††footnotetext:  File: main.tex, printed: 2024-9-1, 10.34

# Solutions for several systems of algebraic differential-difference equations
in $\mathbb{C}^{n}$

Molla Basir Ahamed  Molla Basir Ahamed, Department of Mathematics, Jadavpur
University, Kolkata-700032, West Bengal, India.
<EMAIL_ADDRESS>and  Sanju Mandal  Sanju Mandal,
Department of Mathematics, Jadavpur University, Kolkata-700032, West Bengal,
India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this article, we introduce a new method for solving general quadratic
functional equations in $\mathbb{C}^{n}$. By utilizing Nevanlinna theory in
$\mathbb{C}^{n}$, we explore the existence and form of solutions for the
several systems of general quadratic difference or partial differential-
difference equations of the form $af^{2}+2\alpha fg+bg^{2}+2\beta f+2\gamma
g+C=0$, where $f$ and $g$ are non-constant meromorphic functions in
$\mathbb{C}^{n}$. The results in this paper are some improvements and
generalizations of several existing results on the existence and forms of
solutions for these systems. Specifically, the existing results were related
to Circular-type $f^{2}+g^{2}=1$ and quadratic trinomial functional equations
$f^{2}+2\alpha fg+g^{2}=1$. We provide examples to illustrate the validity of
our conclusions regarding the existence and forms of solutions for such system
of equations.

###### Key words and phrases:

Transcendental entire solutions, Nevanlinna theory, Several complex variables,
Circular-type functional equations, System of functional equations, finite
order, Partial differential-difference equations

###### AMS Subject Classification: Mathematics Subject Classification:

Primary 39A45, 30D35, 35M30, 32W50

##  1\. Introduction

In this paper, we will explore transcendental solutions for several systems of
general quadratic difference or partial differential-difference equations. In
recent years, extensive research has focused on finding finite order entire
and meromorphic solutions in one and several complex variables for the
functional equation

$\displaystyle f^{n}(z)+g^{n}(z)=1\;\mbox{for integer}\;n\geq 1$

regard as the Fermat diophantine equation $x^{n}+y^{n}=1$ over functional
fields, where $n\geq 2$ is an integer. The study of solutions and their
different properties of Circular-type functional equations
$f^{2}(z)+g^{2}(z)=1$ in $\mathbb{C}^{n}$ is an active area of research that
has received significant attention in recent times. Before going into a
detailed information of the study of such equations in $\mathbb{C}^{n}$, let
us revisit a crucial result concerning entire and meromorphic solutions of
Circular-type equations in $\mathbb{C}^{n}$.

######  Theorem A.

([16]) For $h:\mathbb{C}^{n}\rightarrow\mathbb{C}$ entire, the solutions of
the Circular equation $f^{2}(z)+g^{2}(z)=1$ are characterize as follows:

  1. (i)

the entire solutions are $f(z)=\cos(h(z)),\;g(z)=\sin(h(z))$;

  2. (ii)

the meromorphic solutions are of the form
$f(z)=\frac{1-\beta^{2}(z)}{1+\beta^{2}(z)},\;g=\frac{2\beta(z)}{1+\beta^{2}(z)}$,
with $\beta$ being meromorphic on $\mathbb{C}^{n}$.

The study of algebraic equations is likely as ancient as mathematics itself:
Babylonian mathematicians, dating back to around $2000$ BC, demonstrated the
ability to solve certain types of quadratic equations, as evidenced by Old
Babylonian clay tablets. Univariate algebraic equations with rational
coefficients have an extensive history. Ancient mathematicians sought
solutions in the form of radical expressions, such as
$x=\frac{1+\sqrt{5}}{2}$, which represents the positive solution of the
equation $x^{2}-x-1=0$. The ancient Egyptians knew how to solve equations of
degree $2$ in this manner. The study of general quadratic equations holds
significance in mathematics and science because of its applications, making it
a crucial area of exploration. Such equations commonly appear in problems
associated with optimization, control theory, physics, and engineering,
playing an essential role in the development of contemporary technology and
science. Over the functional filed, it is difficult to find entire and
meromorphic solutions for a algebraic $PDE$. Utilizing Nevanlinna theory in
several complex variables and employing complex analysis methods, there was
some literature focusing on solutions of the $PDEs$ and their many variants.

In [10], Khavinson showed that any entire solution of the partial differential
equation $\left(\frac{\partial u}{\partial
z_{1}}\right)^{2}+\left(\frac{\partial u}{\partial z_{2}}\right)^{2}=1$ must
be linear, i.e., $u(z_{1},z_{2})=az_{1}+bz_{2}+c$, where $a,b,c\in\mathbb{C}$
and $a^{2}+b^{2}=1$. Khavinson’s work inspired several researchers on finding
the entire solutions to the equations like $u_{z_{1}}^{2}+u_{z_{2}}^{2}=1$ in
$\mathbb{C}^{2}$. In $2018$, Xu and Cao (see [19, 20]) first obtained the
following results by exploring the finite order solutions of Circular-type
differential-difference equations in $\mathbb{C}^{2}$, and also to that of
difference equations in $\mathbb{C}^{n}$.

######  Theorem B.

([19, Theorem 1.2]) Let $c=(c_{1},c_{2})$ be a constant in $\mathbb{C}^{2}$.
Then any transcendental entire solution with finite order of the partial
difference-differential equation of the Fermat type

$\displaystyle\left(\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+f(z_{1}+c_{1},z_{2}+c_{2})^{2}=1$

has the form of $f(z_{1},z_{2})=\sin(Az_{1}+Bz_{2}+H(z_{2}))$, where $A,B$ are
constants on $\mathbb{C}$ satisfying $A^{2}=1$ and $Ae^{i(Ac_{1}+Bc_{2})}=1$,
$H(z_{2})$ is a polynomial in one variable $z_{2}$ such that
$H(z_{2})=H(z_{2}+c_{2})$. In the special case whenever $c_{2}\neq 0$, we have
$f(z_{1},z_{2})=\sin(Az_{1}+Bz_{2}+constant)$.

######  Theorem C.

(see [19, Theorem 1.4]) Let $c\in\mathbb{C}^{n}\setminus\\{0\\}$. Then any
non-constant entire solution with finite order of the equation

$\displaystyle f^{2}(z)+f^{2}(z+c)=1$

has the form $f(z)=\cos\left(L(z)+B\right)$, where $L$ is a linear function of
the form $L(z)=a_{1}z_{1}+\cdots+a_{n}c_{n}$ on $\mathbb{C}^{n}$ such that
$L(c)=-\frac{\pi}{2}-2k\pi$, $k$ is an integer, and $B$ is a constant in
$\mathbb{C}$.

Motivated by the above discussion, it is natural to raise the question.

######  Question 1.

How to describe the entire solutions for the system of the Circular-type
differential or differential-difference equations in $\mathbb{C}^{2}$?

Nevanlinna theory in one and several complex variables is a useful tool for
the study of existence as well as growth properties of meromorphic solutions
of different complex functional equations. In response to the Question 1,
solution of the following system equations

$\displaystyle\begin{cases}f_{1}(z_{1},z_{2})^{2}+f_{2}(z_{1}+c_{1},z_{2}+c_{2})^{2}=1,\vspace{1.2mm}\\\
f_{2}(z_{1},z_{2})^{2}+f_{1}(z_{1}+c_{1},z_{2}+c_{2})^{2}=1\end{cases}$

and

$\displaystyle\begin{cases}\left(\dfrac{\partial f_{1}(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+[f_{2}(z_{1}+c_{1},z_{2}+c_{2})-f_{1}(z_{1},z_{2})]^{2}=1,\vspace{1.2mm}\\\
\left(\dfrac{\partial f_{2}(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+[f_{1}(z_{1}+c_{1},z_{2}+c_{2})-f_{2}(z_{1},z_{2})]^{2}=1\end{cases}$

are obtained in [23] explicitly using Nevanlinna theory in several complex
variables. However, there have been several studies that focus on finding
solutions of single equations or systems of equations (see [4, 23, 22, 5, 6,
17, 9, 12] and references therein).

Utilizing the difference analogues of Nevanlinna theory in $\mathbb{C}$ by
Halburd and Korhonen [7], Chiang and Feng [3], and subsequently in
$\mathbb{C}^{n}$, respectively, many scholars have tried to characterize all
the meromorphic solutions of Circular-type difference or differential-
difference equations (see [1, 26, 21, 19]). However, Saleeby (see [16]) made
some elementary observation on right factors of meromorphic function to
describe complex analytic solutions to the quadratic trinomial functional
equation $f^{2}(z)+2\alpha f(z)g(z)+g^{2}(z)=1$, $\alpha^{2}\neq 1$ constant
in $\mathbb{C}$ and obtain a result associate with the partial differential
equations $u^{2}_{x}+2\alpha u_{x}u_{y}+u^{2}_{y}=1$. Later, in [13], Liu and
Yang studied the existence and form of solutions of some quadratic trinomial
functional equations and proved that if $\alpha\neq\pm 1,0$, then quadratic
trinomial differential equation $f(z)^{2}+2\alpha
f(z)f^{\prime}(z)+f^{\prime}(z)^{2}=1$ in $\mathbb{C}$ has no transcendental
meromorphic solutions, whereas the difference functional equation
$f(z)^{2}+2\alpha f(z)f(z+c)+f(z+c)^{2}=1$ in $\mathbb{C}$ must have
transcendental entire solutions of order equal to one.

The system of general quadratic equations has been an interesting subject in
the field of complex analysis in connection with the application of Nevanlinna
theory. It appears that the study of systems of the general quadratic
equations which is a general setting of Circular-type as well as trinomial-
type functional equations in several complex variables has not been addressed
in the literature before. In this paper, using Nevanlinna theory, our primary
objective is to explore the existence as well as form of the solutions with
finite order of general quadratic functional equations in $\mathbb{C}^{n}$.

Since a non-constant polynomials in $\mathbb{C}^{n}$ may exhibit periodic
behavior, and we know that polynomials play a crucial role in the solutions of
partial differential or difference equations in $\mathbb{C}^{n}$, the
characterization of solutions in $\mathbb{C}^{n}$ will naturally differ from
that in $\mathbb{C}$. Further, we observe that the Circular equation
$f^{2}+g^{2}=1$ and quadratic trinomial functional equation $f^{2}+2\alpha
fg+g^{2}=1$ are both particular cases of general quadratic functional equation
of the form

$\displaystyle\mathcal{A}(f,g):=af^{2}+2\alpha fg+bg^{2}+2\beta f+2\gamma
g+C=0$

in $\mathbb{C}^{n}$. The most general form of these equations is represented
by quadratic equations with arbitrary coefficients. Therefore, the solutions
for systems of general quadratic equations provide a broader perspective
compared to solutions of these Circular and trinomial equations in
$\mathbb{C}^{n}$.

To the best of our knowledge, the question of what can be said about the
solutions of a system of general quadratic equations in $\mathbb{C}^{n}$ has
not been studied yet. Therefore, the investigation of the solution for systems
involving these general quadratic equations in $\mathbb{C}^{n}$ will be a
novel work. Henceforth, we consider three type system of general quadratic
functional equations in $\mathbb{C}^{n}$ as follows:

$\displaystyle\begin{cases}\mathcal{A}\left(f_{1}(z),\;f_{2}(z+c)\right)=0\\\
\mathcal{A}\left(f_{2}(z),\;f_{1}(z+c)\right)=0,\;\end{cases}$
$\displaystyle\begin{cases}\mathcal{A}\left(f_{1}(z),\;\dfrac{\partial
f_{2}(z)}{\partial z_{1}}\right)=0\vspace{2mm}\\\
\mathcal{A}\left(f_{2}(z),\;\dfrac{\partial f_{1}(z)}{\partial
z_{1}}\right)=0\end{cases}\;\;\;\;\mbox{and}\;\;\;\;\;\;\;\;\begin{cases}\mathcal{A}\left(f_{1}(z+c),\;\dfrac{\partial
f_{2}(z)}{\partial z_{1}}\right)=0\vspace{2mm}\\\
\mathcal{A}\left(f_{2}(z+c),\;\dfrac{\partial f_{1}(z)}{\partial
z_{1}}\right)=0.\end{cases}$

The above discussions lead us to raise a natural question.

######  Question 2.

What can be said about the existence and forms of the solutions of the above
system of general quadratic functional equations in $\mathbb{C}^{n}$?

Motivated by the question above, this article focuses on describing
transcendental solutions for various general quadratic difference equations
and partial differential-difference equations in $\mathbb{C}^{n}$. This paper
aims to explore the existence and precise form of solutions for the mentioned
system of general quadratic functional equations, utilizing Nevanlinna theory
and the difference analog of Nevanlinna theory in $\mathbb{C}^{n}$ (see [2,
11]). These theories will provide a comprehensive answer to Question 2.
Henceforth, throughout the paper, we assume that the corresponding companion
matrices for the equations are non-singular, i.e.,
$\Delta=:abC+2\alpha\beta\gamma-a\gamma^{2}-b\beta^{2}-C\alpha^{2}\neq 0$.
Otherwise, each equation in the system can be linearly factorized, which will
eventually show that the solution $f$ is a constant.

This article is organized as follows: In section 2, we present the main
results of this paper. These main results correspond to Question 2, and we
provide illustrative examples for each result to examine the precise form of
transcendental entire solutions with finite order. We also include several
remarks in support of our some improvements and generalizations results. In
section 3, we provide a comprehensive review of essential Lemmas and present a
detailed proof for each significant result.

##  2\. Main results, corollaries and remarks

In this section, we give the main result of the paper and its immediate
consequences. Throughout this paper, let
$z+c=(z_{1}+c_{1},\ldots,z_{n}+c_{n})$ and $tz=(tz_{1},\ldots,tz_{n})$ for any
$z=(z_{1},\ldots,z_{n}),c=(c_{1},\ldots,c_{n})\in\mathbb{C}^{n}$ and
$t\in\mathbb{C}$. Let us fix several notations which will be used
subsequently:

$\displaystyle\Delta=\begin{vmatrix}a&\alpha&\beta\\\ \alpha&b&\gamma\\\
\beta&\gamma&C\end{vmatrix}=abC+2\alpha\beta\gamma-a\gamma^{2}-b\beta^{2}-C\alpha^{2}\;(\neq
0),\;D=\begin{vmatrix}a&\alpha\\\ \alpha&b\end{vmatrix}=ab-\alpha^{2}\;(\neq
0),$
$\displaystyle\begin{cases}\xi^{\pm}_{1}:=\dfrac{2\alpha}{\sqrt{\left((b-a)\pm\sqrt{(a-b)^{2}+4\alpha^{2}}\right)^{2}+4\alpha^{2}}},\vspace{1.2mm}\\\
\eta^{\pm}_{1}:=\dfrac{(b-a)\pm\sqrt{(a-b)^{2}+4\alpha^{2}}}{\sqrt{\left((b-a)\pm\sqrt{(a-b)^{2}+4\alpha^{2}}\right)^{2}+4\alpha^{2}}}.\end{cases}$

Also, we denote

$\displaystyle
A^{\pm}:=\dfrac{(a+b)\pm\sqrt{(a-b)^{2}+4\alpha^{2}}}{2},\;\;\;\;B^{\mp}:=\dfrac{(a+b)\mp\sqrt{(a-b)^{2}+4\alpha^{2}}}{2}$

and

$\displaystyle\begin{cases}D_{11}=\dfrac{\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;\;D_{12}=\dfrac{\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\;\mbox{and}\;\;T_{1}=\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}},\vspace{1.5mm}\\\
E_{11}=\dfrac{\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;\;E_{12}=\dfrac{\xi^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\;\mbox{and}\;\;T_{2}=\dfrac{\alpha\beta-a\gamma}{ab-\alpha^{2}}.\end{cases}$

Furthermore, for some consequences of the main results, we introduce some
notations:

$\displaystyle\begin{cases}K_{11}=\sqrt{(a+b)\pm\sqrt{(a-b)^{2}+4\alpha^{2}}},\;\;K_{12}=\sqrt{\left((b-a)\pm\sqrt{(a-b)^{2}+4\alpha^{2}}\right)^{2}+4\alpha^{2}},\vspace{1.5mm}\\\
K_{13}=\sqrt{(a+b)\mp\sqrt{(a-b)^{2}+4\alpha^{2}}},\;\;K_{14}=(b-a)\pm\sqrt{(a-b)^{2}+4\alpha^{2})},\vspace{1.5mm}\\\
A_{11}=\dfrac{2\sqrt{2}\alpha}{K_{11}K_{12}},\;\;\;\;\;A_{12}=\dfrac{\sqrt{2}K_{14}}{K_{12}K_{13}},\;\;\;\;B_{11}=\dfrac{\sqrt{2}K_{14}}{K_{11}K_{12}},\;\;\;\;B_{12}=\dfrac{2\sqrt{2}\alpha}{K_{12}K_{13}}.\end{cases}$

The first result of the paper is the following, which answers Question 2
completely in case difference equations in $\mathbb{C}^{n}$.

######  Theorem 2.1.

Let $c=(c_{1},\ldots,c_{n})\in\mathbb{C}^{n}\setminus\\{(0,\ldots,0)\\}$ and
$a,b,C,\alpha,\beta,\gamma\in\mathbb{C}$ such that $ab\neq 0$, $\alpha^{2}\neq
0,ab$ and $\Delta\neq 0$. Then the pair of finite order transcendental entire
solutions in $\mathbb{C}^{n}$ for the system of general quadratic difference
equations

$\displaystyle\begin{cases}\mathcal{A}\left(f_{1}(z),\;f_{2}(z+c)\right)=0,\\\
\mathcal{A}\left(f_{2}(z),\;f_{1}(z+c)\right)=0\;\end{cases}$

must be one of the forms

  1. (i)

$f_{1}(z)=D_{11}\cos[\gamma(z)+b_{1}]-D_{12}\sin[\gamma(z)+b_{1}]+T_{1},\vspace{1.2mm}\\\
f_{2}(z)=D_{11}\cos[\gamma(z)+b_{2}]-D_{12}\sin[\gamma(z)+b_{2}]+T_{1}$,

where $\gamma(z)=L(z)+H(s)$, $L(z)$ is a linear function of the form
$L(z)=\sum_{r=1}^{n}a_{r}z_{r}$, $H(s)$ is a polynomial in
$s(z):=\sum_{r=1}^{n}d_{r}z_{r}$ in $\mathbb{C}$ with
$\sum_{r=1}^{n}d_{r}c_{r}=0$ and $s(z+c)=s(z)$, and $b_{1},b_{2}\in\mathbb{C}$
satisfy

$\displaystyle
e^{2iL(c)}\equiv\frac{R^{2}_{13}}{R^{2}_{11}},\;\;\;\;e^{2i(b_{1}-b_{2})}\equiv\frac{R_{13}}{R_{11}R_{12}}\;\;\;\;\mbox{and}\;\;\;\;T_{1}=T_{2},$

where

$\displaystyle
R_{11}=\frac{iD_{11}-D_{12}}{iE_{11}-E_{12}},\;\;R_{12}=\frac{iD_{11}+D_{12}}{iE_{11}-E_{12}},\;R_{13}=\frac{iE_{11}+E_{12}}{iE_{11}-E_{12}}.$

  1. (ii)

$f_{1}(z)=D_{11}\cos[\gamma(z)+b_{1}]-D_{12}\sin[\gamma(z)+b_{1}]+T_{1},\vspace{1.2mm}\\\
f_{2}(z)=D_{11}\cos[\gamma(z)+b_{2}]+D_{12}\sin[\gamma(z)+b_{2}]+T_{1}$,

where $D_{11},D_{12},\gamma(z)$ are stated as in $(i)$ and
$b_{1},b_{2}\in\mathbb{C}$ satisfy

$\displaystyle
e^{2iL(c)}\equiv\frac{R_{13}}{R_{11}R_{12}},\;\;\;\;e^{2i(b_{1}-b_{2})}\equiv\frac{1}{R^{2}_{11}}\;\;\;\;\mbox{and}\;\;\;\;T_{1}=T_{2}.$

######  Remark 2.1.

It is clear that Theorem 2.1 improves the results from [23, Theorem 1.2], [22,
Theorem 2.1], and [13, Theorem 1.4] for system of general quadratic
differential-difference functional equations in $\mathbb{C}^{n}$.

The examples given below clearly demonstrate the existence of transcendental
entire solutions to the equation system in Theorem 2.1.

######  Example 2.1.

For $c=(3,1,-2)\in\mathbb{C}^{3}$ and $b_{1}=\left(1+\ln 5\right)/(2i)$, and
$b_{2}=1/(2i)$, by a routine computation, it can be easily shown that the pair
$(f_{1},f_{2})$, where

$\displaystyle f_{1}(z)$ $\displaystyle=\frac{-2}{\sqrt{10\pm
6\sqrt{5}}}\cos\left(2z_{1}+\frac{\ln
3}{2i}z_{2}+3z_{3}+(2z_{1}-4z_{2}+z_{3})^{5}+\frac{\left(1+\ln
5\right)}{2i}\right)$ $\displaystyle\quad-\frac{(-1\pm\sqrt{5})}{\sqrt{30\mp
14\sqrt{5}}}\sin\left(2z_{1}+\frac{\ln
3}{2i}z_{2}+3z_{3}+(2z_{1}-4z_{2}+z_{3})^{2}+\frac{\left(1+\ln
5\right)}{2i}\right)-2,$

and

$\displaystyle f_{2}(z)$ $\displaystyle=\frac{-2}{\sqrt{10\pm
6\sqrt{5}}}\cos\left(2z_{1}+\frac{\ln
3}{2i}z_{2}+3z_{3}+(2z_{1}-4z_{2}+z_{3})^{5}+\frac{1}{2i}\right)$
$\displaystyle\quad+\frac{(-1\pm\sqrt{5})}{\sqrt{30\mp
14\sqrt{5}}}\sin\left(2z_{1}+\frac{\ln
3}{2i}z_{2}+3z_{3}+(2z_{1}-4z_{2}+z_{3})^{4}+\frac{1}{2i}\right)-2,$

is transcendental entire solutions with $\rho(f_{1},f_{2})=5>1$ in
$\mathbb{C}^{3}$ of the equations

$\displaystyle\begin{cases}3f_{1}(z)^{2}-4f_{1}(z)f_{2}(z+c)+f_{2}(z+c)^{2}+4f_{1}(z)-4f_{2}(z+c)=1,\\\
3f_{2}(z)^{2}-4f_{2}(z)f_{1}(z+c)+f_{1}(z+c)^{2}+4f_{2}(z)-4f_{1}(z+c)=1.\end{cases}$

######  Remark 2.2.

Theorem 2.1 gives rise to the following result as an immediate corollary that
extends [23, Theorem 1.2] and [22, Theorem 2.1] to include arbitrary constants
as coefficients of the equations, and also generalize from $\mathbb{C}^{2}$ to
$\mathbb{C}^{n}$ for $n\geq 2$.

######  Corollary 2.1.

Let $c=(c_{1},\ldots,c_{n})\in\mathbb{C}^{n}\setminus\\{(0,\ldots,0)\\}$ and
$a,b,\alpha\in\mathbb{C}$ with $ab\neq 0$ and $\alpha^{2}\neq 0,ab$. Then the
pair of finite order transcendental entire solutions in $\mathbb{C}^{n}$ for
the system of general quadratic difference equations

$\displaystyle\begin{cases}af_{1}(z)^{2}+2\alpha
f_{1}(z)f_{2}(z+c)+bf_{2}(z+c)^{2}=1,\\\ af_{2}(z)^{2}+2\alpha
f_{2}(z)f_{1}(z+c)+bf_{1}(z+c)^{2}=1,\end{cases}$

must be one of the forms

  1. (i)

$f_{1}(z)=A_{11}\cos[\gamma(z)+b_{1}]-A_{12}\sin[\gamma(z)+b_{1}],\vspace{1.2mm}\\\
f_{2}(z)=A_{11}\cos[\gamma(z)+b_{2}]-A_{12}\sin[\gamma(z)+b_{2}]$,

where $\gamma(z)=L(z)+H(s)$, $L(z)$ is a linear function of the form
$L(z)=\sum_{r=1}^{n}a_{r}z_{r}$, $H(s)$ is a polynomial in
$s(z):=\sum_{r=1}^{n}d_{r}z_{r}$ in $\mathbb{C}$ with
$\sum_{r=1}^{n}d_{r}c_{r}=0$ and $s(z+c)=s(z)$, and $b_{1},b_{2}\in\mathbb{C}$
satisfy

$\displaystyle
e^{2iL(c)}\equiv\frac{R^{2}_{13}}{R^{2}_{11}}\;\;\;\;\mbox{and}\;\;\;\;e^{2i(b_{1}-b_{2})}\equiv\frac{R_{13}}{R_{11}R_{12}},$

where

$\displaystyle
R_{11}=\frac{iA_{11}-A_{12}}{iB_{11}-B_{12}},\;\;R_{12}=\frac{iA_{11}+A_{12}}{iB_{11}-B_{12}}\;\;\mbox{and}\;\;R_{13}=\frac{iB_{11}+B_{12}}{iB_{11}-B_{12}}.$

  1. (ii)

$f_{1}(z)=A_{11}\cos[\gamma(z)+b_{1}]-A_{12}\sin[\gamma(z)+b_{1}],\vspace{1.2mm}\\\
f_{2}(z)=A_{11}\cos[\gamma(z)+b_{2}]+A_{12}\sin[\gamma(z)+b_{2}]$,

where $A_{11},A_{12},\gamma(z)$ are stated as in $(i)$ and
$b_{1},b_{2}\in\mathbb{C}$ satisfy

$\displaystyle
e^{2iL(c)}\equiv\frac{R_{13}}{R_{11}R_{12}}\;\;\;\;\mbox{and}\;\;\;\;e^{2i(b_{1}-b_{2})}\equiv\frac{1}{R^{2}_{11}},$

where $R_{11},R_{12}\;\mbox{and}\;R_{13}$ are stated in $(i)$ of the corollary
2.1.

The existence of solutions to the general quadratic partial differential
equations is studied in the following result. In fact, this result improved
[22, Theorem 2.2] for more general sense.

######  Theorem 2.2.

Let $a,b,C,\alpha,\beta,\gamma$ in $\mathbb{C}$ such that $ab\neq 0$,
$\alpha^{2}\neq 0,ab$ and $\Delta\neq 0$. Then the system of partial
differential equations

$\displaystyle\begin{cases}\mathcal{A}\left(f_{1}(z),\;\dfrac{\partial
f_{2}(z)}{\partial z_{1}}\right)=0,\vspace{2mm}\\\
\mathcal{A}\left(f_{2}(z),\;\dfrac{\partial f_{1}(z)}{\partial
z_{1}}\right)=0\end{cases}$

has no pair of finite order transcendental entire solutions in
$\mathbb{C}^{n}$.

######  Remark 2.3.

We see that the system of equations in Theorem 2.2 does not admit a pair of
finite transcendental entire solutions. However, by replacing $f_{1}(z)$ and
$f_{2}(z)$ with $f_{1}(z+c)$ and $f_{2}(z+c)$ respectively, a pair of finite
order transcendental entire solutions can be obtain for
$c=(c_{1},\ldots,c_{n})\setminus\\{(0,\ldots,0)\\}$. This can be seen in
Theorem 2.3 below.

In the next result, using Nevanlinna theory, we confirm the existence of
solutions to a system of the general quadratic differential-difference
functional equations in $\mathbb{C}^{n}$ and obtain the precise form of those
solutions, which answers Question 2.

######  Theorem 2.3.

Let $c=(c_{1},\ldots,c_{n})\in\mathbb{C}^{n}\setminus\\{(0,\ldots,0)\\}$,
$d_{1}\in\mathbb{C}\setminus\\{0\\}$ and $a,b,C,\alpha,\beta,\gamma$ in
$\mathbb{C}$ with $ab\neq 0$, $\alpha^{2}\neq 0,ab$ and $\Delta\neq 0$. Then
any pair of finite order transcendental entire solutions in $\mathbb{C}^{n}$
for the system of partial differential-difference equations

$\displaystyle\begin{cases}\mathcal{A}\left(f_{1}(z+c),\;\dfrac{\partial
f_{2}(z)}{\partial z_{1}}\right)=0,\vspace{2mm}\\\
\mathcal{A}\left(f_{2}(z+c),\;\dfrac{\partial f_{1}(z)}{\partial
z_{1}}\right)=0\end{cases}$

must be one of the forms

  1. (i)

$f_{1}(z)=D_{11}\cos[L(z)-L(c)+b_{1}]-D_{12}\sin[L(z)-L(c)+b_{1}]+T_{1}$,
$f_{2}(z)=D_{11}\cos[L(z)-L(c)+b_{2}]-D_{12}\sin[L(z)-L(c)+b_{2}]+T_{1}$,

where $L(z)=\sum_{r=1}^{n}a_{r}z_{r}$ and $b_{1},b_{2}\in\mathbb{C}$ satisfy

$\displaystyle
a^{2}_{1}\equiv\frac{R_{23}}{R_{21}R_{22}},\;\;\;e^{2i(b_{1}-b_{2})}\equiv
1,\;\;\;e^{2iL(c)}\equiv\frac{R_{21}}{R_{22}R_{23}}\;\;\;\mbox{and}\;\;\;T_{2}=0$

for

$\displaystyle
R_{21}=\frac{-D_{11}-iD_{12}}{iE_{11}-E_{12}},\;\;R_{22}=\frac{D_{11}-iD_{12}}{iE_{11}-E_{12}},\;\;R_{23}=\frac{iE_{11}+E_{12}}{iE_{11}-E_{12}}.$

  1. (ii)

$f_{1}(z)=D_{11}\cos[L(z)-L(c)+b_{1}]-D_{12}\sin[L(z)-L(c)+b_{1}]+T_{1},\vspace{1.2mm}\\\
f_{2}(z)=D_{11}\cos[L(z)-L(c)+b_{2}]+D_{12}\sin[L(z)-L(c)+b_{2}]+T_{1}$,

$D_{11},D_{12}$ and $L(z)$ are stated as in $(i)$ and
$b_{1},b_{2}\in\mathbb{C}$ satisfy

$\displaystyle
a^{2}_{1}\equiv\frac{R_{23}}{R_{21}R_{22}},\;\;\;e^{2i(b_{1}-b_{2})}\equiv-\frac{R_{22}}{R_{21}R_{23}},\;\;\;e^{2iL(c)}\equiv-1\;\;\;\mbox{and}\;\;\;T_{2}=0,$

where $R_{21},R_{22}\;\mbox{and}\;R_{23}$ are stated in $(i)$.

######  Remark 2.4.

Theorem 2.3 improves several existing results such as [19, Theorem 1.2], [22,
Theorem 2.3] and [23, Theorem 1.3] for a system of general quadratic
differential-difference equations in $\mathbb{C}^{n}$.

By the following example, we confirm that existence of the transcendental
entire solutions with finite order for the system of equation in Theorem 2.3.

######  Example 2.2.

For $c=(\sqrt{3},-2,1)\in\mathbb{C}^{3}$ and
$b_{1}=\left(1+\ln\left((10i+\sqrt{2})/(10i-\sqrt{2})\right)\right)/(2i)$, and
$b_{2}=1/(2i)$, the following functions $(f_{1},f_{2})$, where

$\displaystyle f_{1}(z)$ $\displaystyle=\frac{4\sqrt{3}}{\sqrt{204\pm
44\sqrt{17}}}\cos\left(\frac{\sqrt{2}}{\sqrt{3}}z_{1}-\frac{\pi}{4}z_{2}-\sqrt{2}z_{3}-\frac{\pi}{2}+\frac{1}{2i}\left(1+\ln\left(\frac{10i+\sqrt{2}}{10i-\sqrt{2}}\right)\right)\right)$
$\displaystyle\quad-\frac{\sqrt{3}(1\pm\sqrt{17})}{\sqrt{136\mp
24\sqrt{17}}}\sin\left(\frac{\sqrt{2}}{\sqrt{3}}z_{1}-\frac{\pi}{4}z_{2}-\sqrt{2}z_{3}-\frac{\pi}{2}+\frac{1}{2i}\left(1+\ln\left(\frac{10i+\sqrt{2}}{10i-\sqrt{2}}\right)\right)\right)+\frac{1}{2}$

and

$\displaystyle f_{2}(z)$ $\displaystyle=\frac{4\sqrt{3}}{\sqrt{204\pm
44\sqrt{17}}}\cos\left(\frac{\sqrt{2}}{\sqrt{3}}z_{1}-\frac{\pi}{4}z_{2}-\sqrt{2}z_{3}-\frac{\pi}{2}+\frac{1}{2i}\right)$
$\displaystyle\quad+\frac{\sqrt{3}(1\pm\sqrt{17})}{\sqrt{136\mp
24\sqrt{17}}}\sin\left(\frac{\sqrt{2}}{\sqrt{3}}z_{1}-\frac{\pi}{4}z_{2}-\sqrt{2}z_{3}-\frac{\pi}{2}+\frac{1}{2i}\right)+\frac{1}{2}$

is a pair of transcendental entire solutions in $\mathbb{C}^{3}$ of the system

$\displaystyle\begin{cases}2f_{1}(z+c)^{2}+4f_{1}(z+c)\dfrac{\partial
f_{2}(z)}{\partial z_{1}}+3\left(\dfrac{\partial f_{2}(z)}{\partial
z_{1}}\right)^{2}-2f_{1}(z+c)-2\dfrac{\partial f_{2}(z)}{\partial
z_{1}}=1,\vspace{1.2mm}\\\ 2f_{2}(z+c)^{2}+4f_{2}(z+c)\dfrac{\partial
f_{1}(z)}{\partial z_{1}}+3\left(\dfrac{\partial f_{1}(z)}{\partial
z_{1}}\right)^{2}-2f_{2}(z+c)-2\dfrac{\partial f_{1}(z)}{\partial
z_{1}}=1.\end{cases}$

As a consequence of Theorem 2.3, we derive the following corollary, which
broadens the scope of [22, Theorem 2.3] by allowing for arbitrary constants in
the equations of the system, and also generalize from $\mathbb{C}^{2}$ to
$\mathbb{C}^{n}$ for $n\geq 2$.

######  Corollary 2.2.

Let $c=(c_{1}\ldots,,c_{n})\setminus\\{(0,\ldots,0)\\}\in\mathbb{C}^{n}$,
$d_{1}\in\mathbb{C}\setminus\\{0\\}$ and $a,b,\alpha\in\mathbb{C}$ with
$\alpha^{2}\neq 0,ab$. Then any pair of finite order transcendental entire
solutions in $\mathbb{C}^{n}$ for the system of partial differential-
difference equations

(2.1) $\displaystyle\begin{cases}af_{1}(z+c)^{2}+2\alpha
f_{1}(z+c)\dfrac{\partial f_{2}(z)}{\partial z_{1}}+b\left(\dfrac{\partial
f_{2}(z)}{\partial z_{1}}\right)^{2}=1,\vspace{1.2mm}\\\
af_{2}(z+c)^{2}+2\alpha f_{2}(z+c)\dfrac{\partial f_{1}(z)}{\partial
z_{1}}+b\left(\dfrac{\partial f_{1}(z)}{\partial
z_{1}}\right)^{2}=1,\end{cases}$

must be one of the forms

  1. (i)

$f_{1}(z)=A_{11}\cos[L(z)-L(c)+b_{1}]-A_{12}\sin[L(z)-L(c)+b_{1}]$,
$f_{2}(z)=A_{11}\cos[L(z)-L(c)+b_{2}]-A_{12}\sin[L(z)-L(c)+b_{2}]$,

where $L(z)=\sum_{r=1}^{n}a_{r}z_{r}$ and $b_{1},b_{2}\in\mathbb{C}$ satisfy

$\displaystyle
a^{2}_{1}\equiv\frac{R_{23}}{R_{21}R_{22}},\;\;e^{2i(b_{1}-b_{2})}\equiv
1\;\;\mbox{and}\;\;e^{2iL(c)}\equiv\frac{R_{21}}{R_{22}R_{23}},$
$\displaystyle
R_{21}=\frac{-A_{11}-iA_{12}}{iB_{11}-B_{12}},\;\;R_{22}=\frac{A_{11}-iA_{12}}{iB_{11}-B_{12}},\;\;R_{23}=\frac{iB_{11}+B_{12}}{iB_{11}-B_{12}}.$

Also $A_{11},A_{12},B_{11}\;\mbox{and}\;B_{12}$ are defined in $(i)$ of
Corollary 2.1.

  1. (ii)

$f_{1}(z)=A_{11}\cos[L(z)-L(c)+b_{1}]-A_{12}\sin[L(z)-L(c)+b_{1}],\vspace{1.2mm}\\\
f_{2}(z)=A_{11}\cos[L(z)-L(c)+b_{2}]+A_{12}\sin[L(z)-L(c)+b_{2}]$,

where $A_{11},A_{12}$ and $L(z)$ are stated as in $(i)$ and
$b_{1},b_{2}\in\mathbb{C}$ satisfy

$\displaystyle
a^{2}_{1}\equiv\frac{R_{23}}{R_{21}R_{22}},\;\;e^{2i(b_{1}-b_{2})}\equiv-\frac{R_{22}}{R_{21}R_{23}}\;\;\mbox{and}\;\;e^{2iL(c)}\equiv-1,$

where $R_{21},R_{22}\;\mbox{and}\;R_{23}$ are stated in $(i)$ of Corollary
2.2.

##  3\. Key lemmas and Proof of the main results

In this study, Nevanlinna’s theory for meromorphic functions plays a
significant role in ensuring the existence and form of solutions. We assume
familiarity with the basic Nevanlinna theory (see [25, 24, 8] for details)
such as the counting function $N(r,f)$, proximity function
$m(r,f)=\int_{S_{n}(r)}\log^{+}|f|\eta_{n}$ and Nevanlinna characteristic
function $T(r,f)=m(r,f)+N(r,f)$ for a meromorphic function $f$ in
$|z|<R\leq\infty$ in $\mathbb{C}^{n}$ where
$\eta_{n}=d^{c}\log|z|^{2}\wedge(dd^{c}\log|z|^{2})^{n-1}$, a positive measure
on the sphere $S_{n}(r)=\\{z\in\mathbb{C}^{n}:|z|=r\\}$ with total measure
$1$. We will also use $\sigma_{n}=(dd^{c}|z|^{2})^{n}$, the Lebesgue measure
on the ball $B_{n}(r)=\\{z\in\mathbb{C}^{n}:|z|\leq r\\}$ normalized so that
the total volume of $B_{n}(r)$ is $r^{2n}$. Define the order of $f$ by

$\displaystyle\rho(f):=\overline{\lim_{r\rightarrow\infty}}\dfrac{\log^{+}T(r,f)}{\log
r}.$

First, we recall here some necessary lemmas which will play key roles in
proving the main results of the paper.

######  Lemma A.

(see [15, 18]) For any entire function $F$ on $\mathbb{C}^{n}$, $F(0)\neq 0$
and put $\rho(n_{F})=\rho<\infty$, where $\rho(n_{F})$ denotes be the order of
the counting function of zeros of $F$. Then there exist a canonical function
$f_{F}$ and a function $g_{F}\in\mathbb{C}^{n}$ such that
$F(z)=f_{F}(z)e^{g_{F}(z)}$. For the special case $n=1$, $f_{F}$ is the
canonical product of Weierstrass.

######  Lemma B.

(see [14]) If $g$ and $h$ are entire functions on the complex plane
$\mathbb{C}$ and $g(h)$ is an entire function of finite order, then there are
only two possible cases: either

  1. (i)

the internal function $h$ is a polynomial and the external function $g$ is of
finite order; or

  2. (ii)

the internal function $h$ is not a polynomial but a function of finite order,
and the external function $g$ is of zero order.

As usual, the notation $``||\;P"$ means that the assertion $P$ holds for all
large $r\in[0,\infty)$ outside a set with finite Lebesgue measure.

######  Lemma C.

(see [8]) Let $f_{j}(\not\equiv 0)$, $j=1,2,3$, be meromorphic functions on
$\mathbb{C}^{m}$ such that $f_{1}$ is not constant and $f_{1}+f_{2}+f_{3}=1$
such that

$\displaystyle||\sum_{j=1}^{3}\left\\{N_{2}\left(r,\frac{1}{f_{j}}\right)+2\overline{N}(r,f_{j})\right\\}<\lambda
T(r,f_{1})+O(\log^{+}T(r,f_{1})),$

for all $r$ outside possibly a set with finite logarithmic measure, where
$\lambda<1$ is a positive number. Then either $f_{2}=1\;\mbox{or}\;f_{3}=1$.

For the convenience of the reader, we will present our proof of Theorems 2.1,
2.2, and 2.3 in all detail here.

###### Proof of Theorem 2.1.

The given system of equations in the result is

(3.1) $\displaystyle\begin{cases}\mathcal{A}(f_{1}(z),\;f_{2}(z+c))=0,\\\
\mathcal{A}(f_{2}(z),\;f_{1}(z+c))=0.\end{cases}$

Assume that $(f_{1},f_{2})$ is a pair of finite order transcendental entire
solution of the system $\eqref{eq-3.1}$. Considering $f_{1}=F+T_{1}$ and
$f_{2}=G+T_{2}$, the first equation of the system (3.1) can be expressed to
the following trinomial form

(3.2) $\displaystyle aF^{2}+2\alpha FG+bG^{2}+\dfrac{\Delta}{D}=0.$

Further, we consider that
$F=\mathcal{U}\xi^{\pm}_{1}-\mathcal{V}\eta^{\pm}_{1}$ and
$G=\mathcal{U}\eta^{\pm}_{1}+\mathcal{V}\xi^{\pm}_{1}$. By a tedious
computation, it becomes evident that (3.2) is simplified to the Circular-type
equation

$\displaystyle\left(\sqrt{\frac{DA^{\pm}}{-\Delta}}\;\mathcal{U}\right)^{2}+\left(\sqrt{\frac{DB^{\mp}}{-\Delta}}\;\mathcal{V}\right)^{2}=1.$

Since $f_{1}$ and $f_{2}$ are finite order entire solutions, it is easy to see
that the functions $F$, $G$, $\mathcal{U}$ and $\mathcal{V}$ are also entire
with finite order in $\mathbb{C}^{n}$. Thus, in view of Theorem A, Lemmas A
and B, it follows that there exists a non-constant polynomial
$h_{1}:\mathbb{C}^{n}\rightarrow\mathbb{C}$ such that
$\sqrt{\frac{DA^{\pm}}{-\Delta}}\;\mathcal{U}=\cos h_{1}(z)$ and
$\sqrt{\frac{DB^{\mp}}{-\Delta}}\;\mathcal{V}=\sin h_{1}(z)$. This turns out
that $f_{1}(z)$ and $f_{2}(z+c)$ can be obtained as

(3.3) $\displaystyle f_{1}(z)=D_{11}\cos h_{1}(z)-D_{12}\sin h_{1}(z)+T_{1},$
(3.4) $\displaystyle f_{2}(z+c)=E_{11}\cos h_{1}(z)+E_{12}\sin
h_{1}(z)+T_{2}.$

By the similar argument, in the case of the second equation of the system
$\eqref{eq-3.1}$, there exists a non-constant polynomial $h_{2}(z)$ in
$\mathbb{C}^{n}$ such that $f_{2}(z)$ and $f_{1}(z+c)$ can be expressed as

(3.5) $\displaystyle f_{2}(z)=D_{11}\cos h_{2}(z)-D_{12}\sin h_{2}(z)+T_{1},$
(3.6) $\displaystyle f_{1}(z+c)=E_{11}\cos h_{2}(z)+E_{12}\sin
h_{2}(z)+T_{2}.$

In view of $\eqref{eq-3.2}\;\mbox{and}\;\eqref{eq-3.5}$, a simple computation
shows that

$\displaystyle D_{11}\cos h_{1}(z+c)-D_{12}\sin h_{1}(z+c)+T_{1}=E_{11}\cos
h_{2}(z)+E_{12}\sin h_{2}(Z)+T_{2}.$

This equation can be written as

(3.7) $\displaystyle
R_{11}e^{i(h_{1}(z+c)+h_{2}(z))}+R_{12}e^{i(h_{2}(z)-h_{1}(z+c))}-[R_{13}e^{ih_{2}(z)}+R_{14}]e^{ih_{2}(z)}=1,$

where $R_{11}$, $R_{12}$, $R_{13}$, and $R_{14}$ are defined by

(3.8)
$\displaystyle\begin{cases}R_{11}=\dfrac{iD_{11}-D_{12}}{iE_{11}-E_{12}},\;\;R_{12}=\dfrac{iD_{11}+D_{12}}{iE_{11}-E_{12}},\vspace{2mm}\\\
R_{13}=\dfrac{iE_{11}+E_{12}}{iE_{11}-E_{12}},\;\;R_{14}=\dfrac{2i(T_{2}-T_{1})}{iE_{11}-E_{12}},\end{cases}$

Clearly, (3.7) id of the form $g_{11}(z)+g_{12}(z)+g_{13}(z)=1$, where

$\displaystyle\begin{cases}g_{11}=R_{11}e^{i(h_{1}(z+c)+h_{2}(z))},\vspace{2mm}\\\
g_{12}=R_{12}e^{i(h_{2}(z)-h_{1}(z+c))},\vspace{2mm}\\\
g_{13}=-\left(R_{13}e^{ih_{2}(z)}+R_{14}\right)e^{ih_{2}(z)}.\end{cases}$

It is easy to see that the function
$g_{13}=-\left(R_{13}e^{ih_{2}(z)}+R_{14}\right)e^{ih_{2}(z)}$ is non-
constant, and the functions $g_{1j}$ $(j=1,2,3)$ have neither zeros nor poles.
Consequently, we have the following estimates

$\displaystyle
N_{2}\left(r,\frac{1}{g_{1j}}\right)+2\overline{N}(r,g_{1j})=o(\log^{+}T(r,g_{1j}))\;\mbox{for
each}\;j=1,2,3.$

It is a simple task to verify that

$\sum_{j=1}^{3}\left\\{N_{2}\left(r,\frac{1}{g_{1j}}\right)+2\overline{N}(r,g_{1j})\right\\}=o(\log^{+}T(r,g_{1j})).$

Because of $\lambda T(r,g_{11})>0$, the strict inequality

$||\sum_{j=1}^{3}\left\\{N_{2}\left(r,\frac{1}{g_{1j}}\right)+2\overline{N}(r,g_{1j})\right\\}<\lambda
T(r,g_{11})+O(\log^{+}T(r,g_{11}))$

of Lemma C is satisfied for all $r$ outside possibly a set with finite
logarithmic measure, where $\lambda<1$ is a positive number. Hence, by Lemma
C, we have

$\displaystyle R_{12}e^{i(h_{2}(z)-h_{1}(z+c))}\equiv
1,\;\;\;\;\mbox{or}\;\;\;\;R_{11}e^{i(h_{1}(z+c)+h_{2}(z))}\equiv 1.$

Otherwise, if

$\displaystyle
e^{ih_{2}(z)}={\left(-R_{14}\pm\sqrt{R^{2}_{14}-4K_{1}R_{13}}\right)}/{2R_{13}}$

for arbitrary constants $K_{1}$, then $h_{2}$ must be a constant, leading to a
contradiction. Similarly, using $\eqref{eq-3.3}\;\mbox{and}\;\eqref{eq-3.4}$,
we see that

(3.9) $\displaystyle D_{11}\cos h_{2}(z+c)-D_{12}\sin
h_{2}(z+c)+T_{1}=E_{11}\cos h_{1}(z)+E_{12}\sin h_{1}(z)+T_{2},$

which can be expressed as

(3.10) $\displaystyle
R_{11}e^{i(h_{2}(z+c)+h_{1}(z))}+R_{12}e^{i(h_{1}(z)-h_{2}(z+c))}-\left(R_{13}e^{ih_{1}(z)}+R_{14}\right)e^{ih_{1}(z)}=1.$

The equation (3.10) is the form of $g_{21}(z)+g_{22}(z)+g_{23}(z)=1$, where

$\displaystyle\begin{cases}g_{21}=R_{11}e^{i(h_{2}(z+c)+h_{1}(z))},\vspace{2mm}\\\
g_{22}=R_{12}e^{i(h_{1}(z)-h_{2}(z+c))},\vspace{2mm}\\\
g_{23}=-\left(R_{13}e^{ih_{1}(z)}+R_{14}\right)e^{ih_{1}(z)}.\end{cases}$

We see that $g_{23}=-\left(R_{13}e^{ih_{1}(z)}+R_{14}\right)e^{ih_{1}(z)}$ is
non constant and the functions $g_{2j}$ $(j=1,2,3)$ have no zeros and poles.
Hence, by the similar argument as applied before, we have

$\displaystyle R_{12}e^{i(h_{1}(z)-h_{2}(z+c))}\equiv
1,\;\;\;\;\mbox{or}\;\;\;\;R_{11}e^{i(h_{2}(z+c)+h_{1}(z))}\equiv 1.$

Otherwise, a reasoning applied as before will lead to a contradiction.

To complete the proof, it is enough to discuss the following four possible
cases.

Case I. First, we suppose that

(3.11) $\displaystyle\begin{cases}R_{12}e^{i(h_{2}(z)-h_{1}(z+c))}\equiv
1,\vspace{2mm}\\\ R_{12}e^{i(h_{1}(z)-h_{2}(z+c))}\equiv 1.\end{cases}$

It follows from (3.11) that
$h_{2}(z)-h_{1}(z+c)=\xi_{1}\;\mbox{and}\;h_{1}(z)-h_{2}(z+c)=\xi_{2}$, where
$\xi_{1}\;\mbox{and}\;\xi_{2}$ are two constants in $\mathbb{C}$. By
combining, we have $h_{k}(z)-h_{k}(z+2c)=\xi_{1}+\xi_{2}$; $k=1,2$. Since
$h_{1}\;\mbox{and}\;h_{2}$ are non-constant polynomials in $\mathbb{C}^{n}$,
we conclude that
$h_{1}(z)=\gamma(z)+b_{1}\;\mbox{and}\;h_{2}(z)=\gamma(z)+b_{2}$, where
$\gamma(z)=L(z)+H(s)$, $L(z)$ is a linear function of the form
$L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$, $H(s)$ is a polynomial in
$s(z):=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in $\mathbb{C}$ with
$d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ such that $s(z+c)=s(z)$, and
$b_{1},b_{2}\in\mathbb{C}$. In other words, $\deg_{s}H(s)\geq 2$. Otherwise,
$L(z)+H(s)+b_{i}$ is still a linear form of $z_{1},z_{2}$. A simple
computation using $\eqref{eq-3.6},\eqref{eq-3.9}\;\mbox{and}\;\eqref{eq-3.10}$
shows that

(3.12) $\displaystyle\begin{cases}R_{12}e^{i(-L(c)-b_{1}+b_{2})}\equiv
1,\vspace{2mm}\\\ R_{12}e^{i(-L(c)+b_{1}-b_{2})}\equiv 1,\vspace{2mm}\\\
R_{11}e^{i(L(c)+b_{1}-b_{2})}\equiv R_{13},\vspace{2mm}\\\
R_{11}e^{i(L(c)-b_{1}+b_{2})}\equiv R_{13}.\end{cases}$

Clearly, it follows from $\eqref{eq-3.11}$ that

(3.13) $\displaystyle
e^{2iL(c)}\equiv\frac{R^{2}_{13}}{R^{2}_{11}}\;\;\;\;\mbox{and}\;\;\;\;e^{2i(b_{1}-b_{2})}\equiv\frac{R_{13}}{R_{11}R_{12}}.$

We claim that $T_{1}=T_{2}$. In fact, using (3.7), (3.11) and
$\eqref{eq-3.11}$, a tedious computation shows that $R_{14}=0$ which implies
that $T_{1}=T_{2}.$

Hence, from $\eqref{eq-3.2}$ and $\eqref{eq-3.4}$, we obtain explicit form of
the solutions

$\displaystyle f_{1}(z)$ $\displaystyle=D_{11}\cos h_{1}(z)-D_{12}\sin
h_{1}(z)+T_{1},$
$\displaystyle=D_{11}\cos(\gamma(z)+b_{1})-D_{12}\sin(\gamma(z)+b_{1})+T_{1}$

and

$\displaystyle f_{2}(z)$ $\displaystyle=D_{11}\cos h_{2}(z)-D_{12}\sin
h_{2}(z)+T_{1},$
$\displaystyle=D_{11}\cos(\gamma(z)+b_{2})-D_{12}\sin(\gamma(z)+b_{2})+T_{1}.$

Case II. Suppose that

$\displaystyle\begin{cases}R_{12}e^{i[h_{2}(z)-h_{1}(z+c)]}\equiv
1,\vspace{2mm}\\\ R_{11}e^{i[h_{2}(z+c)+h_{1}(z)]}\equiv 1.\end{cases}$

The above equations clarify that
$h_{2}(z)-h_{1}(z+c)\equiv\xi_{1}\;\mbox{and}\l\;h_{2}(z+c)+h_{1}(z)\equiv\xi_{2}$,
with $\xi_{1}\;\mbox{and}\;\xi_{2}$ being two constants in $\mathbb{C}$.
Therefore, we obtain $h_{2}(z)+h_{2}(z+2c)\equiv\xi_{1}+\xi_{2}$, which
contradicts the assumption that $h_{2}(z)$ is a non-constant polynomial in
$\mathbb{C}^{n}$.

Case III. Suppose that

$\displaystyle\begin{cases}R_{11}e^{i(h_{1}(z+c)+h_{2}(z))}\equiv
1,\vspace{2mm}\\\ R_{12}e^{i(h_{1}(z)-h_{2}(z+c))}\equiv 1.\end{cases}$

The above two equations lead to
$h_{1}(z+c)+h_{2}(z)\equiv\xi_{1}\;\mbox{and}\;h_{1}(z)-h_{2}(z+c)\equiv\xi_{2}$,
where $\xi_{1}\;\mbox{and}\;\xi_{2}$ denote two constants in $\mathbb{C}$.
Thus, we see that $h_{1}(z)+h_{1}(z+2c)\equiv\xi_{1}+\xi_{2}$, which
contradicts that $h_{1}(z)$ is a non-constant polynomial in $\mathbb{C}^{n}$.

Case IV. Let

(3.14) $\displaystyle\begin{cases}R_{11}e^{i(h_{1}(z+c)+h_{2}(z))}\equiv
1,\vspace{2mm}\\\ R_{11}e^{i(h_{2}(z+c)+h_{1}(z))}\equiv 1.\end{cases}$

From $\eqref{eq-3.13}$ it follows that
$h_{1}(z+c)+h_{2}(z)\equiv\xi_{1}\;\mbox{and}\;h_{2}(z+c)+h_{1}(z)\equiv\xi_{2}$,
where $\xi_{1}\;\mbox{and}\;\xi_{2}$ are two constants in $\mathbb{C}$. Thus
$h_{1}(z+2c)-h_{1}(z)\equiv\xi_{1}-\xi_{2}\;\mbox{and}\;h_{2}(z+2c)-h_{2}(z)\equiv\xi_{2}-\xi_{1}$.
Since $h_{1}\;\mbox{and}\;h_{2}$ are non-constant polynomial in
$\mathbb{C}^{2}$ and similar to the argument as in Case I, we can conclude
that
$h_{1}(z)=\gamma(z)+b^{*}_{1}\;\mbox{and}\;h_{2}(z)=-\gamma(z)+b^{*}_{2}$,
where $\gamma(z)=L(z)+H(s)$, $L(z)$ is a linear function of the form
$L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$, $H(s)$ is a polynomial in
$s(z):=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in $\mathbb{C}$ with
$d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ and $s(z+c)=s(z)$, and
$b^{*}_{1},b^{*}_{2}\in\mathbb{C}$. Using
$\eqref{eq-3.6},\eqref{eq-3.9}\;\mbox{and}\;\eqref{eq-3.13}$, an easy
computation shows that

(3.15) $\displaystyle\begin{cases}R_{11}e^{i(L(c)+b^{*}_{1}+b^{*}_{2})}\equiv
1,\vspace{2mm}\\\ R_{11}e^{i(-L(c)+b^{*}_{1}+b^{*}_{2})}\equiv
1,\vspace{2mm}\\\ R_{12}e^{i(-L(c)-b^{*}_{1}-b^{*}_{2})}\equiv
R_{13},\vspace{2mm}\\\ R_{12}e^{i(L(c)-b^{*}_{1}-b^{*}_{2})}\equiv
R_{13}.\end{cases}$

Hence, it follows from $\eqref{eq-3.14}$ that

(3.16) $\displaystyle
e^{2iL(c)}\equiv\frac{R_{13}}{R_{11}R_{12}}\;\;\;\;\mbox{and}\;\;\;\;e^{2i(b^{*}_{1}+b^{*}_{2})}\equiv\frac{1}{R^{2}_{11}}.$

Using (3.10), (3.14) and $\eqref{eq-3.14}$, a tedious computation shows that
$R_{14}=0\implies T_{1}=T_{2}.$ In view of (3.3) and (3.5), we see that

$\displaystyle f_{1}(z)$ $\displaystyle=D_{11}\cos h_{1}(z)-D_{12}\sin
h_{1}(z)+T_{1},$
$\displaystyle=D_{11}\cos(\gamma(z)+b^{*}_{1})-D_{12}\sin(\gamma(z)+b^{*}_{1})+T_{1}$

and

$\displaystyle f_{2}(z)$ $\displaystyle=D_{11}\cos h_{2}(z)-D_{12}\sin
h_{2}(z)+T_{1},$
$\displaystyle=D_{11}\cos(-\gamma(z)+b^{*}_{2})-D_{12}\sin(-\gamma(z)+b^{*}_{2})+T_{1},$
$\displaystyle=D_{11}\cos(\gamma(z)-b^{*}_{2})+D_{12}\sin(\gamma(z)-b^{*}_{2})+T_{1}.$

Letting $b_{1}=b^{*}_{1}\;\mbox{and}\;b_{2}=-b^{*}_{2}$, the solution
$(f_{1},f_{2})$ takes the form

$\displaystyle
f_{1}(z)=D_{11}\cos(\gamma(z)+b_{1})-D_{12}\sin(\gamma(z)+b_{1})+T_{1},$
$\displaystyle
f_{2}(z)=D_{11}\cos(\gamma(z)+b_{2})+D_{12}\sin(\gamma(z)+b_{2})+T_{1}.$

That completes the proof. ∎

###### Proof of Theorem 2.2.

The system of equations in the result is

(3.17) $\displaystyle\begin{cases}\mathcal{A}\left(f_{1}(z),\;\dfrac{\partial
f_{2}(z)}{\partial z_{1}}\right)=0,\vspace{2mm}\\\
\mathcal{A}\left(f_{2}(z),\;\dfrac{\partial f_{1}(z)}{\partial
z_{1}}\right)=0\end{cases}$

We assume that $(f_{1},f_{2})$ is a pair of finite order transcendental entire
solutions of the system $\eqref{eq-3.16}$. Considering $f_{1}=F+T_{1}$ and
$\frac{\partial f_{2}}{\partial z_{1}}=G+T_{2}$ in the first equation of the
system (3.17), and then by using the analogous reasoning utilized in the proof
of Theorem 2.1, it follows that there is a non-constant polynomial
$h_{1}:\mathbb{C}^{n}\rightarrow\mathbb{C}$ such that

(3.18) $\displaystyle f_{1}(z)=D_{11}\cos h_{1}(z)-D_{12}\sin h_{1}(z)+T_{1},$
(3.19) $\displaystyle\dfrac{\partial f_{2}(z)}{\partial z_{1}}=E_{11}\cos
h_{1}(z)+E_{12}\sin h_{1}(z)+T_{2}.$

Similarly, for the second equation of $\eqref{eq-3.16}$, there exists a non-
constant polynomial $h_{2}(z)$ in $\mathbb{C}^{n}$ such that

(3.20) $\displaystyle f_{2}(z)=D_{11}\cos h_{2}(z)-D_{12}\sin h_{2}(z)+T_{1},$
(3.21) $\displaystyle\dfrac{\partial f_{1}(z)}{\partial z_{1}}=E_{11}\cos
h_{2}(z)+E_{12}\sin h_{2}(z)+T_{2}.$

Using $\eqref{eq-3.17}\;\mbox{and}\;\eqref{eq-3.20}$, an easy computation
shows that

$\displaystyle\left(-D_{11}\sin h_{1}(z)-D_{12}\cos
h_{1}(z)\right)\dfrac{\partial h_{1}(z)}{\partial z_{1}}=E_{11}\cos
h_{2}(z)+E_{12}\sin h_{2}(z)+T_{2}$

which turns out that

(3.22) $\displaystyle R_{21}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{1}(z)+h_{2}(z))}+R_{22}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{2}(z)-h_{1}(z))}-(R_{23}e^{ih_{2}(z)}+R_{24})e^{ih_{2}(z)}=1,$

where

(3.23)
$\displaystyle\begin{cases}R_{21}=\dfrac{-D_{11}-iD_{12}}{iE_{11}-E_{12}},\;\;R_{22}=\dfrac{D_{11}-iD_{12}}{iE_{11}-E_{12}},\vspace{2mm}\\\
R_{23}=\dfrac{iE_{11}+E_{12}}{iE_{11}-E_{12}},\;\;R_{24}=\dfrac{2iT_{2}}{iE_{11}-E_{12}}.\end{cases}$

If $\frac{\partial h_{1}(z)}{\partial z_{1}}\equiv 0$, then it follows from
$\eqref{eq-3.21}$ that
$\left(R_{23}e^{ih_{2}(z)}+R_{24}\right)e^{ih_{2}(z)}\equiv 1$, which implies
that $h_{2}$ is a constant, a contradiction. Therefore, we must have
$\frac{\partial h_{1}(z)}{\partial z_{1}}\not\equiv 0$. We see that equation
(3.22) is of the form $g_{31}+g_{32}+g_{33}=1$, where

$\displaystyle\begin{cases}g_{31}=R_{21}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{1}(z)+h_{2}(z))},\vspace{2mm}\\\ g_{32}=R_{22}\dfrac{\partial
h_{1}(z)}{\partial z_{1}}e^{i(h_{2}(z)-h_{1}(z))},\vspace{2mm}\\\
g_{33}=-(R_{23}e^{ih_{2}(z)}+R_{24})e^{ih_{2}(z)}.\end{cases}$

Then, using (3.22), in view of Lemma C, we obtain

$\displaystyle R_{22}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{2}(z)-h_{1}(z))}\equiv
1,\;\;\;\;\mbox{or}\;\;\;\;R_{21}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{1}(z)+h_{2}(z))}\equiv 1.$

Moreover, from (3.19) and (3.20), an computation shows that

$\displaystyle\left(-D_{11}\sin h_{2}(z)-D_{12}\cos
h_{2}(z)\right)\dfrac{\partial h_{2}(z)}{\partial z_{1}}=E_{11}\cos
h_{1}(z)+E_{12}\sin h_{1}(z)+T_{2},$

which can be written as

(3.24) $\displaystyle R_{21}\dfrac{\partial h_{2}(z)}{\partial
z_{1}}e^{i(h_{2}(z)+h_{1}(z))}+R_{22}\dfrac{\partial h_{2}(z)}{\partial
z_{1}}e^{i(h_{1}(z)-h_{2}(z))}-(R_{23}e^{ih_{1}(z)}+R_{24})e^{ih_{1}(z)}=1.$

If $\frac{\partial h_{2}(z)}{\partial z_{1}}\equiv 0$, then it follows from
$\eqref{eq-3.23}$ that
$\left(R_{23}e^{ih_{1}(z)}+R_{24}\right)e^{ih_{1}(z)}\equiv 1$, which implies
that $h_{2}$ is a constant, a contradiction. Therefore, we must have
$\frac{\partial h_{2}(z)}{\partial z_{1}}\not\equiv 0$. By Lemma C, we easily
obtain from (3.24) that

$\displaystyle R_{22}\dfrac{\partial h_{2}(z)}{\partial
z_{1}}e^{i(h_{1}(z)-h_{2}(z))}\equiv
1,\;\;\;\;\mbox{or}\;\;\;\;R_{21}\dfrac{\partial h_{2}(z)}{\partial
z_{1}}e^{i(h_{2}(z)+h_{1}(z))}\equiv 1.$

To conclude the proof, we only need to consider the subsequent cases.

Case A. Suppose that

(3.25) $\displaystyle\begin{cases}R_{22}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{2}(z)-h_{1}(z))}\equiv 1\vspace{2mm}\\\ R_{22}\dfrac{\partial
h_{2}(z)}{\partial z_{1}}e^{i(h_{1}(z)-h_{2}(z))}\equiv 1\end{cases}$

By using these two equations, we can easily deduce that $h_{1}(z)-h_{2}(z)$ is
equal to a constant in $\mathbb{C}$, which we denote as $\lambda_{1}$. Hence,
both $\frac{\partial h_{1}(z)}{\partial z_{1}}\;\mbox{and}\;\frac{\partial
h_{2}(z)}{\partial z_{1}}$ must be constants. Assume that $\frac{\partial
h_{1}(z)}{\partial z_{1}}\equiv c_{0}$. Since $h_{1}$ and $h_{2}$ are non-
constant polynomials in $\mathbb{C}^{n}$, hence
$h_{1}(z)=c_{0}z_{1}+p(z_{2},\ldots,z_{n})+b_{1}$, where
$p(z_{2},\ldots,z_{n})$ is a polynomial in $z_{2},\ldots,z_{n}$ only in
$\mathbb{C}^{n}$ and $b_{1}\in\mathbb{C}$. Since
$h_{1}(z)-h_{2}(z)=\lambda_{1}$, we see that
$h_{2}(z)=c_{0}z_{1}+p(z_{2},\ldots,z_{n})+b_{2}$, where $b_{2}\in\mathbb{C}$.
Thus, in view of
$\eqref{eq-3.21},\eqref{eq-3.23}\;\mbox{and}\;\eqref{eq-3.24}$, we obtain

(3.26) $\displaystyle\begin{cases}R_{22}c_{0}e^{i(b_{2}-b_{1})}\equiv
1,\vspace{2mm}\\\ R_{22}c_{0}e^{i(b_{1}-b_{2})}\equiv 1,\vspace{2mm}\\\
R_{21}c_{0}e^{i(b_{1}-b_{2})}\equiv R_{23},\vspace{2mm}\\\
R_{21}c_{0}e^{i(b_{2}-b_{1})}\equiv R_{23}.\end{cases}$

A simple computation using the equations in $\eqref{eq-3.25}$ shows that

$\displaystyle
c_{0}^{2}=\frac{R_{23}}{R_{21}R_{22}}\;\;\;\;\;\mbox{and}\;\;\;\;R_{22}R_{23}=R_{21}.$

Thus we have $2i\xi^{\pm}_{1}\eta^{\pm}_{1}(A^{\pm}-B^{\mp})=0$ which implies
that $A^{\pm}=B^{\mp}$, hence

$\displaystyle+\sqrt{(a-b)^{2}+4\alpha^{2}}=-\sqrt{(a-b)^{2}+4\alpha^{2}}\;\;\mbox{or}\;\;-\sqrt{(a-b)^{2}+4\alpha^{2}}=+\sqrt{(a-b)^{2}+4\alpha^{2}}$

Therefore, both the equations yield that $2\sqrt{(a-b)^{2}+4\alpha^{2}}=0$.
Consequently, it follows that $a=b$ and $\alpha=0$, which contradicts
$\alpha^{2}\neq ab$.

Case B. Let

$\displaystyle\begin{cases}R_{22}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{2}(z)-h_{1}(z))}\equiv 1,\vspace{2mm}\\\ R_{21}\dfrac{\partial
h_{2}(z)}{\partial z_{1}}e^{i(h_{2}(z)+h_{1}(z))}\equiv 1.\end{cases}$

By the similar argument used in Case A, we see that
$h_{2}(z)-h_{1}(z)=\lambda_{1}$ and $h_{2}(z)+h_{1}(z)=\lambda_{2}$, where
$\lambda_{1}$ and $\lambda_{2}$ are two constants in $\mathbb{C}$. Thus it
follows that $2h_{2}(z)=\lambda_{1}+\lambda_{2}$, a constant, which
contradicts the fact that $h_{2}(z)$ is non-constant.

Case C. Suppose that

$\displaystyle\begin{cases}R_{21}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{1}(z)+h_{2}(z))}\equiv 1,\vspace{2mm}\\\ R_{22}\dfrac{\partial
h_{2}(z)}{\partial z_{1}}e^{i(h_{1}(z)-h_{2}(z))}\equiv 1.\end{cases}$

The argument used in Case B can be applied similarly to lead to a
contradiction.

Case D. Let

(3.27) $\displaystyle\begin{cases}R_{21}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{1}(z)+h_{2}(z))}\equiv 1,\vspace{2mm}\\\ R_{21}\dfrac{\partial
h_{2}(z)}{\partial z_{1}}e^{i(h_{2}(z)+h_{1}(z))}\equiv 1.\end{cases}$

In view of the equations in $\eqref{eq-3.26}$, it follows that
$h_{1}(z)+h_{2}(z)\equiv\mbox{constant}$, and hence both $\frac{\partial
h_{1}(z)}{\partial z_{1}}\;\mbox{and}\;\frac{\partial h_{2}(z)}{\partial
z_{1}}$ are constants in $\mathbb{C}$. By using the same reasoning as in Case
A, we can readily obtain $h_{1}(z)=d_{0}z_{1}+q(z_{2},\ldots,z_{n})+b_{1}$ and
$h_{2}(z)=-d_{0}z_{1}-q(z_{2},\ldots,z_{n})+b_{2}$, where
$q(z_{2},\ldots,z_{n})$ is a polynomial of $z_{2},\ldots,z_{n}$ only in
$\mathbb{C}^{n}$ and $b_{1},b_{2}\in\mathbb{C}$. Thus, it follows from
$\eqref{eq-3.26}$ that

$\displaystyle\begin{cases}R_{21}re^{i(b_{1}+b_{2})}\equiv 1,\vspace{2mm}\\\
-R_{21}re^{i(b_{1}+b_{2})}\equiv 1,\end{cases}$

from which we obtain that $-1\equiv 1$, which is absurd. This completes the
proof. ∎

###### Proof of Theorem 2.3.

The given system of partial differential-difference equations is

(3.28)
$\displaystyle\begin{cases}\mathcal{A}\left(f_{1}(z+c),\;\dfrac{\partial
f_{2}(z)}{\partial z_{1}}\right)=0,\vspace{2mm}\\\
\mathcal{A}\left(f_{2}(z+c),\;\dfrac{\partial f_{1}(z)}{\partial
z_{1}}\right)=0.\end{cases}$

Assume that $(f_{1},f_{2})$ is a pair of finite order transcendental entire
solutions of the system $\eqref{eq-3.27}$. We consider $f_{1}=F+T_{1}$ and
$\frac{\partial f_{2}}{\partial z_{1}}=G+T_{2}$, in the first equation of the
system (3.28) and by the similar arguments being used in the proof of Theorem
2.1, it follows that there exists a non-constant polynomial
$h_{1}:\mathbb{C}^{n}\rightarrow\mathbb{C}$ such that

(3.29) $\displaystyle f_{1}(z+c)=D_{11}\cos h_{1}(z)-D_{12}\sin
h_{1}(z)+T_{1},$ (3.30) $\displaystyle\dfrac{\partial f_{2}(z)}{\partial
z_{1}}=E_{11}\cos h_{1}(z)+E_{12}\sin h_{1}(z)+T_{2}.$

Similarly, for the second equation of the system $\eqref{eq-3.27}$, there
exists a non-constant polynomial $h_{2}(z)$ in $\mathbb{C}^{n}$ such that

(3.31) $\displaystyle f_{2}(z+c)=D_{11}\cos h_{2}(z)-D_{12}\sin
h_{2}(z)+T_{1},$ (3.32) $\displaystyle\dfrac{\partial f_{1}(z)}{\partial
z_{1}}=E_{11}\cos h_{2}(z)+E_{12}\sin h_{2}(z)+T_{2}.$

From $\eqref{eq-3.28}\;\mbox{and}\;\eqref{eq-3.31}$ we can deduce that

$\displaystyle\left(-D_{11}\sin h_{1}(z)-D_{12}\cos h_{1}(z)\right)$
$\displaystyle\dfrac{\partial h_{1}(z)}{\partial z_{1}}$
$\displaystyle=E_{11}\cos h_{2}(z+c)+E_{12}\sin h_{2}(z+c)+T_{2},$

which can be written as

(3.33) $\displaystyle R_{21}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{1}(z)+h_{2}(z+c))}+R_{22}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}$ $\displaystyle e^{i(h_{2}(z+c)-h_{1}(z))}$
$\displaystyle\quad-(R_{23}e^{ih_{2}(z+c)}+R_{24})e^{ih_{2}(z+c)}=1,$

where

(3.34)
$\displaystyle\begin{cases}R_{21}=\dfrac{-D_{11}-iD_{12}}{iE_{11}-E_{12}},\;\;R_{22}=\dfrac{D_{11}-iD_{12}}{iE_{11}-E_{12}},\vspace{2mm}\\\
R_{23}=\dfrac{iE_{11}+E_{12}}{iE_{11}-E_{12}},\;\;R_{24}=\dfrac{2iT_{2}}{iE_{11}-E_{12}}.\end{cases}$

However, based on $\eqref{eq-3.29}\;\mbox{and}\;\eqref{eq-3.30}$, it is
evident that

$\displaystyle\left(-D_{11}\sin h_{2}(z)-D_{12}\cos h_{2}(z)\right)$
$\displaystyle\dfrac{\partial h_{2}(z)}{\partial z_{1}}$
$\displaystyle=E_{11}\cos h_{1}(z+c)+E_{12}\sin h_{1}(z+c)+T_{2}.$

It follows that

(3.35) $\displaystyle R_{21}\dfrac{\partial h_{2}(z)}{\partial
z_{1}}e^{i(h_{1}(z+c)+h_{2}(z))}+R_{22}\dfrac{\partial h_{2}(z)}{\partial
z_{1}}$ $\displaystyle e^{i(h_{1}(z+c)-h_{2}(z))}$
$\displaystyle\quad-(R_{23}e^{ih_{1}(z+c)}+R_{24})e^{ih_{1}(z+c)}=1,$

where $R_{21},R_{22},R_{23},\;\mbox{and}\;R_{24}$ are non-zero constants as
defined in $\eqref{eq-3.33}$. We see that $\frac{\partial h_{1}}{\partial
z_{1}}\not\equiv 0$ and $\frac{\partial h_{2}}{\partial z_{1}}\not\equiv 0$.
In view of Lemma C and it yields from $\eqref{eq-3.32},\eqref{eq-3.34}$ that

$\displaystyle R_{22}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{2}(z+c)-h_{1}(z))}\equiv
1,\;\;\;\;\mbox{or}\;\;\;\;R_{21}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{1}(z)+h_{2}(z+c))}\equiv 1$

and

$\displaystyle R_{22}\dfrac{\partial h_{2}(z)}{\partial
z_{1}}e^{i(h_{1}(z+c)-h_{2}(z))}\equiv
1,\;\;\;\;\mbox{or}\;\;\;\;R_{21}\dfrac{\partial h_{2}(z)}{\partial
z_{1}}e^{i(h_{1}(z+c)+h_{2}(z))}\equiv 1.$

To complete the proof, we need only to discuss the following possible cases.

Case 1. Let

(3.36) $\displaystyle\begin{cases}R_{22}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{2}(z+c)-h_{1}(z))}\equiv 1\vspace{2mm}\\\ R_{22}\dfrac{\partial
h_{2}(z)}{\partial z_{1}}e^{i(h_{1}(z+c)-h_{2}(z))}\equiv 1\end{cases}$

Hence, it follows that $h_{2}(z+c)-h_{1}(z)\equiv\xi_{1}$ and
$h_{1}(z+c)-h_{2}(z)\equiv\xi_{2}$, where $\xi_{1}$ and $\xi_{2}$ are two
constant in $\mathbb{C}$. This implies that
$h_{k}(z+2c)-h_{k}(z)\equiv\xi_{1}+\xi_{2}$ for $k=1,2$. We claim that
$h_{1}(z)=L(z)+b_{1}$ and $h_{2}(z)=L(z)+b_{2}$, where $L(z)$ is a linear
function as the form $L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$ and $b_{1},b_{2}$ are
two constant in $\mathbb{C}$. In fact, since $h_{1},h_{2}$ are two non-
constant polynomials and $h_{k}(z+2c)-h_{k}(z)$ are two constants, we conclude
that $h_{1}(z)=L(z)+H(s)+b_{1}$ and $h_{2}(z)=L(z)+H(s)+b_{2}$, where $H(s)$
is a polynomial in $s(z):=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in $\mathbb{C}$ with
$d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ and $s(z+c)=s(z)$. Now from $\eqref{eq-3.35}$
it is easy to see that $R_{22}(d_{1}H^{\prime}+a_{1})$ must be constant in
$\mathbb{C}$. By combining with $d_{1}\neq 0$, we obtain $\deg_{s}H\leq 1$.
Thus $L(z)+H(s)$ is still a linear form of $z_{1},\ldots,z_{n}$. Hence
$h_{1}(z)=L(z)+b_{1}$ and $h_{2}(z)=L(z)+b_{2}$. A simple computation using
$\eqref{eq-3.32},\eqref{eq-3.34}\;\mbox{and}\;\eqref{eq-3.35}$ shows that

$\displaystyle\begin{cases}R_{22}a_{1}e^{i(L(c)-b_{1}+b_{2})}\equiv
1,\vspace{2mm}\\\ R_{22}a_{1}e^{i(L(c)+b_{1}-b_{2})}\equiv 1,\vspace{2mm}\\\
R_{21}a_{1}e^{i(-L(c)+b_{1}-b_{2})}\equiv R_{23},\vspace{2mm}\\\
R_{21}a_{1}e^{i(-L(c)-b_{1}+b_{2})}\equiv R_{23},\end{cases}$

which implies that

$\displaystyle
a^{2}_{1}\equiv\frac{R_{23}}{R_{21}R_{22}},\;\;e^{2i(b_{1}-b_{2})}\equiv
1\;\;\mbox{and}\;\;e^{2iL(c)}\equiv\frac{R_{21}}{R_{22}R_{23}},\;\;\mbox{and}\;\;R_{24}=0\implies
T_{2}=0.$

In view of $\eqref{eq-3.28}$ and $\eqref{eq-3.30}$, it is easy to see that

$\displaystyle f_{1}(z)$ $\displaystyle=D_{11}\cos h_{1}(z-c)-D_{12}\sin
h_{1}(z-c)+T_{1},$
$\displaystyle=D_{11}\cos(L(z)-L(c)+b_{1})-D_{12}\sin(L(z)-L(c)+b_{1})+T_{1}$

and

$\displaystyle f_{2}(z)$ $\displaystyle=D_{11}\cos h_{2}(z-c)-D_{12}\sin
h_{2}(z-c)+T_{1},$
$\displaystyle=D_{11}\cos(L(z)-L(c)+b_{2})-D_{12}\sin(L(z)-L(c)+b_{2})+T_{1}.$

Case 2. Suppose that

$\displaystyle\begin{cases}R_{22}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{2}(z+c)-h_{1}(z))}\equiv 1,\vspace{2mm}\\\ R_{21}\dfrac{\partial
h_{2}(z)}{\partial z_{1}}e^{i(h_{1}(z+c)+h_{2}(z))}\equiv 1,\end{cases}$

The above equations yield that $h_{2}(z+c)-h_{1}(z)\equiv\xi_{1}$ and
$h_{1}(z+c)+h_{2}(z)\equiv\xi_{2}$, where $\xi_{1},\xi_{2}$ are two constants
in $\mathbb{C}$. It is easy to see that
$h_{2}(z+2c)+h_{2}(z)\equiv\xi_{1}+\xi_{2}$, which is a contradiction as
$h_{2}(z)$ is a non-constant polynomial in $\mathbb{C}^{n}$.

Case 3. Let

$\displaystyle\begin{cases}R_{21}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{2}(z+c)+h_{1}(z))}\equiv 1,\vspace{2mm}\\\ R_{22}\dfrac{\partial
h_{2}(z)}{\partial z_{1}}e^{i(h_{1}(z+c)-h_{2}(z))}\equiv 1.\end{cases}$

By using the similar argument, we see that $h_{2}(z+c)+h_{1}(z)\equiv\xi_{1}$
and $h_{1}(z+c)-h_{2}(z)\equiv\xi_{2}$, where $\xi_{1},\xi_{2}$ are two
constants in $\mathbb{C}$. It is easy to see that
$h_{1}(z+2c)+h_{1}(z)\equiv\xi_{1}+\xi_{2}$, which is a contradiction as
$h_{1}(z)$ is a non-constant polynomial in $\mathbb{C}^{n}$.

Case 4. Suppose that

(3.37) $\displaystyle\begin{cases}R_{21}\dfrac{\partial h_{1}(z)}{\partial
z_{1}}e^{i(h_{2}(z+c)+h_{1}(z))}\equiv 1,\vspace{2mm}\\\ R_{21}\dfrac{\partial
h_{2}(z)}{\partial z_{1}}e^{i(h_{1}(z+c)+h_{2}(z))}\equiv 1.\end{cases}$

Therefore, it follows that $h_{2}(z+c)+h_{1}(z)\equiv\xi_{1}$ and
$h_{1}(z+c)+h_{2}(z)\equiv\xi_{2}$, where $\xi_{1}$ and $\xi_{2}$ are two
constants in $\mathbb{C}$. This implies that
$h_{1}(z+2c)-h_{1}(z)\equiv\xi_{2}-\xi_{1}$ and
$h_{2}(z+2c)-h_{2}(z)\equiv\xi_{1}-\xi_{2}$. Now we claim that
$h_{1}(z)=L(z)+b^{*}_{1}$ and $h_{2}(z)=-L(z)+b^{*}_{2}$, where $L(z)$ is a
linear function as the form $L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$ and
$b^{*}_{1},b^{*}_{2}$ are two constants in $\mathbb{C}$. In fact, since
$h_{1}(z),h_{2}(z)$ are two non-constant polynomials and
$h_{k}(z+2c)-h_{k}(z)$ are two constant, we conclude that
$h_{1}(z)=L(z)+H(s)+b^{*}_{1}$ and $h_{2}(z)=-L(z)-H(s)+b^{*}_{2}$, where
$H(s)$ is a polynomial in $s(z):=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in $\mathbb{C}$
with $d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ and $s(z+c)=s(z)$. Now from
$\eqref{eq-3.36}$ it is easy to see that $R_{21}(d_{1}H^{\prime}+a_{1})$ must
be constant in $\mathbb{C}$. By combining with $d_{1}\neq 0$, we obtain
$\deg_{s}H\leq 1$. Thus $L(z)+H(s)$ is still a linear form of
$z_{1},\ldots,z_{n}$. Hence $h_{1}(z)=L(z)+b^{*}_{1}$ and
$h_{2}(z)=-L(z)+b^{*}_{2}$. Therefore from
$\eqref{eq-3.32},\eqref{eq-3.34}\;\mbox{and}\;\eqref{eq-3.36}$, it follows
that

$\displaystyle\begin{cases}R_{21}a_{1}e^{i(-L(c)+b^{*}_{1}+b^{*}_{2})}\equiv
1,\vspace{2mm}\\\ -R_{21}a_{1}e^{i(L(c)+b^{*}_{1}+b^{*}_{2})}\equiv
1,\vspace{2mm}\\\ R_{22}a_{1}e^{i(L(c)-b^{*}_{1}-b^{*}_{2})}\equiv
R_{23},\vspace{2mm}\\\ -R_{22}a_{1}e^{i(-L(c)-b^{*}_{1}-b^{*}_{2})}\equiv
R_{23}.\end{cases}$

Thus it follows that

$\displaystyle
a^{2}_{1}\equiv\frac{R_{23}}{R_{21}R_{22}},\;\;e^{2i(b^{*}_{1}+b^{*}_{2})}\equiv-\frac{R_{22}}{R_{21}R_{23}}\;\;\mbox{and}\;\;e^{2iL(c)}\equiv-1,\;\;\mbox{and}\;\;R_{24}=0\implies
T_{2}=0.$

In view of $\eqref{eq-3.28}$ and $\eqref{eq-3.30}$, a simple computation shows
that

$\displaystyle f_{1}(z)$ $\displaystyle=D_{11}\cos h_{1}(z-c)-D_{12}\sin
h_{1}(z-c)+T_{1},$
$\displaystyle=D_{11}\cos(L(z)-L(c)+b^{*}_{1})-D_{12}\sin(L(z)-L(c)+b^{*}_{1})+T_{1}$

and

$\displaystyle f_{2}(z)$ $\displaystyle=D_{11}\cos h_{2}(z-c)-D_{12}\sin
h_{2}(z-c)+T_{1},$
$\displaystyle=D_{11}\cos(-L(z)+L(c)+b^{*}_{2})-D_{12}\sin(-L(z)+L(c)+b^{*}_{2})+T_{1},$
$\displaystyle=D_{11}\cos(L(z)-L(c)-b^{*}_{2})+D_{12}\sin(L(z)-L(c)-b^{*}_{2})+T_{1}.$

Assuming $b_{1}=b^{*}_{1}$ and $b_{2}=-b^{*}_{2}$, we see that
$e^{2i(b_{1}-b_{2})}\equiv-{R_{22}}/{R_{21}R_{23}}$. Hence, the solutions take
the form

$\displaystyle
f_{1}(z)=D_{11}\cos(L(z)-L(c)+b_{1})-D_{12}\sin(L(z)-L(c)+b_{1})+T_{1},$
$\displaystyle
f_{2}(z)=D_{11}\cos(L(z)-L(c)+b_{2})+D_{12}\sin(L(z)-L(c)+b_{2})+T_{1}.$

This completes the proof. ∎

Compliance of Ethical Standards:

Conflict of interest. The authors declare that there is no conflict of
interest regarding the publication of this paper.

Data availability statement. Data sharing is not applicable to this article as
no datasets were generated or analyzed during the current study.

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