case 1, we have $H_{0}(z)=-\frac{a_{1}}{2}$,
$P(z)=-\frac{a_{1}^{2}}{4}$ and
$H_{1}^{2}(z)=-q(z)e^{Q_{t-1}}\mathcal{A}_{1}$. Also, in this case
$L(z,f)=\mathcal{A}_{1}e^{\omega_{1}z^{t}}$.
Case 3. Let $\mathcal{A}_{0}(z)\not\equiv 0$ and $\mathcal{A}_{1}(z)\not\equiv
0$, which implies $H_{0}(z)\not\equiv 0$.
If $v_{t}=-\omega_{1}$ or $v_{t}\neq\pm\omega_{1}$, using Lemma 2.5 on (2), we
have $H_{1}\equiv 0$, a contradiction. If $v_{t}=2\omega_{1}$, by Lemma 2.5,
from (2), we have $\mathcal{A}_{1}\equiv 0$, a contradiction.
If $v_{t}=\omega_{1}$, applying Lemma 2.5 on (2), we have

$\displaystyle
H_{1}(z)(2H_{0}(z)+a_{1})+q(z)e^{Q_{t-1}(z)}\sum_{i=0}^{k}b_{i}H^{(r_{i})}_{0}(z+c_{i})=0,$
(2.18) $\displaystyle
H_{1}^{2}(z)+q(z)e^{Q_{t-1}(z)}\sum_{i=0}^{k}b_{i}\tilde{H}_{1}(z+c_{i})e^{\omega_{1}(z+c_{i})^{t}-\omega_{1}z^{t}}=0,$
(2.19) $\displaystyle P(z)-H_{0}(z)(H_{0}(z)+a_{1})=0.$ (2.20)

Now, we show that $H_{0}(z)$ is a polynomial. If possible let, $H_{0}(z)$ is
transcendental. Then from (2.20), we have

$2T(r,H_{0}(z))+S(r,H_{0}(z))=T(r,P)=O(\log r),$

a contradiction.
Next, from (2.18), we have

$\displaystyle H_{1}(z)=\beta(z)e^{Q_{t-1}(z)},$ (2.21)

where
$\beta(z)=-\frac{q(z)\sum_{i=0}^{k}b_{i}H^{(r_{i})}_{0}(z+c_{i})}{2H_{0}(z)+a_{1}}$.
Since, $f$ is entire, then $\beta(z)$ is a polynomial. Substituting (2.21) in
(2.19), we have

$\displaystyle\beta^{2}(z)+q(z)\sum_{i=0}^{k}b_{i}\tilde{\beta}(z+c_{i})e^{Q_{t-1}(z+c_{i})-Q_{t-1}(z)+\omega_{1}(z+c_{i})^{t}-\omega_{1}z^{t}}=0,$
(2.22)

where $\tilde{\beta}(z+c_{i})$ is a delay-differential polynomial in
$H_{0}(z)$ and $Q_{t-1}(z)$.
Note that $\deg(\omega_{1}(z+c_{i})^{t}-\omega_{1}z^{t})=t-1$ and
$\deg(Q_{t-1}(z+c_{i})-Q_{t-1}(z))\leq t-2$. If $t\geq 2$, applying Lemma 2.5
on (2.22), we have $q(z)=0$, a contradiction. Therefore, $t=1$, i.e.,
$H_{1}(z)$ is a polynomial. Hence, $f(z)$ reduces to the form

$f(z)=H_{0}(z)+H_{1}(z)e^{\omega_{1}z},$

where $H_{0}(z)$ and $H_{1}(z)$ are polynomials. So,
$f\in\Gamma_{1}^{\prime}$. ∎

##  3\. Proofs of Theorems

######  Proof of Theorem 1.1 (i).

Suppose that $f$ be a finite order non-vanishing entire solution of (1.8).
Using Lemma 2.5, we have $f$ is transcendental. Otherwise, we will get
$L(z,f)\equiv 0$, which yields a contradiction. In view of Lemma 2.3, from
(1.8), we get

$\displaystyle n~{}T(r,f)+S(r,f)$ $\displaystyle=$ $\displaystyle
m\left(r,f^{n}(z)+\sum_{i=1}^{n-1}a_{i}f^{i}(z)\right)$ $\displaystyle=$
$\displaystyle m(r,P(z))-q(z)e^{Q(z)}L(z,f))$ $\displaystyle=$ $\displaystyle
m(r,e^{Q(z)})+m(r,L(z,f))+O(1)$ $\displaystyle=$ $\displaystyle
m(r,e^{Q(z)})+m\left(r,\frac{L(z,f)}{f(z)}\right)+m(r,f(z))+O(1)$
$\displaystyle=$ $\displaystyle T(r,e^{Q(z)})+T(r,f(z))+S(r,f)$
$\displaystyle\implies(n-1)~{}T(r,f)$ $\displaystyle\leq$ $\displaystyle
T(r,e^{Q(z)})+S(r,f).$ (3.1)

Therefore, for $n\geq 2$, $\rho(f)\leq\deg{Q(z)}$. If $\rho(f)<\deg{Q(z)}$,
then comparing order of growth of (1.8), we get a contradiction. So,
$\rho(f)=\deg{Q(z)}$. Now, from the definition of type, we have

$\tau(f)=\overline{\lim}_{r\rightarrow\infty}\frac{T(r,f)}{r^{\rho(f)}}=\overline{\lim}_{r\rightarrow\infty}\frac{T(r,f)}{r^{\deg{Q(z)}}}\in(0,\infty),$

i.e., $f$ is of mean type. ∎

######  Proof of Theorem 1.1 (ii).

First, we prove that if zero is a Borel exceptional value of $f(z)$, then we
have $a_{n-1}=\cdots=a_{1}=0\equiv P(z)$. Adopting the similar process as done
in the proof of Theorem 1.2(b) in [10], using Lemmas 2.3, 2.4, 2.6 and
replacing $f(z+c)$ by $L(z,f)$, we can prove our result. In this regards, only
the equation (10) of [10] is replaced by the following lines

$\displaystyle N\left(r,\frac{1}{G(z)}\right)$ $\displaystyle=$ $\displaystyle
N\left(r,\frac{1}{q(z)e^{Q(z)}L(z,f)}\right)$ $\displaystyle\leq$
$\displaystyle
N\left(r,\frac{1}{q(z)}\right)+N\left(r,\frac{1}{L(z,f)}\right)+S(r,f)$
$\displaystyle\leq$
$\displaystyle(k+1)~{}N\left(r,\frac{1}{f(z)}\right)+S(r,f)=S(r,f).$

Next, we prove the converse part of Theorem 1.1 (ii). For this, using Lemmas
2.3, 2.6 and replacing $f(z+c)$ by $L(z,f)$, we proceed similar up to equation
(17) in the proof of Theorem 1.2(c) in [10]. Here, the equations (15), (16)
and (17) of [10], respectively, will be

$\displaystyle T\left(r,\frac{L(z,f)}{f(z)}\right)=S(r,f),$ (3.2)
$\displaystyle\left(f(z)+\frac{a_{n-1}}{n-1}\right)^{n-1}=f^{n-1}(z)+\sum_{i=1}^{n-1}a_{i}f^{i-1}(z)=-q(z)e^{Q(z)}\frac{L(z,f)}{f(z)}$
(3.3)

and

$\displaystyle
a_{i+1}=\frac{(n-1)!}{i!(n-1-i)!}\left(\frac{a_{n-1}}{n-1}\right)^{n-1-i},\;\;i=0,1,...,n-2.$
(3.4)

Case 1. If there exists $i_{0}\in\\{1,\ldots,n-1\\}$ such that $a_{i_{0}}=0$,
then from (3.4), we have all $a_{i}$ must be equal to zero for
$i=1,2,\ldots,n-1$. Therefore, (3.3) becomes

$\displaystyle f(z)^{n-1}=-q(z)e^{Q(z)}\frac{L(z,f)}{f(z)}.$

By using (3.2), for each $\varepsilon>0$, we have

$\displaystyle(n-1)N\left(r,\frac{1}{f(z)}\right)$ $\displaystyle=$
$\displaystyle N\left(r,\frac{1}{q(z)\frac{L(z,f)}{f(z)}}\right)$
$\displaystyle\leq$ $\displaystyle
N\left(r,\frac{1}{q(z)}\right)+N\left(r,\frac{1}{\frac{L(z,f)}{f(z)}}\right)+S(r,f)=S(r,f).$

Therefore, $\lambda(f)<\rho(f)$.

Case 2. If there exists no $i_{0}\in\\{1,\ldots,n-1\\}$ such that
$a_{i_{0}}=0$, then from (3.2) and (3.3), we have

$\displaystyle\overline{N}\left(r,\frac{1}{f(z)+\frac{a_{n-1}}{n-1}}\right)\leq\overline{N}\left(r,\frac{1}{q(z)}\right)+\overline{N}\left(r,\frac{1}{\frac{L(z,f)}{f(z)}}\right)+S(r,f)=S(r,f).$

Using the second main theorem, we have

$\displaystyle T(r,f)$ $\displaystyle\leq$
$\displaystyle\overline{N}\left(r,\frac{1}{f(z)+\frac{a_{n-1}}{n-1}}\right)+\overline{N}\left(r,\frac{1}{f(z)}\right)+\overline{N}\left(r,f(z)\right)$
$\displaystyle=$
$\displaystyle\overline{N}\left(r,\frac{1}{f(z)}\right)+S(r,f)$

Therefore, $\rho(f)\leq\lambda(f)$ but we know that $\lambda(f)\leq\rho(f)$.
Therefore, $\lambda(f)=\rho(f)$. ∎

######  Proof of Theorem 1.1 (iii).

Suppose that $f$ be a non-vanishing finite order entire solution of (1.8).
Similar as Theorem 1.1 (i), $f$ is transcendental.

First suppose that $f$ belongs to $\Gamma_{0}^{\prime}$, which means that $0$
is a Borel exceptional value of $f$. Thus, from Theorem 1.1 (ii), we have
$a_{n-1}=\cdots=a_{1}=0\equiv P(z)$.

Next, we suppose that $P(z)\equiv 0$ and there exists an
$i_{0}\in\\{1,\ldots,n-1\\}$ such that $a_{i_{0}}=0$, then from the converse
part of Theorem 1.1 (ii), we have all of the $a_{i}(i=1,\ldots,n-1)$ must be
zero as well and $\lambda(f)<\rho(f)$. From Hadamard factorization theorem, we
can see that

$\displaystyle f(z)=h(z)e^{\alpha(z)},$ (3.5)

where $\alpha(z)$ is a polynomial and $h(z)$ is the canonical product of zeros
of $f$ with $\deg{\alpha(z)}=\rho(f)=\deg{Q(z)}=t$ and
$\rho(h)=\lambda(h)=\lambda(f)<\rho(f)$.
Substituting (3.5) in (1.8) with all $a_{i}=0$, we have

$\displaystyle
h^{n}(z)e^{n\alpha(z)}+q(z)e^{Q(z)+\alpha(z)}\left(\sum_{i=0}^{k}L_{i}(z,h)e^{\Delta_{c_{i}}\alpha(z)}\right)=0,$
(3.6)

where

$\displaystyle L_{i}(z,h)$ $\displaystyle=$ $\displaystyle
b_{i}\left[h(z+c_{i})M_{k_{i}}(\alpha^{\prime}(z+c_{i}),\alpha^{\prime\prime}(z+c_{i}),\ldots,\alpha^{(k_{i})}(z+c_{i}))\right.$
$\displaystyle\qquad\left.+h^{\prime}(z+c_{i})M_{k_{i}-1}(\alpha^{\prime}(z+c_{i}),\alpha^{\prime\prime}(z+c_{i}),\ldots,\alpha^{(k_{i}-1)}(z+c_{i}))\right.$
$\displaystyle\qquad\left.+\cdots+h^{(k_{i}-1)}(z+c_{i})M_{1}(\alpha^{\prime}(z+c_{i}))+h^{(k_{i})}(z+c_{i})\right].$

Clearly, $\rho(L_{i}(z,h))<t$. Rewriting (3.6), we have

$\displaystyle
h^{n}(z)e^{n\alpha_{t-1}(z)}e^{nu_{t}z^{t}}+q(z)e^{\alpha_{t-1}(z)+Q_{t-1}(z)}\left(\sum_{i=0}^{k}L_{i}(z,h)e^{\Delta_{c_{i}}\alpha(z)}\right)e^{(u_{t}+v_{t})z^{t}}=0$
(3.7)

such that $\alpha(z)=u_{t}z^{t}+\alpha_{t-1}(z)$ and
$Q(z)=v_{t}z^{t}+Q_{t-1}(z)$, where $u_{t}$, $v_{t}$ are non-zero constants
and $\alpha_{t-1}(z)$, $Q_{t-1}(z)$ are of degree $\leq t-1$.
In view of Lemma 2.5, we can easily say that (3.7) is possible only when
$(n-1)u_{t}=v_{t}$. Therefore, (3.7) becomes

$\displaystyle
h^{n}(z)+q(z)e^{(1-n)\alpha_{t-1}(z)+Q_{t-1}(z)}\left(\sum_{i=0}^{k}L_{i}(z,h)e^{\Delta_{c_{i}}\alpha(z)}\right)=0.$
(3.8)

Here, the following cases arise.
Case 1: Let $\rho(h)<t-1>0$. If
$\deg\\{(1-n)\alpha_{t-1}(z)+Q_{t-1}(z)\\}=t-1$, applying Lemma 2.5, we have
$q(z)=0$, a contradiction. If $\deg\\{(1-n)\alpha_{t-1}(z)+Q_{t-1}(z)\\}<t-1$,
from $\deg\\{\Delta_{c_{i}}\alpha(z)\\}=t-1$, by using Lemma 2.5, again we
have $q(z)=0$, a contradiction.
Case 2: Let $\rho(h)\geq t-1>\rho(h)-1$, $t-1>0$. By logarithmic derivative
lemma [5, Corollary 2.5], for each $\varepsilon>0$, we have

$\displaystyle
m\left(r,\frac{L_{i}(z,h)}{h(z)}\right)=O(r^{\rho(h)-1+\varepsilon})+O(\log
r).$ (3.9)

Since, $h$ is entire, using (3.8) and (3.9), we have

$\displaystyle
T\left(r,\sum_{i=0}^{k}\frac{L_{i}(z,h)}{h}e^{\Delta_{c_{i}}\alpha(z)}\right)$
$\displaystyle=$ $\displaystyle
m\left(r,\sum_{i=0}^{k}\frac{L_{i}(z,h)}{h}e^{\Delta_{c_{i}}\alpha(z)}\right)+N\left(r,\sum_{i=0}^{k}\frac{L_{i}(z,h)}{h}e^{\Delta_{c_{i}}\alpha(z)}\right)$
(3.10) $\displaystyle\leq$
$\displaystyle\sum\limits_{i=0}^{k}T\left(r,e^{\Delta_{c_{i}}\alpha(z)}\right)+O(r^{\rho(h)-1+\varepsilon})+O(\log
r).$

Therefore, using (3.8) and (3.10), we obtain for each $\varepsilon>0$

$\displaystyle(n-1)N\left(r,\frac{1}{h(z)}\right)$ $\displaystyle\leq$
$\displaystyle
N\left(r,\frac{1}{\sum_{i=0}^{k}\frac{L_{i}(z,h)}{h}e^{\Delta_{c_{i}}\alpha(z)}}\right)+O(\log
r)$ (3.11) $\displaystyle\leq$ $\displaystyle
T\left(r,\sum_{i=0}^{k}\frac{L_{i}(z,h)}{h}e^{\Delta_{c_{i}}\alpha(z)}\right)+O(\log
r)$ $\displaystyle\leq$
$\displaystyle\sum\limits_{i=0}^{k}T\left(r,e^{\Delta_{c_{i}}\alpha(z)}\right)+O(r^{\rho(h)-1+\varepsilon})+O(\log
r).$

Now, if $c_{i}=c_{j}$ for all $1\leq i,j\leq k$, say, $c$, then (3.11) becomes

$\displaystyle(n-1)N\left(r,\frac{1}{h(z)}\right)\leq
T\left(r,e^{\Delta_{c}\alpha(z)}\right)+O(r^{\rho(h)-1+\varepsilon})+O(\log
r).$

Thus, from the above equation, we have $\lambda(h)\leq t-1$. But in this case,
$\lambda(h)=\rho(h)\geq t-1$. Therefore,
$\lambda(f)=\lambda(h)=t-1=\rho(f)-1$.

If $t-1=0$, we get $\rho(h)=\lambda(h)<\rho(f)=t=1$. If $h(z)$ is
transcendental, then $h(z)$ has infinitely many zeros. Now noting that
$\Delta_{c_{i}}\alpha(z)$ is of degree $t-1$, from (3.11), we get
$N\left(r,\frac{1}{h(z)}\right)=O(\log r)$, a contradiction. Therefore, $h(z)$
will be a polynomial. So, $f$ belongs to $\Gamma_{0}^{\prime}$.
Case 3: Let $\rho(h)\geq t$. Then from (3.11), we obtain $\lambda(h)<\rho(h)$,
a contradiction. So, $h(z)$ will be a polynomial. Therefore, $f$ belongs to
$\Gamma_{0}^{\prime}$. ∎

######  Proof of Theorem 1.1 (iv).

Suppose that $f$ is a non-vanishing finite order entire solution of (1.8).
Similarly, $f$ is transcendental. If possible let, $P(z)\not\equiv 0$. By the
assumption, $card\\{z:p(z)=p^{\prime}(z)=p^{\prime\prime}(z)=0\\}\geq 1$ or
$card\\{z:p(z)=p^{\prime}(z)=0\\}\geq 2$, where
$p(z)=z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z$, we mean that $p(z)$ has at least
one zero with multiplicity at least three or at least $2$ zeros with
multiplicities at least two. In view of Lemma 2.3 and the second main theorem,
we have

$\displaystyle n~{}T(r,f)$ $\displaystyle=$ $\displaystyle
T(r,f^{n}(z)+a_{n-1}f^{n-1}(z)+\cdots+a_{1}f(z))$ $\displaystyle\leq$
$\displaystyle\overline{N}\left(r,\frac{1}{f^{n}+a_{n-1}f^{n-1}+\cdots+a_{1}f-P(z)}\right)$
$\displaystyle+\overline{N}\left(r,\frac{1}{f^{n}+a_{n-1}f^{n-1}+\cdots+a_{1}f}\right)+\overline{N}(r,f^{n}+a_{n-1}f^{n-1}+\cdots+a_{1}f)$
$\displaystyle\leq$
$\displaystyle\overline{N}\left(r,\frac{1}{q(z)L(z,f)}\right)+(n-2)~{}T(r,f)+S(r,f)$
$\displaystyle\leq$ $\displaystyle
T\left(r,q(z)L(z,f)\right)+(n-2)~{}T(r,f)+S(r,f)$ $\displaystyle\leq$
$\displaystyle
N\left(r,q(z)L(z,f)\right)+m\left(r,q(z)L(z,f)\right)+(n-2)~{}T(r,f)+S(r,f)$
$\displaystyle\leq$ $\displaystyle
m\left(r,q(z)\frac{L(z,f)}{f(z)}\right)+m(r,f(z))+(n-2)~{}T(r,f)+S(r,f)$
$\displaystyle\leq$ $\displaystyle T(r,f)+(n-2)~{}T(r,f)+S(r,f)$
$\displaystyle\leq$ $\displaystyle(n-1)~{}T(r,f)+S(r,f),$

a contradiction. Therefore, $P(z)\equiv 0$.

Since, at least one $a_{i_{0}}=0$ $(i_{0}=1,2,\ldots,n-1)$, using Theorem 1.1
(iii), we have $f\in\Gamma_{0}^{\prime}$.
Note that, since, $P(z)\equiv 0$ and at least one $a_{i_{0}}=0$
$(i_{0}=1,2,\ldots,n-1)$, from Theorem 1.1 (ii), we have all of $a_{j}$’s
$(j=1,\ldots,n-1)$ must be zero. Therefore $p(z)$ is of the form $z^{n}$,
which implies $card\\{z:p(z)=p^{\prime}(z)=0\\}\geq 2$ is not possible. ∎

######  Proof of Theorem 1.1 (v).

From Theorem 1.1 (i), we have $\rho(f)=\deg(Q(z))$. Using Lemmas 2.8 \- 2.10,
we can prove the result. ∎

## References

  * [1] T. Biswas and A. Banerjee, Finite-order transcendental entire solutions of generalized non-linear shift equations, Mediterr. J. Math., 18(2021), Art. 130, DOI: 10.1007/s00009-021-01743-3.
  * [2] Z. X. Chen, Complex Differences and Difference Equations, Science Press, 2014.
  * [3] M. F. Chen, Z. S. Gao and J. L. Zhang, Entire solutions of certain type of non-linear difference equations, Comput. Methods Funct. Theo., 19(1)(2019), 17-36.
  * [4] W. Chen, P. Hu and Q. Wang, Entire solutions of two certain types of non-linear differential-difference equations, Comput. Methods Funct. Theo., (2020), DOI: 10.1007/s40315-020-00343-8.
  * [5] Y. M. Chiang and S. J. Feng, On the Nevanlinna characteristic of $f(z+\eta)$ and difference equations in the Complex plane, Ramanujan J., 16(1)(2008), 105-129.
  * [6] W. K. Hayman, Meromorphic Functions, Clarendon Press, Oxford, 1964.
  * [7] R. Korhonen, An extension of Picard’s theorem for meromorphic functions of small hyper-order, J. Math. Anal. Appl., 357(2009), 244-253.
  * [8] I. Laine, Nevanlinna Theory and Complex Differential Equations, Walter de Gruyter, Berlin-New York, 1993.
  * [9] C. P. Li, F. Lü and J. F. Xu, Entire solutions of non-linear differential‑difference equations, SpringerPlus, 5, 609(2016), DOI: 10.1186/s40064-016-2255-9.
  * [10] N. Li and L. Z. Yang, Solutions of non-linear difference equations, J. Math. Anal. Appl., 452(2017), 1128-1144.
  * [11] L. W. Liao, C. C. Yang and J. J. Zhang, On meromorphic solutions of certain type of non-linear differential equations, Ann. Acad. Sci. Fenn. Math., 38(2013), 581-593.
  * [12] K. Liu, Exponential polynomials as solutions of differential-difference equations of certain types, Mediterr. J. Math., 13(2016), 3015-3027.
  * [13] H. F. Liu, Z. Q. Mao and D. Zheng, Meromorphic solutions of certain non-linear difference equations, Open Math., 18(2020), 1292-1301.
  * [14] N. Steinmetz, Zur Wertverteilung von Exponentialpolynomen, Manuscr. Math., 26(1978), 155-167.
  * [15] Z. T. Wen, J. Heittokangas and I. Laine, Exponential polynomials as solutions of certain non-linear difference equations, Acta Math. Sin. 28(2012), 1295-1306.
  * [16] J. F. Xu and J. X. Rong, Exponential polynomials and non-linear differential-difference equations, Journal of Function Spaces, Vol. 2020, Art. 6901270, 1-10.
  * [17] C. C. Yang, On entire solution of a certain type of non-linear differential equation, Bull. Austral. Math. Soc., 64(3)(2001), 377-380.
  * [18] C. C. Yang and I. Laine, On analogies between non-linear difference and differential equations, Proc. Jpn. Acad., Ser. A, 86(2010), 10-14.
  * [19] C. C. Yang and H. X. Yi, Uniqueness theory of meromorphic functions, Kluwer Acad. Publishers, Norwell, (2003).
  * [20] J. Zhang and L. W. Liao, On entire solutions of a certain type of non-linear differential and difference equations, Taiwan. J. Math., 15(2011), 2145-2157.