††footnotetext:  File: main.tex, printed: 2024-8-28, 3.28

# Characterization of solutions of refined Fermat-type functional 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>
###### Abstract.

The main purpose of this article is concerned with the existence and the
precise forms of the transcendental solutions of several refined versions of
Fermat-type functional equations with polynomial coefficients in several
complex variables by utilizing the Nevanlinna theory of meromorphic functions
in several complex variables. In fact, we investigate the existence and forms
of the transcendental solutions of non-linear quadratic trinomial equations in
$\mathbb{C}^{n}$. As a consequence of our result, we show that solutions of
binomial equations in $\mathbb{C}^{n}$ can be explored and this exploration
broaden the scope of the study of functional equations in $\mathbb{C}^{n}$.
The results we obtained are improvements over certain recent findings, noted
as remarks. In addition, some examples relevant to the content of the paper
have been exhibited in support of the validation of each results. Further, we
discuss situations when solutions of such equations does not exist.

###### Key words and phrases:

Transcendental entire solutions, Nevanlinna theory, Fermat-type functional
equations, functions in several complex variables, finite order, Partial
differential-difference equations

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

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

###### Contents

  1. 1 Introduction
  2. 2 Description of the solutions of Fermat-type functional equations in $\mathbb{C}^{n}$
  3. 3 Main results, corollaries and remarks
    1. 3.1 Some examples in support of Remark 3.5
  4. 4 Key lemmas and Proof of the main results
    1. 4.1 Proof of Theorem 3.1
    2. 4.2 Proof of Theorem 3.2
    3. 4.3 Proof of Theorem 3.3
  5. 5 Declaration

##  1\. Introduction

It is well known that, the complex oscillation theory of meromorphic solutions
of differential equations is an important topic in complex analysis. Some
results can be found in [22], where the Nevanlinna theory is an effective
research tool. Recently, many results on meromorphic solutions of complex
difference equations were rapidly obtained, such as [8, 9] and so on. In
recent years, an extensive study of finding finite order entire and
meromorphic solutions both in one and several complex variables of the
functional equation

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

what is called Fermat-type functional equations. In this study, Nevanlinna
value distribution theory for meromorphic functions plays a significant role.
In fact, the study of Fermat-type functional equations has a long history. In
$1939$, Iyer [20] studied the entire solutions of (1.1) for $n=2$ and showed
that the only solution is of the form $f(z)=\cos(\varphi(z))$ and
$g(z)=\sin(\varphi(z))$, where $\varphi$ is an entire function. Inspired by
the results in [20], many scholars paid considerable attention to
investigating the existence and form of entire and meromorphic solutions of
some variations of equations. We refer to the articles [43, 40, 42] and
references therein for some recent development on the topic. In the past
decade, with the establishment of difference analogues of Nevanlinna theory
(see e.g., [4, 8, 16]), many researchers studied the solutions of Fermat type
difference and also Fermat-type differential-difference equations (see [28,
34, 18]). Here, an equation is called differential-difference equation, if it
includes derivatives, shifts or differences of $f(z)$, which can be called
$DDE$ for short (see [31]). As far as we know, there are few results on
solutions of complex partial differential-difference equations.

The Fermat type equation (1.1) appears in particle mechanics, in particular,
it appears as the Lagrange function in the Lagrangian functional describing
the “action” of a system. Finding the precise form of the solution with the
help of Nevanlinna theory is an active area of research in function theory and
researchers are paid considerable attention to this topic in recent years. For
example, Liu _et al._ [27] have investigated the Fermat-type difference
equation $f^{2}(z)+f^{2}(z+c)=1$ in $\mathbb{C}$ and obtained the
transcendental entire solutions of finite order $f(z)=\sin(Az+B)$, where $B$
is a constant and $A=((4k+1)\pi)/(2c)$, where $k$ is an integer. Later, Han
and Lü [17] established the solution to the more general complex difference
equation $f^{n}(z)+g^{n}(z)=e^{\alpha z+\beta}$. Moreover, Liu _et al._ [27]
showed that the existence of solutions for the complex differential-difference
equations ${f^{\prime}}^{2}(z)+f(z+c)^{2}=1$ and
${f^{\prime}}^{2}(z)+(f(z+c)-f(z))^{2}=1$ in $\mathbb{C}$.

We know that the partial differential equations (PDEs) are occurring in
various areas of applied mathematics, such as fluid mechanics, nonlinear
acoustics, gas dynamics, and traffic flow (see [10, 12]). In general, it is
difficult to find entire and meromorphic solutions for a nonlinear PDE. By
employing Nevanlinna theory and the method of complex analysis, there were a
number of literature focusing on the solutions of some PDEs and their many
variants, readers can refer to [6, 21, 25, 15, 38]. The solutions of Fermat-
type $PDEs$ were investigated by [23, 36, 18]. Most noticeably, in $1995$,
Khavinson [21] derived that any entire solution of the partial differential
equation in $\mathbb{C}^{2}$,

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

is necessarily linear, i.e., $u(z_{1},z_{2})=az_{1}+bz_{2}+c$, where
$a,b,c\in\mathbb{C}$, and $a^{2}+b^{2}=1$. This $PDE$ in the real variable
case occurs in the study of characteristic surfaces and wave propagation
theory, and it is the two-dimensional eiconal equation, one of the main
equations of geometric optics (see [10]). A more general equation of the form
$F^{2}+G^{2}=e^{p(z)}$, where $p(z)$ is a polynomial has been studied recently
to find the solutions. Due to the development of the logarithmic derivative
lemma, in recent years, functional equations involving derivatives with shift
or differences operators of meromorphic functions are studied extensively.
Furthermore, Li [24, 25] have continued the research and discussed solutions
of a series of $PDEs$ with more general forms including $\left({\partial
f}/{\partial z_{1}}\right)^{2}+\left({\partial f}/{\partial
z_{2}}\right)^{2}=e^{g}$, $\left({\partial f}/{\partial
z_{1}}\right)^{2}+\left({\partial f}/{\partial z_{2}}\right)^{2}=p$, etc.,
where $g,p$ are polynomials in $\mathbb{C}^{2}$.

In [27], it is established that there is no finite order transcendental entire
solution of Fermat-type difference equation $f^{n}(z)+f^{m}(z+c)=1$ when
$n>m>1$ or $n=m>2$. As a continuation of the research, later, the following
result considering with polynomial coefficients in the equation in
$\mathbb{C}$ is established in [29].

######  Theorem 1.1.

[29] Let $P(z)$, $Q(z)$ be non-zero polynomials. If the difference equation

(1.2) $\displaystyle f^{2}(z)+P^{2}(z)f^{2}(z+c)=Q(z)$

admits a transcendental entire solution of finite order, then $P(z)\equiv\pm
1$ and $Q(z)$ reduces to a constant $q$. Thus, $f(z)=\sqrt{q}\sin(Az+B)$,
where $B$ is a constant and $A={(4k+1)\pi}/{2c}$, where $k$ is a integer.

In the next section, we give details on the study of fermat-type functional
equations in several complex variables.

##  2\. Description of the solutions of Fermat-type functional equations in
$\mathbb{C}^{n}$

Over the past few years, Nevanlinna theory has made significant strides in the
study of several complex variables. Notable advancements include the second
fundamental theory of Nevanlinna and the application of difference analogues
to logarithmic derivative lemmas, which have contributed substantially to the
literature on holomorphic functions and their associated functional equations.
Furthermore, through the application of the Nevanlinna theory for several
complex variables with a difference perspective, Xu and Cao [41] investigated
and characterized the solution of the Fermat-type difference equation
$f^{2}(z)+f^{2}(z+c)=1$ in $\mathbb{C}^{n}$. A significant characteristic of
polynomials in $\mathbb{C}^{n}$ (for $n\geq 2$) is their ability to exhibit
periodicity, which is a fundamentally distinct behavior compared to
polynomials in $\mathbb{C}$. As a result, the solutions of functional
equations in $\mathbb{C}^{n}$ (for $n\geq 2$) with polynomial coefficients
will possess distinct characteristics.

The study of solutions to Fermat-type equations in $\mathbb{C}^{n}$ is a
captivating area of research that has recently garnered significant attention.
Before delving into a detailed examination of such equations in
$\mathbb{C}^{n}$, let us revisit a crucial result concerning Fermat-type
equations in $\mathbb{C}^{n}$. In order to achieve symmetry, similar to the
case of single-variable equations, we can summarize the results on Fermat-type
functional equations in $\mathbb{C}^{n}$ for $n\geq 3$. By utilizing [38,
Theorems 1.1 and 1.2], along with the findings presented in [13, 14] and [3],
we obtain the following theorem, which characterizes the entire and
meromorphic solutions in $\mathbb{C}^{n}$ for the equation $f^{m}+g^{m}=1$
(possibly up to a fractional linear transformation of the uniformizing
variable). It is important to mention 19th-century results on the global
parameterization of elliptic curves, such as Picard’s work ([37]), Dixon’s
contributions to $sm$ and $cm$ functions (refer to [11]), and Liouville’s
result stating that any elliptic function $f$ can be expressed as
$F(\wp)+\wp^{\prime}G(\wp)$, where $\wp$ represents the Weierstrass elliptic
function and $F$ and $G$ are rational functions of their arguments (see [32]).

######  Theorem 2.1.

[38] For $h:\mathbb{C}^{n}\rightarrow\mathbb{C}$ entire, the solutions of the
equation $f^{m}+g^{m}=1$, $m>1$ are characterize as follows:

  1. (i)

for $m=2$, the entire solutions are $f(z)=\cos(h(z)),\;g(z)=\sin(h(z))$;

  2. (ii)

for $m>2$, there are no non-constant entire solutions;

  3. (iii)

for $m=2$, the meromorphic solutions are
$f(z)={(1-\beta^{2}(z))}/{(1+\beta^{2}(z))}$ and
$g(z)={2\beta(z)}/{(1+\beta^{2}(z))}$, with $\beta$ being meromorphic on
$\mathbb{C}^{n}$.

  4. (iv)

for $m=3$, the meromorphic solutions are of the form

$\displaystyle
f(z)=\frac{1}{2\wp(h)}\left(1+\frac{\wp^{\prime}(h)}{\sqrt{3}}\right)\;\;\mbox{and}\;\;g(z)=\frac{\omega}{2\wp(h)}\left(1-\frac{\wp^{\prime}(h)}{\sqrt{3}}\right),$

where $\left(\wp^{\prime}\right)^{2}=4\wp^{3}-1$;

  5. (v)

for $m>3$, there are no non-constant meromorphic solutions.

With the introduction of the difference analogue of Nevanlinna theory for
meromorphic functions with several complex variables, particularly the
logarithmic derivative lemma proposed by Cao and Korhonen [5], numerous
researchers have focused on exploring complex difference equations and complex
Fermat-type difference equations with constant coefficients. This can be seen
in various studies (e.g., [40, 44] and their references). To gain a
comprehensive understanding of Nevanlinna theory in $\mathbb{C}^{n}$, we
suggest referring to [47] and the literature mentioned therein. It is worth
noting that non-constant polynomials in $\mathbb{C}^{n}$ may exhibit
periodicity when $n\geq 2$. Therefore, investigating solutions to Fermat-type
difference equations in $\mathbb{C}^{n}$ is an intriguing avenue of research,
as these solutions possess fundamentally different characteristics compared to
those in a single complex variable.

Motivated by the above discussion and to continue the research for further
investigations on Fermat-type functional equations in $\mathbb{C}^{n}$, we
raised the following questions:

######  Question 2.1.

Can we derive a result corresponding to Theorem 1.1 with polynomial
coefficients in $\mathbb{C}^{n}$?

######  Question 2.2.

What can be said about the solutions of equations when $f(z)$ is replaced by
$\frac{\partial f(z)}{\partial z_{i}}$ and $Q(z)$ is replaced by
$Q(z)e^{g(z)}$ in Theorem 1.1, where $g$ is a polynomial in $\mathbb{C}^{n}$?

We now recall here an interesting result in $\mathbb{C}^{2}$, where an
exploration presented in [48] delved into the existence and precise
characteristics of transcendental entire solutions in $\mathbb{C}^{2}$.

######  Theorem 2.2.

[48] Let $c=(c_{1},c_{2})\in\mathbb{C}^{2}\setminus\\{(0,0)\\}$ and
$a_{1},a_{2},a_{3}$ be nonzero constant in $\mathbb{C}$. If the Fermat-type
functional equation

(2.1) $\displaystyle a_{1}f(z)^{2}+(a_{2}f(z+c)+a_{3}f(z))^{2}=1$

has a transcendental entire solution $f(z)$ with finite order. Then
$a^{2}_{1}+a^{2}_{3}=a^{2}_{2}$ and $f(z)$ is of the form

$\displaystyle f(z)=\dfrac{1}{a_{1}}\sin\left(L(z)+H(s)+B\right),$

where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$\alpha_{1},\alpha_{2}\in\mathbb{C}$, $H(s)$ is a polynomial in
$s:=c_{2}z_{1}-c_{1}z_{2}$ in $\mathbb{C}^{2}$, and $L(z)$ satisfies
$L(c)=\alpha_{1}c_{1}+\alpha_{2}c_{2}=\theta+k\pi\pm{\pi}/{2}$,
$\tan\theta={a_{3}}/{a_{1}}$.

After in depth study of Theorem 2.2 and its proof, we understand that two fold
generalizations is possible. One is that the result can be generalized in
$\mathbb{C}^{n}$ ($n\geq 2$) instead of $\mathbb{C}^{2}$, and the other one is
that the functional equation in (2.1) can be chosen with a more general
setting. Henceforth, Theorem 2.2 lead us to ask the following question.

######  Question 2.3.

What had happened about the solutions if the constant $1$ in R.H.S of (2.1) is
replaced by $e^{g}$ in Theorem 2.2, where $g$ is a polynomial in
$\mathbb{C}^{n}$ for $n\geq 2$?

In this paper, one of our aim is to answer Question 2.3. Before delving into
our investigations, we would like to highlight some results obtained by
Saleeby regarding trinomial quadratic equations. In [38], Saleeby made
important observations on right factors of meromorphic functions to describe
complex analytic solutions to the quadratic functional equation $f^{2}+2\alpha
fg+g^{2}=1$, where $\alpha^{2}\neq 1$ is a constant in $\mathbb{C}$. Saleeby’s
work also associated with partial differential equations of the form
$u^{2}_{x}+2\alpha u_{x}u_{y}+u^{2}_{y}=1$. Building upon this, Liu and Yang
[30] further investigated the existence and form of solutions for quadratic
trinomial functional equations. It is proved in [30] that if $\alpha\neq\pm
1,0$, then the equation $f(z)^{2}+2\alpha
f(z)f^{\prime}(z)+{f^{\prime}}^{2}(z)=1$ has no transcendental meromorphic
solutions. However, the equation $f(z)^{2}+2\alpha f(z)f(z+c)+f(z+c)^{2}=1$
must have transcendental entire solutions of order equal to one. In recent
years, there has been extensive research on trinomial partial differential-
difference equations for functions of two complex variables. Motivated by the
techniques used by Saleeby in [38] to solve trinomial equations, we raised the
following questions corresponding to Theorems 2.2.

######  Question 2.4.

Do solutions exist for equations (2.1) if the constant $1$ is replaced by a
function $e^{g}$, where $g$ is a polynomial in $\mathbb{C}^{n}$? Moreover,
with this setting, can we find solutions if the binomial equation is replaced
by a quadratic trinomial equation in $\mathbb{C}^{n}$ with arbitrary
coefficients?

Motivated by the above questions, this article is concerned with the
description of entire solutions for several difference equations and partial
differential-difference of Fermat type with more general forms in
$\mathbb{C}^{n}$. Our main purpose of this paper is to discuss the finite
order transcendental entire solutions of a Fermat-type binomial and quadratic
trinomial functional equation with polynomial coefficients and find the
precise form of the solutions. Henceforth, we consider the following equations

(2.2) $\displaystyle
af^{2}(z)+bP^{2}(z)[a_{1}f(z+c)+a_{0}f(z)]^{2}=Q(z)e^{g(z)},$

(2.3) $\displaystyle af^{2}(z+c)+bP^{2}(z)\left(\frac{\partial f(z)}{\partial
z_{i}}\right)^{2}=Q(z)e^{g(z)}$

and

(2.4) $\displaystyle af^{2}(z)+2\omega
f(z)\left(\gamma_{1}f(z+c)+\gamma_{2}f(z)\right)+b\left(\gamma_{1}f(z+c)+\gamma_{2}f(z)\right)^{2}=e^{g(z)},$

where $P(z)$, $Q(z)$ and $g(z)$ are non-zero polynomials in $\mathbb{C}^{n}$.
Moreover, we show that the precise form of the polynomial $g(z)$ in
$\mathbb{C}^{n}$ can be obtained.

This article is organized as follows: In section 3, we presents the main
results of this paper, which address Fermat-type binomial and trinomial
differential-difference equations with polynomial coefficients in
$\mathbb{C}^{n}$. In fact, these main results correspond to the Questions 2.1,
2.2, 2.3, and 2.4, 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 improved results. In section
4, we provide a comprehensive review of essential Lemmas and present a
detailed proof for each significant result.

##  3\. Main results, corollaries and remarks

The main tools used in this paper are the Nevanlinna theory and difference
analogue of Nevanlinna theory with several complex variables. Throughout this
paper, we assume that $z+c=(z_{1}+c_{1},\ldots,z_{n}+c_{n})$, for any
$z=(z_{1},\ldots,z_{n})$ and $c=(c_{1},\ldots,c_{n})$ are in $\mathbb{C}^{n}$.
In fact, we know that any polynomial $P(z)$ in $\mathbb{C}^{n}$ of degree $p$
can be expressed as
$P(z)=\sum_{|I|=0}^{p}a_{\alpha_{1},\ldots,\alpha_{n}}z^{\alpha_{1}}_{1}\cdots
z^{\alpha_{n}}_{n}$, where $I=(\alpha_{1},\ldots,\alpha_{n})$ be a multi-index
with $|I|=\sum_{j=0}^{n}\alpha_{j}$ and $\alpha_{j}$ are non-negative
integers. For solving quadratic trinomial equations, we define

$\displaystyle\omega_{1}:=-\frac{\omega}{\sqrt{ab}}\pm\frac{\sqrt{\omega^{2}-ab}}{\sqrt{ab}}\;\;\mbox{and}\;\;\omega_{2}:=-\frac{\omega}{\sqrt{ab}}\mp\frac{\sqrt{\omega^{2}-ab}}{\sqrt{ab}}.$

Our key contribution in this paper pertains to Question 2.1 and demonstrates
that the polynomial $P(z)$ discussed in (2.2) eventually attains a constant
value, thereby enabling the determination of the solutions’ precise form.

######  Theorem 3.1.

Let $c=(c_{1},\ldots,c_{n})\in\mathbb{C}^{n}\setminus\\{(0,\ldots,0)\\}$ and
$ab\neq 0$. If $f$ is a transcendental entire solution with finite order of
the equation (2.2), then the polynomial $P(z)$ reduces to non-zero constant
${\sqrt{a}}/{i\sqrt{b}\left(a_{0}\pm
a_{1}e^{\frac{L_{1}(c)+L_{2}(c)}{2}}\right)}$. Moreover, $f$ and $g$ must be
one of the following to forms:

  1. (i)

$g(z)=\psi(s)+K+\sum_{i=1}^{n}(\alpha_{i}+\beta_{i})z_{i}$, where $\psi$ is a
polynomial in $s=\sum_{i=1}^{n}d_{i}z_{i}$ with $\sum_{i=1}^{n}d_{i}c_{i}=0$
with $\psi(z+c)=\psi(z)$; Moreover,

$\displaystyle
f(z)=\dfrac{Q_{1}(z)e^{L_{1}(z)+\psi_{1}(s)+k_{1}}+Q_{2}(z)e^{L_{2}(z)+\psi_{2}(s)+k_{2}}}{2\sqrt{a}},$

where $Q_{1}(z)=\sum_{i=1}^{n}m_{1i}z_{i}+H_{1}(s_{1})+B_{1}$ with
$\sum_{i=1}^{n}m_{1i}c_{i}=0$ and
$Q_{2}(z)=\sum_{i=1}^{n}m_{2i}z_{i}+H_{2}(s_{1})+B_{2}$ with
$\sum_{i=1}^{n}m_{2i}c_{i}=0$; $H_{1}(s_{1}),H_{2}(s_{1})$ are polynomials in
$s_{1}=\sum_{i=1}^{n}e_{i}z_{i}$ with $\sum_{i=1}^{n}e_{i}c_{i}=0$,
$H_{1}(z+c)=H_{1}(z)$, $H_{2}(z+c)=H_{2}(z)$ and $B_{1},B_{2}$ are constants.
Furthermore, $L_{1}(z)=\sum_{i=1}^{n}\alpha_{i}z_{i}$,
$L_{2}(z)=\sum_{i=1}^{n}\beta_{i}z_{i}$, $\psi_{1}(s)$ and $\psi_{2}(s)$ are
polynomials in $s$ in $\mathbb{C}^{n}$ with $\psi_{1}(z+c)=\psi_{1}(z)$ and
$\psi_{2}(z+c)=\psi_{2}(z)$, $k_{1},k_{2}$ are constants.

  2. (ii)

$\displaystyle f(z)=\beta(z)e^{L_{21}(z)}\;\mbox{and}\;g(z)=2L_{21}(z)+B,$

where $\beta(z)$ satisfies
$a\beta^{2}(z)+bP^{2}(z)\left(a_{1}\beta(z+c)+a_{0}\beta(z)\right)^{2}=e^{B}Q(z)$
and $L_{21}(z)=\sum_{i=1}^{n}\gamma_{i}z_{i}$ and
$\gamma_{1},\ldots,\gamma_{n},B$ are constants

Theorem 3.1 yields several noteworthy observations, all of which we
meticulously examine in the subsequent remark.

######  Remark 3.1.

We note that

  1. (i)

Theorem 3.1 improves that in [27, Theorem 2.1, Theorem 2.3] and [18, Theorem
1.1], and [7, Theorem 1.3, Theorem 1.5] in the sense that in this result
equations are considered with polynomial coefficients in $\mathbb{C}^{n}$.

  2. (ii)

In particular, if $n=1,m$; $a=1,b=1,a_{1}=1,a_{0}=0$ and $g(z)=2k\pi i$,
$k\in\mathbb{Z}$, one can obtain the conclusions of Theorem 1.1[29] in
$\mathbb{C}$ and [18, Theorem 1.1] in $\mathbb{C}^{n}$. As a result, we obtain
a more improved result.

  3. (iii)

In particular, if $a_{1}=1,a_{0}=-1$ and $g(z)=2k\pi i$, $k\in\mathbb{Z}$,
then it is easy to see that (2.2) reduces to the equation
$af^{2}(z)+bP^{2}(z)\left(\Delta_{c}f(z)\right)^{2}=Q(z)$, where $P(z),Q(z)$
are two non-zero polynomials in $\mathbb{C}^{n}$. Solving this equation,
finally it can be obtained that $bP^{2}(z)\left(Q(z+c)-Q(z)\right)=aQ(z)$.
Clearly, if $\deg P(z)=p$ and $\deg Q(z)=q$, then we see that $2p+q-1=q$, we
obtain $p={1}/{2}$, which is a contradiction. Therefore, this equation has no
transcendental entire solutions with finite order. Consequently, our result is
an improvement of that in [26, Proposition 5.3] and [27, Theorem 2.3].

  4. (iv)

The conclusion of Theorem 3.1 reveals also one crucial fact: if one considers
the polynomial $P(z)$ in equation (2.2) as non-constant, then (2.2) does not
admit any solutions.

######  Remark 3.2.

In $\mathbb{C}$, it can be observed that when a polynomial $p(z)$ and a non-
zero constant $c$ satisfy the equation $p(z+c)-p(z)=\xi$, with $\xi$ as a
constant, the polynomial $p(z)$ is must be linear and can be written in the
form $p(z)=az+b$, where $b$ represents a constant belonging to $\mathbb{C}$.
But, if $p(z)$ is a polynomial in $\mathbb{C}^{n}$ such that
$p(z+c)-p(z)=\eta$, where $\eta$ is a constant in $\mathbb{C}$, then $p(z)$
must be of the form $p(z)=L(z)+H(s)+B$, where $L(z)=\sum_{i=1}^{n}a_{n}z_{n}$
and $H(s)$ is a polynomial in $s=\sum_{i=1}^{n}d_{i}z_{i}$ with
$\sum_{i=1}^{n}d_{i}c_{i}=0$. In this case, it is worth pointing out that
$p(z)$ is not necessarily linear, i.e., degree of $p(z)$ grater than or equal
to $1$. Consequently, taking into account this property of a polynomial, it
becomes clear that to verify the accuracy of this statement, the following
examples demonstrate instances where the transcendental solutions of certain
difference equations in $\mathbb{C}^{n}$ may order $\rho(f)\geq 1$.

######  Example 3.1.

Let $c=(2,-1,3)\in\mathbb{C}^{3}$, the transcendental entire solutions in
$\mathbb{C}^{3}$ of the difference equation

$\displaystyle
7f^{2}(z)-\frac{{7}}{\left(5\pm\sqrt{3}e^{(4+i)+\frac{\pi}{2}}\right)^{2}}($
$\displaystyle\sqrt{3}f(z+c)+5f(z))^{2}$
$\displaystyle\quad=Q_{1}(z)Q_{2}(z)e^{4z_{1}+(2\pi+i)z_{2}+(\pi+i)z_{3}+\psi(s)+\frac{\pi
i}{3}}$

where $Q_{1}(z)=(4z_{1}+2z_{2}-2z_{3})+(2z_{1}+z_{2}-z_{3})^{5}+{\pi i}/{12}$,
$Q_{2}(z)=(5z_{1}+z_{2}-3z_{3})+(2z_{1}+z_{2}-z_{3})^{8}+{\pi i}/{13}$ and
$\psi(s)$ is a polynomial in $s=z_{1}+(2+3i)z_{2}+iz_{3}$; must be of the form

$\displaystyle f(z)$
$\displaystyle=\dfrac{1}{2\sqrt{7}}\left(Q_{1}(z)e^{3z_{1}+iz_{2}+\pi
z_{3}+(z_{1}+(2+3i)z_{2}+iz_{3})^{10}+\frac{\sqrt{3}\pi i}{7}}\right.$
$\displaystyle\quad\left.+Q_{2}(z)e^{z_{1}+2\pi
z_{2}+iz_{3}+(z_{1}+(2+3i)z_{2}+iz_{3})^{7}+\frac{\sqrt{7}\pi}{11}}\right).$

It is easy to see that $\rho(f)=\max\\{10,7\\}=10>1$.

######  Example 3.2.

Let $c=(5,-2,3)\in\mathbb{C}^{3}$, the transcendental entire solutions in
$\mathbb{C}^{3}$ of the difference equation

$\displaystyle
13f^{2}(z)-\frac{{13}}{\left(\frac{\sqrt{7}}{2}\pm\sqrt{5}e^{\frac{17-5i}{2}}\right)^{2}}\bigg{(}$
$\displaystyle\sqrt{5}f(z+c)+\frac{\sqrt{7}}{2}f(z)\bigg{)}^{2}$
$\displaystyle\quad=Q_{1}(z)Q_{2}(z)e^{5z_{1}+(1+i)z_{2}-(2+i)z_{3}+\psi(s)+\frac{\sqrt{17}\pi
i}{15}},$

where $Q_{1}(z)=(4z_{1}+z_{2}-6z_{3})+(z_{1}+z_{2}-z_{3})^{7}+{\pi i}/{4}$,
$Q_{2}(z)=(6z_{1}+3z_{2}-8z_{3})+(z_{1}+z_{2}-z_{3})^{6}+{\pi i}/{9}$ and
$\psi(s)$ is a polynomial in $s=3z_{1}+6z_{2}-z_{3}$; must be of the form

$\displaystyle f(z)$
$\displaystyle=\frac{1}{2\sqrt{13}}\left(Q_{1}(z)e^{2z_{1}+z_{2}-iz_{3}+(3z_{1}+6z_{2}-z_{3})^{13}+\frac{\sqrt{5}\pi
i}{2}}\right.$
$\displaystyle\left.\quad+Q_{2}(z)e^{3z_{1}+iz_{2}-2z_{3}+(3z_{1}+6z_{2}-z_{3})^{5}+\frac{\sqrt{7}\pi
i}{5}}\right).$

Clearly, we have $\rho(f)=\max\\{13,5\\}=13>1$.

The subsequent corollary for functional equations in $\mathbb{C}^{2}$, derived
from Theorem 3.1, is as follows.

######  Corollary 3.1.

Let $c=(c_{1},c_{2})\in\mathbb{C}^{2}\setminus\\{(0,0)\\}$ and $a,b\neq 0$. If
$f$ is a transcendental entire solution with finite order of the equation
(2.2), then the polynomials $P(z)$ reduce to non-zero constant
${\sqrt{a}}/{i\sqrt{b}\left(a_{0}\pm
a_{1}e^{\frac{L_{1}(c)+L_{2}(c)}{2}}\right)}$. Moreover, $f$ and $g$ must be
one of the following to forms:

  1. (i)

$g(z)=(\alpha_{1}+\beta_{1})z_{1}+(\alpha_{2}+\beta_{2})z_{2}+\psi(s)+K$,
where $\psi$ is a polynomial in $s=c_{2}z_{1}-c_{1}z_{2}$. Further,

$\displaystyle
f(z)=\dfrac{Q_{1}(z)e^{L_{1}(z)+\psi_{1}(s)+k_{1}}+Q_{2}(z)e^{L_{2}(z)+\psi_{2}(s)+k_{2}}}{2\sqrt{a}},$

where $Q_{1}(z)=\sum_{i=1}^{2}m_{1i}z_{i}+H_{1}(s_{1})+B_{1}$, with
$\sum_{i=1}^{2}m_{1i}c_{i}=0$ and
$Q_{2}(z)=\sum_{i=1}^{2}m_{2i}z_{i}+H_{2}(s_{1})+B_{2}$ with
$\sum_{i=1}^{2}m_{2i}c_{i}=0$; where $H_{1}(s_{1}),H_{2}(s_{1})$ is a
polynomial is $s_{1}=c_{2}z_{1}-c_{1}z_{2}$, $B_{1},B_{2}$ are constants.
Furthermore, $L_{1}(z)=\sum_{i=1}^{2}\alpha_{i}z_{i}$,
$L_{2}(z)=\sum_{i=1}^{2}\beta_{i}z_{i}$, $\psi_{1}(s)$ and $\psi_{2}(s)$ are
polynomials in $s$ in $\mathbb{C}^{2}$, $k_{1},k_{2}$ are constants.

  2. (ii)

$\displaystyle f(z)=\beta(z)e^{L_{21}(z)}\;\mbox{and}\;g(z)=2L_{21}(z)+B,$

where $\beta(z)$ satisfies
$a\beta^{2}(z)+bP^{2}(z)\left(a_{1}\beta(z+c)+a_{0}\beta(z)\right)^{2}=e^{B}Q(z)$
and $L_{21}(z)=\gamma_{1}z_{1}+\gamma_{2}z_{2}$ and $\gamma_{1},\gamma_{2},B$
are constants.

The second result we obtain in this paper is the following, which answers to
the Question 2.2. We show that if $f$ is a transcendental entire solution with
finite order of the equation (2.3), then the polynomials $P(z)$ and $Q(z)$
reduce to non-zero constants.

######  Theorem 3.2.

Let $c=(c_{1},\ldots,c_{n})\in\mathbb{C}^{n}\setminus\\{(0,\ldots,0)\\}$ and
$ab\neq 0$. If $f$ is a transcendental entire solution with finite order of
the equation (2.3), then the polynomials $P(z)$ and $Q(z)$ reduce to non-zero
constants. Moreover, $f$ and $g$ must be one of the following forms:

  1. (i)

$\displaystyle
f(z)=\dfrac{\alpha_{1}e^{h_{1}(z-c)}+\alpha_{2}e^{h_{2}(z-c)}}{2\sqrt{a}}\;\mbox{and}\;g(z)=(r_{1}+r_{2})+\sum_{i=1}^{n}(a_{i}+b_{i})z_{i}$

where $h_{1}(z)=r_{1}+\sum_{i=1}^{n}a_{i}z_{i}$ and
$h_{2}(z)=r_{2}+\sum_{i=1}^{n}b_{i}z_{i}$, and satisfy the relations
$e^{\sum_{i=1}^{n}(a_{i}-b_{i})c_{i}}=-a_{i}/b_{i}$; where
$r_{1},r_{2},a_{i},b_{i}\in\mathbb{C}$ for $i=1,2,\ldots,n$.

  2. (ii)

$\displaystyle
f(z)=\gamma(z-c)e^{L_{1}(z-c)+H(s)}\;\mbox{and}\;g(z)=2L_{1}(z)+2H(s)+r_{5},$

where $L(z)=\sum_{i=1}^{n}\xi_{i}z_{i}$ and $H(s)$ is a polynomial in
$s:=\sum_{i=1}^{n}d_{i}z_{i}$ in $\mathbb{C}^{n}$ such that
$\sum_{i=1}^{n}d_{i}c_{i}=0$ with $H(z+c)=H(z)$,
$\xi_{1},\ldots,\xi_{n},r_{5}$ are constants and $\gamma(z)$ satisfies

$\displaystyle
a\gamma^{2}(z)+bP^{2}(z)e^{-2L_{1}(c)}\left(\frac{\partial\gamma(z-c)}{\partial
z_{i}}+(\xi_{i}+d_{i})\gamma(z-c)\right)^{2}=e^{r_{5}}Q(z).$

######  Remark 3.3.

We conclude some remark corresponding to Theorem 3.2.

  1. (i)

We note that Theorem 3.2 improves the results of [27, Theorem 1.3] and [18,
Theorem 1.3].

  2. (ii)

If $n=1$, $P(z)=1$, $Q(z)=1$, $a=1,b=1$ and $g(z)=2k\pi i$, $k\in\mathbb{Z}$,
then we obtain the conclusion of [27, Theorem 1.3]. Thus, our result exhibits
a higher level of improvement compared to previous results.

  3. (iii)

In particular, if $n=2$, $a=1,b=1$ and $g(z)=2k\pi i$, $k\in\mathbb{Z}$, one
can obtain the conclusions of [18, Theorem 1.3] in $\mathbb{C}^{2}$.
Accordingly, our result can be regarded as a more refined and improved version
in comparison to the existing ones.

Our third result of this paper is the following and we obtained the precise
form of the solutions to a quadratic trinomial difference equations in
$\mathbb{C}^{n}$, which answers to the Question 2.4. In fact, we obtain the
following result under the assumption $\omega^{2}\neq 0,ab$.

######  Theorem 3.3.

Let $c=(c_{1},\ldots,c_{n})\in\mathbb{C}^{n}\setminus\\{(0,\ldots,0)\\}$,
$ab\neq 0$ and $\gamma_{1},\gamma_{2}$ are nonzero constants, and
$\omega^{2}\neq 0,ab$. If the difference equation (2.4) admits a
transcendental entire solution of finite order, then $g(z)$ must be a
polynomial of the form $g(z)=L(z)+H(s)+B_{3}$, where $L(z)$ is a linear
function of the form $L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$ and $H(s)$ is a
polynomial in $s:=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in $\mathbb{C}^{n}$ with
$d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ with $H(z+c)=H(z)$,
$a_{1},\ldots,a_{n}\in\mathbb{C}$. Further, $f(z)$ must satisfy one of the
following cases:

  1. (i)

$\displaystyle
f(z)=\frac{\omega_{2}\xi^{2}-\omega_{1}}{\xi\sqrt{a}(\omega_{2}-\omega_{1})}e^{\frac{1}{2}(L(z)+H(s)+B_{3})},$

where

$\displaystyle
e^{\frac{a_{1}c_{1}+\cdots+a_{n}c_{n}}{2}}=\dfrac{(\omega_{1}\gamma_{2}\sqrt{b}-\sqrt{a})-(\omega_{2}\gamma_{2}\sqrt{b}-\sqrt{a})\xi^{2}}{\gamma_{1}\sqrt{b}(\omega_{2}\xi^{2}-\omega_{1})}.$

  2. (ii)

$\displaystyle
f(z)=\dfrac{\omega_{2}e^{L_{1}(z)+H_{1}(s)+D_{1}}-\omega_{1}e^{L_{2}(z)+H_{2}(s)+D_{2}}}{\sqrt{a}(\omega_{2}-\omega_{1})},$

where $L_{l}(z)=a_{l1}z_{1}+\cdots+a_{ln}z_{n}$ and $H_{l}(s)$ for $l=1,2$ are
polynomial in $s$ in $\mathbb{C}^{n}$ and
$a_{l1},\ldots,a_{ln},D_{1},D_{2}\in\mathbb{C}$ for $l=1,2$. Also, satisfying

$\displaystyle L_{1}(z)+H_{1}(s)\neq L_{2}(z)+H_{2}(s),\;\;g(z)=L(z)+H(s)+D,$

where $L(z)=L_{1}(z)+L_{2}(z)$, $H(s)=H_{1}(s)+H_{2}(s)$ and $D=D_{1}+D_{2}$
and

$\displaystyle-\frac{(\omega_{2}\gamma_{2}\sqrt{b}-\sqrt{a})}{\omega_{2}\gamma_{1}\sqrt{b}}e^{-L_{1}(c)}\equiv
1\;\;\;\;\mbox{and}\;\;\;\;-\frac{(\omega_{1}\gamma_{2}\sqrt{b}-\sqrt{a})}{\omega_{1}\gamma_{1}\sqrt{b}}e^{-L_{2}(c)}\equiv
1.$

######  Remark 3.4.

Theorem 3.3 strengthened the result for the quadratic trinomial equation with
arbitrary coefficients of [48, Theorem 2.2] in $\mathbb{C}^{n}$.

By the following examples, we show that forms of the solutions in part (i) and
(ii) of Theorem 3.3 are precise.

######  Example 3.3.

For $c=(5,2,-3)\in\mathbb{C}^{3}$, by a routine computation, it can be easily
shown that the transcendental entire solutions in $\mathbb{C}^{3}$ of the
difference equation

$\displaystyle 2f^{2}(z)+8f(z)[5f(z+c)-3f(z)]$
$\displaystyle+3[5f(z+c)-3f(z)]^{2}$
$\displaystyle\quad=e^{3z_{1}+\ln\left(\frac{10\pm 9\sqrt{10}}{5(4\pm
3\sqrt{10})}\right)z_{2}+5z_{3}+H(s)+\frac{5\pi i}{6}},$

where $H(s)$ is a polynomial in $s:=4z_{1}-z_{2}+6z_{3}$ of degree $n\geq 1$;
must be of the form

$\displaystyle f(z_{1},z_{2},z_{3})=\frac{(-4\mp 3\sqrt{10})}{\mp
4\sqrt{10}}e^{\frac{1}{2}\left(3z_{1}+\ln\left(\frac{10\pm 9\sqrt{10}}{5(4\pm
3\sqrt{10})}\right)z_{2}+5z_{3}+(4z_{1}-z_{2}+6z_{3})^{n}+\frac{5\pi
i}{6}\right)},$

and the order of this transcendental entire solution is $\rho(f)=n\geq 1$.

######  Example 3.4.

For $c=(2,-3,1)\in\mathbb{C}^{3}$, the transcendental entire solutions in
$\mathbb{C}^{3}$ of the difference equation

$\displaystyle 3f^{2}(z)$
$\displaystyle+10f(z)[3f(z+c)-2f(z)]+[3f(z+c)-2f(z)]^{2}$
$\displaystyle\quad=e^{11z_{1}+\left(4-\frac{1}{3}\ln\left(\frac{7\mp
2\sqrt{22}}{3(5\mp\sqrt{22})}\right)\right)z_{2}+\left(\ln\left(\frac{7\pm
2\sqrt{22}}{3(5\pm\sqrt{22})}\right)-10\right)z_{3}+H(s)+\frac{181\pi
i}{88}},$

where $H(s)$ is a polynomial in $s:=5z_{1}+4z_{2}+2z_{3}$; must be of the form

$\displaystyle f(z_{1},z_{2},z_{3})$
$\displaystyle=\dfrac{(-5\mp\sqrt{22})}{\mp
2\sqrt{66}}e^{6z_{1}+4z_{2}+\ln\left(\frac{7\pm
2\sqrt{22}}{3(5\pm\sqrt{22})}\right)z_{3}+(5z_{1}+4z_{2}+2z_{3})^{n}+\frac{7\pi
i}{8}}$ $\displaystyle\quad-\dfrac{(-5\pm\sqrt{22})}{\mp
2\sqrt{66}}e^{5z_{1}-\frac{1}{3}\ln\left(\frac{7\mp
2\sqrt{22}}{3(5\mp\sqrt{22})}\right)z_{2}-10z_{3}+(5z_{1}+4z_{2}+2z_{3})^{m}+\frac{13\pi
i}{11}},$

and the order of this transcendental entire solution is
$\rho(f)=\max\\{n,m\\}\geq 1$.

In Theorem 3.3, the complete solutions of (2.4) are thoroughly explored in the
case when $\omega^{2}\neq 0,ab$. However, the study of finding the solutions
would be considered complete and comprehensive if we also determine the
solutions of (2.4) when $\omega^{2}=0,ab$. Therefore, it is natural to answer
the following question.

######  Question 3.1.

What can we say about the solutions of (2.4) when $\omega=0$ or
$\omega^{2}=ab$?

We answer this Question 3.1 in the subsequent discussion. However, as a
consequence of Theorem 3.3, we obtain the result which encompasses the scope
of Theorem 2.2 in case of trinomial equations in $\mathbb{C}^{2}$.

######  Corollary 3.2.

Let $c=(c_{1},c_{2})\in\mathbb{C}^{2}\setminus\\{(0,0)\\}$, and
$\gamma_{1},\gamma_{2}$ are nonzero constants and $ab\neq 0$, and
$\omega^{2}\neq ab$. If the difference equation

$\displaystyle af^{2}(z)+2\omega
f(z)\left(\gamma_{1}f(z+c)+\gamma_{2}f(z)\right)+b\left(\gamma_{1}f(z+c)+\gamma_{2}f(z)\right)^{2}=e^{g(z)}$

admits a transcendental entire solution of finite order, then $g(z)$ must be a
polynomial function of the form $g(z)=L(z)+H(s)+B_{3}$, where $L(z)$ is a
linear function of the form $L(z)=a_{1}z_{1}+a_{2}z_{2}$ and $H(s)$ is a
polynomial in $s:=c_{2}z_{1}-c_{1}z_{2}$ in $\mathbb{C}^{2}$. Further, $f(z)$
must satisfy one of the following cases:

  1. (i)

$\displaystyle
f(z)=\dfrac{\omega_{2}\xi^{2}-\omega_{1}}{\xi\sqrt{a}(\omega_{2}-\omega_{1})}e^{\frac{1}{2}(L(z)+H(s)+B_{3})},$

and satisfying

$\displaystyle
e^{\frac{a_{1}c_{1}+a_{2}c_{2}}{2}}=\dfrac{(\omega_{1}\gamma_{2}\sqrt{b}-\sqrt{a})-(\omega_{2}\gamma_{2}\sqrt{b}-\sqrt{a})\xi^{2}}{\gamma_{1}\sqrt{b}(\omega_{2}\xi^{2}-\omega_{1})}.$

  2. (ii)

$\displaystyle
f(z)=\dfrac{\omega_{2}e^{L_{1}(z)+H_{1}(s)+D_{1}}-\omega_{1}e^{L_{2}(z)+H_{2}(s)+D_{2}}}{\sqrt{a}(\omega_{2}-\omega_{1})},$

where $L_{l}(z)=a_{l1}z_{1}+a_{l2}z_{2}$ and $H_{l}(s)$ for $l=1,2$ are
polynomial in $s:=c_{2}z_{1}-c_{1}z_{2}$ in $\mathbb{C}^{2}$ and
$D_{1},D_{2}\in\mathbb{C}$. Also, satisfying

$\displaystyle L_{1}(z)+H_{1}(s)\neq L_{2}(z)+H_{2}(s),\;\;g(z)=L(z)+H(s)+D,$

where $L(z)=L_{1}(z)+L_{2}(z)$, $H(s)=H_{1}(s)+H_{2}(s)$ and $D=D_{1}+D_{2}$,
and

$\displaystyle-\dfrac{(\omega_{2}\gamma_{2}\sqrt{b}-\sqrt{a})}{\omega_{2}\gamma_{1}\sqrt{b}}e^{-L_{1}(c)}\equiv
1\;\;\;\;\mbox{and}\;\;\;\;-\dfrac{(\omega_{1}\gamma_{2}\sqrt{b}-\sqrt{a})}{\omega_{1}\gamma_{1}\sqrt{b}}e^{-L_{2}(c)}\equiv
1.$

What could be the precise form of the solutions of (2.4) when $\omega^{2}=ab$
or $\omega^{2}=0$ _i.e., if_ $\omega=\pm\sqrt{a}\sqrt{b}$ or $\omega=0$, we
now discuss in details in the following remark.

######  Remark 3.5.

For $\omega=\pm\sqrt{a}\sqrt{b}$ and $\omega=0$, equation (2.4) can be
respectively written as

(3.1)
$\displaystyle\sqrt{a}f(z)\pm\sqrt{b}\left(\gamma_{1}f(z+c)+\gamma_{2}f(z)\right)=\pm
e^{g(z)/2}$

and

(3.2) $\displaystyle
af^{2}(z)+b\left(\gamma_{1}f(z+c)+\gamma_{2}f(z)\right)^{2}=e^{g(z)}.$

It suffices to discuss the following three cases:

Case 1: Assume that $g(z)$ is a constant. However, in this case, it is easy to
see that $e^{g(z)/2}$ is also a constant, say $E(\neq 0)$. Then by the
standard method of finding solutions of complex difference equations with
constant coefficients (see [2, 1]), it can be shown that the precise form of
the transcendental entire solutions in $\mathbb{C}^{n}$ of (3.1) will be of
the form

$\displaystyle
f(z)=\left(\mp\frac{(\sqrt{a}\pm\gamma_{2}\sqrt{b})}{\gamma_{1}\sqrt{b}}\right)^{z/c}\pi(z)\pm\frac{E}{\left(\sqrt{a}\pm(\gamma_{1}+\gamma_{2})\sqrt{b}\right)},$

where $\pi$ is a transcendental entire function with $\pi(z+c)=\pi(z)$.

Case 2: If $g$ is non-constant and $c$-periodic, _i.e._ $g(z+c)=g(z)$, then
the precise form of the transcendental entire solutions in $\mathbb{C}^{n}$ of
(3.1) will be of the form

$\displaystyle
f(z)=\left(\mp\frac{(\sqrt{a}\pm\gamma_{2}\sqrt{b})}{\gamma_{1}\sqrt{b}}\right)^{z/c}\pi(z)\pm\frac{e^{g(z)/2}}{\left(\sqrt{a}\pm(\gamma_{1}+\gamma_{2})\sqrt{b}\right)},$

where $\pi$ is a transcendental entire function with $\pi(z+c)=\pi(z)$ and $g$
is a polynomial with $g(z+c)=g(z)$.

Case 3: If $g$ is non-constant and non-periodic. Then the precise form of the
transcendental entire solutions in $\mathbb{C}^{n}$ of (3.1) will be of the
form

$\displaystyle
f(z)=\left(\mp\frac{(\sqrt{a}\pm\gamma_{2}\sqrt{b})}{\gamma_{1}\sqrt{b}}\right)^{z/c}\pi(z)\pm\frac{e^{g(z)/2}}{\left(\sqrt{a}\pm(\gamma_{1}+\gamma_{2})\sqrt{b}\right)},$

where $\pi$ is a transcendental entire function with $\pi(z+c)=\pi(z)$ and
$g(z)$ is a polynomial of the form $g(z)=L(z)+H_{1}(s)+B$, where
$L(z)=\alpha_{1}z_{1}+\cdots+\alpha_{n}z_{n}$, $H_{1}(s_{1})$ is a polynomial
in $s_{1}=\sum_{i=1}^{n}e_{i}z_{i}$, with $\sum_{i=1}^{n}e_{i}c_{i}=0$,
$H_{1}(z+c)=H_{1}(z)$ and $B$ is a constant, and satisfying

$\displaystyle\pm
e^{\frac{L(c)}{2}}=\pm\left(\sqrt{a}\pm\sqrt{b(\gamma_{1}+\gamma_{2})}\right)\mp
1.$

###  3.1. Some examples in support of Remark 3.5

######  Example 3.5.

The transcendental entire solutions in $\mathbb{C}^{n}$ of the difference
equation

$\displaystyle 3f(z)+2\left(\sqrt{3}f(z+c)+\sqrt{5}f(z)\right)=e^{g(z)/2}$

must be of the form

$\displaystyle
f(z)=\left(-\frac{(3+2\sqrt{5})}{2\sqrt{3}}\right)^{z/c}\sin\left(\dfrac{2\pi
z}{c}\right)+\frac{e^{g(z)/2}}{\left(3+2(\sqrt{5}+\sqrt{3})\right)},$

where $g$ is a polynomial with $g(z+c)=g(z)$, i.e., $g$ is of the form
$g(z)=\sum_{i=1}^{n}\gamma_{i}z_{i}+H(s)+A$, where
$\sum_{i=1}^{n}\gamma_{i}c_{i}=0$, $H(s)$ is a polynomial is
$s=\sum_{i=1}^{n}d_{i}z_{i}$ with $\sum_{i=1}^{n}d_{i}c_{i}=0$, $H(z+c)=H(z)$
and $A$ is a constant.

######  Example 3.6.

The transcendental entire solutions in $\mathbb{C}^{n}$ of the difference
equation

$\displaystyle\sqrt{7}f(z)+\sqrt{3}\left(2f(z+c)-f(z)\right)=-e^{g(z)/2}$

must be of the form

$\displaystyle
f(z)=\left(-\frac{(\sqrt{7}-\sqrt{3})}{2\sqrt{3}}\right)^{z/c}\cos\left(\dfrac{2\pi
z}{c}\right)-\frac{e^{g(z)/2}}{\left(\sqrt{7}+\sqrt{3}\right)},$

where $g$ is a polynomial with $g(z+c)=g(z)$, i.e., $g$ is of the form
$g(z)=\sum_{i=1}^{n}\gamma_{i}z_{i}+H(s)+A$, where
$\sum_{i=1}^{n}\gamma_{i}c_{i}=0$, $H(s)$ is a polynomial is
$s=\sum_{i=1}^{n}d_{i}z_{i}$ with $\sum_{i=1}^{n}d_{i}c_{i}=0$, $H(z+c)=H(z)$
and $A$ is a constant.

######  Example 3.7.

The transcendental entire solutions in $\mathbb{C}^{n}$ of the difference
equation

$\displaystyle\sqrt{5}f(z)-\sqrt{2}\left(3f(z+c)-2f(z)\right)=e^{g(z)/2}$

must be of the form

$\displaystyle
f(z)=\left(\frac{(\sqrt{5}+2\sqrt{2})}{3\sqrt{2}}\right)^{z/c}\sin\left(\dfrac{2\pi
z}{c}\right)+\frac{e^{g(z)/2}}{\left(\sqrt{5}-\sqrt{2}\right)},$

where $g$ is a polynomial with $g(z+c)=g(z)$, i.e., $g$ is of the form
$g(z)=\sum_{i=1}^{n}\gamma_{i}z_{i}+H(s)+A$, where
$\sum_{i=1}^{n}\gamma_{i}c_{i}=0$, $H(s)$ is a polynomial is
$s=\sum_{i=1}^{n}d_{i}z_{i}$ with $\sum_{i=1}^{n}d_{i}c_{i}=0$, $H(z+c)=H(z)$
and $A$ is a constant.

######  Example 3.8.

The transcendental entire solutions in $\mathbb{C}^{n}$ of the difference
equation

$\displaystyle\sqrt{2}f(z)-\sqrt{3}\left(\sqrt{7}f(z+c)+\sqrt{11}f(z)\right)=-e^{g(z)/2}$

must be of the form

$\displaystyle
f(z)=\left(\frac{(\sqrt{2}-\sqrt{33})}{\sqrt{21}}\right)^{z/c}\cos\left(\dfrac{2\pi
z}{c}\right)-\frac{e^{g(z)/2}}{\left(\sqrt{2}-(\sqrt{7}+\sqrt{11})\sqrt{3}\right)},$

where $g$ is a polynomial with $g(z+c)=g(z)$, i.e., $g$ is of the form
$g(z)=\sum_{i=1}^{n}\gamma_{i}z_{i}+H(s)+A$, where
$\sum_{i=1}^{n}\gamma_{i}c_{i}=0$, $H(s)$ is a polynomial is
$s=\sum_{i=1}^{n}d_{i}z_{i}$ with $\sum_{i=1}^{n}d_{i}c_{i}=0$, $H(z+c)=H(z)$
and $A$ is a constant.

It is worth noticing that if we allow $\omega=0$ in (2.4), then it reduces to
binomial equation (3.2) of the form $F^{2}+G^{2}=e^{g}$. We need to explore
the solutions in this case too. Hence, we obtain the following result for the
case $\omega=0$, i.e., corresponding to the equation (3.2). Consequently, the
following result is an answer to Question 2.3

######  Theorem 3.4.

Let $c=(c_{1},\ldots,c_{n})\in\mathbb{C}^{n}\setminus\\{(0,\ldots,0)\\}$,
$ab\neq 0$ and $\gamma_{1},\gamma_{2}$ are nonzero constants. If the
difference equation (3.2) admits a transcendental entire solution of finite
order, then $g(z)$ must be a polynomial of the form $g(z)=L(z)+H(s)+A$, where
$L(z)$ is a linear function of the form $L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$
and $H(s)$ is a polynomial in $s:=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in
$\mathbb{C}^{n}$ with $d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ with $H(z+c)=H(z)$,
$a_{1},\ldots,a_{n},A\in\mathbb{C}$. Further, $f(z)$ must satisfy one of the
following cases:

  1. (i)

$\displaystyle
f(z)=\dfrac{(\xi^{2}+1)}{2\xi\sqrt{a}}e^{\frac{1}{2}(L(z)+H(s)+A)},$

where $g(z)=L(z)+H(s)+A$, where $L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$;
$a_{1},\ldots,a_{n},A\in\mathbb{C}$ satisfying

$\displaystyle
e^{\frac{L(c)}{2}}=\dfrac{\sqrt{a}(\xi^{2}-1)-i\gamma_{2}\sqrt{b}(\xi^{2}+1)}{i\gamma_{1}\sqrt{b}(\xi^{2}+1)}.$

  2. (ii)

$\displaystyle
f(z)=\dfrac{e^{L_{1}(z)+H_{1}(s)+B_{1}}+e^{L_{2}(z)+H_{2}(s)+B_{2}}}{2\sqrt{a}},$

where $L_{l}(z)=a_{l1}z_{1}+\cdots+a_{ln}z_{n}$ and $H_{l}(s)$ for $l=1,2$ are
polynomial in $s$ in $\mathbb{C}^{n}$ and
$a_{l1},\ldots,a_{ln},B_{1},B_{2}\in\mathbb{C}$ for $l=1,2$. Also, satisfying

$\displaystyle L_{1}(z)+H_{1}(s)\neq L_{2}(z)+H_{2}(s),\;\;g(z)=L(z)+H(s)+B,$

where $L(z)=L_{1}(z)+L_{2}(z)$, $H(s)=H_{1}(s)+H_{2}(s)$ and $B=B_{1}+B_{2}$
and

$\displaystyle\dfrac{(i\gamma_{2}\sqrt{b}-\sqrt{a})}{-i\gamma_{1}\sqrt{b}}e^{-L_{1}(c)}\equiv
1\;\;\;\;\mbox{and}\;\;\;\;\dfrac{(i\gamma_{2}\sqrt{b}+\sqrt{a})}{-i\gamma_{1}\sqrt{b}}e^{-L_{2}(c)}\equiv
1.$

An interesting observation is that, in particular, if
$\sqrt{a}=a_{1},\sqrt{b}\gamma_{1}=a_{2},\sqrt{b}\gamma_{2}=a_{3}$ and
$g(z)=2k\pi i$, $k\in\mathbb{Z}$ in (3.2), one can obtain the conclusions of
[48, Theorem 2.2] in $\mathbb{C}^{2}$. Accordingly, our result can be regarded
as a more refined and improved version in comparison to [48, Theorem 2.2].

######  Corollary 3.3.

Let $c=(c_{1},c_{2})\in\mathbb{C}^{2}\setminus\\{(0,0)\\}$, $ab\neq 0$ and
$\gamma_{1},\gamma_{2}$ are nonzero constants. If the difference equation
(3.2) admits a transcendental entire solution of finite order, then $g(z)$
must be a polynomial of the form $g(z)=L(z)+H(s)+A$, where $L(z)$ is a linear
function of the form $L(z)=a_{1}z_{1}+a_{2}z_{2}$ and $H(s)$ is a polynomial
in $s:=c_{2}z_{1}-c_{1}z_{2}$ in $\mathbb{C}^{2}$ with $H(z+c)=H(z)$,
$a_{1},a_{2},A\in\mathbb{C}$. Further, $f(z)$ must satisfy one of the
following cases:

  1. (i)

$\displaystyle
f(z)=\dfrac{(\xi^{2}+1)}{2\xi\sqrt{a}}e^{\frac{1}{2}(L(z)+H(s)+A)},$

where

$\displaystyle
e^{\frac{L(c)}{2}}=\dfrac{\sqrt{a}(\xi^{2}-1)-i\gamma_{2}\sqrt{b}(\xi^{2}+1)}{i\gamma_{1}\sqrt{b}(\xi^{2}+1)}.$

  2. (ii)

$\displaystyle
f(z)=\dfrac{e^{L_{1}(z)+H_{1}(s)+B_{1}}+e^{L_{2}(z)+H_{2}(s)+B_{2}}}{2\sqrt{a}},$

where $L_{l}(z)=a_{l1}z_{1}+a_{l2}z_{2}$ and $H_{l}(s)$ for $l=1,2$ are
polynomial in $s$ in $\mathbb{C}^{2}$ and
$a_{l1},\ldots,a_{ln},B_{1},B_{2}\in\mathbb{C}$ for $l=1,2$, satisfying

$\displaystyle L_{1}(z)+H_{1}(s)\neq
L_{2}(z)+H_{2}(s),\;\;g(z)=L_{3}(z)+H_{3}(s)+B_{3},$

where $L_{3}(z)=L_{1}(z)+L_{2}(z)$, $H_{3}(s)=H_{1}(s)+H_{2}(s)$ and
$B_{3}=B_{1}+B_{2}$ and

$\displaystyle\frac{(i\gamma_{2}\sqrt{b}-\sqrt{a})}{-i\gamma_{1}\sqrt{b}}e^{-L_{1}(c)}\equiv
1\;\;\;\;\mbox{and}\;\;\;\;\frac{(i\gamma_{2}\sqrt{b}+\sqrt{a})}{-i\gamma_{1}\sqrt{b}}e^{-L_{2}(c)}\equiv
1.$

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

We shall use the following standard notations of value distribution theory
(see details [19, 46]): $T(r,f),\;m(r,f),\;N(r,f),\;\overline{N}(r,f),\ldots$
in $\mathbb{C}^{n}$. We denote by $S(r,f)$, any function satisfying
$S(r,f)=\circ\\{T(r,f)\\}$ as $r\rightarrow\infty$, possibly outside a set of
finite measure. 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.

######  Lemma 4.1.

[35, 39] 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 4.2.

[33] 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.

######  Lemma 4.3.

[19] Let $f_{j}(\not\equiv 0)\;(j=1,2,\ldots,m;\;m\geq 3)$ be meromorphic
functions on $\mathbb{C}^{n}$ such that $f_{1},\ldots,f_{m-1}$ are non-
constant and $f_{1}+\ldots+f_{m}=1$, and such that

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

holds for $j=1,\ldots,m-1$, where $\lambda<1$ is a positive number. Then
$f_{m}=1$.

######  Lemma 4.4.

[19] Suppose that $a_{0}(z),a_{1}(z),\ldots,a_{m}(z)\;(m\geq 1)$ are
meromorphic functions on $\mathbb{C}^{n}$ and
$g_{0}(z),g_{1}(z),\ldots,g_{m}(z)$ are entire functions on $\mathbb{C}^{n}$
such that $g_{i}(z)-g_{j}(z)$ are not constants for $0\leq i<j\leq m$. If

$\displaystyle\sum_{i=0}^{m}a_{i}(z)e^{g_{i}(z)}\equiv 0$

and $||T(r,a_{i})=o(T(r)),\;i=0,1,\ldots,m$ holds, where
$T(r):=\displaystyle\min_{0\leq i<j\leq m}T(r,e^{g_{i}-g_{j}})$, then
$a_{i}(z)\equiv 0$ for $i=0,1,\ldots,m$.

For the convenience of the reader, we will present our proof of Theorems 3.1,
3.2, and 3.3 in all detail.

###  4.1. Proof of Theorem 3.1

###### Proof of Theorem 3.1.

For the sake of simplicity, first we define
$L_{c}(f(z)):=a_{1}f(z+c)+a_{0}f(z)$, where $a_{1},a_{0}\in\mathbb{C}$. It is
easy to see that the given equation (2.2) can be expressed as

(4.1)
$\displaystyle\left(\frac{\sqrt{a}f(z)+i\sqrt{b}P(z)\left(L_{c}(f(z))\right)}{e^{\frac{g(z)}{2}}}\right)\left(\frac{\sqrt{a}f(z)-i\sqrt{b}P(z)\left(L_{c}(f(z))\right)}{e^{\frac{g(z)}{2}}}\right)=Q(z).$

In view of (4.1) and Lemma 4.1, the following pair of equations can be
obtained

(4.2)
$\displaystyle\begin{cases}\sqrt{a}f(z)+iP(z)\left(a_{1}f(z+c)+a_{0}f(z)\right)=Q_{1}(z)e^{\frac{g(z)}{2}+\alpha(z)}\vspace{1.2mm}\\\
\sqrt{a}f(z)-iP(z)\left(a_{1}f(z+c)+a_{0}f(z)\right)=Q_{2}(z)e^{\frac{g(z)}{2}-\alpha(z)},\end{cases}$

where $\alpha(z)$ is an non-constant entire function on $\mathbb{C}^{n}$ and
$Q(z)=Q_{1}(z)Q_{2}(z)$. For brevity, we define

(4.3) $\displaystyle
h_{1}(z):=\dfrac{g(z)}{2}+\alpha(z)\;\mbox{and}\;h_{2}(z):=\dfrac{g(z)}{2}-\alpha(z).$

Since $f$ is a transcendental entire function with finite order, in view of
Lemma 4.2, we see that $\alpha(z)$ must be a polynomial in $\mathbb{C}^{n}$.
Solving the pair of equations in (4.2), we easily obtain

(4.4)
$\displaystyle\begin{cases}f(z)=\dfrac{Q_{1}(z)e^{h_{1}(z)}+Q_{2}(z)e^{h_{2}(z)}}{2\sqrt{a}}\vspace{2mm}\\\
a_{1}f(z+c)+a_{0}f(z)=\dfrac{Q_{1}(z)e^{h_{1}(z)}-Q_{2}(z)e^{h_{2}(z)}}{2i\sqrt{b}P(z)}.\end{cases}$

Combining both the expressions in (4.4), a simple computation shows that

(4.5)
$\displaystyle\dfrac{a_{1}}{\sqrt{a}}\dfrac{Q_{1}(z+c)}{P_{1}(z)Q_{1}(z)}e^{h_{1}(z+c)-h_{1}(z)}+\dfrac{a_{1}}{\sqrt{a}}\dfrac{Q_{2}(z+c)}{P_{1}(z)Q_{1}(z)}e^{h_{2}(z+c)-h_{1}(z)}-\dfrac{Q_{2}(z)}{Q_{1}(z)}e^{h_{2}(z)-h_{1}(z)}=1,$

where $P_{1}(z)=\frac{1}{i\sqrt{b}P(z)}-\frac{a_{0}}{\sqrt{a}}$.

Moreover, the equation (4.5) can be written as $g_{1}(z)+g_{2}(z)+g_{3}(z)=1$,
where

$\displaystyle\begin{cases}g_{1}(z)=\dfrac{a_{1}}{\sqrt{a}}\dfrac{Q_{1}(z+c)}{P_{1}(z)Q_{1}(z)}e^{h_{1}(z+c)-h_{1}(z)}\vspace{2mm}\\\
g_{2}(z)=\dfrac{a_{1}}{\sqrt{a}}\dfrac{Q_{2}(z+c)}{P_{1}(z)Q_{1}(z)}e^{h_{2}(z+c)-h_{1}(z)}\vspace{2mm}\\\
g_{3}(z)=-\dfrac{Q_{2}(z)}{Q_{1}(z)}e^{h_{2}(z)-h_{1}(z)}\end{cases}$

In order to complete the proof, it is enough to discuss the following two
possible cases.

Case 1. Suppose that $e^{h_{2}(z)-h_{1}(z)}$ is not a constant. Then, it is
easy to see that polynomial $h_{2}(z)-h_{1}(z)$ is a non-constant. Our aim is
to show that both the functions $g_{2}$ and $g_{3}$ are non-constants.
Otherwise, suppose that $g_{2}(z)=D_{1}$ and $g_{3}(z)=-D_{2}$, where
$D_{1},D_{2}$ are both constants. Then, an easy computation leads to

(4.6) $\displaystyle
e^{h_{2}(z+c)-h_{1}(z)}=\frac{D_{1}\sqrt{a}}{a_{1}}\frac{P_{1}(z)Q_{1}(z)}{Q_{2}(z+c)}\;\mbox{and}\;e^{h_{2}(z)-h_{1}(z)}=\frac{D_{2}Q_{1}(z)}{Q_{2}(z)}.$

It is easy to see from (4.6) that the left side is a transcendental entire
function, whereas the right side of it is a rational function, which is a
contradiction. Consequently, we see that

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

as $r$ sufficiently large outside possibly a set $E$ of $r$ with finite
logarithmic measure. Thus, in view of Lemma 4.3 (with $m=3$), we must have
$g_{1}(z)\equiv 1$ i.e.,

(4.7) $\displaystyle
a_{1}Q_{1}(z+c)e^{h_{1}(z+c)-h_{1}(z)}\equiv\sqrt{a}P_{1}(z)Q_{1}(z).$

We observe that left side of (4.7) is transcendental entire, whereas right
side is polynomial, and this shows that $h_{1}(z+c)-h_{1}(z)=\xi$, where
$\xi\in\mathbb{C}$ is a constant. An easy computation shows that
$h_{1}(z)=L_{1}(z)+\psi_{1}(s)+k_{1}$, where
$L_{1}(z)=\alpha_{1}z_{1}+\cdots+\alpha_{n}z_{n}$ and $\psi_{1}(s)$ is a
polynomial in $s:=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in $\mathbb{C}^{n}$ with
$d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ and $\psi_{1}(z+c)=\psi_{1}(z)$;
$\alpha_{1},\ldots,\alpha_{n},k_{1}\in\mathbb{C}$.

Further, from (4.7), we see that

(4.8) $\displaystyle
a_{1}Q_{1}(z+c)e^{L_{1}(c)}\equiv\sqrt{a}P_{1}(z)Q_{1}(z).$

Since $Q_{1}(z)$ is a non-zero polynomial, from (4.8), we must have

(4.9) $\displaystyle P_{1}(z)=\frac{a_{1}e^{L_{1}(c)}}{\sqrt{a}}.$

Thus, it follows from (4.8) that $Q_{1}(z+c)=Q_{1}(z)$ _i.e._ the polynomial
$Q_{1}$ is $c$-periodic. By a simple computation, it can be easily shown that
$Q_{1}(z)=m_{11}z_{1}+\cdots+m_{1n}z_{n}+H_{1}(s_{1})+B_{1}$, where
$m_{11}c_{1}+\cdots+m_{1n}c_{n}=0$, $H_{1}(s_{1})$ is a polynomial is
$s_{1}=\sum_{i=1}^{n}e_{i}z_{i}$ with $\sum_{i=1}^{n}e_{i}c_{i}=0$,
$H_{1}(z+c)=H_{1}(z)$ and $B_{1}$ is a constant.

Therefore, in view of (4.5) and (4.7), we obtain

(4.10) $\displaystyle
a_{1}Q_{2}(z+c)e^{h_{2}(z+c)-h_{2}(z)}=\sqrt{a}P_{1}(z)Q_{2}(z).$

By the similar argument being used for $h_{1}(z)$, we obtain
$h_{2}(z)=L_{2}(z)+\psi_{2}(s)+k_{2}$, where
$L_{2}(z)=\beta_{1}z_{1}+\cdots+\beta_{n}z_{n}$ and $\psi_{2}(s)$ is a
polynomial in $s:=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in $\mathbb{C}^{n}$ with
$d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ with $\psi_{2}(z+c)=\psi_{2}(z)$;
$\beta_{1},\ldots,\beta_{n},k_{2}\in\mathbb{C}$. Hence, the equation (4.10)
reduces to

(4.11) $\displaystyle a_{1}Q_{2}(z+c)e^{L_{2}(c)}=\sqrt{a}P_{1}(z)Q_{2}(z).$

Since $Q_{2}(z)$ is a non-zero polynomial, (4.10) shows that

(4.12) $\displaystyle P_{1}(z)=\frac{a_{1}e^{L_{2}(c)}}{\sqrt{a}}.$

As a matter of fact, from (4.11), we see that $Q_{2}(z+c)=Q_{2}(z)$. This
shows that $Q_{2}(z)=m_{21}z_{1}+\cdots+m_{2n}z_{n}+H_{2}(s_{1})+B_{2}$, where
$m_{21}c_{1}+\cdots+m_{2n}c_{n}=0$, $H_{2}(s_{1})$ is a polynomial is
$s_{1}=\sum_{i=1}^{n}e_{i}z_{i}$ with $\sum_{i=1}^{n}e_{i}c_{i}=0$,
$H_{2}(z+c)=H_{2}(z)$ and $B_{2}$ is a constant. Multiplying (4.9) and (4.12),
it is easy to see that

(4.13) $\displaystyle aP^{2}_{1}(z)=a^{2}_{1}e^{L_{1}(z)+L_{2}(c)}.$

Now plugging $P_{1}(z)=\frac{1}{i\sqrt{b}P(z)}-\frac{a_{0}}{\sqrt{a}}$ into
(4.13), we see that $P(z)$ becomes a constant of the form

$\displaystyle P(z)=\dfrac{\sqrt{a}}{i\sqrt{b}\left(a_{0}\pm
a_{1}e^{\frac{L_{1}(c)+L_{2}(c)}{2}}\right)}=\mbox{constant}.$

Moreover, from (4.3), it is easy to see that

$\displaystyle
g(z)=L(z)+\psi(s)+K=(\alpha_{1}+\beta_{1})z_{1}+\cdots+(\alpha_{n}+\beta_{n})z_{n}+\psi(s)+K,$

where $L(z)=L_{1}(z)+L_{2}(z)$, $\psi(s)=\psi_{1}(s)+\psi_{2}(s)$ and
$K=k_{1}+k_{2}$.

Case 2. Suppose that $e^{h_{2}(z)-h_{1}(z)}$ is a constant. Clearly,
$h_{2}(z)-h_{1}(z)$ must be a constant. In fact, we see that
$h_{2}(z+c)-h_{1}(z+c)$ is also a constant. Thus it follows from (4.5) that
$e^{h_{1}(z+c)-h_{1}(z)}$ is a constant, otherwise, we will arrive at a
contradiction. Therefore, we conclude that $h_{1}(z)=L_{21}(z)+k_{3}$ and
$h_{2}(z)=L_{21}(z)+k_{4}$, where
$L_{21}(z)=\gamma_{1}z_{1}+\gamma_{2}z_{2}+\cdots+\gamma_{n}z_{n}$ and
$\gamma_{1},\ldots,\gamma_{n},k_{3},k_{4}\in\mathbb{C}$. Thus, from (4.3) it
follows that

$\displaystyle
g(z)=2(\gamma_{1}z_{1}+\gamma_{2}z_{2}+\cdots+\gamma_{n}z_{n})+B,\;\;\mbox{where}\;\;B=k_{3}+k_{4}.$

However, we see that the solution $f$ takes the following form

$\displaystyle
f(z)=\dfrac{Q_{1}(z)e^{L_{21}(z)+k_{3}}+Q_{2}(z)e^{L_{21}(z)+k_{4}}}{2\sqrt{a}}=\beta(z)e^{L_{21}(z)},$

where $\beta(z)$ satisfies

$\displaystyle
a\beta^{2}(z)+bP^{2}(z)\left(a_{1}\beta(z+c)+a_{0}\beta(z)\right)^{2}=e^{B}Q(z).$

This completes the proof. ∎

###  4.2. Proof of Theorem 3.2

###### Proof of Theorem 3.2.

The equation (2.3) can be written as

$\displaystyle\left(\frac{\sqrt{a}f(z+c)}{e^{\frac{g(z)}{2}}}\right)^{2}+\left(\frac{\sqrt{b}P(z)\left(\frac{\partial
f(z)}{\partial z_{i}}\right)}{e^{\frac{g(z)}{2}}}\right)^{2}=Q(z).$

By the similar argument being used in case of proof of Theorem 3.1, it can be
easily shown that

(4.14)
$\displaystyle\begin{cases}f(z+c)=\dfrac{Q_{1}(z)e^{h_{1}(z)}+Q_{2}(z)e^{h_{2}(z)}}{2\sqrt{a}}\vspace{2mm}\\\
\dfrac{\partial f(z)}{\partial
z_{i}}=\dfrac{Q_{1}(z)e^{h_{1}(z)}-Q_{2}(z)e^{h_{2}(z)}}{2i\sqrt{b}P(z)},\end{cases}$

where

(4.15) $\displaystyle
h_{1}(z)=\dfrac{g(z)}{2}+\alpha(z)\;\;\;\;\;\mbox{and}\;\;\;\;\;h_{2}(z)=\dfrac{g(z)}{2}-\alpha(z).$

Combining both the equations in (4.10), a simple computation shows that

(4.16)
$\displaystyle\sqrt{a}Q_{1}(z+c)e^{h_{1}(z+c)-h_{1}(z)}-\sqrt{a}Q_{2}(z+c)e^{h_{2}(z+c)-h_{1}(z)}$
$\displaystyle-i\sqrt{b}P(z+c)\left(Q_{2}(z)\frac{\partial h_{2}}{\partial
z_{i}}+\frac{\partial Q_{2}}{\partial
z_{i}}\right)e^{h_{2}(z)-h_{1}(z)}=i\sqrt{b}P(z+c)\left(Q_{1}(z)\frac{\partial
h_{1}(z)}{\partial z_{i}}+\frac{\partial Q_{1}}{\partial z_{i}}\right).$

However, equation (4.16) can be expressed as

(4.17) $\displaystyle g_{1}(z)+g_{2}(z)+g_{3}(z)=1,$

where

(4.18)
$\displaystyle\begin{cases}g_{1}(z)=\dfrac{\sqrt{a}Q_{1}(z+c)}{i\sqrt{b}P(z+c)\left(Q_{1}(z)\frac{\partial
h(z)}{\partial z_{i}}+\frac{\partial Q_{1}}{\partial
z_{i}}\right)}e^{h_{1}(z+c)-h_{1}(z)}\vspace{2mm}\\\
g_{2}(z)=-\dfrac{\sqrt{a}Q_{2}(z+c)}{i\sqrt{b}P(z+c)\left(Q_{1}(z)\frac{\partial
h(z)}{\partial z_{i}}+\frac{\partial Q_{1}}{\partial
z_{i}}\right)}e^{h_{2}(z+c)-h_{1}(z)}\vspace{2mm}\\\
g_{3}(z)=-\dfrac{Q_{2}(z)\frac{\partial h_{2}}{\partial z_{i}}+\frac{\partial
Q_{2}}{\partial z_{i}}}{Q_{1}(z)\frac{\partial h_{2}}{\partial
z_{i}}+\frac{\partial Q_{1}}{\partial
z_{i}}}e^{h_{2}(z)-h_{1}(z)}.\end{cases}$

To complete the proof, it suffices to discuss the following cases:

Case I. If $e^{h_{2}(z)-h_{1}(z)}$ is non-constant, then it follows that the
polynomial $h_{2}(z)-h_{1}(z)$ must be non-constant. We claim that $g_{2}$ and
$g_{3}$ are non-constants. Otherwise, if $g_{2}(z)=C_{2}$ and
$g_{3}(z)=C_{3}$, where $C_{2}$ and $C_{3}$ are constants, then from (4.18),
it follows that

$\displaystyle\begin{cases}-\dfrac{\sqrt{a}Q_{2}(z+c)}{iC_{2}\sqrt{b}P(z+c)\left(Q_{1}(z)\frac{\partial
h(z)}{\partial z_{i}}+\frac{\partial Q_{1}}{\partial
z_{i}}\right)}=e^{h_{1}(z)-h_{2}(z+c)}\vspace{2mm}\\\
\mbox{and}\;\;\;\;-\dfrac{Q_{2}(z)\frac{\partial h_{2}}{\partial
z_{i}}+\frac{\partial Q_{2}}{\partial
z_{i}}}{C_{3}\left(Q_{1}(z)\frac{\partial h_{2}}{\partial
z_{i}}+\frac{\partial Q_{1}}{\partial
z_{i}}\right)}=e^{h_{1}(z)-h_{2}(z)}.\end{cases}$

It is easy to see that right side is a transcendental entire function but left
side is a rational functions; both these equations leads us to a
contradiction.

In view of Lemma 4.3, we must have $g_{1}(z)\equiv 1$ i.e., we have

(4.19)
$\displaystyle\dfrac{\sqrt{a}Q_{1}(z+c)}{i\sqrt{b}P(z+c)\left(Q_{1}(z)\frac{\partial
h(z)}{\partial z_{i}}+\frac{\partial Q_{1}}{\partial z_{i}}\right)}\equiv
e^{h_{1}(z)-h_{1}(z+c)}.$

Evidently, it follows from (4.19) that $h_{1}(z)-h_{1}(z+c)$ must be constant
and hence $e^{h_{1}(z)-h_{1}(z+c)}$ becomes a constant, say $A_{11}$,
otherwise, we arrive at a contradiction. Assume that

(4.20)
$\displaystyle\dfrac{\sqrt{a}Q_{1}(z+c)}{i\sqrt{b}P(z+c)\left(Q_{1}(z)\frac{\partial
h(z)}{\partial z_{i}}+\frac{\partial Q_{1}}{\partial z_{i}}\right)}=A_{11}.$

In view of (4.17) and (4.19), a simple computation shows that

(4.21)
$\displaystyle-\dfrac{\sqrt{a}Q_{2}(z+c)}{i\sqrt{b}P(z+c)\left(Q_{2}(z)\frac{\partial
h_{2}}{\partial z_{i}}+\frac{\partial Q_{2}}{\partial
z_{i}}\right)}=e^{h_{2}(z)-h_{2}(z+c)}.$

Similarly, it follows from (4.21) that $h_{2}(z)-h_{2}(z+c)$ must be a
constant, Hence, $e^{h_{2}(z)-h_{2}(z+c)}$ must be a constant, say $A_{12}$.
Suppose that

(4.22)
$\displaystyle-\dfrac{\sqrt{a}Q_{2}(z+c)}{i\sqrt{b}P(z+c)\left(Q_{2}(z)\frac{\partial
h_{2}}{\partial z_{i}}+\frac{\partial Q_{2}}{\partial z_{i}}\right)}=A_{12}.$

Rewriting (4.20), we get

(4.23) $\displaystyle
P(z+c)=\frac{\sqrt{a}Q_{1}(z+c)}{A_{11}i\sqrt{b}\left(Q_{1}(z)\frac{\partial
h_{1}}{\partial z_{i}}+\frac{\partial Q_{1}}{\partial z_{i}}\right)}$

which shows that $Q_{1}(z)$ and $\frac{\partial h_{1}}{\partial z_{i}}$ must
be constant, say $\alpha_{1}$ and $\beta_{1}$, respectively. Then
$h_{1}(z)=\beta_{1}z_{i}+\phi(w)$, where
$w=(z_{1},\ldots,z_{i-1},z_{i+1},\ldots,z_{n})$. Thus, it follows from (4.22)
that
$P(z+c)=\frac{\sqrt{a}\alpha_{1}}{i\sqrt{b}A_{11}\alpha_{1}\beta_{1}}=P(z)$.
Moreover, from (4.22), we obtain

$\displaystyle
P(z+c)=-\frac{\sqrt{a}Q_{2}(z+c)}{i\sqrt{b}A_{12}\left(Q_{2}(z)\frac{\partial
h_{2}}{\partial z_{i}}+\frac{\partial Q_{2}}{\partial z_{i}}\right)}$

and from this, we conclude that $Q_{2}(z)$ and $\frac{\partial h_{2}}{\partial
z_{i}}$ are non-zero constant, say $\alpha_{2}$ and $\beta_{2}$, respectively.
Thus $h_{2}(z)=\beta_{2}z_{i}+\psi(w)$ and this shows that
$P(z+c)=-\frac{\sqrt{a}\alpha_{2}}{i\sqrt{b}\alpha_{2}\beta_{2}A_{12}}=P(z).$
Consequently, we have

$\displaystyle\frac{\sqrt{a}\alpha_{1}}{i\sqrt{b}\alpha_{1}\beta_{1}A_{11}}=-\frac{\sqrt{a}\alpha_{2}}{i\sqrt{b}\alpha_{2}\beta_{2}A_{12}}$

and this implies that $A_{11}\beta_{1}+A_{12}\beta_{2}=0$. Moreover,
$e^{h_{1}(z)-h_{1}(z+c)}=A_{11}$ implies that $h_{1}(z)-h_{1}(z+c)=\ln
A_{11}+2k\pi i$; $k\in\mathbb{Z}$. That is

$\displaystyle\phi(w+c)-\phi(w)=-(\beta_{1}c_{i}+\ln A_{11}+2k\pi
i)=\mbox{constant}.$

Hence,
$\phi(w)=a_{1}z_{1}+\cdots+a_{i-1}z_{i-1}+a_{i+1}z_{i+1}+\cdots+a_{n}z_{n}+r_{1}$.
Therefore,
$h_{1}(z)=a_{1}z_{1}+\cdots+a_{i-1}z_{i-1}+\beta_{1}z_{i}+a_{i+1}z_{i+1}+\cdots+a_{n}z_{n}+r_{1}$.
Similarly, we obtain that
$h_{2}(z)=b_{1}z_{1}+\cdots+b_{i-1}z_{i-1}+\beta_{2}z_{i}+b_{i+1}z_{i+1}+\cdots+b_{n}z_{n}+r_{2}$.
Consequently,

$\displaystyle\beta_{2}e^{(a_{1}-b_{1})c_{1}+\cdots+(a_{i-1}-b_{i-1})c_{i-1}+(\beta_{1}-\beta_{2})c_{i}+(a_{i+1}-b_{i+1})c_{i+1}+\cdots+(a_{n}-b_{n})c_{n}}+\beta_{1}=0.$

Case-II: If $e^{h_{2}(z)-h_{1}(z)}$ is a constant, then $h_{2}(z)-h_{1}(z)$
must be a constant. Then by (4.17), $e^{h_{1}(z+c)-h_{1}(z)}$ is also a
constant, otherwise, we get contradiction. Hence, we deduce that
$h_{1}(z)=L_{1}(z)+H(s)+r_{3}$ and $h_{2}(z)=L_{1}(z)+H(s)+r_{4}$, where
$L_{1}(z)=\xi_{1}z_{1}+\cdots+\xi_{n}z_{n}$, $H(s)$ is a polynomial in
$s=d_{1}z_{1}+\cdots+d_{n}z_{n}$ such that $d_{1}c_{1}+\cdots+d_{n}c_{n}=0$
with $H(z+c)=H(z)$ and $r_{3},r_{4}\in\mathbb{C}$. Therefore, from (4.15), it
is easy to see that

$\displaystyle g(z)=2L_{1}(z)+2H(s)+r_{5},\;\mbox{where}\;r_{3}+r_{4}=r_{5}.$

Indeed, we have the form of the solution as

$\displaystyle f(z)$
$\displaystyle=\frac{Q_{1}(z-c)e^{L_{1}(z-c)+H(s)+r_{3}}+Q_{2}(z-c)e^{L_{1}(z-c)+H(s)+r_{4}}}{2\sqrt{a}}$
$\displaystyle=\gamma(z-c)e^{L_{1}(z-c)+H(s)},$

where $\gamma(z)$ satisfies

$\displaystyle
a\gamma^{2}(z)+bP^{2}(z)e^{-2L_{1}(c)}\left(\frac{\partial\gamma(z-c)}{\partial
z_{i}}+(\xi_{i}+d_{i})\gamma(z-c)\right)^{2}=e^{r_{5}}Q(z).$

This completes the proof. ∎

###  4.3. Proof of Theorem 3.3

###### Proof of Theorem 3.3.

Suppose that $f(z)$ is a transcendental entire solution of finite order of the
equation (2.4). First of all, we write (2.4) as

(4.24) $\displaystyle aF^{2}+2\omega FG+bG^{2}=1,$

where $F$ and $G$ are defined by

(4.25) $\displaystyle
F=\dfrac{f(z)}{e^{\frac{g(z)}{2}}}\;\;\;\;\mbox{and}\;\;\;\;G=\dfrac{\gamma_{1}f(z+c)+\gamma_{2}f(z)}{e^{\frac{g(z)}{2}}}$

Further, we can transform (4.1) to be

$\displaystyle(\sqrt{a}F-\omega_{1}\sqrt{b}G)(\sqrt{a}F-\omega_{2}\sqrt{b}G)=1.$

Since $f$ is a finite order transcendental entire function and $g$ is a
polynomial, by Lemmas 4.1 and 4.2, we see that there exists a polynomial $p$
in $\mathbb{C}^{n}$ such that

(4.26)
$\displaystyle\sqrt{a}F-\omega_{1}\sqrt{b}G=e^{p}\;\;\;\;\mbox{and}\;\;\;\;\sqrt{a}F-\omega_{2}\sqrt{b}G=e^{-p}.$

Using (4.25) and (4.26), a simple computation shows that

(4.27) $\displaystyle
f(z)=\dfrac{\omega_{2}e^{p(z)}-\omega_{1}e^{-p(z)}}{\sqrt{a}(\omega_{2}-\omega_{1})}e^{\frac{g(z)}{2}},$
(4.28)
$\displaystyle\gamma_{1}f(z+c)+\gamma_{2}f(z)=\dfrac{e^{p(z)}-e^{-p(z)}}{\sqrt{b}(\omega_{2}-\omega_{1})}e^{\frac{g(z)}{2}}.$

For brevity, let

(4.29) $\displaystyle
h_{1}(z)=\dfrac{g(z)}{2}+p(z)\;\;\;\;\mbox{and}\;\;\;\;h_{2}(z)=\dfrac{g(z)}{2}-p(z).$

The equations (4.27) and (4.28) can be written as

(4.30) $\displaystyle
f(z)=\dfrac{\omega_{2}e^{h_{1}(z)}-\omega_{1}e^{h_{2}(z)}}{\sqrt{a}(\omega_{2}-\omega_{1})}$
(4.31)
$\displaystyle\gamma_{1}f(z+c)+\gamma_{2}f(z)=\dfrac{e^{h_{1}(z)}-e^{h_{2}(z)}}{\sqrt{b}(\omega_{2}-\omega_{1})}.$

Combining (4.30) and (4.31), an easy computation shows that

(4.32) $\displaystyle
R_{11}e^{h_{1}(z)-h_{1}(z+c)}+R_{12}e^{h_{2}(z)-h_{1}(z+c)}+R_{13}e^{h_{2}(z+c)-h_{1}(z+c)}\equiv
1,$

where

$\displaystyle
R_{11}=-\dfrac{(\omega_{2}\gamma_{2}\sqrt{b}-\sqrt{a})}{\omega_{2}\gamma_{1}\sqrt{b}},\;\;R_{12}=\dfrac{(\omega_{1}\gamma_{2}\sqrt{b}-\sqrt{a})}{\omega_{2}\gamma_{1}\sqrt{b}}\;\;\mbox{and}\;\;R_{13}=\dfrac{\omega_{1}}{\omega_{2}}.$

Case A: If $e^{h_{2}(z+c)-h_{1}(z+c)}$ is a constant, then
$h_{2}(z+c)-h_{1}(z+c)$ must be a constant. We set $h_{2}(z+c)-h_{1}(z+c)=k$,
where $k\in\mathbb{C}$. It follows from (4.29), that $p(z)$ is a constant. Let
$\xi=e^{p(z)}$, then from (4.27) and (4.28), it is easy to see that

(4.33) $\displaystyle
f(z)=B_{1}e^{\frac{g(z)}{2}}\;\;\mbox{and}\;\;\gamma_{1}f(z+c)+\gamma_{2}f(z)=B_{2}e^{\frac{g(z)}{2}},$

where

$\displaystyle
B_{1}=\dfrac{\omega_{2}\xi-\omega_{1}\xi^{-1}}{\sqrt{a}(\omega_{2}-\omega_{1})}\;\;\mbox{and}\;\;B_{2}=\dfrac{\xi-\xi^{-1}}{\sqrt{b}(\omega_{2}-\omega_{1})}.$

It is easy see that $B_{1}\neq 0$ and $B_{2}\neq 0$. From (4.33), we see that

(4.34)
$\displaystyle\frac{\gamma_{1}B_{1}}{(B_{2}-\gamma_{2}B_{1})}e^{\frac{g(z+c)-g(z)}{2}}=1.$

Since $g(z)$ is a polynomial, then (4.34) implies $g(z+c)-g(z)=\eta$, where
$\eta$ is a constant in $\mathbb{C}$. Therefore, it follows that
$g(z)=L(z)+H(s)+B_{3}$, where $L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$ and $H(s)$
is a polynomial in $s:=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in $\mathbb{C}^{n}$ with
$d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ with $H(z+c)=H(z)$, $B_{3}\in\mathbb{C}$.
Substituting $g(z)$ into (4.34) a simple computation shows that

$\displaystyle
e^{\frac{a_{1}c_{1}+\cdots+a_{n}c_{n}}{2}}=\frac{(\omega_{1}\gamma_{2}\sqrt{b}-\sqrt{a})-(\omega_{2}\gamma_{2}\sqrt{b}-\sqrt{a})\xi^{2}}{\gamma_{1}\sqrt{b}(\omega_{2}\xi^{2}-\omega_{1})}.$

Therefore, in view of first equation of (4.33), we obtain

$\displaystyle
f(z)=\frac{\omega_{2}\xi^{2}-\omega_{1}}{\xi\sqrt{a}(\omega_{2}-\omega_{1})}e^{\frac{1}{2}(L(z)+H(s)+B_{3})}.$

Case B: If $e^{h_{2}(z+c)-h_{1}(z+c)}$ is not a constant, then obviously,
$R_{11}\equiv 0$ and $R_{12}\equiv 0$ cannot hold at the same time. Otherwise,
from (4.32) we see that $R_{13}e^{h_{2}(z+c)-h_{1}(z+c)}\equiv 1$, a
contradiction. If $R_{11}\equiv 0$ and $R_{12}\not\equiv 0$, then in view of
(4.32), we obtain

(4.35) $\displaystyle
R_{12}e^{h_{2}(z)-h_{1}(z+c)}+R_{13}e^{h_{2}(z+c)-h_{1}(z+c)}\equiv 1.$

Since $e^{h_{2}(z+c)-h_{1}(z+c)}$ is not a constant, it follows that
$e^{h_{2}(z)-h_{1}(z+c)}$ is not a constant. Moreover,
$e^{h_{2}(z+c)-h_{2}(z)}$ is not a constant. Otherwise,
$h_{2}(z+c)-h_{2}(z)=\xi_{1}$, where $\xi_{1}\in\mathbb{C}$. From (4.35), we
see that $(R_{12}e^{-\xi_{1}}+R_{13})e^{h_{2}(z+c)-h_{1}(z+c)}\equiv 1$, which
is a contradiction as $e^{h_{2}(z+c)-h_{1}(z+c)}$ is non-constant. Hence,
(4.35) can be written as

(4.36) $\displaystyle
R_{12}e^{h_{2}(z)}+R_{13}e^{h_{2}(z+c)}-e^{h_{1}(z+c)}\equiv 0.$

In view of Lemma 4.4, from (4.36), we get a contradiction. Similarly, if
$R_{11}\not\equiv 0$ and $R_{12}\equiv 0$, we get a contradiction. Thus, we
conclude that $R_{11}\not\equiv 0$ and $R_{12}\not\equiv 0$.

Since $h_{1}(z),h_{2}(z)$ are polynomials and $e^{h_{2}(z+c)-h_{1}(z+c)}$ is
non-constant, then by Lemma 4.3 and with the help of (4.32), we obtain

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

Sub-case B1: Assume that $R_{11}e^{h_{1}(z)-h_{1}(z+c)}\equiv 1$. From (4.32),
it yields that $-\frac{R_{12}}{R_{13}}e^{h_{2}(z)-h_{2}(z+c)}\equiv 1$. Since
$h_{1}(z),h_{2}(z)$ are polynomials, we see that
$h_{1}(z)-h_{1}(z+c)=\lambda_{1}$ and $h_{2}(z)-h_{2}(z+c)=\lambda_{2}$, where
$\lambda_{1},\lambda_{2}\in\mathbb{C}$. Thus, it follows that
$h_{1}(z)=L_{1}(z)+H_{1}(s)+D_{1}$ and $h_{2}(z)=L_{2}(z)+H_{2}(s)+D_{2}$,
where $L_{l}(z)=a_{l1}z_{1}+\cdots+a_{ln}z_{n}$ and $H_{l}(s)$ for $l=1,2$ are
polynomial in $s:=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in $\mathbb{C}^{n}$,
$d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ with $H_{l}(z+c)=H_{l}(z)$ for $l=1,2$, and
$D_{1},D_{2}\in\mathbb{C}$. Obviously, $L_{1}(z)+H_{1}(s)\neq
L_{2}(z)+H_{2}(s)$. Otherwise, we have $h_{2}(z+c)-h_{1}(z+c)$ is a constant,
which shows that $e^{h_{2}(z+c)-h_{1}(z+c)}$ is a constant, a contradiction.
Substituting $h_{1}(z)$ and $h_{2}(z)$ into
$R_{11}e^{h_{1}(z)-h_{1}(z+c)}\equiv 1$ and
$-\frac{R_{12}}{R_{13}}e^{h_{2}(z)-h_{2}(z+c)}\equiv 1$, we obtain

$\displaystyle-\frac{(\omega_{2}\gamma_{2}\sqrt{b}-\sqrt{a})}{\omega_{2}\gamma_{1}\sqrt{b}}e^{-L_{1}(c)}\equiv
1\;\;\;\;\mbox{and}\;\;\;\;-\frac{(\omega_{1}\gamma_{2}\sqrt{b}-\sqrt{a})}{\omega_{1}\gamma_{1}\sqrt{b}}e^{-L_{2}(c)}\equiv
1.$

Therefore, from (4.7) it follows that $f$ takes the form

$\displaystyle
f(z)=\dfrac{\omega_{2}e^{L_{1}(z)+H_{1}(s)+D_{1}}-\omega_{1}e^{L_{2}(z)+H_{2}(s)+D_{2}}}{\sqrt{a}(\omega_{2}-\omega_{1})}.$

Moreover, from (4.6) we see that

$\displaystyle g(z)=h_{1}(z)+h_{2}(z)=L(z)+H(s)+D,$

where $L(z):=L_{1}(z)+L_{2}(z)$, $H(s):=H_{1}(s)+H_{2}(s)$ and
$D:=D_{1}+D_{2}$.

Sub-case B2: Assume that $R_{12}e^{h_{2}(z)-h_{1}(z+c)}\equiv 1$. Then, from
(4.32) it is easy to see that
$-\frac{R_{11}}{R_{13}}e^{h_{1}(z)-h_{2}(z+c)}\equiv 1$. Since
$h_{1}(z),h_{2}(z)$ is a polynomial, it follows that
$h_{2}(z)-h_{1}(z+c)=\lambda_{3}$ and $h_{1}(z)-h_{2}(z+c)=\lambda_{4}$, where
$\lambda_{3},\lambda_{4}\in\mathbb{C}$. A simple computation shows that
$h_{1}(z+2c)-h_{1}(z)=-\lambda_{3}-\lambda_{4}$ and
$h_{2}(z+2c)-h_{2}(z)=-\lambda_{3}-\lambda_{4}$. Thus, we deduce that
$h_{1}(z)=L(z)+H(s)+D_{3}$ and $h_{2}(z)=L(z)+H(s)+D_{4}$, where
$L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$ and $H(s)$ is a polynomial in
$s:=d_{1}z_{1}+\cdots+d_{n}z_{n}$ in $\mathbb{C}^{n}$ with
$d_{1}c_{1}+\cdots+d_{n}c_{n}=0$ with $H(z+c)=H(z)$, and
$D_{3},D_{4}\in\mathbb{C}$. Now, we see that
$h_{2}(z+c)-h_{1}(z+c)=D_{4}-D_{3}$, which shows that
$e^{h_{2}(z+c)-h_{1}(z+c)}$ is a constant, a contradiction. This completes the
proof. ∎

##  5\. Declaration

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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# The Laplacian of The Integral Of The Logarithmic Derivative of the Riemann-
Siegel-Hardy Z-function

Stephen Crowley

###### Abstract

The integral
$R(t)=\pi^{-1}\int\frac{\mathrm{d}}{\mathrm{d}t}\operatorname{lnZ}(t)\mathrm{d}t=\pi^{-1}\left(\ln\zeta\left(\frac{1}{2}+it\right)+i\vartheta(t)\right)$
of the logarithmic derivative of the Hardy Z function
$Z(t)=e^{i\vartheta(t)}\zeta\left(\frac{1}{2}+it\right)$, where $\vartheta(t)$
is the Riemann-Siegel theta function, and $\zeta(t)$ is the Riemann zeta
function, is used as a basis for the construction of a pair of transcendental
entire functions $\nu(t)=-\nu(1-t)=-\left(\Delta
R\left(\frac{i}{2}-it\right)\right)^{-1}=-G\left(\frac{i}{2}-it\right)$ where
$G=-(\Delta R(t))^{-1}$ is the derivative of the additive inverse of the
reciprocal of the Laplacian $\Delta f(t)=\ddot{f}(t)$ of $R(t)$ and
$\chi(t)=-\chi(1-t)=\dot{\nu}(t)=-iH\left(\frac{i}{2}-it\right)$ where
$H(t)$=$\dot{G}(t)$ has roots at the local minima and maxima of $G(t)$. When
$H(t)=0$ and $\dot{H}(t)=\ddot{G}(t)=\Delta G(t)>0$, the point $t$ marks a
minimum of $G(t)$ where it coincides with a Riemann zero, i.e.,
$\zeta\left(\frac{1}{2}+it\right)=0$, otherwise when $H(t)=0$ and
$\dot{H}(t)=\Delta G(t)<0$, the point $t$ marks a local maximum $G(t)$,
marking midway points between consecutive minima. Considered as a sequence of
distributions or wave functions, $\nu_{n}(t)=\nu(1+2n+2t)$ converges to
$\nu_{\infty}(t)=\lim_{n\rightarrow\infty}\nu_{n}(t)=\sin^{2}(\pi t)$ and
$\chi_{n}(t)=\chi(1+2n+2t)$ to
$\chi_{\infty}(t)=\lim_{n\rightarrow\infty}\chi_{n}(t)=-8\cos(\pi t)\sin(\pi
t)$

###### Contents

  1. 1 Derivations
    1. 1.1 Standard Definitions
    2. 1.2 The Logarithmic Derivative of $Z(t)$ and its Integral
    3. 1.3 The Laplacian
  2. 2 Appendix
    1. 2.1 Wave Mechanics
    2. 2.2 The One-Dimensional Wave Equation
    3. 2.3 The Poisson Bracket and Lagrangian Mechanics
      1. 2.3.1 The Euler-Lagrange equation
      2. 2.3.2 Quantum Mechanics of General Lagrangian Systems
      3. 2.3.3 Noether’s Theorem and Lie Groups
    4. 2.4 Schroedinger’s Time-Dependent Equation and Nonstationary Wave Motion
      1. 2.4.1 Operators and Observables: Dirac’s Time-Dependent Theory
      2. 2.4.2 The String Theoretic Partition Function

##  1 Derivations

###  1.1 Standard Definitions

Let $\zeta(t)$ be the Riemann zeta function

$\begin{array}[]{lll}\zeta(t)&=\sum_{n=1}^{\infty}n^{-s}&\forall\operatorname{Re}(s)>1\\\
&=(1-2^{1-s})\sum_{n=1}^{\infty}n^{-s}(-1)^{n-1}&\forall\operatorname{Re}(s)>0\end{array}$
(1)

and $\vartheta(t)$ be Riemann-Siegel vartheta function $\vartheta(t)$

$\vartheta(t)=-\frac{i}{2}\left(\ln\Gamma\left(\frac{1}{4}+\frac{it}{2}\right)-\ln\Gamma\left(\frac{1}{4}-\frac{it}{2}\right)\right)-\frac{\ln(\pi)t}{2}$
(2)

where $\arg(z)=\frac{\ln(z)-\ln(\bar{z})}{2i}$ and
$\overline{\Gamma(z)}=\Gamma(\bar{z})$. The Hardy $Z$ function[Ivi13] can then
be written as

$Z(t)=e^{i\vartheta(t)}\zeta\left(\frac{1}{2}+it\right)$ (3)

which can be mapped isometrically back to the $\zeta$ function

$\zeta(t)=e^{-i\vartheta\left(\frac{i}{2}-it\right)}Z\left(\frac{i}{2}-it\right)$
(4)

due to the isometry

$t=\frac{1}{2}+i\left(\frac{i}{2}-it\right)$ (5)

of the Mobius transforms111Thanks to Matti Pitkänen for pointing out this is a
Mobius transform pair, among other things
$f(t)=\frac{at+b}{ct+d}=\left(\begin{array}[]{cc}a&b\\\ c&d\end{array}\right)$
with

$\left(\begin{array}[]{cc}-i&\frac{i}{2}\\\
0&1\end{array}\right)\operatorname{and}\operatorname{its}\operatorname{inverse}\left(\begin{array}[]{cc}i&\frac{1}{2}\\\
0&1\end{array}\right)$ (6)

making possible the Riemann-Siegel-Hardy correspondence. Furthermore, let
$S(t)$ be argument of $\zeta$ normalized by $\pi$ defined by

$\begin{array}[]{ll}S(t)&=\pi^{-1}\arg\left(\zeta\left(\frac{1}{2}+it\right)\right)\\\
&=-\frac{i}{2\pi}\left(\ln\zeta\left(\frac{1}{2}+it\right)-\ln\zeta\left(\frac{1}{2}-it\right)\right)\end{array}$
(7)

The Bäcklund counting formula gives the exact number of zeros on the critical
strip up to level $t$, not just on the criticial line,

$N(t)=\operatorname{Im}(R(t))=\frac{\vartheta(t)}{\pi}+1+S(t)$ (8)

The relationship between the functions $N(t)$, $S(t)$, and $Z(t)$ is
demonstrated by

$\ln\zeta\left(\frac{1}{2}+it\right)=\ln|Z(t)|+i\pi S(t)$ (9)

These formulas are true independent of the Riemann hypothesis which posits
that all complex zeros of $\zeta(s+it)$ have real part $s=\frac{1}{2}$.
[Ivi13, Corrollary 1.8 p.13]

###  1.2 The Logarithmic Derivative of $Z(t)$ and its Integral

Let $Q(t)$ be the logarithmic derivative of $Z(t)$ given by

$\begin{array}[]{ll}Q(t)&=\frac{\dot{\mathrm{Z}}(t)}{Z(t)}\\\
&=\frac{\mathrm{d}}{\mathrm{d}t}\operatorname{lnZ}(t)\\\
&=i\left(\frac{\dot{\zeta}\left(\frac{1}{2}+it\right)}{\zeta\left(\frac{1}{2}+it\right)}+\frac{\Psi\left(\frac{1}{4}-\frac{i}{2t}\right)+\Psi\left(\frac{1}{4}+\frac{i}{2t}\right)-2\ln(\pi)}{4}\right)\\\
&=i\left(\frac{\dot{z}(t)}{z(t)}+\frac{g^{-}(t)+g^{+}(t)-2\ln(\pi)}{4}\right)\end{array}$
(10)

where

$\Psi(x)=\frac{\mathrm{d}}{\mathrm{d}x}\ln(\Gamma(x))=\frac{\frac{\mathrm{d}}{\mathrm{d}x}\Gamma(x)}{\Gamma(x)}$
(11)

is the digamma function, the logarithmic derivative of the $\Gamma$ function
and $z(t)$ and $g^{\pm}(t)$ have been introduced to simplify the expressions.
The function $Q(t)$ has singularities at $\pm\frac{i}{2}(4n-3)$ with residues

$\underset{t=\frac{i}{2}(4n-3)}{\operatorname{Res}(Q(t))}=\frac{-8\zeta(2-2n)n^{2}+4\zeta(2-2n)n}{16\zeta(2-2n)n^{2}-8\zeta(2-2n)n}=-\frac{1}{2}$
(12)

and

$\underset{t=-\frac{i}{2}(4n-3)}{\operatorname{Res}(Q(t))}=\frac{-2\zeta(2n-1)+4\zeta(2n-1)n}{8\zeta(2n-1)n-4\zeta(2n-1)}=\frac{1}{2}$
(13)

Now, the integral of the logarithmic derivative of Z is defined by

$\begin{array}[]{cl}R(t)&=\pi^{-1}\int Q(t)\mathrm{d}t\\\
&=\pi^{-1}\int\frac{\dot{\mathrm{Z}}(t)}{Z(t)}\mathrm{d}t\\\
&=\pi^{-1}\int\frac{\mathrm{d}}{\mathrm{d}t}\operatorname{lnZ}(t)\mathrm{d}t\\\
&=\pi^{-1}\int
i\left(\frac{\dot{z}(t)}{z(t)}+\frac{g^{-}(t)+g^{+}(t)-2\ln(\pi)}{4}\right)\mathrm{d}t\\\
&=\pi^{-1}\int
i\left(\frac{\dot{\zeta}\left(\frac{1}{2}+\operatorname{it}\right)}{\zeta\left(\frac{1}{2}+it\right)}+\frac{\Psi\left(\frac{1}{4}-\frac{i}{2t}\right)+\Psi\left(\frac{1}{4}+\frac{i}{2t}\right)-2\ln(\pi)}{4}\right)\mathrm{d}t\\\
&=\pi^{-1}\left(\ln\zeta\left(\frac{1}{2}+it\right)+i\vartheta(t)\right)\\\
&=\pi^{-1}\left(\ln\zeta\left(\frac{1}{2}+it\right)+\frac{i}{2}\left(i\left(\ln\Gamma\left(\frac{1}{4}-\frac{it}{2}\right)-\ln\Gamma\left(\frac{1}{4}+\frac{it}{2}\right)\right)-\ln(\pi)t\right)\right)\end{array}$
(14)

###  1.3 The Laplacian

The Laplacian $=\nabla\cdot\nabla=\nabla^{2}$ is a differential operator which
corresponds to the divergence of the gradient of a function $f(x)$ on a
Euclidean space $x\in X\underline{\subset}\mathbbm{R}^{d}$ and is denoted by

$f(x)=\sum_{i=1}^{d}\frac{\partial}{\partial x^{2}_{i}}f(x_{i})\forall x\in
X\subseteq\mathbbm{R}^{d}$ (15)

and is simply the second derivative of a function when $d=1$

$f(x)=\ddot{f}(x)=\frac{\mathrm{d}}{\mathrm{d}x^{2}}f(x)\forall x\in
X\subseteq\mathbbm{R}$ (16)

Let $G(t)$ be the additive inverse of the reciprocal(also known as
multiplicative inverse) of the Laplacian of $R(t)$ interpreted as a partition
function

$\begin{array}[]{ll}G(t)&=-\frac{1}{\Delta R(t)}\\\
&=\frac{8z(t)^{2}\pi}{z(t)^{2}(\dot{g}^{-}(t)-\dot{g}^{+}(t))-8(z(t)\ddot{z}(t)+\dot{z}(t)^{2})}\end{array}$
(17)

---
Figure 1: The Real and Imaginary Parts of $R(t)$ compared with $G(t)$

Now, let $H(t)$ be the derivative of $G(t)$ given by

$\begin{array}[]{ll}H(t)&=\dot{G}(t)\\\
&=\frac{\mathrm{d}}{\mathrm{d}t}\left(-\frac{1}{\Delta R(t)}\right)\\\
&=-4i\pi z(t)\frac{z(t)^{3}\Delta
g(t)+8(2z(t)^{2}\dddot{z}(t)-6z(t)\dot{z}(t)\ddot{z}(t)-4\dot{z}(t)^{3})}{(z(t)^{2}+8(z(t)\ddot{z}(t)-\dot{z}(t)^{2}))^{2}}\end{array}$
(18)

When $H(t)=0$ and $\dot{H}(t)=\ddot{G}(t)=\Delta G(t)>0$, the point $t$ marks
a minimum of $G(t)$ where it coincides with a Riemann zero, i.e.,
$\zeta\left(\frac{1}{2}+it\right)=0$, otherwise when $H(t)=0$ and
$\dot{H}(t)=\Delta G(t)<0$, the point $t$ marks a local maximum $G(t)$,
marking midway points between consecutive minima.

Figure 2: $G(t),H(t)\operatorname{and}\dot{H}(t)$

With the identity $t=\frac{1}{2}+i\left(\frac{i}{2}-it\right)$, define
$\nu(t)$ as the Mobius transform of the partition function

$\nu(t)=-\left(\Delta
R\left(\frac{i}{2}-it\right)\right)^{-1}=-G\left(\frac{i}{2}-it\right)$ (19)

then $\nu(t)$ has zeros at the positive odd integers, zero, and negative even
integers. In the same way, let

$\begin{array}[]{ll}\chi(t)&=\dot{\nu}(t)\\\
&=-iH\left(\frac{i}{2}-it\right)\\\
&=-4i\pi\zeta(t)\frac{\zeta(t)^{3}\left(\ddot{\Psi}\left(\frac{t}{2}\right)-\ddot{\Psi}\left(\frac{1}{2}-\frac{t}{2}\right)\right)-8(6\zeta(t)\dot{\zeta}(t)\ddot{\zeta}(t)+2\zeta(t)^{{}^{2}}\dddot{\zeta}(t)+4\dot{\zeta}(t)^{3})}{\zeta(t)^{2}\left(\left(\dot{\Psi}\left(\frac{t}{2}\right)-\dot{\Psi}\left(\frac{1}{2}-\frac{t}{2}\right)\right)+8(\zeta(t)\ddot{\zeta}(t)-\dot{\zeta}(t)^{2})\right)^{2}}\end{array}$
(20)

which also has zeros on the real line at the positive odd integers, zero, and
the negative even integers.

$\begin{array}[]{rll}\lim_{t\rightarrow 2n-1}\nu(t)&=\lim_{t\rightarrow
2n-1}\chi(t)&=0\forall n>0\\\ \lim_{t\rightarrow 0}\nu(t)&=\lim_{t\rightarrow
0}\chi(t)&=0\\\
\lim_{t\rightarrow-2n}\nu(t)&=\lim_{t\rightarrow-2n}\chi(t)&=0\forall
n<0\end{array}$ (21)

Both $\nu(t)$ and $\chi(t)$ satisify similiar functional equations

$\nu(t)=\nu(1-t)$ (22)

and

$\chi(t)=-\chi(1-t)$ (23)

So the even $\mu(t)$ and odd $\psi(t)$ transcendental entire functions can be
defined

$\mu(t)=\nu\left(t+\frac{1}{2}\right)$ (24)
$\psi(t)=\chi\left(t+\frac{1}{2}\right)$ (25)

which satisify the functional symmetries

$\mu(t)=\mu(-t)$ (26)

and

$\psi(t)=-\psi(-t)$ (27)

The function $\chi(t)$ has as a subset of its roots, the roots of the Rieman
zeta function $\zeta(t)$, the converse is not true, since $\chi(t)$ is a
function of $\zeta(t)$ and its first, second, and third derivatives.

$\\{t:\zeta(t)=0\\}\subset\\{t:\chi(t)=0\\}$ (28)

Let

$\chi_{n}(t)=\chi(1+2n+2t)$ (29)

and

$\nu_{n}(t)=\nu(1+2n+2t)$ (30)

which both satisify the symmetries

$\chi_{-\frac{1}{2}}\left(\frac{t}{2}\right)=\chi(t)$ (31)
$\nu_{-\frac{1}{2}}\left(\frac{t}{2}\right)=\nu(t)$ (32)

as well as

$\chi_{n}\left(-\frac{1}{2}\right)=\chi(2n)$ (33)
$\nu_{{}_{n}}\left(-\frac{1}{2}\right)=\nu(2n)$ (34)

The sequence of wave functions $\nu_{n}(t)$ converges, thanks to quantum
ergodicity, to

$\nu_{\infty}(t)=\lim_{n\rightarrow\infty}\nu_{n}(t)=\sin^{2}(\pi t)$ (35)

which has a limiting maximum

$\begin{array}[]{ll}\lim_{n\rightarrow\infty}\max_{0<t<1}\nu_{n}(t)&=\max_{0<t<1}\nu_{\infty}(t)\\\
&=\lim_{n\rightarrow\infty}\nu(2n)\\\ &=\frac{8}{\pi}\\\ &\cong
2.546479089470\ldots\end{array}$ (36)

and the associated differential is

$\begin{array}[]{cl}\chi_{\infty}(t)&=\lim_{n\rightarrow\infty}\chi_{n}(t)\\\
&=\frac{4}{\pi}\frac{\mathrm{d}}{\mathrm{d}t}\nu_{\infty}(t)\\\
&=\frac{4}{\pi}\frac{\mathrm{d}}{\mathrm{d}t}\sin^{2}(\pi t)\\\ &=-8\sin(\pi
t)\cos(\pi t)\end{array}$ (37)

It is worth mentioning that the pair-correlation function for the zeros of
$\zeta(s)$ is

$r_{2}(x)=1-\frac{\sin^{2}(\pi x)^{2}}{\pi^{2}x^{2}}$

if the Riemann hypothesis is true. [11] The point $n=1$ is where
$\chi_{n}\left(-\frac{1}{2}\right)$ and $\nu_{n}\left(-\frac{1}{2}\right)$
attain their greatest values among the integers.

---
$n$ | $\nu_{n}\left(-\frac{1}{2}\right)$ | $\chi_{n}\left(-\frac{1}{2}\right)$
---|---|---
$1$ | 117.43532857805377782 | 9447.7593604718560
$2$ | 003.03320654562255410 | 0000.2816402783351
$3$ | 002.77176105375846239 | 0000.0589526080385
$4$ | 002.69653220844944185 | 0000.0240373109008
5 | 002.66095375057354766 | 0000.0132398305603
$6$ | 002.63970550787458574 | 0000.0088555925215
$7$ | 002.62532107269483772 | 0000.0060558883634
… | $\ldots\ldots\ldots\ldots\ldots\ldots$ | $\ldots\ldots\ldots\ldots\ldots\ldots\ldots$
$\infty$ | $\frac{8}{\pi}$ | 0

Table 1: Even values of $\nu(n)$ and $\chi(n)$

This phenomena of having the first wave function having a much larger size
than all of the remainders is mentioned in [Kna99, Theorem 22] The convergence
of $\lim_{n\rightarrow\infty}\nu_{n}(t)\rightarrow\nu_{\infty}(t)$ and
$\lim_{n\rightarrow\infty}\chi_{n}(t)\rightarrow\chi_{\infty}(t)$ might be
interpreted as a manifestation of quantum ergodicity [11, C p.16]. There is an
essential singularity[CB90, 54 p.169] at $s_{0}\cong 1.98757823\ldots$

$\lim_{s\rightarrow s_{0}}\chi(s)=\infty$ (38)

We also have the limits

$\lim_{s\rightarrow\\{0,1\\}}\dot{\chi}(s)=4\pi$ (39)

The integral over this sine-squared kernel is

$\int_{0}^{1}\sin^{2}(\pi t)\mathrm{d}t=\frac{1}{2}$ (40)

whereas

$\frac{}{}\int_{0}^{1}\frac{\nu(t)}{\nu\left(\frac{1}{2}\right)}\mathrm{d}t\cong.46693755153559653755\ldots$
(41)

Figure 3: Top Left: The function $\nu(s)$ normalized by its maximum value on
$[0,1]$ Top Right: The function $\nu(s)$ normalized by its maximum value and
subtracted from $\sin^{2}(\pi x)$ Bottom Left: Convegence of
$\chi_{n}(t)\rightarrow-8\cos(\pi t)\sin(\pi t)$ Bottom Right: Convegence of
$\nu_{n}(t)\rightarrow\sin^{2}(\pi t)\frac{\pi}{8}$

##  2 Appendix

###  2.1 Wave Mechanics

###  2.2 The One-Dimensional Wave Equation

The one-dimensional wave equation is

$-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{x^{2}}u(x,t)+V(x,t)u(t,x)=-\frac{\hbar}{i}\frac{\partial}{\partial
t}u(x,t)$ (42)

[Flü94, II.A.4 p.25][Gan06, 4.2.1 p.][Coo03, Ch. 8 p.147]

###  2.3 The Poisson Bracket and Lagrangian Mechanics

“The” Poisson bracket, expressed with Einstein summation convention (for the
repeated index $i$)

$\\{A,B\\}\equiv\frac{\partial A}{\partial p_{i}}\frac{\partial B}{\partial
q_{i}}-\frac{\partial B}{\partial p_{i}}\frac{\partial A}{\partial q_{i}}$
(43)

has the antisymmetry property

$\\{A,B\\}=-\\{B,A\\}$ (44)

and the so-called Jacobi identity

$\\{A,\\{B,C\\}\\}-\\{B,\\{A,C\\}\\}+\\{C,\\{A,B\\}\\}=0$ (45)

Two quantitities $A$ and $B$ are said to commute if their
$\operatorname{Poisson}\operatorname{bracket}$ $\\{A,B\\}$ vanishes, that is,
$\\{A,B\\}=0$. Hamilton’s equations of motion for the system

$\dot{p}_{i}=-\frac{H}{q_{i}}$ (46) $\dot{q}_{i}=\frac{\partial H}{\partial
q_{i}}$ (47)

where $H$ is a Legendre-transformed function of the Lagrangian called the
Hamiltonian

$H\equiv\frac{\partial
L}{\partial\dot{q}_{i}}\dot{q}_{i}-L(q_{i},\dot{q}_{i},t)$ (48)

whose value for any given time $t$ gives the energy

$\mathcal{A}[q_{i}]=\int_{t_{a}}^{t_{b}}L(q_{i},\dot{q}_{i},t)\mathrm{d}t$
(49)

of the system where $L(q_{i},\dot{q}_{i},t)$ is the Lagrangian of the system
and

$q_{i}(t)=q_{i}^{\operatorname{cl}}(t)+\delta q_{i}(t)$ (50)

is an arbitrary path where $q_{i}^{\operatorname{cl}}(t)$ is the classical
orbit or classical path of the system and

$\delta q_{i}(t_{a})\equiv q_{i}(t)-q^{\operatorname{cl}}_{i}(t)$ (51)

[Kle04, 1.1]

####  2.3.1 The Euler-Lagrange equation

The Euler-Lagrange equation

$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial
L}{\partial\dot{q}_{i}}\right)=\frac{\partial L}{\partial q_{i}}$ (52)

indicates that the action $S$ given by

$S=\int_{t_{1}}^{t_{2}}L(t)\mathrm{d}t$ (53)

where

$L(t)=T-V(x(t))$ (54)

is the Lagrangian and

$T(t)=\frac{1}{2}m\dot{x}(t)^{2}$ (55)

is the kinetic energy which is stationary for the physical solutions
$q_{i}(t)$.[Gan06, 4.2.1]

####  2.3.2 Quantum Mechanics of General Lagrangian Systems

The coordinate transformation

$x^{i}=x^{i}(q^{\mu})$ (56)

implies the relation

$\partial_{\mu}=\frac{\partial}{\partial q^{\mu}}=e^{i}_{\mu}(q)\partial_{i}$
(57)

between the derivatives $\partial_{\mu}$ and

$\partial_{i}\equiv\frac{\partial}{\partial x^{i}}$ (58)

where

$e^{i}_{\mu}(q)\equiv\partial_{\mu}x^{i}(q)$ (59)

is a transformation matrix called the basis p-ad where $p$ is the prefix
corresponding to $n$, the dimension of $x$, monad when $n=1$, dyad when $n=2$,
triad when $n=$3, and so on. Let

$e_{i}^{\mu}(q)=\frac{\partial q^{\mu}}{\partial x^{i}}$ (60)

be the inverse matrix called the reciprocal p-ad. The
$\operatorname{basis}p-\operatorname{ad}$ and its recpiprocal satisify the
orthogonality and completeness relations

$e^{i}_{\mu}e_{i}^{\nu}=\delta_{\mu}^{\nu}$ (61)

and

$e_{\mu}^{i}e_{j}^{\mu}=\delta_{j}^{i}$ (62)

The inverse of $\partial_{\mu}$ is

$\partial_{i}=e_{i}^{\mu}(q)\partial_{\mu}$ (63)

which is related to the curvilinear transform of the Cartesian quantum-
mechanical momentum operators by

$\hat{p}_{i}=-i\hbar\partial_{i}=-i\hbar e_{i}^{\mu}(q)\partial_{\mu}$ (64)

The Hamiltonian operator for free particles is defined by

$\hat{H}_{0}=\hat{T}=\frac{1}{2M}\hat{p}^{2}=-\frac{\hbar^{2}}{2M}\Delta$ (65)

where its metric tensor is given by

$g^{\mu\nu}(q)\equiv e^{\mu}_{i}(q)e^{\nu}_{i}(q)$ (66)

and its inverse by

$g_{\mu\nu}(q)\equiv e^{i}_{\mu}(q)e^{i}_{\nu}(q)$ (67)

The Laplacian of a metric tensor is then expressed

$\begin{array}[]{ll}\Delta&=\partial_{i}^{2}\\\
&=e^{i\mu}\partial_{\mu}e^{i\nu}\partial_{\nu}\\\
&=e^{i\mu}e^{i\nu}\partial_{\mu}\partial_{\nu}+(e^{i\mu}\partial_{\mu}e^{i\nu})\partial_{\nu}\\\
&=g^{\mu\nu}(q)\partial_{\mu}\partial_{\nu}-\Gamma_{\mu}^{\mu\nu}(q)\partial_{\nu}\end{array}$
(68)

where

$\Gamma_{\mu\nu}^{\lambda}(q)=-e^{i}_{v}(q)\partial_{\mu}e_{i}^{\lambda}(q)=e_{i}^{\lambda}(q)\partial_{u}e^{i}_{\nu}(q)$
(69)

is the affine-connection [Kle04, 1.13][Pod28]

####  2.3.3 Noether’s Theorem and Lie Groups

From Noether’s theorem it is known that continuous symmetries have
corresponding conservation laws.[Gan06, 4.2.1] Let

$\alpha_{s}$ (70)

be a continuous family of symmetries which is a $1$-parameter subgroup
$s\mapsto\alpha_{s}$ in the Lie group of symmetries. A Lie group is a group
whose operations are compatible with the smooth structure. A smooth structure
on a manifold allows for an unambiguous notion of smooth function.

###  2.4 Schroedinger’s Time-Dependent Equation and Nonstationary Wave Motion

The function $\cos^{2}(t)=1-\sin^{2}(t)$ arises in the Dirac phase averaving
method of calculating transition probabilities of non-stationary states in the
time-dependent Schroedinger equation. [Gol61, 15.1] 3

####  2.4.1 Operators and Observables: Dirac’s Time-Dependent Theory

Let

$y(x,t)=A\cos\left(\frac{2\pi px}{h}\right)\cos\left(\frac{2\pi\epsilon
t}{h}\right)$

where

$\epsilon=hv$ (71)

which relates the energies of a “matter wave” system, whatever that is, maybe
a fermionic system, to the frequencies of quanta it emits or absorbs. There is
always some arbitriness with choice of units. This wave function has the
symmetry

$\frac{\partial^{2}y(x,t)}{x^{2}}=-\left(\frac{2\pi}{h}\right)^{2}p^{2}y(x,t)$
(72)

so that the kinetic energy of a particle is obtained from the wave function

$-\frac{(2\pi)^{2}}{2m}\frac{\partial^{2}y(x,t)}{x^{2}}=\frac{p^{2}}{2m}y(x,t)$
(73)

which means formally that

$\frac{1}{2m}(p)^{2}y(x,t)=-\frac{1}{2m}\left(\frac{2\pi}{i}\frac{\partial}{\partial
x}\right)^{2}$ (74)

which suggests the identifiation of the Schroedinger momentum operator

$\left(\frac{2\pi}{i}\frac{\partial}{\partial x}\right)\leftrightarrow p$ (75)

The equation for “nonfree” particles is augmented by

$\frac{m}{2}\left(\frac{2\pi}{i}\frac{\partial}{\partial
x}\right)^{2}y(x,t)=[\epsilon-V(x)]y(x,t)$ (76)

where $\epsilon$ is the total energy of the particle and $V(x)$ is its
potential energy. The first Schroedinger equation is then expressed

$\epsilon
y(x,t)=\left[\frac{1}{2m}\left(\frac{2\pi}{i}\frac{\partial}{\partial
x}\right)^{2}+V(x)\right]y(x,t)=\left[\frac{1}{2m}(p)^{2}+V(x)\right]y(x,t)$
(77)

or

$H_{y(x,t)}=\epsilon y(x,t)$

where $H$ is the operator corresponding to the Hamiltonian of the (point)
particle. [Gol61, 11.5] The time-dependent Schroedinger equation is written

$\frac{\partial^{2}(x,t)}{\partial
t^{2}}=-\frac{1}{(2\pi)^{2}}\epsilon^{2}y(x,t)=-\left(\frac{H}{2\pi}\right)^{2}y(x,t)$
(78)

or equivalently as

$0=\left((2\pi)^{2}\frac{\partial^{2}}{\partial
t^{2}}+H^{2}\right)y(x,t)=\left(H+i2\pi\frac{\partial}{\partial
t}\right)\left(H-i2\pi\frac{\partial}{\partial t}\right)y(x,t)$ (79)

where

$Hy(x,t)=i2\pi\frac{\partial}{\partial t}y(x,t)$ (80)

It is also worth mentioning that in Dirac’s theory of the time-dependent
Schroedinger there is a Hamiltonian of the form

$H=H_{0}+\lambda H_{1}(t)$ (81)

which has a transition probability per unit time of

$w(p\rightarrow-p)=\frac{V_{0}^{2}\sin^{2}\left(2l\sqrt{\frac{2m\epsilon}{{}^{\hbar^{2}}}}\right)}{2\sqrt{8m}\epsilon^{\frac{3}{2}}L}$

where $V_{0},\epsilon,h,\operatorname{and}L$ are suitable constants. [Gol61,
15.5]

####  2.4.2 The String Theoretic Partition Function

The string theoretic partition function is defined as

$Z_{st}(\omega)=Z_{st,R}(\omega)=\operatorname{Tr}[e^{2\pi
i\omega_{1}P-2\pi\omega_{2}H}]=(q\bar{q})^{-\frac{1}{24}}\operatorname{Tr}(q^{L_{0}^{+}}\bar{q}^{L^{-}_{0}})$
(82)

where $L^{+}_{0}$ and $L^{-}_{0}$ are the Virasoro generators defined by

$L_{0}^{\pm}=\frac{(p^{+})^{2}+(p^{-})^{2}}{2}+\sum_{n=1}^{\infty}\alpha_{-n}^{\pm}\alpha_{n}^{\pm}$
(83)

and $\alpha_{-n}^{\pm}\alpha_{n}$ are related to something called a Fubini-
Veneziano field and $p^{{}_{+}}$ is the left-momentum operator and $p^{-}$ is
right-momentum operator. [Lap08, 2.2.1]

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