††footnotetext:  File: main.tex, printed: 2024-8-28, 0.40

# Transcendental entire solutions of several general quadratic type PDEs and
PDDEs in $\mathbb{C}^{2}$

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 functional equations $f^{2}+g^{2}=1$ and $f^{2}+2\alpha fg+g^{2}=1$ are
respectively called Fermat-type binomial and trinomial equations. It is of
interest to know about the existence and form of the solutions of general
quadratic functional equations. Utilizing Nevanlinna’s theory for several
complex variables, in this paper, we study the existence and form of the
solutions to the general quadratic partial differential or partial
differential-difference equations of the form $af^{2}+2\alpha fg+bg^{2}+2\beta
f+2\gamma g+C=0$ in $\mathbb{C}^{2}$. Consequently, we obtain certain
corollaries of the main results of this paper concerning binomial equations
which generalize many results in [Rocky Mountain J. Math. 51(6) (2021),
2217-2235] in the sense of arbitrary coefficients.

###### Key words and phrases:

Transcendental entire solutions, Nevanlinna theory, Several complex variables,
Fermat-type equations, finite order, Partial differential-difference equations

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

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

##  1\. Introduction

The study of the Fermat type functional equation has been the interesting
subject in the field of complex analysis in connection with extensions of
Nevanlinna theory. Finding the precise solutions of different Fermat-type
equations in several complex variables now-a-days becomes a central objectives
in the value distribution theory of Nevanlinna. As we know that, partial
differential equations are occurring in various areas of applied mathematics,
such as fluid mechanics, nonlinear acoustic, gas dynamics and traffic flow
(see [7],[8]). However, in general, it is difficult to find entire and
meromorphic solutions for a nonlinear partial differential equations (PDEs)
and differential-difference equations (DDEs). The study of complex DDEs can be
traced back to Naftalevich’s research [29, 30]. Naftalevich have investigated
the meromorphic solutions on complex DDEs with one complex variable by
employing the operator theory and iteration method. Moreover, by employing
Nevanlinna theory and the method of complex analysis, there were a number of
literature focusing on the solutions of the PDEs and DDEs and many variants of
them.

In a number of articles, it is dealt with the finite order solutions to the
Fermat-type binomial and trinomial equations in $\mathbb{C}$ and
$\mathbb{C}^{2}$. Moreover, it appears recently in the literature that the
transcendental entire solutions of the Fermat-type $f^{2}+g^{2}=1$ (binomial
type) or $f^{2}+2\alpha fg+g^{2}=1$ (trinomial type) equations os system of
equations in several complex variables have been studied (see [40, 41, 43] and
references therein). However, no attempt has been made here to develop
methodologies on finding solutions of the general quadratic functional
equations of the form $af^{2}+2\alpha fg+bg^{2}+2\beta f+2\gamma g+C=0$ (a
general setting of binomial and trinomial type equations) in $\mathbb{C}$ and
$\mathbb{C}^{2}$. In this paper, our main objective is to find solutions to
the system of general quadratic equations in $\mathbb{C}^{2}$. In this paper,
we develop methods using some transformations and discuss how to solve
reducing this general quadratic equations into a binomial form. The estimate
we obtain in the course of proof seems to be independent interest. In fact,
our main results provide a natural characterization of the solutions of the
equations of our investigation.

It has always been a well-known and interesting problem to investigate the
existence and form of solution for Fermat-type equations

(1.1) $\displaystyle f^{m}(z)+g^{m}(z)=1$

regard as the Fermat diophantine equation $x^{m}+y^{m}=1$ over functional
fields, where $n\geq 2$ is an integer. In $1995$, A. Wiles [39], A. Wiles and
R. Taylor [37] pointed out that the Fermat equation $x^{m}+y^{m}=1$ does not
admit nontrivial solutions in rational numbers for $m\geq 3$, and does admit
nontrivial rational solutions for $m=2$. In fact, more earlier, Montel [28],
Iyer [18] and Gross [11] had established results discussing the existence of
solutions for Fermat type functional equation (1.1) and pointed out the
following:

  1. (i)

for $m=2$, the entire solutions of (1.1) are $f(z)=\cos(\varphi(z))$ and
$g(z)=\sin(\varphi(z))$, where $\varphi$ is an entire function;

  2. (ii)

for $m>2$, there are no non-constant entire solutions of (1.1);

  3. (iii)

for $m=2$, the meromorphic solutions of (1.1) are of the form

$\displaystyle
f(z)=\frac{2\varphi(z)}{1+\varphi^{2}(z)}\;\;\mbox{and}\;\;g(z)=\frac{1-\varphi^{2}(z)}{1+\varphi^{2}(z)},$

where $\varphi$ is a meromorphic function;

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

  5. (v)

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

In $1933$, Cartan [3] first considered the Fermat Diophantine functional
equations $f^{m}+g^{n}=1$, and observed that all the entire solutions $f$ and
$g$ must be constants in case of when $mn>m+n$ ; a similar observations are
made also Gundersen and Hayman in [13]. Since then, researchers have paid
their considerable attention to this study and investigated the existence of
entire or meromorphic solutions for Fermat-type functional equation of the
form $f^{n}+g^{m}=1$, where $f,g$ are in general, meromorphic functions and
$m,n\in\mathbb{N}$ (see e.g., [11, 28, 36, 47, 48]). Continuing the study, in
$2004$, Yang and Li [48] investigated to find the form of solutions to the
equation (1.1) in view of replacing $g(z)$ by $f^{\prime}(z)$ when $m=2$, and
showed that the transcendental entire solutions of
$f(z)^{2}+f^{\prime}(z)^{2}=1$ must of the form
$f(z)=\frac{1}{2}\left(Ae^{\alpha z}+\frac{1}{A}e^{-\alpha z}\right),$ where
$A,\alpha$ are non-zero complex constants. In [48], the authors also studied
the existence of meromorphic solutions to the equation
$f^{2}(z)+\left(a_{n}f^{(n)}(z)+a_{n+1}f^{(n+1)}(z)\right)^{2}=1$ for $a_{n}$
and $a_{n+1}$ be non-zero constants and proved that it has no transcendental
meromorphic solutions. In $2015$, Liu and Dong [24] further studied the
existence of solutions of the following Fermat-type differential equation

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

and proved that equation (1.2) does not admit any transcendental meromorphic
solution. For different aspects of the solutions of Fermat-type functional
equation, we refer to the articles [4, 12, 14, 27] and references therein.
Moreover, for other aspects of Fermat-type functional equations and several
questions on it, we refer the article [12] by Gundersen

For analogue study of Fermat-type functional equations in several complex
variables, we recall here the classic result which plays a significant role.

######  Theorem 1.1.

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

  1. (i)

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

  2. (ii)

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

The first order non-linear partial differential equation
$u_{z_{1}}^{2}+u_{z_{2}}^{2}=1$ is a special form of the functional equation
$f^{2}+g^{2}=1$ in $\mathbb{C}^{2}$ that admits non-constant complex analytic
solutions such as

$\displaystyle\begin{cases}f(z_{1},z_{2})=\cos(h(z_{1},z_{2}))\\\
g(z_{1},z_{2})=\sin(h(z_{1},z_{2}))\end{cases}\mbox{and}\;\;\begin{cases}f(z_{1},z_{2})=\sec(h(z_{1},z_{2}))\\\
g(z_{1},z_{2})=i\tan(h(z_{1},z_{2}))\end{cases}$

for an entire function $h$ in $\mathbb{C}^{2}$.

From now on, we will understand by $z+w=(z_{1}+w_{1},z_{2}+w_{2})$ for any
$z=(z_{1},z_{2})$, $w=(w_{1},w_{2})$ are in $\mathbb{C}^{2}$, the shift of
$f(z)$ is defined by $f(z+c)$, whereas the difference of $f(z)$ is defined by
$\Delta_{c}f(z)=f(z+c)-f(z)$ (see [20]).

Let us also recall here the definitions of difference equations and partial
differential-difference equations in $\mathbb{C}^{n}$.

######  Definition 1.1.

[50] An equation is called a difference equation, if the equation includes
shifts or difference of $f(z)$ in in $\mathbb{C}^{n}$, which can be called
$DE$ for short. An equation is called a complex partial differential-
difference equation, if this equation includes partial derivatives, shift or
difference of $f(z)$ in $\mathbb{C}^{n}$, which can be called $PDDE$ for
short.

Due to the development of the difference analogues of Nevanlinna theory in
$\mathbb{C}$ by Halburd and Korhonen [15], Chiang and Feng [6] as well as in
$\mathbb{C}^{n}$ [1, 2, 9, 38, 49], many scholars paid their considerable
attention to exploring several properties of the solutions to some complex
difference equations both in $\mathbb{C}$ and $\mathbb{C}^{n}$. Most
noticeably, $1995$, Khavinson [19] (also, Saleby [33]) derived 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., the solution takes the form
$u(z_{1},z_{2})=az_{1}+bz_{2}+c$, where $a,b,c\in\mathbb{C}$, and
$a^{2}+b^{2}=1$. In fact, Khavinson’s work inspired researchers on finding the
entire solutions to the equations $u_{z_{1}}^{2}+u_{z_{2}}^{2}=1$ in
$\mathbb{C}^{2}$. For example, Li [21, 22, 23] investigated on the partial
differential equations with more general forms such as
$u_{z_{1}}^{2}+u_{z_{2}}^{2}=p$, $u_{z_{1}}^{2}+u_{z_{2}}^{2}=e^{q}$, etc,
where $p,q$ are polynomials in $\mathbb{C}^{2}$. With the help of the
difference Nevanlinna theory for several complex variables (see [2, 20])
obtained some interesting results on the existence of the solutions for some
Fermat-type partial differential-difference equations from one complex
variable to several several complex variables.

In the recent years, Xu and his coauthors made some vital contribution
establishing results on the Fermat-type equations with various settings in two
complex variables. For extensive research work of Xu, the readers are referred
to the articles [40, 45, 41, 43, 44] and references therein. Moreover, for
entire solutions of several quadratic binomial and trinomial partial
differential-difference equations in $\mathbb{C}^{2}$, we refer to the article
[16] and references there. In addition, entire solutions of certain type of
nonlinear differential-difference equations are discussed in [5]. Our
interests in this paper to explore the solutions of equations considered in
[44] with a more general settings, namely for general quadratic functional
equations. Henceforth, we first recall here a series of results obtained by Xu
_et al._ [44] who have investigated the form of entire solutions to quadratic
binomial functional equations and obtained the following result.

######  Theorem 1.2.

[44] Let $f(z_{1},z_{2})$ be a transcendental entire solution with finite
order of the partial differential equations $\left(f(z)+\frac{\partial
f(z)}{\partial z_{1}}\right)^{2}+\left(f(z)+\frac{\partial f(z)}{\partial
z_{2}}\right)^{2}=1.$ Then $f(z_{1},z_{2})$ is of the form
$f(z_{1},z_{2})=\pm\frac{\sqrt{2}}{2}+\eta e^{-(z_{1}+z_{2})}$ or
$f(z_{1},z_{2})=\frac{1}{2}\sin(z_{2}-z_{1}+\eta_{1})+\frac{1}{2}\cos(z_{2}-z_{1}+\eta_{1})+\eta_{2}e^{-(z_{1}+z_{2})}$,
where $\eta,\eta_{1},\eta_{2}\in\mathbb{C}$.

######  Theorem 1.3.

[44] The PDE $\left(f(z)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+\left(f(z)+\frac{\partial^{2}f(z)}{\partial
z^{2}_{1}}\right)^{2}=1.$ does not admit any transcendental entire solution
with finite order.

######  Theorem 1.4.

[44] Let $f(z_{1},z_{2})$ be a transcendental entire solution with finite
order of the PDE $\left(f(z)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+\left(f(z)+\frac{\partial^{2}f(z)}{\partial z_{1}\partial
z_{2}}\right)^{2}=1.$ Then $f(z_{1},z_{2})$ is of the form
$f(z_{1},z_{2})=\pm\frac{\sqrt{2}}{2}+\eta e^{z_{2}-z_{1}},$ where
$\eta\in\mathbb{C}$.

######  Theorem 1.5.

[44] Let $c=(c_{1},c_{2})(\neq(0,0))\in\mathbb{C}^{2}$ and
$s_{0}=c_{1}+c_{2}$, and let $f(z_{1},z_{2})$ be a transcendental entire
solution with finite order of the PDE $\left(f(z+c)+\frac{\partial
f(z)}{\partial z_{1}}\right)^{2}+\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{2}}\right)^{2}=1.$ Then $f(z_{1},z_{2})$ is of the form

  1. (i)

$f(z_{1},z_{2})=\phi(z_{1}+z_{2}),$ where $\phi(s)$ is a transcendental entire
function with finite order in $s:=z_{1}+z_{2}$ satisfying
$\phi(s+s_{0})+\phi^{\prime}(s)=\pm\frac{\sqrt{2}}{2};$

  2. (ii)

$f(z_{1},z_{2})=\frac{1+i}{2(\alpha_{1}-\alpha_{2})}e^{L(z)+B}-\frac{1-i}{2(\alpha_{1}-\alpha_{2})}e^{-L(z)-B}+\phi(z_{1}+z_{2}),$
where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$\alpha_{1},\alpha_{2},B\in\mathbb{C}$ and $\phi(s)$ satisfies
$\alpha^{2}_{1}+\alpha^{2}_{2}=-2,\;e^{L(c)}=-\frac{i\alpha_{1}+\alpha_{2}}{1+i}=-\frac{1-i}{i\alpha_{1}-\alpha_{2}},\;\phi^{\prime}(s)+\phi(s+s_{0})=0.$

######  Theorem 1.6.

[44] Let $c=(c_{1},c_{2})(\neq(0,0))\in\mathbb{C}^{2}$. Suppose that
$f(z_{1},z_{2})$ be a transcendental entire solution with finite order of the
PDE $\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+\left(f(z+c)+\frac{\partial^{2}f(z)}{\partial
z^{2}_{1}}\right)^{2}=1.$ Then $f(z_{1},z_{2})$ is of the form
$f(z_{1},z_{2})=\pm\frac{\sqrt{2}}{2}+\eta e^{z_{1}+\frac{2k\pi i\pm\pi
i-c_{1}}{c_{2}}z_{2}}$, where $k\in\mathbb{Z}_{+}$ and $\eta\in\mathbb{C}$; or
$f(z_{1},z_{2})=-\frac{1+i}{2\alpha_{1}(\alpha_{1}-1)}e^{L(z)+B}-\frac{1-i}{2\alpha_{1}(\alpha_{1}-1)}e^{-L(z)-B}+\eta
e^{z_{1}+\frac{2k\pi i\pm\pi i-c_{1}}{c_{2}}z_{2}},$ where
$L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$, and $\eta\in\mathbb{C}$,
$\alpha_{1},\alpha_{2},c_{1},c_{2}$ satisfies
$(i\alpha_{1}+\alpha^{2}_{1})^{2}=2,\;e^{2L(c)}=-i.$

We see that all the equations considered in Theorems 1.2 to 1.6 are of the
form $F^{2}+G^{2}=1$ (binomial type) and it is easy to see that $F^{2}+2\alpha
FG+G^{2}=1$ (trinomial type) is a general setting of binomial, and moreover,
the general quadratic equation of the form $aF^{2}+2\alpha FG+bG^{2}+2\beta
F+2\gamma G+C=0$ is the natural extensions of both the binomial and trinomial
equations. To the best of knowledge of the auhtors, since no investigation is
done on exploring the existence and form of the solutions with this general
quadratic functional equations. Thus, for further investigations, it is
natural to raise the following question.

######  Question 1.1.

Can we derive results corresponding to the equations in Theorems 1.2 to 1.6
considering general quadratic equations? If so, then can we find the precise
form of the solutions of such equations?

In this paper, our main aim is to answer Question 1.1 completely. Before start
our investigations, we want to mention here some results of Saleeby obtained
for trinomial quadratic equations. In fact, Saleeby [34] made some elementary
observation on right factors of meromorphic function to describe complex
analytic solutions to the quadratic functional equation $f^{2}+2\alpha
fg+g^{2}=1$, $\alpha^{2}\neq 1$ 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 $2016$, Liu and Yang [26] studied the
existence and the form of solutions of some quadratic trinomial functional
equations and proved that if $\alpha\neq\pm 1,0$, then equation
$f(z)^{2}+2\alpha f(z)f^{\prime}(z)+f^{\prime}(z)^{2}=1$ has no transcendental
meromorphic solutions, whereas the equation $f(z)^{2}+2\alpha
f(z)f^{\prime}(z+c)+f^{\prime}(z+c)^{2}=1$ must have transcendental entire
solutions of order equal to one. In the recent years, trinomial $PDDEs$ for
functions of two complex variables are studied extensively.

Motivated by the techniques solving trinomial equations used by Saleeby in
[34], our primary concern in this paper to establish results corresponding to
Theorems 1.2 to 1.6 considering arbitrary coefficients in the equations. More
precisely, we aim to generalize these results not only for trinomial cases but
for complete general quadratic equations. In order to establish results
corresponding to the equations in Theorems 1.2 to 1.6 for simplicity, we
consider the general quadratic difference equations $P(L_{1}(f),L_{2}(f))=0$,
$P(L_{1}(f),L_{3}(f))=0$ and $P(L_{1}(f),L_{4}(f))=0$ or partial differential-
difference equations $P(M_{1}(f),M_{2}(f))=0$ and $P(M_{1}(f),M_{3}(f))=0$ in
$\mathbb{C}^{2}$. Our main interest will be to find the solution of equation
$P(x,y)=0$, where $P(x,y)=ax^{2}+2\alpha xy+by^{2}+2\beta x+2\gamma y+C=0$
which is irreducible i.e., equivalent to saying that the corresponding curve
is non-singular. We define the following notations

$\displaystyle\begin{cases}L_{1}(f(z))=f(z)+\dfrac{\partial f(z)}{\partial
z_{1}},L_{2}(f(z))=f(z)+\dfrac{\partial f(z)}{\partial z_{2}},\\\
L_{3}(f(z))=f(z)+\dfrac{\partial^{2}f(z)}{\partial
z^{2}_{1}},L_{4}(f(z))=f(z)+\dfrac{\partial^{2}f(z)}{\partial z_{1}\partial
z_{2}},\\\ M_{1}(f(z))=f(z+c)+\dfrac{\partial f(z)}{\partial
z_{1}},M_{2}(f(z))=f(z+c)+\dfrac{\partial f(z)}{\partial
z_{2}},M_{3}(f(z))=f(z+c)+\dfrac{\partial^{2}f(z)}{\partial
z^{2}_{1}}\end{cases}$

Employing elementary linear transformations and utilizing Nevanlinna’s theory,
in this paper, we will examine the existence and the precise form of the
solutions of the following general quadratic $PDDEs$.

$\displaystyle\begin{cases}aL_{1}(f)^{2}+2\alpha
L_{1}(f)L_{2}(f)+bL_{2}(f)^{2}+2\beta L_{1}(f)+2\gamma L_{2}(f)+C=0,\\\
aL_{1}(f)^{2}+2\alpha L_{1}(f)L_{3}(f)+bL_{3}(f)^{2}+2\beta L_{1}(f)+2\gamma
L_{3}(f)+C=0,\\\ aL_{1}(f)^{2}+2\alpha L_{1}(f)L_{4}(f)+bL_{4}(f)^{2}+2\beta
L_{1}(f)+2\gamma L_{4}(f)+C=0,\\\ aM_{1}(f)^{2}+2\alpha
M_{1}(f)M_{2}(f)+bM_{2}(f)^{2}+2\beta M_{1}(f)+2\gamma M_{2}(f)+C=0,\\\
aM_{1}(f)^{2}+2\alpha M_{1}(f)M_{3}(f)+bM_{3}(f)^{2}+2\beta M_{1}(f)+2\gamma
M_{3}(f)+C=0.\end{cases}$

In the above-mentioned equations, we assume the corresponding companion
matrices are no-singular, i.e., $\Delta\neq 0$ , otherwise, each equation can
be factorized linearly, which will eventually lead to the fact that the
function $f$ is a constant. Before we state the main results of this paper,
below we discuss some techniques regarding how to transform a general
quadratic equation into a binomial equation using transformations.

##  2\. Finding solutions to the general quadratic equation with the help of
some transformations

The most general quadratic first order partial differential equation
$P(u_{x},u_{y})=0$ over $\mathbb{C}$ can be written as

(2.1) $P(u_{x},u_{y}):=au_{x}^{2}+2\alpha u_{x}u_{y}+bu_{y}^{2}+2\beta
u_{x}+2\gamma u_{y}+C=0,$

where $a,\alpha,b,\beta,\gamma,c$ are complex constants and
$(x,y)\in\mathbb{C}^{2}$. It is easy to see that (2.1) is associated with the
algebraic equation

(2.2) $\displaystyle ax^{2}+2\alpha xy+by^{2}+2\beta x+2\gamma y+C=0.$

This will represent a pair of straight lines only if the L.H.S of (2.2) can be
resolve into two linear factors. Arranging the equation as a quadratic in $x$,
we have

$\displaystyle ax^{2}+2(\alpha y+\beta)x+(by^{2}+2\gamma y+C)=0.$

Therefore, if $a\neq 0$ a simple computation shows that

$\displaystyle x=\frac{-(\alpha y+\beta)\pm\sqrt{(\alpha
y+\beta)^{2}-a(by^{2}+2\gamma y+C)}}{a}.$

Hence left hand side of (2.2) can be resolve into two linear factors, if
$(\alpha y+\beta)^{2}-a(by^{2}+2\gamma y+C)$, i.e,
$y^{2}(\alpha^{2}-ab)+2(\alpha\beta-a\gamma)y+(\beta^{2}-aC)$ would be a
perfect square. This is a quadratic in $y$, and hence is a perfect square, if
its discriminant be zero, i.e, if
$4(\alpha\beta-a\gamma)^{2}-4(\alpha^{2}-ab)(\beta^{2}-aC)=0$ or

(2.3) $\displaystyle
a(abC+2\alpha\beta\gamma-a\gamma^{2}-b\beta^{2}-C\alpha^{2})=0.$

The companion matrix of the quadratic equation (2.2) is the following

$A=\begin{bmatrix}a&\alpha&\beta\\\ \alpha&b&\gamma\\\
\beta&\gamma&C\end{bmatrix}.$

Let $\Delta:=\det(A)$, where $\Delta$ is known to be an invariant of the conic
as $\Delta=0$ in invariant under linear transformations (see [10]). A simple
computation shows that
$\Delta=abC+2\alpha\beta\gamma-a\gamma^{2}-b\beta^{2}-C\alpha^{2}$. Now since
$a\neq 0$, from (2.3), we see that $\Delta=0$. Therefore, when $\Delta=0$ the
equation (2.2) represent a pair of straight lines. These pair of straight
lines will be parallel if $\alpha^{2}=ab$ and be intersecting if
$\alpha^{2}\neq ab$. In addition, when $\Delta\neq 0$, the equation (2.2) is
transformed into a circle $x^{2}+y^{2}=1$, or, an ellipse
$x^{2}/a^{2}+y^{2}/b^{2}=1$, or, a parabola, $y^{2}=x$. Define,
$D=\det\begin{bmatrix}a&\alpha\\\ \alpha&b\end{bmatrix}=ab-\alpha^{2}$. Let
$\alpha^{2}\neq ab$. Then under the transformation

(2.4) $\displaystyle
x=X+x_{1}\;\mbox{and}\;y=Y+y_{1},\;\mbox{where}\;x_{1}=\frac{\alpha\gamma-b\beta}{ab-\alpha^{2}}\;\mbox{and}\;y_{1}=\frac{\alpha\beta-a\gamma}{ab-\alpha^{2}}$

a straightforward computation shows that equation $\eqref{eq-2.4}$ reduces to

(2.5) $\displaystyle aX^{2}+2\alpha XY+bY^{2}+C_{1}=0,$

where $C_{1}=ax^{2}_{1}+2\alpha x_{1}y_{1}+by^{2}_{1}+2\beta x_{1}+2\gamma
y_{1}+C.$ In view of the values of $x_{1}$ and $y_{1}$, we see that $C_{1}$
can be written as
$C_{1}=\frac{abC+2\alpha\beta\gamma-a\gamma^{2}-b\beta^{2}-C\alpha^{2}}{ab-\alpha^{2}}=\frac{\Delta}{D}.$
Hence, the equation $\eqref{eq-2.5}$ can be written as

(2.6) $\displaystyle aX^{2}+2\alpha XY+bY^{2}+\frac{\Delta}{D}=0,$

where $D\neq 0$ as $\alpha^{2}\neq ab$. To remove the $XY$-term from (2.6), we
take the transformation
$(X,Y)=\left(\mathcal{U}\xi^{\pm}_{1}-\mathcal{V}\eta^{\pm}_{1},\;\mathcal{U}\eta^{\pm}_{1}+\mathcal{V}\xi^{\pm}_{1}\right),$
where

$\displaystyle\begin{cases}\xi^{\pm}_{1}:=\dfrac{2\alpha}{\sqrt{\left((b-a)\pm\sqrt{(a-b)^{2}+4\alpha^{2}}\right)^{2}+4\alpha^{2}}},\vspace{1.2mm}\\\
\eta^{\pm}_{1}:=\dfrac{(b-a)\pm\sqrt{(a-b)^{2}+4\alpha^{2}}}{\sqrt{\left((b-a)\pm\sqrt{(a-b)^{2}+4\alpha^{2}}\right)^{2}+4\alpha^{2}}}.\end{cases}$

Under this transformation, the equation $\eqref{eq-2.6}$ is transformed to

(2.7)
$\displaystyle\left(\frac{\mathcal{U}}{1/\sqrt{A^{\pm}}}\right)^{2}+\left(\frac{\mathcal{V}}{1/\sqrt{B^{\mp}}}\right)^{2}+\frac{\Delta}{D}=0,$

where

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

It is easy to see that (2.7) can be written as

(2.8)
$\displaystyle\left(\sqrt{\frac{DA^{\pm}}{-\Delta}}\;\mathcal{U}\right)^{2}+\left(\sqrt{\frac{DB^{\mp}}{-\Delta}}\;\mathcal{V}\right)^{2}=1,$

which is associated with a circle in $\mathbb{C}^{2}$. Therefore, in view of
part -(i) of Theorem 1.1, we must have

$\displaystyle(\mathcal{U},\mathcal{V})=\left(\frac{\cos
h(z)}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\frac{\sin
h(z)}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\right),$

where $h:\mathbb{C}^{2}\rightarrow\mathbb{C}$ is an entire function. In view
of the above observations, it follows from $\eqref{eq-2.4}$ that the
transcendental solutions $(x,y)$ to the general quadratic equation (2.2) take
the form

(2.9) $\displaystyle(x,y)=\left(D_{11}\cos h(z)-D_{12}\sin
h(z)+T_{1},E_{11}\cos h(z)+E_{12}\sin h(z)+T_{2}\right)$

where

(2.10)
$\displaystyle\begin{cases}D_{11}=\dfrac{\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;\;D_{12}=\dfrac{\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\;\mbox{and}\;\;T_{1}=\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}},\vspace{1.5mm}\\\
E_{11}=\dfrac{\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;\;E_{12}=\dfrac{\xi^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\;\mbox{and}\;\;T_{2}=\dfrac{\alpha\beta-a\gamma}{ab-\alpha^{2}}.\end{cases}$

##  3\. Main results and their consequences

In this paper, we investigate the existence and the form of entire solutions
for general quadratic PDEs and PDDES with two complex variables. We obtain
Theorems 3.1 to 3.5 corresponding to Theorems 1.2 to 1.6 respectively, and
hence Question 1.1 is answered completely. The first main result of the paper
is the following which extends Theorem 1.4 for general quadratic PDEs.

######  Theorem 3.1.

Let $a,b,C,\alpha,\beta,\gamma\in\mathbb{C}$ with $\alpha^{2}\neq ab$,
$\Delta\neq 0$ and $f(z_{1},z_{2})$ be a finite order transcendental entire
solutions for the general quadratic partial differential equation

(3.1) $\displaystyle aL_{1}(f)^{2}+2\alpha
L_{1}(f)L_{2}(f)+bL_{2}(f)^{2}+2\beta L_{1}(f)+2\gamma L_{2}(f)+C=0.$

Then $f(z_{1},z_{2})$ is of the forms

  1. (i)

$f(z_{1},z_{2})=A_{0}+R_{4}e^{-(z_{1}+z_{2})},$ where $K_{1},K_{2},R_{4}$ are
constants with $K^{2}_{1}+K^{2}_{2}=1$,

$\displaystyle\begin{cases}A_{0}=\dfrac{K_{1}\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}-\dfrac{K_{2}\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}},\;K_{2}=\dfrac{-R_{2}R_{3}\pm
R_{1}\sqrt{R^{2}_{1}+R^{2}_{2}-R^{2}_{3}}}{R^{2}_{1}+R^{2}_{2}},\vspace{2mm}\\\
R_{1}=\dfrac{\xi^{\pm}_{1}-\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;R_{2}=\dfrac{\xi^{\pm}_{1}+\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\mbox{and}\;R_{3}=\dfrac{\beta(\alpha+b)-\gamma(\alpha+a)}{ab-\alpha^{2}}.\end{cases}$

  2. (ii)

$\displaystyle f(z_{1},z_{2})$
$\displaystyle=\left(\dfrac{E_{11}+E_{12}}{2}\right)\sin\left(\dfrac{R_{12}}{R_{11}}z_{1}+z_{2}+R_{5}\right)$
$\displaystyle+\left(\dfrac{E_{11}-E_{12}}{2}\right)\cos\left(\dfrac{R_{12}}{R_{11}}z_{1}+z_{2}+R_{5}\right)+R_{6}e^{-(z_{1}+z_{2})}+T_{2}.$

where $R_{5}$ is a constant and $T_{1}=T_{2}$.

Before, we state some consequences of the main result as the precise form of
the solutions for trinomial and binomial equations, we introduce some
notations below. Denote

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

We obtain the following result from Theorem 3.1 showing the precise form of
the solutions in case of trinomial quadratic PDEs.

######  Corollary 3.1.

Let $a,b,C,\alpha\in\mathbb{C}$ with $\alpha^{2}\neq ab$, $\Delta\neq 0$ and
$f(z_{1},z_{2})$ be a finite order transcendental entire solutions for the
general quadratic partial differential equation

$\displaystyle a\left(f(z)+\dfrac{\partial f(z)}{\partial
z_{1}}\right)^{2}+2\alpha\left(f(z)+\dfrac{\partial f(z)}{\partial
z_{1}}\right)\left(f(z)+\dfrac{\partial f(z)}{\partial
z_{2}}\right)+b\left(f(z)+\dfrac{\partial f(z)}{\partial z_{2}}\right)^{2}=1.$

Then $f(z_{1},z_{2})$ is of the forms

  1. (i)

$f(z_{1},z_{2})=A_{31}+R_{4}e^{-(z_{1}+z_{2})},$ where $K_{1},K_{2},R_{4}$ are
constants with $K^{2}_{1}+K^{2}_{2}=1$, $A_{31}=A_{11}K_{1}-A_{12}K_{2}$,
$K_{2}=\pm\dfrac{S_{11}}{\sqrt{S^{2}_{11}+S^{2}_{12}}}$, where
$S_{11}=\frac{\sqrt{2}(2\alpha-
K_{14})}{K_{11}K_{12}}\;\mbox{and}\;S_{12}=\frac{\sqrt{2}(2\alpha+K_{14})}{K_{11}K_{12}}.$

  2. (ii)

$\displaystyle f(z_{1},z_{2})$
$\displaystyle=\left(\dfrac{B_{11}+B_{12}}{2}\right)\sin\left(\dfrac{S_{14}}{S_{13}}z_{1}+z_{2}+R_{5}\right)$
$\displaystyle+\left(\dfrac{B_{11}-B_{12}}{2}\right)\cos\left(\dfrac{S_{14}}{S_{13}}z_{1}+z_{2}+R_{5}\right)+R_{6}e^{-(z_{1}+z_{2})}.$

where $R_{5}$ is a constant,
$S_{13}=B_{11}(A_{11}-B_{11})-B_{12}(A_{12}+B_{12})$ and
$S_{14}=A_{11}(A_{11}-B_{11})+A_{12}(A_{12}+B_{12})$.

We obtain the following corollary of Theorem 3.1 and it gives the solutions in
generalized form of the same in Theorem 1.2.

######  Corollary 3.2.

Let $a,b\in\mathbb{C}$ with $\Delta\neq 0$ and $f(z_{1},z_{2})$ be a finite
order transcendental entire solutions for the general quadratic partial
differential equation

$\displaystyle a\left(f(z)+\dfrac{\partial f(z)}{\partial
z_{1}}\right)^{2}+b\left(f(z)+\dfrac{\partial f(z)}{\partial
z_{2}}\right)^{2}=1.$

Then $f(z_{1},z_{2})$ is of the forms

  1. (i)

$f(z_{1},z_{2})=A_{0}+R_{4}e^{-(z_{1}+z_{2})},$ where $K_{1*},K_{2*},R_{4}$
are constants with $K^{2}_{1*}+K^{2}_{2*}=1$, $A_{0}=-A_{21}K_{2}$ and
$K_{2*}=\mp\frac{K_{21}}{\sqrt{K^{2}_{21}+K^{2}_{22}}}$

  2. (ii)

$\displaystyle f(z_{1},z_{2})$
$\displaystyle=\dfrac{A_{22}}{2}\sin\left(-\dfrac{A^{2}_{21}}{A^{2}_{22}}z_{1}+z_{2}+R_{5}\right)$
$\displaystyle+\dfrac{A_{22}}{2}\cos\left(-\dfrac{A^{2}_{21}}{A^{2}_{22}}z_{1}+z_{2}+R_{5}\right)+R_{6}e^{-(z_{1}+z_{2})},\mbox{where}\;R_{5},R_{6}\;\mbox{are
constants}.$

######  Theorem 3.2.

Let $a,b,C,\alpha,\beta,\gamma\in\mathbb{C}$ with $\alpha^{2}\neq ab$ and
$\Delta\neq 0$. The partial differential equation

(3.2) $\displaystyle aL_{1}(f)^{2}+2\alpha
L_{1}(f)L_{3}(f)+bL_{3}(f)^{2}+2\beta L_{1}(f)+2\gamma L_{3}(f)+C=0$

does not admit any transcendental entire solution with finite order.

######  Remark 3.1.

In view of the proof of Theorem 3.2, it can be easily shown that the trinomial
quadratic PDE, $a\left(f(z)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+2\alpha\left(f(z)+\frac{\partial f(z)}{\partial
z_{1}}\right)\left(f(z)+\frac{\partial^{2}f(z)}{\partial
z^{2}_{1}}\right)+b\left(f(z)+\frac{\partial^{2}f(z)}{\partial
z^{2}_{1}}\right)^{2}=1$ and also the binomial quadratic PDE,
$a\left(f(z)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+b\left(f(z)+\frac{\partial^{2}f(z)}{\partial
z^{2}_{1}}\right)^{2}=1$ do not admit any transcendental entire solutions.
Consequently, the Theorem 1.3 is established with more general setting of the
equation.

######  Theorem 3.3.

Let $a,b,C,\alpha,\beta,\gamma\in\mathbb{C}$ with $\alpha^{2}\neq ab$,
$\Delta\neq 0$ and $f(z_{1},z_{2})$ be a finite order transcendental entire
solutions for the general quadratic partial differential equation

(3.3) $\displaystyle aL_{1}(f)^{2}+2\alpha
L_{1}(f)L_{4}(f)+bL_{4}(f)^{2}+2\beta L_{1}(f)+2\gamma L_{4}(f)+C=0.$

Then $f(z_{1},z_{2})$ is of the forms
$f(z_{1},z_{2})=-A_{0}+B_{11}e^{(z_{1}-z_{2})}$, where $K_{5},K_{6},B_{11}$
are constant with $K^{2}_{5}+K^{2}_{6}=1$,

$\displaystyle\begin{cases}A_{0}=\dfrac{K_{5}\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}-\dfrac{K_{6}\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}},\;K_{6}=\dfrac{-R_{2}R_{3}\pm
R_{1}\sqrt{R^{2}_{1}+R^{2}_{2}-R^{2}_{3}}}{R^{2}_{1}+R^{2}_{2}},\\\
R_{1}=\dfrac{\xi^{\pm}_{1}-\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;R_{2}=\dfrac{\xi^{\pm}_{1}+\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\mbox{and}\;R_{3}=\dfrac{\beta(\alpha+b)-\gamma(\alpha+a)}{ab-\alpha^{2}}.\end{cases}$

In case of trinomial quadratic PDEs, we obtain the following result as a
consequence of Theorem 3.3.

######  Corollary 3.3.

Let $a,b,C,\alpha\in\mathbb{C}$ with $\alpha^{2}\neq ab$, $\Delta\neq 0$ and
$f(z_{1},z_{2})$ be a finite order transcendental entire solutions for the
general quadratic trinomial PDE

$\displaystyle a\left(f(z)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+2\alpha\left(f(z)+\frac{\partial f(z)}{\partial
z_{1}}\right)\left(f(z)+\dfrac{\partial^{2}f(z)}{\partial z_{1}\partial
z_{2}}\right)+b\left(f(z)+\dfrac{\partial^{2}f(z)}{\partial z_{1}\partial
z_{2}}\right)^{2}=1.$

Then $f(z_{1},z_{2})$ is of the forms
$f(z_{1},z_{2})=-A_{31}+B_{11}e^{(z_{1}-z_{2})}$, where $K_{5*},K_{6*},B_{11}$
are constant with $K^{2}_{5*}+K^{2}_{6*}=1$,
$A_{31}=A_{11}K_{5*}-A_{12}K_{6*}$ and
$K_{6*}=\pm\frac{S_{11}}{\sqrt{S^{2}_{11}+S^{2}_{12}}},$ where
$S_{11}=\frac{\sqrt{2}(2\alpha-
K_{14})}{K_{11}K_{12}}\;\mbox{and}\;S_{12}=\frac{\sqrt{2}(2\alpha+K_{14})}{K_{11}K_{12}}.$

We obtain the following corollary of Theorem 3.3 which generalizes Theorem
1.4.

######  Corollary 3.4.

Let $a,b\in\mathbb{C}$ with $\Delta\neq 0$ and $f(z_{1},z_{2})$ be a finite
order transcendental entire solutions for the general quadratic binomial PDE

$\displaystyle a\left(f(z)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+b\left(f(z)+\dfrac{\partial^{2}f(z)}{\partial z_{1}\partial
z_{2}}\right)^{2}=1.$

Then $f(z_{1},z_{2})$ is of the forms
$f(z_{1},z_{2})=-A_{0}+B_{11}e^{(z_{1}-z_{2})}$ where $K_{11*},K_{22*},B_{11}$
are constant with $K^{2}_{11*}+K^{2}_{22*}=1$, $A_{0}=-A_{21}K_{22*}$ and
$K_{22*}=\mp\frac{K_{21}}{\sqrt{K^{2}_{21}+K^{2}_{22}}}.$

######  Theorem 3.4.

Let $a,b,C,\alpha,\beta,\gamma\in\mathbb{C}$ with $\alpha^{2}\neq ab$,
$\Delta\neq 0$, $c=(c_{1},c_{2})(\neq(0,0))\in\mathbb{C}^{2}$,
$s_{0}=c_{1}+c_{2}$ and $f(z_{1},z_{2})$ be a finite order transcendental
entire solutions for the general quadratic partial differential equation

(3.4) $\displaystyle aM_{1}(f)^{2}+2\alpha
M_{1}(f)M_{2}(f)+bM_{2}(f)^{2}+2\beta M_{1}(f)+2\gamma M_{2}(f)+C=0.$

Then $f(z_{1},z_{2})$ is one of the following forms

  1. (i)

$f(z_{1},z_{2})=\psi(z_{1}+z_{2})$, where $\psi(s)$ is a transcendental entire
function with finite order in $s:=z_{1}+z_{2}$, satisfying
$\psi^{\prime}(s)+\psi(s+s_{0})=A_{0},$ where $K_{7},K_{8}$ are constants with
$K^{2}_{7}+K^{2}_{8}=1$, where

$\begin{cases}A_{0}=\dfrac{K_{7}\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}-\dfrac{K_{8}\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}},\;K_{6}=\dfrac{-R_{2}R_{3}\pm
R_{1}\sqrt{R^{2}_{1}+R^{2}_{2}-R^{2}_{3}}}{R^{2}_{1}+R^{2}_{2}},\vspace{2mm}\\\
R_{1}=\dfrac{\xi^{\pm}_{1}-\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;R_{2}=\dfrac{\xi^{\pm}_{1}+\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\mbox{and}\;R_{3}=\dfrac{\beta(\alpha+b)-\gamma(\alpha+a)}{ab-\alpha^{2}}.\end{cases}$

  2. (ii)

$\displaystyle
f(z_{1},z_{2})=\dfrac{(D_{11}-E_{11})}{(\alpha_{1}-\alpha_{2})}\sin(L(z)+B)+\dfrac{(D_{12}+E_{12})}{(\alpha_{1}-\alpha_{2})}\cos(L(z)+B)+(T_{1}-T_{2})z_{1}+\psi_{1}(s).$

where

$\displaystyle
e^{2iL(c)}=\dfrac{(D_{11}+iD_{12})\alpha_{2}-(E_{11}-iE_{12})\alpha_{1}}{(E_{11}+iE_{12})\alpha_{1}-(D_{11}-iD_{12})\alpha_{2}}\;\mbox{and}\;L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2},\;\alpha_{1},\alpha_{2},B\in\mathbb{C}$

and

$\displaystyle(E^{2}_{11}+E^{2}_{12})\alpha^{2}_{1}$
$\displaystyle+(D^{2}_{11}+D^{2}_{12})\alpha^{2}_{2}-2(D_{11}E_{11}-D_{12}E_{12})\alpha_{1}\alpha_{2}$
$\displaystyle+[i(D_{11}-E_{11})+(D_{12}+E_{12})]^{2}=0$

and $\psi_{1}(s)$ satisfies

$\displaystyle\psi^{\prime}_{1}(s)+\psi_{1}(s+s_{0})$
$\displaystyle=\left(\dfrac{E_{11}\alpha_{1}-D_{11}\alpha_{2}}{\alpha_{1}-\alpha_{2}}\right)\cos(L(z)+B)+\left(\dfrac{E_{12}\alpha_{1}-D_{12}\alpha_{2}}{\alpha_{1}-\alpha_{2}}\right)\sin(L(z)+B)$
$\displaystyle-\left(\dfrac{D_{11}-E_{11}}{\alpha_{1}-\alpha_{2}}\right)\sin(L(z)+L(c)+B)-\left(\dfrac{D_{12}-E_{12}}{\alpha_{1}-\alpha_{2}}\right)\cos(L(z)+L(c)+B)$
$\displaystyle-(T_{1}-T_{2})(z_{1}+c)+T_{2}.$

For trinomial quadratic PDEs, the following result is a corollary of Theorem
3.4.

######  Corollary 3.5.

Let $a,b,C,\alpha\in\mathbb{C}$ with $\alpha^{2}\neq ab$, $\Delta\neq 0$,
$c=(c_{1},c_{2})(\neq(0,0))\in\mathbb{C}^{2}$, $s_{0}=c_{1}+c_{2}$ and
$f(z_{1},z_{2})$ be a finite order transcendental entire solutions for the
general quadratic partial differential equation

$\displaystyle a\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+2\alpha\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{1}}\right)\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{2}}\right)+b\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{2}}\right)^{2}=1.$

Then $f(z_{1},z_{2})$ is one of the following forms

  1. (i)

$f(z_{1},z_{2})=\psi(z_{1}+z_{2})$, where $\psi(s)$ is a transcendental entire
function with finite order in $s:=z_{1}+z_{2}$, satisfying
$\psi^{\prime}(s)+\psi(s+s_{0})=A_{31},$ where $K_{7*},K_{8*}$ are constant
with $K^{2}_{7*}+K^{2}_{8*}=1$, $A_{31}=A_{11}K_{7*}-A_{12}K_{8*}$,
$K_{8*}=\pm\frac{S_{11}}{\sqrt{S^{2}_{11}+S^{2}_{12}}}$,
$S_{11}=\frac{\sqrt{2}(2\alpha-
K_{14})}{K_{11}K_{12}}\;\mbox{and}\;S_{12}=\frac{\sqrt{2}(2\alpha+K_{14})}{K_{11}K_{12}}.$

  2. (ii)

$\displaystyle
f(z_{1},z_{2})=\dfrac{(A_{11}-B_{11})}{(\alpha_{1}-\alpha_{2})}\sin(L(z)+B)+\dfrac{(A_{12}+B_{12})}{(\alpha_{1}-\alpha_{2})}\cos(L(z)+B)+\psi_{1}(s).$

where

$\displaystyle\begin{cases}L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$\alpha_{1},\alpha_{2},B\in\mathbb{C},\;e^{2iL(c)}=\dfrac{(A_{11}+iA_{12})\alpha_{2}-(B_{11}-iB_{12})\alpha_{1}}{(B_{11}+iB_{12})\alpha_{1}-(A_{11}-iA_{12})\alpha_{2}},\vspace{2mm}\\\
(B^{2}_{11}+B^{2}_{12})\alpha^{2}_{1}+(A^{2}_{11}+A^{2}_{12})\alpha^{2}_{2}-2(A_{11}B_{11}-A_{12}B_{12})\alpha_{1}\alpha_{2}\\\
+[i(A_{11}-B_{11})+(A_{12}+B_{12})]^{2}=0.\end{cases}$

and $\psi_{1}(s)$ satisfies

$\displaystyle\psi^{\prime}_{1}(s)+\psi_{1}(s+s_{0})$
$\displaystyle=\left(\dfrac{B_{11}\alpha_{1}-A_{11}\alpha_{2}}{\alpha_{1}-\alpha_{2}}\right)\cos(L(z)+B)+\left(\dfrac{B_{12}\alpha_{1}-A_{12}\alpha_{2}}{\alpha_{1}-\alpha_{2}}\right)\sin(L(z)+B)$
$\displaystyle-\left(\dfrac{A_{11}-B_{11}}{\alpha_{1}-\alpha_{2}}\right)\sin(L(z)+L(c)+B)-\left(\dfrac{A_{12}-B_{12}}{\alpha_{1}-\alpha_{2}}\right)\cos(L(z)+L(c)+B).$

The following corollary of Theorem 3.4 generalizes Theorem 1.5.

######  Corollary 3.6.

Let $a,b\in\mathbb{C}$ with $\Delta\neq 0$,
$c=(c_{1},c_{2})(\neq(0,0))\in\mathbb{C}^{2}$, $s_{0}=c_{1}+c_{2}$ and
$f(z_{1},z_{2})$ be a finite order transcendental entire solutions for the
general quadratic partial differential equation

$\displaystyle a\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+b\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{2}}\right)^{2}=1.$

Then $f(z_{1},z_{2})$ is one of the following forms

  1. (i)

$f(z_{1},z_{2})=\psi(z_{1}+z_{2})$, where $\psi(s)$ is a transcendental entire
function with finite order in $s:=z_{1}+z_{2}$, satisfying
$\psi^{\prime}(s)+\psi(s+s_{0})=A_{0},$ where $K_{7**},K_{8**}$ are constant
with $K^{2}_{7**}+K^{2}_{8**}=1$, $A_{0}=-A_{21}K_{8**}$ and
$K_{8**}=\mp\frac{K_{21}}{\sqrt{K^{2}_{21}+K^{2}_{22}}}.$

  2. (ii)

$\displaystyle
f(z_{1},z_{2})=\dfrac{-A_{22}}{(\alpha_{1}-\alpha_{2})}\sin(L(z)+B)+\dfrac{A_{21}}{(\alpha_{1}-\alpha_{2})}\cos(L(z)+B)+\psi_{1}(s).$

where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$\alpha_{1},\alpha_{2},B\in\mathbb{C}$ and
$e^{2iL(c)}=\dfrac{iA_{21}\alpha_{2}-A_{22}\alpha_{1}}{iA_{21}\alpha_{2}+A_{22}\alpha_{1}}$
and
$A^{2}_{22}\alpha^{2}_{1}+A^{2}_{21}\alpha^{2}_{2}+(A_{21}-iA_{22})^{2}=0,$
and $\psi_{1}(s)$ satisfies

$\displaystyle\psi^{\prime}_{1}(s)+\psi_{1}(s+s_{0})$
$\displaystyle=\left(\dfrac{A_{22}\alpha_{1}}{\alpha_{1}-\alpha_{2}}\right)\cos(L(z)+B)+\left(\dfrac{-A_{21}\alpha_{2}}{\alpha_{1}-\alpha_{2}}\right)\sin(L(z)+B)$
$\displaystyle-\left(\dfrac{-A_{22}}{\alpha_{1}-\alpha_{2}}\right)\sin(L(z)+L(c)+B)-\left(\dfrac{A_{21}}{\alpha_{1}-\alpha_{2}}\right)\cos(L(z)+L(c)+B).$

######  Theorem 3.5.

Let $a,b,C,\alpha,\beta,\gamma\in\mathbb{C}$ with $\alpha^{2}\neq ab$,
$\Delta\neq 0$, $c=(c_{1},c_{2})(\neq(0,0))\in\mathbb{C}^{2}$,
$s_{0}=c_{1}+c_{2}$ and $f(z_{1},z_{2})$ be a finite order transcendental
entire solutions for the general quadratic partial differential equation

(3.5) $\displaystyle aM_{1}(f)^{2}+2\alpha
M_{1}(f)M_{3}(f)+bM_{3}(f)^{2}+2\beta M_{1}(f)+2\gamma M_{3}(f)+C=0.$

Then $f(z_{1},z_{2})$ is one of the following forms

  1. (i)

$f(z_{1},z_{2})=A_{0}+e^{(z_{1}+Az_{2}+B)},$ where $K_{9},K_{10},B$ are
constant with $K^{2}_{9}+K^{2}_{10}=1$, where

$\displaystyle\begin{cases}A_{0}=\dfrac{K_{9}\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}-\dfrac{K_{10}\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}},\;A=\dfrac{2k\pi
i\pm\pi i-c_{1}}{c_{2}},K_{10}=\dfrac{-R_{2}R_{3}\pm
R_{1}\sqrt{R^{2}_{1}+R^{2}_{2}-R^{2}_{3}}}{R^{2}_{1}+R^{2}_{2}},\vspace{2mm}\\\
R_{1}=\dfrac{\xi^{\pm}_{1}-\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;R_{2}=\dfrac{\xi^{\pm}_{1}+\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\mbox{and}\;R_{3}=\dfrac{\beta(\alpha+b)-\gamma(\alpha+a)}{ab-\alpha^{2}}.\end{cases}$

  2. (ii)

$\displaystyle
f(z_{1},z_{2})=(-e^{-c_{1}})^{\frac{z_{2}}{c_{2}}}\pi(z_{2})e^{z_{1}}+R_{11}\sin[L(z)-L(c)+B]+R_{12}\cos[L(z)-L(c)+B],$

where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$\alpha_{1},\alpha_{2},B\in\mathbb{C}$ and

$\displaystyle\begin{cases}e^{2iL(c)}=\dfrac{(iD_{11}-D_{12})\alpha_{1}-(E_{11}-iE_{12})}{(iD_{11}+D_{12})\alpha_{1}+(iE_{12}+E_{11})},\vspace{2mm}\\\
(D^{2}_{11}+D^{2}_{12})\alpha^{4}_{1}+2(D_{11}E_{12}+D_{12}E_{11})\alpha^{3}_{1}+(E^{2}_{11}+E^{2}_{12})\alpha^{2}_{1}\\\
\quad+[(iD_{11}+D_{12})-(iE_{11}-E_{12})]=0,\;\mbox{where}\;\pi(z_{2}+c_{2})=\pi(z_{2}).\end{cases}$

The existence and form of trinomial quadratic PDEs is shown by the following
result which is a consequence of Theorem 3.5.

######  Corollary 3.7.

Let $a,b,C,\alpha\in\mathbb{C}$ with $\alpha^{2}\neq ab$, $\Delta\neq 0$,
$c=(c_{1},c_{2})(\neq(0,0))\in\mathbb{C}^{2}$, $s_{0}=c_{1}+c_{2}$ and
$f(z_{1},z_{2})$ be a finite order transcendental entire solutions for the
general quadratic partial differential equation

$\displaystyle a\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+2\alpha\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{1}}\right)\left(f(z+c)+\dfrac{\partial^{2}f(z)}{\partial
z^{2}_{1}}\right)+b\left(f(z+c)+\dfrac{\partial^{2}f(z)}{\partial
z^{2}_{1}}\right)^{2}=1.$

Then $f(z_{1},z_{2})$ is one of the following forms

  1. (i)

$f(z_{1},z_{2})=A_{31}+e^{(z_{1}+Az_{2}+B)},$ where $K_{9*},K_{10*},B$ are
constants with $K^{2}_{9*}+K^{2}_{10*}=1$, where

$\begin{cases}A_{31}=A_{11}K_{9*}-A_{12}K_{10*},\;A=\dfrac{2k\pi i\pm\pi
i-c_{1}}{c_{2}}\;\mbox{and}\;K_{10*}=\pm\dfrac{S_{11}}{\sqrt{S^{2}_{11}+S^{2}_{12}}},\vspace{2mm}\\\
S_{11}=\dfrac{\sqrt{2}(2\alpha-
K_{14})}{K_{11}K_{12}}\;\mbox{and}\;S_{12}=\dfrac{\sqrt{2}(2\alpha+K_{14})}{K_{11}K_{12}}.\end{cases}$

  2. (ii)

$\displaystyle
f(z_{1},z_{2})=(-e^{-c_{1}})^{\frac{z_{2}}{c_{2}}}\pi(z_{2})e^{z_{1}}+S_{15}\sin[L(z)-L(c)+B]+S_{16}\cos[L(z)-L(c)+B].$

where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$\alpha_{1},\alpha_{2},B\in\mathbb{C}$ and

$\displaystyle
e^{2iL(c)}=\dfrac{(iA_{11}-A_{12})\alpha_{1}-(B_{11}-iB_{12})}{(iA_{11}+A_{12})\alpha_{1}+(iB_{12}+B_{11})},$
$\displaystyle(A^{2}_{11}+A^{2}_{12})\alpha^{4}_{1}+$ $\displaystyle
2(A_{11}B_{12}+A_{12}B_{11})\alpha^{3}_{1}+(B^{2}_{11}+B^{2}_{12})\alpha^{2}_{1}$
$\displaystyle+[(iA_{11}+A_{12})-(iB_{11}-B_{12})]=0,\;\mbox{where}\;\pi(z_{2}+c_{2})=\pi(z_{2})$

and
$S_{15}=\frac{(A_{11}\alpha_{1}+B_{12})-\alpha_{1}(A_{12}\alpha_{1}+B_{11})}{\alpha^{2}_{1}+1}$,
$S_{16}=\frac{(A_{12}\alpha_{1}+B_{11})+\alpha_{1}(A_{11}\alpha_{1}+B_{12})}{\alpha^{2}_{1}+1}.$

We obtain the following corollary of Theorem 3.5 which generalizes Theorem
1.6.

######  Corollary 3.8.

Let $a,b\in\mathbb{C}$ with $\Delta\neq 0$,
$c=(c_{1},c_{2})(\neq(0,0))\in\mathbb{C}^{2}$, $s_{0}=c_{1}+c_{2}$ and
$f(z_{1},z_{2})$ be a finite order transcendental entire solutions for the
general quadratic partial differential equation

$\displaystyle a\left(f(z+c)+\frac{\partial f(z)}{\partial
z_{1}}\right)^{2}+b\left(f(z+c)+\dfrac{\partial^{2}f(z)}{\partial
z^{2}_{1}}\right)^{2}=1.$

Then $f(z_{1},z_{2})$ is one of the following forms

  1. (i)

$f(z_{1},z_{2})=A_{0}+e^{(z_{1}+Az_{2}+B)},$ where $K_{9**},K_{10**},B$ are
constant with $K^{2}_{9**}+K^{2}_{10**}=1$, $A_{0}=-A_{21}K_{10**}$,
$A=\dfrac{2k\pi i\pm\pi i-c_{1}}{c_{2}}$,
$K_{10**}=\mp\frac{K_{21}}{\sqrt{K^{2}_{21}+K^{2}_{22}}}.$

  2. (ii)

$\displaystyle
f(z_{1},z_{2})=(-e^{-c_{1}})^{\frac{z_{2}}{c_{2}}}\pi(z_{2})e^{z_{1}}+S_{17}\sin[L(z)-L(c)+B]+S_{18}\cos[L(z)-L(c)+B].$

where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$\alpha_{1},\alpha_{2},B\in\mathbb{C}$, $e^{2iL(c)}=-1$ and
$A^{2}_{21}\alpha^{4}_{1}+2A_{21}A_{22}\alpha^{3}_{1}+A^{2}_{22}\alpha^{2}_{1}+(A_{21}-iA_{22})^{2}=0,$
where $\pi(z_{2}+c_{2})=\pi(z_{2})$ and
$S_{17}=\frac{-\alpha_{1}(A_{21}\alpha_{1}+A_{22})}{\alpha^{2}_{1}+1},$
$S_{18}=\frac{(A_{21}\alpha_{1}+A_{22})}{\alpha^{2}_{1}+1}.$

##  4\. Proof of the main results

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

######  Lemma 4.1.

[32, 35] For any entire function $F$ on $\mathbb{C}^{n}$, $F(0)\neq 0$ and put
$\rho(n_{F})=\rho<\infty$. 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.

######  Remark 4.1.

Here, denote $\rho(n_{F})$ to be the order of the counting function of zeros
of $F$.

######  Lemma 4.2.

[31] 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.

[17] 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

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

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

Now we discuss the proof of the main results of the paper.

###### Proof of Theorem 3.1.

Suppose that $f$ is a transcendental entire solution of (3.1) with finite
order. Under the transformations discussed in Section 2, we that the equation
(3.1) can be written as

(4.1)
$\displaystyle\left(\dfrac{DA^{\pm}}{-\Delta}\right)u^{2}+\left(\dfrac{DB^{\mp}}{-\Delta}\right)v^{2}=1$

We divide the proof into two cases.

Case 1: If $\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is a constant, then it follows
from (4.1) that $\sqrt{\frac{DB^{\mp}}{-\Delta}}v$ is also constant. Denoting

$\displaystyle\sqrt{\dfrac{DA^{\pm}}{-\Delta}}u=K_{1}\;\mbox{and}\;\sqrt{\dfrac{DB^{\mp}}{-\Delta}}v=K_{2},$

in view of (4.1), it is easy to see that $K^{2}_{1}+K^{2}_{2}=1$. This clearly
forces to

(4.2) $\displaystyle L_{1}(f)=f(z)+\dfrac{\partial f}{\partial
z_{1}}=\dfrac{K_{1}\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}-\dfrac{K_{2}\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}}:=A_{1},$
(4.3) $\displaystyle L_{2}(f)=f(z)+\dfrac{\partial f}{\partial
z_{2}}=\dfrac{K_{1}\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}+\dfrac{K_{2}\xi^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\beta-a\gamma}{ab-\alpha^{2}}:=B_{1}.$

In view of (4.2) and (4.3), we see that

(4.4) $\displaystyle\dfrac{\partial f}{\partial z_{1}}-\dfrac{\partial
f}{\partial z_{2}}=A_{1}-B_{1}$

The characteristic equations of (4.4) are
$\dfrac{dz_{1}}{dt}=1,\;\dfrac{dz_{2}}{dt}=-1,\;\dfrac{df}{dt}=A_{1}-B_{1}.$
Using the initial conditions: $z_{1}=0$, $z_{2}=s$ and we thus get
$f=f(0,s):=\phi(s)$ with a parameter $s$. Therefore, a simple computation
shows that $z_{1}=t,\;z_{2}=-t+s,$ and

$\displaystyle
f(t,s)=\int_{0}^{t}(A_{1}-B_{1})dt+\phi(s)=(A_{1}-B_{1})t+\phi(s),$

where $\phi(s)$ is a transcendental entire function with finite order in $s$.
Then we obtain $z_{1}=t$ and $s=z_{1}+z_{2}$, the solution of the equation
(4.4) is of the form

(4.5) $\displaystyle f(z_{1},z_{2})=(A_{1}-B_{1})z_{1}+\phi(z_{1}+z_{2}).$

Since $f$ is an entire function, we have $\frac{\partial^{2}f}{\partial
z_{1}\partial z_{2}}=\frac{\partial^{2}f}{\partial z_{2}\partial z_{1}}$,
hence from (4.2) and (4.3), we obtain $\frac{\partial f}{\partial
z_{1}}=\frac{\partial f}{\partial z_{2}}$. Therefore from (4.4) it follows
that $A_{1}=B_{1}$. Furthermore, from (4.2), (4.3) and using $A_{1}=B_{1}$,
$K^{2}_{1}+K^{2}_{2}=1$, a simple computation shows that

$\displaystyle K_{2}=\dfrac{-R_{2}R_{3}\pm
R_{1}\sqrt{R^{2}_{1}+R^{2}_{2}-R^{2}_{3}}}{R^{2}_{1}+R^{2}_{2}}$

where

$\displaystyle
R_{1}=\dfrac{\xi^{\pm}_{1}-\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;R_{2}=\dfrac{\xi^{\pm}_{1}+\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\mbox{and}\;R_{3}=\dfrac{\beta(\alpha+b)-\gamma(\alpha+a)}{ab-\alpha^{2}}.$

Thus, the equation (4.5) becomes

(4.6) $\displaystyle f(z_{1},z_{2})=\phi(z_{1}+z_{2}).$

Substituting (4.6) into (4.3), we obtain
$\phi^{\prime}(z_{1}+z_{2})+\phi(z_{1}+z_{2})=A_{1}$ and an easy computation
shows that

(4.7)
$\displaystyle\phi(z_{1}+z_{2})=A_{1}+R_{4}e^{-(z_{1}+z_{2})},\;R_{4}\in\mathbb{C}$

Hence, from (4.6) and (4.7), we see that the precise form of the solution is
$f(z_{1},z_{2})=A_{1}+R_{4}e^{-(z_{1}+z_{2})}.$

Case 2: If $\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is not a constant, then in view
of the fact that the entire solutions of the functional equation
$f^{2}+g^{2}=1$ are $f=\cos h(z)$, $g=\sin h(z)$, where $h(z)$ is an entire
function, it follows from (4.1) and ….

(4.8) $\displaystyle L_{1}(f)=f(z)+\dfrac{\partial f}{\partial
z_{1}}=D_{11}\cos h(z)-D_{12}\sin h(z)+T_{1},$ (4.9) $\displaystyle
L_{2}(f)=f(z)+\dfrac{\partial f}{\partial z_{2}}=E_{11}\cos h(z)+E_{12}\sin
h(z)+T_{2}.$

Subtracting (4.9) from (4.8), we obtain

(4.10) $\displaystyle\dfrac{\partial f}{\partial z_{1}}-\dfrac{\partial
f}{\partial z_{2}}=(D_{11}-E_{11})\cos h(z)-(D_{12}+E_{12})\sin
h(z)+(T_{1}-T_{2}).$

Using the fact that $\frac{\partial^{2}f}{\partial z_{1}\partial
z_{2}}=\frac{\partial^{2}f}{\partial z_{2}\partial z_{1}}$, from (4.8) and
(4.9), a routine computation shows that

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

From (4.10) and (4.11), we obtain

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

Sub-case 2.1: Suppose that

$\displaystyle D_{11}\dfrac{\partial h}{\partial z_{2}}-E_{11}\dfrac{\partial
h}{\partial z_{1}}+(D_{12}+E_{12})\neq 0,$ $\displaystyle
D_{12}\dfrac{\partial h}{\partial z_{2}}+E_{12}\dfrac{\partial h}{\partial
z_{1}}-(D_{11}-E_{11})\neq 0\;\mbox{and}\;(T_{1}-T_{2})\neq 0.$

It follows from (4.12) that

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

Since $f$ is a finite order transcendental entire solution of the equation
(3.1), by Lemmas 4.1, 4.2, (4.8) and (4.9), we conclude that $h(z)$ is a non-
constant polynomial in $\mathbb{C}^{2}$. As $h(z)$ is a non-constant
polynomial, we assume that $h(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$. Using
Nevanlinna theory we deduce that $T(r,\tan h)=O\\{T(r,h)+\log r+r^{n}\\}$. But
$\lim_{r\rightarrow\infty}\frac{T(r,\tan h)}{T(r,h)+\log r+r^{n}}=\infty$ when
$h$ is a non-constant polynomial. Therefore $h$ must be a constant, a
contradiction.

Sub-case 2.2: Suppose that

$\displaystyle D_{11}\dfrac{\partial h}{\partial z_{2}}-E_{11}\dfrac{\partial
h}{\partial z_{1}}+(D_{12}+E_{12})=0,D_{12}\dfrac{\partial h}{\partial
z_{2}}+E_{12}\dfrac{\partial h}{\partial z_{1}}-(D_{11}-E_{11})\neq
0,T_{1}-T_{2}\neq 0.$

Thus the equation (4.12) can be written as

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

Since $f$ is a finite order transcendental entire solution of the equation
(3.1), by Lemmas 4.1, 4.2, (4.8) and (4.9), we conclude that $h(z)$ is a non-
constant polynomial in $\mathbb{C}^{2}$. Similarly as in the sub-case $2.1$,
we obtain that $\lim_{r\rightarrow\infty}\frac{T(r,\tan h)}{T(r,h)+\log
r+r^{n}}=\infty$ when $h$ is a non-constant polynomial. Therefore $h$ must be
a constant, a contradiction.

Sub-case 2.3: Assume

$\displaystyle D_{11}\dfrac{\partial h}{\partial z_{2}}-E_{11}\dfrac{\partial
h}{\partial z_{1}}+(D_{12}+E_{12})\neq 0,D_{12}\dfrac{\partial h}{\partial
z_{2}}+E_{12}\dfrac{\partial h}{\partial
z_{1}}-(D_{11}-E_{11})=0,T_{1}-T_{2}\neq 0,$

By the similar argument used in Sub-case $1$, we arrive a contradiction.

Sub-case 2.4: Suppose that

$\displaystyle D_{11}\dfrac{\partial h}{\partial z_{2}}-E_{11}\dfrac{\partial
h}{\partial z_{1}}+(D_{12}+E_{12})\neq 0,D_{12}\dfrac{\partial h}{\partial
z_{2}}+E_{12}\dfrac{\partial h}{\partial z_{1}}-(D_{11}-E_{11})\neq
0,T_{1}-T_{2}=0.$

By the similar argument used in Sub-case $1$, we arrive a contradiction.

Sub-case 2.5: Suppose that

$\displaystyle D_{11}\dfrac{\partial h}{\partial z_{2}}-E_{11}\dfrac{\partial
h}{\partial z_{1}}+(D_{12}+E_{12})=0,D_{12}\dfrac{\partial h}{\partial
z_{2}}+E_{12}\dfrac{\partial h}{\partial
z_{1}}-(D_{11}-E_{11})=0,\;T_{1}-T_{2}\neq 0.$

In view of (4.12), we see that $T_{1}-T_{2}=0$ which contradicts
$T_{1}-T_{2}\neq 0$.

Sub-case 2.6: Assume that

$\displaystyle D_{11}\dfrac{\partial h}{\partial z_{2}}-E_{11}\dfrac{\partial
h}{\partial z_{1}}+(D_{12}+E_{12})=0,D_{12}\dfrac{\partial h}{\partial
z_{2}}+E_{12}\dfrac{\partial h}{\partial z_{1}}-(D_{11}-E_{11})\neq
0,\;T_{1}-T_{2}=0.$

Using (4.12), we obtain $\cos h(z)=0$ and this shows that $h(z)$ is a
constant, a contradiction.

Sub-case 2.7: Let

$\displaystyle D_{11}\dfrac{\partial h}{\partial z_{2}}-E_{11}\dfrac{\partial
h}{\partial z_{1}}+(D_{12}+E_{12})\neq 0,D_{12}\dfrac{\partial h}{\partial
z_{2}}+E_{12}\dfrac{\partial h}{\partial
z_{1}}-(D_{11}-E_{11})=0,\;T_{1}-T_{2}=0.$

In view of (4.12) it is easy to see that $\sin h(z)=0$, this implies that
$h(z)$ is a constant, a contradiction.

Sub-case 2.8: Suppose that

$\displaystyle D_{11}\dfrac{\partial h}{\partial z_{2}}-E_{11}\dfrac{\partial
h}{\partial z_{1}}+(D_{12}+E_{12})=0,D_{12}\dfrac{\partial h}{\partial
z_{2}}+E_{12}\dfrac{\partial h}{\partial
z_{1}}-(D_{11}-E_{11})=0,T_{1}-T_{2}=0.$

Since $T_{1}=T_{2}$, a simple computation shows that

(4.13) $\displaystyle R_{11}\dfrac{\partial h}{\partial
z_{1}}-R_{12}\dfrac{\partial h}{\partial z_{2}}=0$

where
$R_{11}=E_{11}(D_{11}-E_{11})-E_{12}(D_{12}+E_{12}),\;R_{12}=D_{11}(D_{11}-E_{11})+D_{12}(D_{12}+E_{12}).$
Thus, it follows from (4.13) that

(4.14) $\displaystyle
h(z)=z_{2}+\dfrac{R_{12}}{R_{11}}z_{1}+R_{5},\;\;\mbox{where}\;R_{5}\in\mathbb{C}.$

Substituting this $h(z)$ into (4.10), we obtain

(4.15) $\displaystyle\dfrac{\partial f}{\partial z_{1}}-\dfrac{\partial
f}{\partial z_{2}}$
$\displaystyle=(D_{11}-E_{11})\cos\left(z_{2}+\dfrac{R_{12}}{R_{11}}z_{1}+R_{5}\right)-(D_{12}+E_{12})\sin\left(z_{2}+\dfrac{R_{12}}{R_{11}}z_{1}+R_{5}\right).$

The characteristic equations of (4.15) are
$\dfrac{dz_{1}}{dt}=1,\;\dfrac{dz_{2}}{dt}=-1$, and hence

$\displaystyle\dfrac{df}{dt}=(D_{11}-E_{11})\cos\left(z_{2}+\dfrac{R_{12}}{R_{11}}z_{1}+R_{5}\right)-(D_{12}+E_{12})\sin\left(z_{2}+\dfrac{R_{12}}{R_{11}}z_{1}+R_{5}\right).$

Using the initial conditions: $z_{1}=0$, $z_{2}=s$ and $f=f(0,s):=\phi_{1}(s)$
with a parameter $s$, wee see that $z_{1}=t,\;z_{2}=-t+s,$ and

$\displaystyle f(t,s)$ $\displaystyle=\phi_{1}(s)$
$\displaystyle+\int_{0}^{t}\left((D_{11}-E_{11})\cos\left(z_{2}+\dfrac{R_{12}}{R_{11}}z_{1}+R_{5}\right)-(D_{12}+E_{12})\sin\left(z_{2}+\dfrac{R_{12}}{R_{11}}z_{1}+R_{5}\right)\right)dt$
$\displaystyle=\phi_{1}(s)+(D_{11}-E_{11})\int_{0}^{t}\cos\left(\dfrac{R_{12}-R_{11}}{R_{11}}t+s+R_{5}\right)dt$
$\displaystyle\quad-(D_{12}+E_{12})\int_{0}^{t}\sin\left(\dfrac{R_{12}-R_{11}}{R_{11}}t+s+R_{5}\right)dt$
$\displaystyle=\phi_{2}(s)+\dfrac{(D_{11}-E_{11})R_{11}}{(R_{12}-R_{11})}\sin\left(\dfrac{R_{12}-R_{11}}{R_{11}}t+s+R_{5}\right)$
$\displaystyle\quad+\dfrac{(D_{12}+E_{12})R_{11}}{(R_{12}-R_{11})}\cos\left(\dfrac{R_{12}-R_{11}}{R_{11}}t+s+R_{5}\right)$

where
$\phi_{2}(s)=\phi_{1}(s)-\frac{(D_{11}-E_{11})R_{11}}{(R_{12}-R_{11})}\sin\left(s+R_{5}\right)-\frac{(D_{12}+E_{12})R_{11}}{(R_{12}-R_{11})}\cos\left(s+R_{5}\right)$
is a transcendental entire function with finite order in $s$. Using $t=z_{1}$,
$s=z_{1}+z_{2}$, we easily obtain

(4.16) $\displaystyle f(z_{1},z_{2})$
$\displaystyle=\dfrac{(D_{11}-E_{11})R_{11}}{(R_{12}-R_{11})}\sin\left(\dfrac{R_{12}}{R_{11}}z_{1}+z_{2}+R_{5}\right)$
$\displaystyle+\dfrac{(D_{12}+E_{12})R_{11}}{(R_{12}-R_{11})}\cos\left(\dfrac{R_{12}}{R_{11}}z_{1}+z_{2}+R_{5}\right)+\phi_{2}(z_{1}+z_{2})$

Substituting (4.16) into (4.9) gives that

$\displaystyle\phi^{\prime}_{2}(z_{1}+z_{2})+\phi_{2}(z_{1}+z_{2})=R_{21}\sin\left(\dfrac{R_{12}}{R_{11}}z_{1}+z_{2}+R_{5}\right)+R_{22}\cos\left(\dfrac{R_{12}}{R_{11}}z_{1}+z_{2}+R_{5}\right)+T_{2},$

where

$\displaystyle
R_{21}=\dfrac{E_{12}R_{12}+(E_{11}-D_{11}+D_{12})R_{11}}{(R_{12}-R_{11})}\;\mbox{and}\;R_{22}=\dfrac{E_{11}R_{12}-(E_{12}+D_{11}+D_{12})R_{11}}{(R_{12}-R_{11})}.$

It follows easily that

(4.17) $\displaystyle\phi_{2}(z_{1}+z_{2})$
$\displaystyle=\dfrac{(R_{21}+R_{22})}{2}\sin\left(\dfrac{R_{12}}{R_{11}}z_{1}+z_{2}+R_{5}\right)$
$\displaystyle-\dfrac{(R_{21}-R_{22})}{2}\cos\left(\dfrac{R_{12}}{R_{11}}z_{1}+z_{2}+R_{5}\right)+R_{6}e^{-(z_{1}+z_{2})}+T_{2}.$

Therefore, from (4.16) and (4.17), we obtain the precise form of the solution
$f$ as

$\displaystyle f(z_{1},z_{2})$
$\displaystyle=\left(\dfrac{E_{11}+E_{12}}{2}\right)\sin\left(\dfrac{R_{12}}{R_{11}}z_{1}+z_{2}+R_{5}\right)$
$\displaystyle+\left(\dfrac{E_{11}-E_{12}}{2}\right)\cos\left(\dfrac{R_{12}}{R_{11}}z_{1}+z_{2}+R_{5}\right)+R_{6}e^{-(z_{1}+z_{2})}+T_{2}.$

This completes the proof. ∎

###### Proof of Theorem 3.2.

Suppose that $f$ is a transcendental entire solution of (3.2) with finite
order. We see that the equation (3.2) can be written as

(4.18)
$\displaystyle\left(\dfrac{DA^{\pm}}{-\Delta}\right)u^{2}+\left(\dfrac{DB^{\mp}}{-\Delta}\right)v^{2}=1$

We discuss the following two possible cases.

Case 1: If $\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is a constant, then it follows
from (4.18) that $\sqrt{\frac{DB^{\mp}}{-\Delta}}v$ is also constant. Denote

$\displaystyle\sqrt{\dfrac{DA^{\pm}}{-\Delta}}u=K_{3}\;\mbox{and}\;\sqrt{\dfrac{DB^{\mp}}{-\Delta}}v=K_{4}.$

In view of (4.18), it is easy to see that $K^{2}_{3}+K^{2}_{4}=1$, hence

(4.19) $\displaystyle L_{1}(f)=f(z)+\dfrac{\partial f}{\partial
z_{1}}=\dfrac{K_{3}\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}-\dfrac{K_{4}\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}}:=A_{2},$
(4.20) $\displaystyle L_{3}(f)=f(z)+\dfrac{\partial^{2}f}{\partial
z^{2}_{1}}=\dfrac{K_{1}\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}+\dfrac{K_{2}\xi^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\beta-a\gamma}{ab-\alpha^{2}}:=B_{2}.$

Subtracting (4.20) from (4.19), we see that

(4.21) $\displaystyle\dfrac{\partial f}{\partial
z_{1}}-\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=A_{2}-B_{2}$

Differentiating (4.19) with respect to $z_{1}$ we obtain

(4.22) $\displaystyle\dfrac{\partial f}{\partial
z_{1}}+\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=0$

Therefore, it follows from (4.20) and (4.22) that

(4.23) $\displaystyle f(z)-\dfrac{\partial f}{\partial z_{1}}=B_{2}$

Differentiating (4.23) with respect to $z_{1}$ we obtain

(4.24) $\displaystyle\dfrac{\partial f}{\partial
z_{1}}-\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=0$

In view of (4.21) and (4.24), it is clear that $A_{2}=B_{2}$. Using
$K^{3}_{1}+K^{4}_{2}=1$, $A_{2}=B_{2}$ and from (4.2) and (4.3), a simple
computation shows that

$\displaystyle K_{3}=\dfrac{-R_{2}R_{3}\pm
R_{1}\sqrt{R^{2}_{1}+R^{2}_{2}-R^{2}_{3}}}{R^{2}_{1}+R^{2}_{2}}$

where

$\displaystyle
R_{1}=\dfrac{\xi^{\pm}_{1}-\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;R_{2}=\dfrac{\xi^{\pm}_{1}+\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\mbox{and}\;R_{3}=\dfrac{\beta(\alpha+b)-\gamma(\alpha+a)}{ab-\alpha^{2}}.$

Therefore, from (4.24) it follows that

(4.25) $\displaystyle
f(z_{1},z_{2})=e^{z_{1}+\phi_{1}(z_{2})}+\phi_{2}(z_{2}),$

where $\phi_{1}(z_{2})$, $\phi_{2}(z_{2})$ are two functions in $z_{2}$.
Substituting (4.25) into (4.19) we obtain
$2e^{z_{1}+\phi_{1}(z_{2})}+\phi_{2}(z_{2})=A_{2}.$ Thus, it follows that
$\phi_{2}(z_{2})=A_{2}$ and $2e^{z_{1}+\phi_{1}(z_{2})}=0$, which is
impossible.

Case 2: If $\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is not a constant. Similar to
the argument as in the proof of Theorem 3.1, there exists an entire function
$h(z)$ such that

(4.26) $\displaystyle L_{1}(f)=f(z)+\dfrac{\partial f}{\partial
z_{1}}=D_{11}\cos h(z)-D_{12}\sin h(z)+T_{1},$ (4.27) $\displaystyle
L_{3}(f)=f(z)+\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=E_{11}\cos
h(z)+E_{12}\sin h(z)+T_{2}.$

By the similar reasoning applied previously, we obtain

(4.28) $\displaystyle\dfrac{\partial f}{\partial
z_{1}}-\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=(D_{11}-E_{11})\cos
h(z)-(D_{12}+E_{12})\sin h(z)+(T_{1}-T_{2}).$

Differentiating (4.26) with respect to $z_{1}$ it follows that

(4.29) $\displaystyle\dfrac{\partial f}{\partial
z_{1}}+\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=-D_{11}\sin
h(z)\dfrac{\partial h}{\partial z_{1}}-D_{12}\cos h(z)\dfrac{\partial
h}{\partial z_{1}}$

From (4.27) and (4.29) it is easy to see that

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

Differentiating (4.30) with respect to $z_{1}$ and using (4.28) we obtain

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

Since $f$ is a finite order transcendental entire solution of the equation
(3.2), by the Lemmas 4.1, 4.2, (4.26) and (4.27), we conclude that $h(z)$ is a
non-constant polynomial in $\mathbb{C}^{2}$. Similar arguments as in the proof
of Theorem 3.1 gives that

(4.31) $\displaystyle(D_{12}+E_{12})-E_{11}\dfrac{\partial h}{\partial
z_{1}}-D_{12}\left(\dfrac{\partial h}{\partial
z_{1}}\right)^{2}+D_{11}\dfrac{\partial^{2}h}{\partial z^{2}_{1}}=0,$ (4.32)
$\displaystyle(D_{11}-E_{11})-E_{12}\dfrac{\partial h}{\partial
z_{1}}-D_{11}\left(\dfrac{\partial h}{\partial
z_{1}}\right)^{2}-D_{12}\dfrac{\partial^{2}h}{\partial z^{2}_{1}}=0,$ (4.33)
$\displaystyle\mbox{and}\;(T_{1}-T_{2})=0.$

In view of (4.31) and (4.32), we see that

(4.34) $\displaystyle\dfrac{\partial^{2}h}{\partial
z^{2}_{1}}-S_{1}\dfrac{\partial h}{\partial z_{1}}=S_{2},$

where
$S_{1}=\frac{D_{11}E_{11}-D_{12}E_{12}}{D^{2}_{11}+D^{2}_{12}}\;\;\mbox{and}\;\;S_{2}=-\frac{D_{11}E_{12}+D_{12}E_{11}}{D^{2}_{11}+D^{2}_{12}}$
which implies that
$h(z_{1},z_{2})=e^{S_{1}z_{1}+\phi_{1}(z_{2})}+e^{\phi_{2}(z_{2})}-\frac{S_{2}}{S_{1}}z_{1}$,
is a transcendental polynomial, which is a contradiction as $h(z)$ is a non-
constant polynomial. This completes the proof of Theorem 3.2. ∎

###### Proof of Theorem 3.3.

Suppose that $f$ is a transcendental entire solution of (3.3) with finite
order. The equation (3.3) can be written as

(4.35)
$\displaystyle\left(\dfrac{DA^{\pm}}{-\Delta}\right)u^{2}+\left(\dfrac{DB^{\mp}}{-\Delta}\right)v^{2}=1$

Now we will discuss two cases below.

Case 1: If $\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is a constant, then it follows
from (4.35) that $\sqrt{\frac{DB^{\mp}}{-\Delta}}v$ is also constant. Denote

$\displaystyle\sqrt{\dfrac{DA^{\pm}}{-\Delta}}u=K_{5}\;\mbox{and}\;\sqrt{\dfrac{DB^{\mp}}{-\Delta}}v=K_{6}$

In view of (4.35), it is easy to see that $K^{2}_{5}+K^{2}_{6}=1$. Thus, we
have

(4.36) $\displaystyle L_{1}(f)=f(z)+\dfrac{\partial f}{\partial
z_{1}}=\dfrac{K_{5}\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}-\dfrac{K_{6}\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}}:=A_{3},$
(4.37) $\displaystyle L_{4}(f)=f(z)+\dfrac{\partial^{2}f}{\partial
z_{1}\partial
z_{2}}=\dfrac{K_{5}\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}+\dfrac{K_{6}\xi^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\beta-a\gamma}{ab-\alpha^{2}}:=B_{3}.$

An easy computation using (4.36) and (4.37) shows that

(4.38) $\displaystyle\dfrac{\partial f}{\partial
z_{1}}-\dfrac{\partial^{2}f}{\partial z_{1}\partial z_{2}}=A_{3}-B_{3}$

Differentiating (4.36) with respect to $z_{2}$ we obtain

(4.39) $\displaystyle\dfrac{\partial f}{\partial
z_{2}}+\dfrac{\partial^{2}f}{\partial z_{2}\partial z_{1}}=0$

From (4.37) and (4.39), it follows that

(4.40) $\displaystyle f(z)-\dfrac{\partial f}{\partial z_{2}}=B_{3}$

Differentiating (4.40) with respect to $z_{1}$, we obtain

(4.41) $\displaystyle\dfrac{\partial f}{\partial
z_{1}}-\dfrac{\partial^{2}f}{\partial z_{1}\partial z_{2}}=0$

In view of (4.39) and (4.41) it is easy to see that $A_{3}=B_{3}$. Using
$K^{2}_{5}+K^{2}_{6}=1$, $A_{3}=B_{3}$ and from (4.36), (4.37) a simple
computation shows that

$\displaystyle K_{6}=\dfrac{-R_{2}R_{3}\pm
R_{1}\sqrt{R^{2}_{1}+R^{2}_{2}-R^{2}_{3}}}{R^{2}_{1}+R^{2}_{2}}$

where

$\displaystyle
R_{1}=\dfrac{\xi^{\pm}_{1}-\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;R_{2}=\dfrac{\xi^{\pm}_{1}+\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\mbox{and}\;R_{3}=\dfrac{\beta(\alpha+b)-\gamma(\alpha+a)}{ab-\alpha^{2}}.$

Combining (4.39) and (4.41) yields that

(4.42) $\displaystyle\dfrac{\partial f}{\partial z_{1}}+\dfrac{\partial
f}{\partial z_{2}}=0,$

which shows that $f(z_{1},z_{2})=\phi(z_{2}-z_{1})$. Substituting
$f(z_{1},z_{2})$ into (4.36), we obtain
$\phi^{\prime}(z_{2}-z_{1})-\phi(z_{2}-z_{1})=-A_{3}.$ Of course,
$\phi(z_{2}-z_{1})=-A_{3}+B_{11}e^{(z_{1}-z_{2})}$, $B_{11}\in\mathbb{C}$.
Hence, $f$ takes the form $f(z_{1},z_{2})=-A_{3}+B_{11}e^{(z_{1}-z_{2})}$.

Case 2: If $\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is not a constant, similar
argument being used in the proof of Theorem 3.1 that, there exists an entire
function $h(z)$ such that

(4.43) $\displaystyle L_{1}(f)=f(z)+\dfrac{\partial f}{\partial
z_{1}}=D_{11}\cos h(z)-D_{12}\sin h(z)+T_{1},$ (4.44) $\displaystyle
L_{4}(f)=f(z)+\dfrac{\partial^{2}f}{\partial z_{1}\partial z_{2}}=E_{11}\cos
h(z)+E_{12}\sin h(z)+T_{2}.$

In view of (4.43) and (4.44), a simple computation shows that

(4.45) $\displaystyle\dfrac{\partial f}{\partial
z_{1}}-\dfrac{\partial^{2}f}{\partial z_{1}\partial z_{2}}=(D_{11}-E_{11})\cos
h(z)-(D_{12}+E_{12})\sin h(z)+(T_{1}-T_{2}).$

Differentiating (4.43) with respect to $z_{2}$, we obtain

(4.46) $\displaystyle\dfrac{\partial f}{\partial
z_{2}}+\dfrac{\partial^{2}f}{\partial z_{2}\partial z_{1}}=-D_{11}\sin
h(z)\dfrac{\partial h}{\partial z_{2}}-D_{12}\cos h(z)\dfrac{\partial
h}{\partial z_{2}}$

From (4.44) and (4.46), it follows that

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

Differentiating (4.47) with respect to $z_{1}$, and using (4.45), the
following can be obtained

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

However, since $f$ is a finite order transcendental entire solution of the
equation (3.3), by Lemmas 4.1, 4.2, (4.43) and (4.44), we conclude that $h(z)$
is a non-constant polynomial in $\mathbb{C}^{2}$. The similar argument being
used in the proof of Theorem 3.1 will help us to obtain the following

(4.49) $\displaystyle D_{11}\dfrac{\partial^{2}h}{\partial z_{1}\partial
z_{2}}-D_{12}\dfrac{\partial h}{\partial z_{1}}\dfrac{\partial h}{\partial
z_{2}}-E_{11}\dfrac{\partial h}{\partial z_{1}}+(D_{12}+E_{12})=0,$ (4.50)
$\displaystyle(D_{11}-E_{11})-D_{12}\dfrac{\partial^{2}h}{\partial
z_{1}\partial z_{2}}-D_{11}\dfrac{\partial h}{\partial z_{1}}\dfrac{\partial
h}{\partial z_{2}}-E_{12}\dfrac{\partial h}{\partial z_{1}}=0$ (4.51)
$\displaystyle\mbox{and}\;(T_{1}-T_{2})=0.$

In view of (4.49) and (4.50), a simple computation shows that $\frac{\partial
h}{\partial z_{1}}\frac{\partial h}{\partial z_{2}}+S_{3}\frac{\partial
h}{\partial z_{1}}=S_{4},$ where

$\displaystyle
S_{3}=\dfrac{D_{11}E_{12}+D_{12}E_{11}}{D^{2}_{11}+D^{2}_{12}}\;\;\mbox{and}\;\;S_{4}=\dfrac{D_{11}(D_{11}-E_{11})+D_{12}(D_{12}+E_{12})}{D^{2}_{11}+D^{2}_{12}}$

which implies that
$h(z_{1},z_{2})=\phi(z_{1})e^{-S_{3}z_{2}}+\phi(z_{2})+\frac{S_{4}}{S_{3}}z_{1}$,
is a transcendental polynomial, which is a contradiction as $h(z)$ is a non-
constant polynomial. This completes the proof of Theorem 3.3. ∎

###### Proof of Theorem 3.4.

Suppose that $f$ is a transcendental entire solution of (3.4) with finite
order. Under the transformation mentioned in Section 2, (3.4) can be written
as

(4.52)
$\displaystyle\left(\dfrac{DA^{\pm}}{-\Delta}\right)u^{2}+\left(\dfrac{DB^{\mp}}{-\Delta}\right)v^{2}=1$

Case 1: If $\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is a constant, then it follows
from (4.52) that $\sqrt{\frac{DB^{\mp}}{-\Delta}}v$ is also constant. Denote

$\displaystyle\sqrt{\dfrac{DA^{\pm}}{-\Delta}}u=K_{7}\;\mbox{and}\;\sqrt{\dfrac{DB^{\mp}}{-\Delta}}v=K_{8}.$

From (4.52), it is clear that $K^{2}_{7}+K^{2}_{8}=1$. As a matter of fact, we
see that

(4.53) $\displaystyle M_{1}(f)=f(z+c)+\dfrac{\partial f}{\partial
z_{1}}=\dfrac{K_{7}\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}-\dfrac{K_{8}\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}}:=A_{4},$
(4.54) $\displaystyle M_{2}(f)=f(z+c)+\dfrac{\partial f}{\partial
z_{2}}=\dfrac{K_{7}\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}+\dfrac{K_{8}\xi^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\beta-a\gamma}{ab-\alpha^{2}}:=B_{4}.$

From (4.53) and (4.54), we obtain

(4.55) $\displaystyle\dfrac{\partial f}{\partial z_{1}}-\dfrac{\partial
f}{\partial z_{2}}=A_{4}-B_{4}.$

The characteristic equations of (4.55) are
$\frac{dz_{1}}{dt}=1,\;\frac{dz_{2}}{dt}=-1,\;\frac{df}{dt}=A_{4}-B_{4}.$
Using the initial conditions: $z_{1}=0$, $z_{2}=s$ and $f=f(0,s):=\psi(s)$
with a parameter $s$. Therefore, a simple computation shows that
$z_{1}=t,\;z_{2}=-t+s,$ and
$f(t,s)=\int_{0}^{t}(A_{4}-B_{4})dt+\psi(s)=(A_{4}-B_{4})t+\psi(s),$ where
$\psi(s)$ is a transcendental entire function with finite order in $s$. Then
we obtain $z_{1}=t$ and $s=z_{1}+z_{2}$, the solution of the equation (4.55)
is of the form

(4.56) $\displaystyle f(z_{1},z_{2})=(A_{4}-B_{4})z_{1}+\psi(z_{1}+z_{2}).$

In view of $\frac{\partial^{2}f}{\partial z_{1}\partial
z_{2}}=\frac{\partial^{2}f}{\partial z_{2}\partial z_{1}}$, using (4.2) and
(4.3), we obtain that $\frac{\partial f(z+c)}{\partial z_{1}}=\frac{\partial
f(z+c)}{\partial z_{2}}$. Thus, it follows from (4.55) that $A_{4}=B_{4}$.
Also, from (4.53), (4.54) and using $A_{0}=B_{0}$, $K^{2}_{7}+K^{2}_{8}=1$, a
simple computation shows that $K_{8}=\frac{-R_{2}R_{3}\pm
R_{1}\sqrt{R^{2}_{1}+R^{2}_{2}-R^{2}_{3}}}{R^{2}_{1}+R^{2}_{2}}$, where

$\displaystyle
R_{1}=\dfrac{\xi^{\pm}_{1}-\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;R_{2}=\dfrac{\xi^{\pm}_{1}+\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\mbox{and}\;R_{3}=\dfrac{\beta(\alpha+b)-\gamma(\alpha+a)}{ab-\alpha^{2}}.$

From (4.56) and using $A_{4}=B_{4}$ we obtain
$f(z_{1},z_{2})=\psi(z_{1}+z_{2})$, where $\psi(s)$ is a transcendental entire
function with finite order in $s$. Also $\psi(s)$ satisfying the relation
$\psi^{\prime}(s)+\psi(s+s_{0})=A_{4}$, where $s_{0}=c_{1}+c_{2}$.

Case 2: If $\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is not a constant. By the
similar argument being used previously, we obtain

(4.57) $\displaystyle M_{1}(f)=f(z+c)+\dfrac{\partial f}{\partial
z_{1}}=D_{11}\cos h(z)-D_{12}\sin h(z)+T_{1},$ (4.58) $\displaystyle
M_{2}(f)=f(z+c)+\dfrac{\partial f}{\partial z_{2}}=E_{11}\cos h(z)+E_{12}\sin
h(z)+T_{2}.$

Subtracting (4.58) from (4.57), we see that

(4.59) $\displaystyle\dfrac{\partial f}{\partial z_{1}}-\dfrac{\partial
f}{\partial z_{2}}=(D_{11}-E_{11})\cos h(z)-(D_{12}+E_{12})\sin
h(z)+(T_{1}-T_{2}).$

Since $\frac{\partial^{2}f}{\partial z_{1}\partial
z_{2}}=\frac{\partial^{2}f}{\partial z_{2}\partial z_{1}}$, using (4.57) and
(4.58), a simple computation shows that

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

Therefore, it follows from (4.59) and (4.60) that

(4.61)
$\displaystyle\left(\dfrac{\\{(D_{12}-iD_{11})+(E_{12}+iE_{11})\\}e^{ih(z+c)}+2i(T_{2}-T_{1})}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{ih(z+c)}$
$\displaystyle+\left(\dfrac{(D_{11}+iD_{12})\dfrac{\partial h}{\partial
z_{2}}-(E_{11}-iE_{12})\dfrac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{i[h(z+c)+h(z)]}$
$\displaystyle+\left(\dfrac{(E_{11}+iE_{12})\dfrac{\partial h}{\partial
z_{1}}-(D_{11}-iD_{12})\dfrac{\partial h}{\partial
z_{2}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{i[h(z+c)-h(z)]}=1.$

If $\left(\frac{(D_{11}+iD_{12})\frac{\partial h}{\partial
z_{2}}-(E_{11}-iE_{12})\frac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)=0$, then we have
$\left(\frac{(E_{11}+iE_{12})\frac{\partial h}{\partial
z_{1}}-(D_{11}-iD_{12})\frac{\partial h}{\partial
z_{2}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)\neq 0$. Otherwise,
$\left(\frac{\\{(D_{12}-iD_{11})+(E_{12}+iE_{11})\\}e^{ih(z+c)}+2i(T_{2}-T_{1})}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{ih(z+c)}=1$,
thus, it follows that $e^{ih(z+c)}=\frac{i(T_{1}-T_{2})\pm
i\sqrt{(T_{2}-T_{1})^{2}+[(D_{12}-iD_{11})+(E_{12}+iE_{11})][(iE_{11}-E_{12})-(iD_{11}+D_{12})]}}{(D_{12}-iD_{11})+(E_{12}+iE_{11})}$,
which implies that $h(z)$ must be a constant, and hence,
$\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is a constant, which is a contradiction.
Therefore, the equation (4.61) is reduced to the following form

$\displaystyle\left(\dfrac{\\{(D_{12}-iD_{11})+(E_{12}+iE_{11})\\}e^{ih(z+c)}+2i(T_{2}-T_{1})}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{ih(z+c)}$
$\displaystyle+\left(\dfrac{(E_{11}+iE_{12})\dfrac{\partial h}{\partial
z_{1}}-(D_{11}-iD_{12})\dfrac{\partial h}{\partial
z_{2}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{i[h(z+c)-h(z)]}=1.$

By the similar argument being used in [44], using Second Fundamental Theorem
of Nevanlinna for several complex variables, we obtain
$T(r,e^{ih(z+c)})=o\left(T\left(r,e^{h}\right)\right)$, which is a
contradiction. Therefore, $\left(\frac{(D_{11}+iD_{12})\frac{\partial
h}{\partial z_{2}}-(E_{11}-iE_{12})\frac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)\neq 0$. Similarly, it can be
shown that $\left(\frac{(E_{11}+iE_{12})\frac{\partial h}{\partial
z_{1}}-(D_{11}-iD_{12})\frac{\partial h}{\partial
z_{2}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)\neq 0$. Thus, in view of
Lemma 4.3 and (4.61), we easily obtain

(4.62) $\displaystyle\left(\dfrac{(E_{11}+iE_{12})\dfrac{\partial h}{\partial
z_{1}}-(D_{11}-iD_{12})\dfrac{\partial h}{\partial
z_{2}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{i[h(z+c)-h(z)]}\equiv 1$

Also, from (4.61), it follows that

(4.63) $\displaystyle\left(\dfrac{(D_{11}+iD_{12})\dfrac{\partial h}{\partial
z_{2}}-(E_{11}-iE_{12})\dfrac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{i[h(z)-h(z+c)]}\equiv 1.$

Since $h(z)$ is a polynomial, from (4.62)(or (4.63)), it follows that
$h(z+c)-h(z)=\xi$, where $\xi$ is a constant in $\mathbb{C}$. Hence, we see
that $h(z)=L(z)+H(z)+B$, where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$H(z):=H(s_{1})$, $H(s_{1})$ is a polynomial in $s_{1}=c_{2}z_{1}-c_{1}z_{2}$.
We show that $H(z)\equiv 0$. Using (4.62) and (4.63), a simple computation
shows that

(4.64)
$\displaystyle\dfrac{(E_{11}+iE_{12})\alpha_{1}-(D_{11}-iD_{12})\alpha_{2}+[(D_{11}-iD_{12})c_{1}+(E_{11}+iE_{12})c_{2}]H^{\prime}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}=e^{-iL(c)}$

and

(4.65)
$\displaystyle\dfrac{(D_{11}+iD_{12})\alpha_{2}-(E_{11}-iE_{12})\alpha_{1}-[(D_{11}+iD_{12})c_{1}+(E_{11}-iE_{12})c_{2}]H^{\prime}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}=e^{iL(c)}.$

In view of (4.64) and (4.65), now it follows that
$[(D_{11}-iD_{12})c_{1}+(E_{11}+iE_{12})c_{2}]H^{\prime}$ and
$[(D_{11}+iD_{12})c_{1}+(E_{11}-iE_{12})c_{2}]H^{\prime}$ are constants. Since
$(c_{1},c_{2})\neq(0,0)$, it follows that $H^{\prime}$ is a constant, that is
$\deg_{s_{1}}H\leq 1$. Hence, it is easy to see that $L(z)+H(z)+B$ is still
linear form of $\alpha_{1}z_{1}+\alpha_{2}z_{2}+B$, which means that
$H(z)\equiv 0$. Therefore it is easy to see that
$h(z)=L(z)+B=\alpha_{1}z_{1}+\alpha_{2}z_{2}+B$. Now, in view of (4.64) and
(4.65), we see that

$\begin{cases}\dfrac{(E_{11}+iE_{12})\alpha_{1}-(D_{11}-iD_{12})\alpha_{2}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}e^{iL(c)}=1;\vspace{2mm}\\\
\dfrac{(D_{11}+iD_{12})\alpha_{2}-(E_{11}-iE_{12})\alpha_{1}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}e^{-iL(c)}=1,\par\end{cases}$

which implies that

$\begin{cases}e^{2iL(c)}=\dfrac{(D_{11}+iD_{12})\alpha_{2}-(E_{11}-iE_{12})\alpha_{1}}{(E_{11}+iE_{12})\alpha_{1}-(D_{11}-iD_{12})\alpha_{2}},\vspace{2mm}\\\
(E^{2}_{11}+E^{2}_{12})\alpha^{2}_{1}+(D^{2}_{11}+D^{2}_{12})\alpha^{2}_{2}-2(D_{11}E_{11}-D_{12}E_{12})\alpha_{1}\alpha_{2}\\\
\quad+[i(D_{11}-E_{11})+(D_{12}+E_{12})]^{2}=0.\end{cases}$

Substituting $h(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}+B$ into (4.59), we obtain

(4.66) $\displaystyle\dfrac{\partial f}{\partial z_{1}}-\dfrac{\partial
f}{\partial z_{2}}$
$\displaystyle=(D_{11}-E_{11})\cos(\alpha_{1}z_{1}+\alpha_{2}z_{2}+B)$
$\displaystyle-(D_{12}+E_{12})\sin(\alpha_{1}z_{1}+\alpha_{2}z_{2}+B)+(T_{1}-T_{2}).$

The characteristic equations of (4.66) are
$\frac{dz_{1}}{dt}=1,\;\frac{dz_{2}}{dt}=-1$ and we see that

(4.67)
$\displaystyle\dfrac{df}{dt}=(D_{11}-E_{11})\cos(\alpha_{1}z_{1}+\alpha_{2}z_{2}+B)-(D_{12}+E_{12})\sin(\alpha_{1}z_{1}+\alpha_{2}z_{2}+B)+(T_{1}-T_{2}).$

Using the initial conditions: $z_{1}=0$, $z_{2}=s$ and $f=f(0,s):=\psi_{0}(s)$
with a parameter $s$. Therefore, it is easy to see that
$z_{1}=t,\;z_{2}=-t+s,$ and

$\displaystyle f(t,s)$
$\displaystyle=\int_{0}^{t}[(D_{11}-E_{11})\cos(\alpha_{1}z_{1}+\alpha_{2}z_{2}+B)-(D_{12}+E_{12})\sin(\alpha_{1}z_{1}+\alpha_{2}z_{2}+B)]dt$
$\displaystyle\quad+\int_{0}^{t}(T_{1}-T_{2})dt+\psi_{0}(s)$
$\displaystyle=\int_{0}^{t}((D_{11}-E_{11})\cos[(\alpha_{1}-\alpha_{2})t+\alpha_{2}s+B]-(D_{12}+E_{12})\sin[(\alpha_{1}-\alpha_{2})t+\alpha_{2}s+B])dt$
$\displaystyle\quad+\int_{0}^{t}(T_{1}-T_{2})dt+\psi_{0}(s)$
$\displaystyle=\dfrac{(D_{11}-E_{11})}{(\alpha_{1}-\alpha_{2})}\sin[(\alpha_{1}-\alpha_{2})t+\alpha_{2}s+B]+\dfrac{(D_{12}+E_{12})}{(\alpha_{1}-\alpha_{2})}\cos[(\alpha_{1}-\alpha_{2})t+\alpha_{2}s+B]$
$\displaystyle\quad+(T_{1}-T_{2})t+\psi_{1}(s)$

where $\psi_{1}(s)$ is a transcendental entire function with finite order in
$s$ such that

$\displaystyle\psi_{1}(s)=\psi_{0}(s)-\dfrac{(D_{11}-E_{11})}{(\alpha_{1}-\alpha_{2})}\sin(\alpha_{2}s+B)-\dfrac{(D_{12}+E_{12})}{(\alpha_{1}-\alpha_{2})}\cos(\alpha_{2}s+B).$

Therefore, we have

(4.68) $\displaystyle f(z_{1},z_{2})$
$\displaystyle=\dfrac{(D_{11}-E_{11})}{(\alpha_{1}-\alpha_{2})}\sin(\alpha_{1}z_{1}+\alpha_{2}z_{2}+B)+\dfrac{(D_{12}+E_{12})}{(\alpha_{1}-\alpha_{2})}\cos(\alpha_{1}z_{1}+\alpha_{2}z_{2}+B)$
$\displaystyle+(T_{1}-T_{2})z_{1}+\psi_{1}(s).$

which can be written as

$\displaystyle
f(z_{1},z_{2})=\dfrac{(D_{11}-E_{11})}{(\alpha_{1}-\alpha_{2})}\sin(L(z)+B)+\dfrac{(D_{12}+E_{12})}{(\alpha_{1}-\alpha_{2})}\cos(L(z)+B)+(T_{1}-T_{2})z_{1}+\psi_{1}(s).$

Substituting (4.68) into (4.57), we can obtain that $\psi_{1}(s)$ satisfies

$\displaystyle\psi^{\prime}_{1}(s)+\psi_{1}(s+s_{0})$
$\displaystyle=\left(\dfrac{E_{11}\alpha_{1}-D_{11}\alpha_{2}}{\alpha_{1}-\alpha_{2}}\right)\cos(L(z)+B)+\left(\dfrac{E_{12}\alpha_{1}-D_{12}\alpha_{2}}{\alpha_{1}-\alpha_{2}}\right)\sin(L(z)+B)$
$\displaystyle-\left(\dfrac{D_{11}-E_{11}}{\alpha_{1}-\alpha_{2}}\right)\sin(L(z)+L(c)+B)-\left(\dfrac{D_{12}-E_{12}}{\alpha_{1}-\alpha_{2}}\right)\cos(L(z)+L(c)+B)$
$\displaystyle-(T_{1}-T_{2})(z_{1}+c)+T_{2}.$

This completes the proof of the Theorem 3.4. ∎

###### Proof of Theorem 3.5.

Suppose that $f$ is a transcendental entire solution of (3.5) with finite
order. The equation (3.5) can be written as

(4.69)
$\displaystyle\left(\dfrac{DA^{\pm}}{-\Delta}\right)u^{2}+\left(\dfrac{DB^{\mp}}{-\Delta}\right)v^{2}=1$

Now we will discuss two cases below.

Case 1: If $\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is a constant, then it follows
from (4.69) that $\sqrt{\frac{DB^{\mp}}{-\Delta}}v$ is also constant. Denote

$\displaystyle\sqrt{\dfrac{DA^{\pm}}{-\Delta}}u=K_{9},\;\;\;\;\sqrt{\dfrac{DB^{\mp}}{-\Delta}}v=K_{10}$

In view of (4.69), it is easy to see that $K^{2}_{9}+K^{2}_{10}=1$. So we have

(4.70) $\displaystyle M_{1}(f)=f(z+c)+\dfrac{\partial f}{\partial
z_{1}}=\dfrac{K_{9}\xi^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}-\dfrac{K_{10}\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\gamma-b\beta}{ab-\alpha^{2}}:=A_{5},$
(4.71) $\displaystyle M_{3}(f)=f(z+c)+\dfrac{\partial^{2}f}{\partial
z^{2}_{1}}=\dfrac{K_{9}\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}}+\dfrac{K_{10}\xi^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}+\dfrac{\alpha\beta-a\gamma}{ab-\alpha^{2}}:=B_{5}.$

A simple computation using (4.70) and (4.71) shows that

(4.72) $\displaystyle\dfrac{\partial f}{\partial
z_{1}}-\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=A_{5}-B_{5}.$

Differentiating (4.70) with respect to $z_{1}$ we obtain

(4.73) $\displaystyle\dfrac{\partial f(z+c)}{\partial
z_{1}}+\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=0$

Thus it follows from (4.71) and (4.73) that

(4.74) $\displaystyle f(z+c)-\dfrac{\partial f(z+c)}{\partial z_{1}}=B_{5}$

Differentiating (4.74) with respect to $z_{1}$, we obtain

(4.75) $\displaystyle\dfrac{\partial f(z+c)}{\partial
z_{1}}-\dfrac{\partial^{2}f(z+c)}{\partial z^{2}_{1}}=0$

In view of (4.72) and (4.75) it is easy to see that $A_{5}=B_{5}$. Using
$K^{2}_{9}+K^{2}_{10}=1$, $A_{5}=B_{5}$ and from (4.70), (4.71), a simple
computation shows that $K_{2}=\frac{-R_{2}R_{3}\pm
R_{1}\sqrt{R^{2}_{1}+R^{2}_{2}-R^{2}_{3}}}{R^{2}_{1}+R^{2}_{2}}$, where

$\displaystyle
R_{1}=\dfrac{\xi^{\pm}_{1}-\eta^{\pm}_{1}}{\sqrt{\frac{DA^{\pm}}{-\Delta}}},\;R_{2}=\dfrac{\xi^{\pm}_{1}+\eta^{\pm}_{1}}{\sqrt{\frac{DB^{\mp}}{-\Delta}}}\;\mbox{and}\;R_{3}=\dfrac{\beta(\alpha+b)-\gamma(\alpha+a)}{ab-\alpha^{2}}.$

In view of (4.72), using $A_{5}=B_{5}$, we obtain

(4.76) $\displaystyle
f(z_{1},z_{2})=e^{z_{1}+\psi_{1}(z_{2})}+\psi_{2}(z_{2}).$

where $\psi_{1}(z_{2})$, $\psi_{2}(z_{2})$ are two function in $z_{2}$. From
(4.70) and (4.76), it follows that
$e^{(z_{1}+c_{1})+\psi_{1}(z_{2}+c_{2})}+e^{z_{1}+\psi_{1}(z_{2})}+\psi_{2}(z_{2}+c_{2})=A_{5}$
which implies that
$\psi_{2}(z_{2})=A_{5},\;\;\psi_{1}(z_{2})=Az_{2}+B,\;\;\mbox{where}\;\;A=\frac{2k\pi
i\pm\pi i-c_{1}}{c_{2}}$ Therefore, from (4.76) we obtain
$f(z_{1},z_{2})=A_{5}+e^{(z_{1}+Az_{2}+B)}.$
Case 2: If $\sqrt{\frac{DA^{\pm}}{-\Delta}}u$ is not a constant. By the
similar arguments being used in the proof of Theorem 3.1, there exists an
entire function $h(z)$ such that

(4.77) $\displaystyle M_{1}(f)=f(z+c)+\dfrac{\partial f}{\partial
z_{1}}=D_{11}\cos h(z)-D_{12}\sin h(z)+T_{1},$ (4.78) $\displaystyle
M_{3}(f)=f(z+c)+\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=E_{11}\cos
h(z)+E_{12}\sin h(z)+T_{2}.$

In view of (4.77) and (4.78), we obtain

(4.79) $\displaystyle\dfrac{\partial f}{\partial
z_{1}}-\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=(D_{11}-E_{11})\cos
h(z)-(D_{12}+E_{12})\sin h(z)+(T_{1}-T_{2}).$

Differentiating (4.77) with respect to $z_{1}$, we obtain

(4.80) $\displaystyle\dfrac{\partial f(z+c)}{\partial
z_{1}}+\dfrac{\partial^{2}f}{\partial z^{2}_{1}}=-D_{11}\sin
h(z)\dfrac{\partial h}{\partial z_{1}}-D_{12}\cos h(z)\dfrac{\partial
h}{\partial z_{1}}.$

Thus it follows from (4.78) and (4.80) that

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

Differentiating (4.81) with respect to $z_{1}$ and using (4.79), we obtain

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

This can be expressed as the following

(4.82)
$\displaystyle\left(\dfrac{\\{(D_{12}-iD_{11})+(E_{12}+iE_{11})\\}e^{ih(z+c)}+2i(T_{2}-T_{1})}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{ih(z+c)}$
$\displaystyle+\left(\dfrac{(D_{11}+iD_{12})\dfrac{\partial^{2}h}{\partial
z^{2}_{1}}+(iD_{11}-D_{12})\left(\dfrac{\partial h}{\partial
z_{1}}\right)^{2}-(E_{11}-iE_{12})\dfrac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{i[h(z+c)+h(z)]}$
$\displaystyle+\left(\dfrac{(iD_{12}-D_{11})\dfrac{\partial h}{\partial
z_{2}}+(iD_{11}+D_{12})\left(\dfrac{\partial h}{\partial
z_{1}}\right)^{2}+(iE_{12}+E_{11})\dfrac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{i[h(z+c)-h(z)]}=1.$

If $\left(\frac{(D_{11}+iD_{12})\frac{\partial^{2}h}{\partial
z^{2}_{1}}+(iD_{11}-D_{12})\left(\frac{\partial h}{\partial
z_{1}}\right)^{2}-(E_{11}-iE_{12})\frac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)=0$, then from (4.82) it is
easy to see that $\left(\frac{(iD_{12}-D_{11})\frac{\partial h}{\partial
z_{2}}+(iD_{11}+D_{12})\left(\frac{\partial h}{\partial
z_{1}}\right)^{2}+(iE_{12}+E_{11})\frac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)\neq 0$. Otherwise, it
follows that
$\left(\frac{\\{(D_{12}-iD_{11})+(E_{12}+iE_{11})\\}e^{ih(z+c)}+2i(T_{2}-T_{1})}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)=1$,
thus, a simple computation shows that $e^{ih(z+c)}=\frac{i(T_{1}-T_{2})\pm
i\sqrt{(T_{2}-T_{1})^{2}+[(D_{12}-iD_{11})+(E_{12}+iE_{11})][(iE_{11}-E_{12})-(iD_{11}+D_{12})]}}{(D_{12}-iD_{11})+(E_{12}+iE_{11})}$,
which implies $h(z)$ is a constant. Thus, the equation (4.82) reduces to

$\displaystyle\left(\dfrac{\\{(D_{12}-iD_{11})+(E_{12}+iE_{11})\\}e^{ih(z+c)}+2i(T_{2}-T_{1})}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{ih(z+c)}$
$\displaystyle+\left(\dfrac{(iD_{12}-D_{11})\dfrac{\partial h}{\partial
z_{2}}+(iD_{11}+D_{12})\left(\dfrac{\partial h}{\partial
z_{1}}\right)^{2}+(iE_{12}+E_{11})\dfrac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{i[h(z+c)-h(z)]}=1.$

By the similar argument being used previously, in view of the Second
Fundamental Theorem of Nevanlinna for several complex variables, we obtain
$T(r,e^{2ih(z+c)})=o\left(T(r,e^{h})\right)$ which is a contradiction. Hence,
$\frac{(D_{11}+iD_{12})\frac{\partial^{2}h}{\partial
z^{2}_{1}}+(iD_{11}-D_{12})\left(\frac{\partial h}{\partial
z_{1}}\right)^{2}-(E_{11}-iE_{12})\frac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\neq 0$. Similarly, we obtain
$\frac{(iD_{12}-D_{11})\frac{\partial h}{\partial
z_{2}}+(iD_{11}+D_{12})\left(\frac{\partial h}{\partial
z_{1}}\right)^{2}+(iE_{12}+E_{11})\frac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\neq 0$. Therefore, in view of the
Lemma 4.3 using (4.82), we obtain

(4.83) $\displaystyle\left(\dfrac{(iD_{12}-D_{11})\dfrac{\partial h}{\partial
z_{2}}+(iD_{11}+D_{12})\left(\dfrac{\partial h}{\partial
z_{1}}\right)^{2}+(iE_{12}+E_{11})\dfrac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{i[h(z+c)-h(z)]}\equiv 1.$

From (4.82) and (4.83) it follows that

(4.84)
$\displaystyle\left(\dfrac{(D_{11}+iD_{12})\dfrac{\partial^{2}h}{\partial
z^{2}_{1}}+(iD_{11}-D_{12})\left(\dfrac{\partial h}{\partial
z_{1}}\right)^{2}-(E_{11}-iE_{12})\dfrac{\partial h}{\partial
z_{1}}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}\right)e^{i[h(z)-h(z+c)]}\equiv 1.$

Since $h(z)$ is a polynomial, then equation (4.83)(or (4.84)) implies
$h(z+c)-h(z)=\eta$, where $\eta$ is a constant in $\mathbb{C}$. Therefore, it
follows that $h(z)=L(z)+H(z)+B$, where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$H(z):=H(s_{1})$, $H(s_{1})$ is a polynomial in $s_{1}=c_{2}z_{1}-c_{1}z_{2}$.
Now, we show that $H(z)\equiv 0$. From (4.62) and (4.63), it follows that

(4.85)
$\displaystyle(iD_{11}+D_{12})\alpha^{2}_{1}+(iE_{12}+E_{11})\alpha_{1}+c_{2}[2\alpha_{1}(iD_{11}+D_{12})+(iE_{12}+E_{11})]H^{\prime}$
$\displaystyle+c^{2}_{2}(iD_{11}+D_{12})(H^{\prime})^{2}=e^{-iL(c)}[(iD_{11}+D_{12})-(iE_{11}-E_{12})]$

and

(4.86)
$\displaystyle(iD_{11}-D_{12})\alpha^{2}_{1}-(E_{11}-iE_{12})\alpha_{1}+c_{2}[2\alpha_{1}(iD_{11}-D_{12})-(E_{11}-iE_{12})]H^{\prime}$
$\displaystyle+c^{2}_{2}(iD_{11}-D_{12})(H^{\prime})^{2}=e^{iL(c)}[(iD_{11}+D_{12})-(iE_{11}-E_{12})].$

Combining with the fact that $(c_{1},c_{2})\neq(0,0)$, it follows that
$H^{\prime}$ is a constant, that is $\deg_{s_{1}}H\leq 1$. Hence,
$L(z)+H(z)+B$ is still linear form of $\alpha_{1}z_{1}+\alpha_{2}z_{2}+B$,
which means that $H(z)\equiv 0$. Therefore it is easy to see that
$h(z)=L(z)+B=\alpha_{1}z_{1}+\alpha_{2}z_{2}+B$. Thus, the equations (4.85)
and (4.86) become

(4.87)
$\displaystyle\dfrac{(iD_{11}+D_{12})\alpha^{2}_{1}+(iE_{12}+E_{11})\alpha_{1}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}e^{iL(c)}=1,$

and

(4.88)
$\displaystyle\dfrac{(iD_{11}-D_{12})\alpha^{2}_{1}-(E_{11}-iE_{12})\alpha_{1}}{(iD_{11}+D_{12})-(iE_{11}-E_{12})}e^{-iL(c)}=1.$

From (4.87) and (4.88), a simple computation shows that

$\displaystyle
e^{2iL(c)}=\dfrac{(iD_{11}-D_{12})\alpha_{1}-(E_{11}-iE_{12})}{(iD_{11}+D_{12})\alpha_{1}+(iE_{12}+E_{11})}$

and

$\displaystyle(D^{2}_{11}+D^{2}_{12})\alpha^{4}_{1}$
$\displaystyle+2(D_{11}E_{12}+D_{12}E_{11})\alpha^{3}_{1}+(E^{2}_{11}+E^{2}_{12})\alpha^{2}_{1}$
$\displaystyle+[(iD_{11}+D_{12})-(iE_{11}-E_{12})]=0.$

The equation (4.81) can be written as

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

that is

(4.89) $\displaystyle\dfrac{\partial f}{\partial z_{1}}-f(z)$
$\displaystyle=-\left(D_{12}\alpha_{1}+E_{11}\right)\cos[L(z)-L(c)+B]$
$\displaystyle-\left(D_{11}\alpha_{1}+E_{12}\right)\sin[L(z)-L(c)+B]-T_{2},$

Therefore, a simple computation shows that

(4.90) $\displaystyle
f(z_{1},z_{2})=\phi(z_{2})e^{z_{1}}+R_{11}\sin[L(z)-L(c)+B]+R_{12}\cos[L(z)-L(c)+B],$

where $\phi(z_{2})$ is an entire function in $z_{2}$ and
$R_{11}=\frac{(D_{11}\alpha_{1}+E_{12})-\alpha_{1}(D_{12}\alpha_{1}+E_{11})}{\alpha^{2}_{1}+1},\;\;R_{12}=\frac{(D_{12}\alpha_{1}+E_{11})+\alpha_{1}(D_{11}\alpha_{1}+E_{12})}{\alpha^{2}_{1}+1}$.
Substituting (4.90) into (4.77), it is easy to see that

$\displaystyle\phi(z_{2}+c_{2})e^{c_{1}}e^{z_{1}}+(R_{11}+D_{12})\sin(L(z)+B)+(R_{12}-D_{11})\cos(L(z)+B)$
$\displaystyle+R_{11}\alpha_{1}\cos[L(z)-L(c)+B]-R_{12}\alpha_{1}\sin[L(z)-L(c)+B]=T_{1}-\phi(z_{2})e^{z_{1}}.$

Comparing both side the coefficient of $e^{z_{1}}$ we obtain
$\phi(z_{2}+c_{2})=-e^{-c_{1}}\phi(z_{2}),$ which implies that
$\phi(z_{2})=(-e^{-c_{1}})^{\frac{z_{2}}{c_{2}}}\pi(z_{2}),\;\;\mbox{where}\;\;\pi(z_{2}+c_{2})=\pi(z_{2}).$
Therefore, from (4.90), we see that the precise form of the solution is

$\displaystyle
f(z_{1},z_{2})=(-e^{-c_{1}})^{\frac{z_{2}}{c_{2}}}\pi(z_{2})e^{z_{1}}+R_{11}\sin[L(z)-L(c)+B]+R_{12}\cos[L(z)-L(c)+B].$

This completes the proof of the Theorem 3.5. ∎

Acknowledgment: The authors would like to thank the referee for their helpful
suggestions and comments to improve the exposition of the paper.

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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