1. Introduction
The Fermat type functional equation given by
\begin{equation}
f^2+g^2=1
\end{equation}can also be called the Pythagorean functional equation. In 1966, Gross [Reference Gross5] proved the classical result for equation (1.1) which is that the non-constant entire solutions of (1.1) are as follows:
\begin{equation*}f = \cos a(z) \quad \text{and} \quad g = \sin a(z),\end{equation*}where a(z) is an entire function. Actually, the study of the Fermat type functional equations such as (1.1) can be traced back to Montel [Reference Montel20] and Pólya [Reference Pólya21]. The Pythagorean functional equation also defined the eikonal equation:
\begin{equation*}[u_{z_1}]^2+[u_{z_2}]^2=1,\end{equation*} which was discussed by Li in [Reference Li17] and Khavinson in [Reference Khavinson7]. They proved that any entire solution of the above eikonal equation in 
$\mathbb{C}^2$ must be linear. We can see that the eikonal equation is a typical partial differential equation. Subsequently, Chang and Li [Reference Chang and Li4] further considered the following more general partial differential equation (PDE):
\begin{equation}
\big[X_1(u)\big])^2+\big[X_2(u)\big]^2=1
\end{equation} in 
$\mathbb{C}^2$, where
\begin{equation*}
X_1=p_1\frac{\partial}{\partial z_1}+p_2 \frac{\partial}{\partial z_2}
\quad \text{and} \quad X_2=p_3\frac{\partial}{\partial z_1}+p_4 \frac{\partial}{\partial z_2},\end{equation*} are linearly independent operators with pj being polynomials in 
$\mathbb{C}^2$, and gave the forms of the entire solution of (1.2).
In general, it is not a very simple problem to find the special entire and meromorphic solutions for a nonlinear PDE. For nearly a hundred years, Nevanlinna’s value distribution theory of meromorphic function has been developed rapidly ([Reference Cao and Korhonen2, Reference Korhonen8, Reference Ronkin22, Reference Stoll26, Reference Vitter27]), which plays an important role in studying the properties of solutions of (partial) differential equations, (partial) difference equations, functions with several variables, holomorphic curves and so on. It should be pointed out that the method of complex analysis and the classic Nevanlinna theory have attracted considerable attention in exploring the solutions of a large number of PDEs and their variants can be found in [Reference Cao and Xu3, Reference Khavinson7, Reference Lü9, Reference Lü, Lü, Li and Xu11–Reference Li14, Reference Li16, Reference Liu, Laine and Yang18, Reference Liu and Yang19, Reference Xu and Jiang28–Reference Xu, Xu and Liu31]. For example, Hu and Yang [Reference Hu and Yang6] obtained the forms of transcendental meromorphic solutions of a nonlinear PDE, and discussed the growth and uniqueness of the solutions; Yuan in [Reference Yuan, Huang and Shang34] gave all travelling meromorphic exact solutions of two nonlinear physical equation with the help of the method of complex analysis; Lü [Reference Lü9] studied the entire and meromorphic solutions of the nonlinear PDE 
$u_t-Cu^mu_x=Q,$ where 
$C\;(\neq 0)$ is a constant, 
$m\in \mathbb{N}_+$ and Q is a polynomial; Xu and Cao [Reference Xu and Cao32, Reference Xu and Cao33] studied the Fermat type partial differential equation 
$u^2+u_{z_1}^2=1$, and pointed out that any transcendental entire solution with finite order of this equation has the form given by 
$u(z_1, z_2) = \sin \big(z_1 + g(z_2)\big)$, where 
$g(z_2)$ is a polynomial in one variable z2. In 2005, Li [Reference Li15] further discussed a nonlinear PDE in the following form:
\begin{equation*}\left[u_{z_1}\right]^2+\left[u_{z_2}\right]^2=e^g,\end{equation*}which is more general than the Pythagorean functional equation. In fact, this equation in the real-variable case appears frequently in geometrical optics and wave propagation, which can be used to describe the wave fronts of light in an inhomogeneous medium with a variable index of refraction eg.
Theorem A (see [Reference Li15])
Let g be a polynomial in 
$\mathbb{C}^2$. Then u is an entire solution of the following partial differential equation:
\begin{equation}
\left[u_{z_1}\right]^2+\left[u_{z_2}\right]^2=e^g
\end{equation} in 
$\mathbb{C}^2$ if and only if
(i) 
$u=f(c_1z_1+c_2z_2)$ or
(ii) 
$u=\phi_1(z_1+iz_2)+\phi_2(z_1-iz_2),$
where f is an entire function in 
$\mathbb{C}$ satisfying
\begin{equation*}
f'(c_1z_1+c_2z_2)=\pm e^{g/2},\end{equation*} c1 and c2 are two constants satisfying 
$c_1^2+c_2^2=1$, and ϕ1 and ϕ2 are entire functions in 
$\mathbb{C}$ satisfying the condition is given by
\begin{equation*}
\phi'_1(z_1+iz_2)\phi'_2(z_1-iz_2)=\frac{1}{4}e^g.\end{equation*} Moreover, the forms of 
$f, \phi_1$ and ϕ2 can be found in [Reference Li14].
In addition, Saleeby [Reference Saleeby23–Reference Saleeby25] extended the result about the Pythagorean functional equation (the eikonal equation) by exploring the solutions of the quadratic trinomial functional equations 
$ f^2+2\alpha fg+g^2=1$, when 
$\alpha^2\neq 1$, 
$\alpha\in \mathbb{C}$ and
\begin{equation}
\left(u_{z_1}\right)^2+2\alpha u_{z_1}u_{z_2}+\left(u_{z_2}\right)^2=1,~~\alpha^2\neq 1,
\end{equation} and obtained that equation (1.4) can admit the nonlinear entire and meromorphic solutions in the case when 
$\alpha^2\neq 1$. Very recently, Lü [Reference Lü10] generalised equation (1.4) when the constant 1 on the right-hand side is replaced by eg, and demonstrated the following result.
Theorem B (see [Reference Lü10] Proposition 2.1)
Let g be a polynomial in 
$\mathbb{C}^2$. Let t1 and t2 be two distinct roots of the equation 
$1+2Bt+t^2=0 \;\;(B\neq \pm1)$. Then u is an entire solution of the following partial differential equation:
\begin{equation}
\left(u_{z_1}\right)^2+2B u_{z_1}u_{z_2}+\left(u_{z_2}\right)^2=e^g
\end{equation} in 
$\mathbb{C}^2$ if and only if
(i) 
$u(z_1,z_2)=F(z_1+t_2z_2+A(z_1+t_1z_2))$ and
(ii) 
$u(z_1,z_2)=\phi_1(z_1+t_1z_2)+\phi_2(z_1+t_2z_2),$
where F is an entire function in 
$\mathbb{C}$ satisfying the condition is given by
\begin{equation*}
AF'^2(z_1+t_2z_2+A(z_1+t_1z_2))=e^{g-\log \tau},\end{equation*} where A is a non-zero constant, 
$\tau=4(1-B^2),$ and ϕ1 and ϕ2 are entire functions in 
$\mathbb{C}$ satisfying the following conditions:
\begin{equation*}
\phi_1'(z_1+t_1z_2)=e^{\alpha(z_1+t_1z_2)}\quad \text{and} \quad
\phi_2'(z_1+t_2z_2)=e^{\beta(z_1+t_2z_2)},\end{equation*}where α and β are two polynomials such that
\begin{equation*}
\alpha(z_1+t_1z_2)+\beta(z_1+t_2z_2)=g(z_1,z_2)-\log \tau.\end{equation*}Inspired by the ideas of Chang and Li [Reference Chang and Li4], Li [Reference Li15] and Lü [Reference Lü10], one could naturally ask the following questions:
Question 1.1.
How to give the specific expressions for the solutions of equations (1.3) and (1.5) when 
$u_{z_1}$ and 
$u_{z_2}$ are replaced by some partial differential polynomials?
Our motivation in this article is to answer the above question. More precisely, this paper is devoted to giving a description of the finite-order transcendental entire solutions of the following quadratic binomial and trinomial partial differential equations:
\begin{equation}
\mathcal {P}_1(u)^2+\mathcal {P}_2(u)^2=e^{g}
\end{equation}and
\begin{equation}
\mathcal {P}_1(u)^2+2\alpha \mathcal {P}_1(u)\mathcal {P}_2(u)+\mathcal {P}_2(u)^2=e^{g},
\end{equation} where 
$\alpha^2\neq 0,1$, 
$g(z_1,z_2)=\alpha_1z_1+\alpha_2z_2+\beta_0$, 
$a_j,b_j,c_j(j=1,2)$ are constants in 
$\mathbb{C}$ and
\begin{equation*}\mathcal {P}_1(u)=a_1 u+b_1 u_{z_1}+c_1u_{z_2}
\quad \text{and} \quad \mathcal {P}_2(u)=a_2 u+b_2u_{z_1}+c_2u_{z_2}.\end{equation*} Obviously, such PDEs are the deformation equations of equations (1.3) to (1.5). When 
$\alpha=\pm1$, equation (1.7) can assume the following form:
\begin{equation}
au+bu_{z_1}+cu_{z_2}=e^{g/2}.
\end{equation}By a simple calculation, one can obtain that equation (1.8) has the transcendental entire solutions of the forms given by
\begin{equation*}
u=\Phi\left(z_2-\frac{c}{b}z_1\right)e^{-\frac{a}{b}z_1}+\frac{2}{2a+\alpha_1b+\alpha_2c}e^{g/2},\end{equation*} if 
$2a+\alpha_1b+\alpha_2c\neq0$, and
\begin{equation*}
u=\Phi\left(z_2-\frac{c}{b}z_1\right)e^{-\frac{a}{b}z_1}+\frac{z_1}{b}e^{g/2},\end{equation*} if 
$2a+\alpha_1b+\alpha_2c=0$, where 
$\Phi(x)$ is an entire function in x in 
$\mathbb{C}$. So, in this paper, we only deal with the case when 
$\alpha^2\neq 1$ for equation (1.7).
In this article, the Nevanlinna theory with several complex variables and the characteristic equations for quasi-linear PDEs are the main theoretical tools. Let us now introduce the structure of our article briefly. We devote Section 2 to exhibit our main results concerning the forms of solutions of Eqs. (1.6) and (1.7) and some variations, and list a series of examples to show the correctness of the forms of solutions in every case and the significant difference in the order of solutions for the equations involving a single variable to those involving several variables. The proof of the main results (Theorem 2.1 and Theorem 2.5) will be presented in Section 3 and Section 4, respectively.
2. Results and examples
The first main theorem is about the existence and the forms of the solutions for Equ. (1.6). Throughout the article, unless otherwise specified, we always assume that 
$\beta_1,\beta_2,\eta, \eta_0, \eta_1, \eta_2$ are constants, which can be different in each of their occurrences. Let
\begin{equation*}
\begin{aligned}
&A_1=-\frac{i}{2D}\left[(b_1^2+b_2^2)\alpha_1+(b_2+ib_1)(c_2-ic_1)\alpha_2+2(a_1b_1+a_2b_2)\right],\\
&A_2=\frac{i}{2D}\left[(b_1^2+b_2^2)\alpha_1+(b_2-ib_1)(c_2+ic_1)\alpha_2+2(a_1b_1+a_2b_2)\right],\\
&\mu_1=(b_1+ib_2)\alpha_1+(c_1+ic_2)\alpha_2+2(a_1+ia_2),\\
&\mu_2=(b_1-ib_2)\alpha_1+(c_1-ic_2)\alpha_2+2(a_1-ia_2).
\end{aligned}
\end{equation*}Theorem 2.1. Let 
$g(z_1,z_2)=\alpha_1z_1+\alpha_2z_2+\beta_0$, 
$\alpha_1,\alpha_2,\beta_0\in \mathbb{C}$, 
$D:=b_1c_2-b_2c_1\neq 0$, 
$D_1:=a_1b_2-a_2b_1\neq0$ and 
$D_2:=a_1c_2-a_2c_1\neq0$. Also let 
$u(z_1,z_2)$ be the finite-order transcendental entire solution of equation (1.6).
(i) If 
$\alpha_1\neq -\frac{2D_2}{D}$ or 
$\alpha_2\neq -\frac{2D_1}{D},$ then
\begin{equation*} u(z_1,z_2)=\pm\frac{2}{A}e^{g/2}+\eta_0e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},\end{equation*}where
\begin{equation*}
A=\sqrt{(2a_1+\alpha_1b_1+\alpha_2c_1)^2+(2a_2+\alpha_1b_2+\alpha_2c_2)^2};\end{equation*} (ii) If 
$\mu_1\mu_2\neq 0,$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{1}{\mu_1}e^{Q_1(z_1,z_2)}
+\frac{1}{\mu_2}e^{Q_2(z_1,z_2)}+\eta_0e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},
\end{aligned}
\end{equation*} (iii) If 
$\mu_1=0, \mu_2\neq 0$, then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\left(\eta_1z_1+\eta_0\right)e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]}
+\frac{1}{\mu_2}e^{Q_2(z_1,z_2)},
\end{aligned}
\end{equation*} (iv) If 
$\mu_1\neq 0$ and 
$\mu_2=0,$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{1}{\mu_1}e^{Q_1(z_1,z_2)}
+\left(\eta_2z_1+\eta_0\right)e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]};
\end{aligned}
\end{equation*} (v) If 
$\mu_1=0$ and 
$\mu_2=0,$ that is, 
$\alpha_1=\frac{2(a_2c_1-a_1c_2)}{D}$ and 
$\alpha_2=\frac{2(a_1b_2-a_2b_1)}{D},$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\left(\eta z_1
+\eta_0\right)e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},
\end{aligned}
\end{equation*} where 
$\eta, \eta_0, \eta_1, \eta_2, \beta_1, \beta_2$ are constants satisfying 
$e^{\beta_1+\beta_2}=e^{\beta_0}$ and
\begin{align*}
Q_1(z_1,z_2)&=-\frac{a_2+ia_1}{b_2+ib_1}z_1+A_1\left(z_2-\frac{c_2+ic_1}{b_2+ib_1}z_1\right)+\beta_1,\\
Q_2(z_1,z_2)&=-\frac{a_2-ia_1}{b_2-ib_1}z_1+A_2\left(z_2-\frac{c_2-ic_1}{b_2-ib_1}z_1\right)+\beta_2.
\end{align*}The following examples show the existence of transcendental entire solutions of equation (1.6) for every case in Theorem 2.1.
Example 2.1. Let
\begin{equation*}
u(z_1,z_2)=\frac{1}{\sqrt{26}}e^{z_1+z_2}+e^{-3z_1+4z_2}.\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and u is a transcendental entire solution of equation (1.6) with 
$a_1=2, b_1=2, c_1=1$, 
$a_2=-1$, 
$b_2=c_2=1$ and 
$g(z)=2z_1+2z_2$. Obviously, we can find that
\begin{equation*}\alpha_1(a_2b_1-a_1b_2)+\alpha_2(a_2c_1-a_1c_2)=-14\neq 0
\quad \text{and} \quad A=2\sqrt{26}.\end{equation*}This shows that the forms of solutions for Equ. (1.6) the case (i) in Theorem 2.1 are correct.
Example 2.2. Let 
$\eta_0\in \mathbb{C}$ and
\begin{equation*}
u(z_1,z_2)=\frac{1}{5}e^{\frac{1+5i}{2}z_1+\frac{1-10i}{2}z_2}+\frac{1}{5}e^{\frac{1-5i}{2}z_1+\frac{1+10i}{2}z_2}+\eta_0 e^{-2z_1+3z_2}.
\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and 
$u(z_1,z_2)$ is a transcendental entire solution of (1.6) with 
$a_1=1, b_1=2, c_1=1$, 
$a_2=-1, b_2=1, c_2=1$ and 
$g(z)=z_1+z_2$. Obviously, 
$\mu_1=\mu_2=5 \neq 0$. Hence, this example corresponds to the case 
$\mu_1=\mu_2\neq 0$ in Theorem 2.1.
Example 2.3. Let 
$\eta_0\in \mathbb{C}$ and
\begin{equation*}
u(z_1,z_2)=(-\frac{i}{2}z_1+\eta_0)e^{z_1-(1+i)z_2}+\frac{1+i}{4}e^{iz_1+(1-i)z_2}.
\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and u is a transcendental entire solution of (1.6) with 
$a_1=1, b_1=i, c_1=1$, 
$a_2=i, b_2=1, c_2=1$ and 
$g(z)=(1+i)z_1-2iz_2$. Obviously, 
$\mu_1=0$ and 
$\mu_2=2-2i\neq 0$. Hence, this example corresponds to the case 
$\mu_1=0$, 
$\mu_2\neq 0$ in Theorem 2.1.
Example 2.4. Let
\begin{equation*}
u(z_1,z_2)=\left(\frac{1}{\sqrt{5}}z_1+1\right)e^{-\frac{2}{3}z_1-\frac{1}{3}z_2}.\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and u is a transcendental entire solution of equation (1.6) with 
$a_1=1, b_1=2, c_1=-1$, 
$a_2=1, b_2=1$, 
$c_2=1$ and 
$g(z)=-\frac{4}{3}z_1-\frac{2}{3}z_2$. Obviously, 
$\mu_1=\mu_2= 0$. Hence, this example corresponds to the case 
$\mu_1=\mu_2=0$ in Theorem 2.1.
From Theorem 2.1, one can easily deduce the following corollary.
Corollary 2.1. Let
\begin{equation*}g(z_1,z_2)=\alpha_1z_1+\alpha_2z_2+\beta_0\quad \text{and} \quad
\alpha_1,\alpha_2,\beta_0\in \mathbb{C}.\end{equation*} Then the finite order transcendental entire solution 
$f(z_1,z_2)$ of the following equation:
\begin{equation}
\left(u+u_{z_1}\right)^2+\left(u+u_{z_2}\right)^2=e^g
\end{equation}possesses one of the following forms:
(i)
\begin{equation*} u(z_1,z_2)=\pm\frac{2e^{g/2}}{\sqrt{(\alpha_1+2)^2+(\alpha_2+2)^2}}+\eta_0e^{-(z_1+z_2)},\end{equation*} for 
$\alpha_1\neq -2$ and 
$\alpha_2=-2$;
(ii) If 
$\alpha_1+i\alpha_2\neq -2(1+i)$ and 
$\alpha_1-i\alpha_2\neq 2(i-1),$ then
\begin{equation*}
u(z_1,z_2)=\frac{1}{\alpha_1+i\alpha_2+2(1+i)}e^{\gamma_1(z)}+\frac{1}{\alpha_1-i\alpha_2+2(1-i)}e^{\gamma_2(z)}+\eta_0 e^{-(z_1+z_2)};\end{equation*} (iii) If 
$\alpha_1+i\alpha_2=-2(1+i)$ and 
$\alpha_1-i\alpha_2\neq 2(i-1),$ then
\begin{equation*}
u(z_1,z_2)=\frac{1}{\alpha_1-i\alpha_2+2(1-i)}e^{\gamma_2(z)}+(\eta_1z_1+\eta_0)e^{-(z_1+z_2)};\end{equation*} (iv) If 
$\alpha_1+i\alpha_2\neq -2(1+i)$ and 
$\alpha_1-i\alpha_2=2(i-1),$ then
\begin{equation*}
u(z_1,z_2)=\frac{1}{\alpha_1+i\alpha_2+2(1+i)}e^{\gamma_1(z)}+(\eta_2z_1+\eta_0)e^{-(z_1+z_2)};\end{equation*} (v) If 
$\alpha_1=\alpha_2=-2,$ then
\begin{equation*}
u(z_1,z_2)=(\eta z_1+\eta_0) e^{-(z_1+z_2)},\end{equation*} where 
$\beta_1, \beta_2, \eta, \eta_j \;\;(j=0,1,2,3)$ satisfy 
$e^{\beta_1+\beta_2}=e^{\beta_0}$ and
\begin{equation*}
\begin{aligned}
& \gamma_1(z)=\frac{\alpha_1+i(\alpha_2+2)}{2}z_1+\frac{\alpha_2-i(\alpha_1+2)}{2}z_2+\beta_1,\\
&\gamma_2(z)=\frac{\alpha_1-i(\alpha_2+2)}{2}z_1+\frac{\alpha_2+i(\alpha_1+2)}{2}z_2+\beta_2.
\end{aligned}
\end{equation*}Here we give some examples to show that the forms of solutions in Corollary 2.1 are precise.
Example 2.5. Let 
$\eta_0\in \mathbb{C}$ and
\begin{equation*}
u(z_1,z_2)=\frac{1}{\sqrt{13}}e^{z_1+2z_2}+\eta_0e^{-(z_1+z_2)}.\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and 
$u(z_1,z_2)$ is a transcendental entire solution of equation (2.1) with 
$g(z)=2z_1+4z_2$.
Example 2.6. Let 
$\eta_0\in \mathbb{C}$ and
\begin{equation*}
u(z_1,z_2)=(z_1+\eta_0)e^{-(z_1+z_2)}.\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and 
$u(z_1,z_2)$ is a transcendental entire solution of equation (2.1) with 
$g(z)=-2z_1-2z_2$.
When 
$a_1=a_2=0$, that is, 
$D_1=0$ in (1.6), by using an argument similar to that in Theorem 2.1, we have the following result.
Theorem 2.2. Let
\begin{equation*}g(z_1,z_2)=\alpha_1z_1+\alpha_2z_2+\beta_0\quad \text{and} \quad
\alpha_1,\alpha_2,\beta_0\in \mathbb{C}\end{equation*} and D ≠ 0. Also let 
$u(z_1,z_2)$ be the finite-order transcendental entire solution of the following equation:
\begin{equation*}
\left[b_1u_{z_1}+c_1u_{z_2}\right]^2+\left[b_2u_{z_1}+c_2u_{z_2}\right]^2=e^g.
\end{equation*} (i) If 
$\alpha_1\neq 0$ or 
$\alpha_2\neq 0,$ then
\begin{equation*}
u(z_1,z_2)=\pm\frac{2}{\sqrt{(b_1\alpha_1+c_1\alpha_2)^2+(b_2\alpha_1+c_2\alpha_2)^2}}e^{g/2}+\eta_0;\end{equation*} (ii) If 
$A_3A_4\neq 0,$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{-ie^{A_3\left(z_2-\frac{c_2+ic_1}{b_2+ib_1}z_1\right)+\beta_1}}{\alpha_1(b_2-b_1i)+\alpha_2(c_2-c_1i)}
+\frac{i e^{A_4\left(z_2-\frac{c_2-ic_1}{b_2-ib_1}z_1\right)+\beta_2}}{\alpha_1(b_2+b_1i)+\alpha_2(c_2+c_1i)}+\eta_0;
\end{aligned}
\end{equation*} (iii) If 
$A_3=0,$ then 
$A_4=\alpha_2$ and
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{ie^{\alpha_1z_1+\alpha_2z_2+\beta_2}}{2(\alpha_1b_2+\alpha_2c_2)}+\frac{e^{\beta_1}}{2D}\left[(c_2+c_1i)z_1-(b_2+b_1i)z_2\right]
+\eta_0,
\end{aligned}
\end{equation*} (iv) If 
$A_4=0,$ then 
$A_3=\alpha_2$ and
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{-ie^{\alpha_1z_1+\alpha_2z_2+\beta_1}}{2(\alpha_1b_2+\alpha_2c_2)}+\frac{e^{\beta_2}}{2D}\left[(c_2-c_1i)z_1-(b_2-b_1i)z_2\right]
+\eta_0,
\end{aligned}
\end{equation*} where 
$\eta_0, \beta_1, \beta_2$ are constants satisfying 
$e^{\beta_1+\beta_2}=e^{\beta_0}$ and
\begin{equation*}
\begin{aligned}
&A_3=\frac{b_1-b_2i}{2D}[\alpha_1(b_2-b_1i)+\alpha_2(c_2-c_1i)],\\
&A_4=\frac{b_1+b_2i}{2D}[\alpha_1(b_2+b_1i)+\alpha_2(c_2+c_1i)].
\end{aligned}
\end{equation*}For the case when D = 0, we consider the solutions of the following PDEs:
\begin{equation}
\left(a_1u+b_1u_{z_1}\right)^2+\left(a_2u+b_2u_{z_1}\right)^2=e^g
\end{equation}and
\begin{equation}
\mathcal {L}_1(u)^2+\mathcal {L}_2(u)^2=e^g,
\end{equation}where
\begin{equation*}\mathcal {L}_1(u)= a_1u+b_1\left(u_{z_1}+u_{z_2}\right)
\quad \text{and} \quad \mathcal {L}_2(u)=a_2u+b_2\left(u_{z_1}+u_{z_2}\right),\end{equation*}and demonstrate the following theorem.
Theorem 2.3. Let 
$g(z_1,z_2)=\alpha_1z_1+\alpha_2z_2+\beta_0$, 
$\alpha_1,\alpha_2,\beta_0\in \mathbb{C}$, 
$a_j,b_j\in \mathbb{C}, j=1,2$ and 
$D_1\neq0$. If equation (2.2) admits the finite-order transcendental entire solution 
$u(z_1,z_2),$ then 
$u(z_1,z_2)$ possesses the following forms:
\begin{equation*}
u(z_1,z_2)=\pm\frac{2e^{g/2}}{\sqrt{(2a_1+\alpha_1b_1)^2+(2a_2+\alpha_1b_2)^2}}\end{equation*}or
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{1}{2D_1}\left[(b_2+ib_1)e^{-\frac{a_1-ia_2}{b_1-ib_2}z_1+\phi_1(z_2)}
+(b_2-ib_1)e^{-\frac{a_1+ia_2}{b_1+ib_2}z_1+\phi_2(z_2)}\right],
\end{aligned}
\end{equation*}and
\begin{equation*}2(a_1b_1+a_2b_2)+\alpha_1(b_1^2+b_2^2)=0,\end{equation*} where 
$\phi_1(z_2)$ and 
$\phi_2(z_2)$ are polynomials in z2 satisfying the condition is given by
\begin{equation*}\phi_1(z_2)+\phi_2(z_2)=\alpha_2z_2+\beta_0.\end{equation*}Theorem 2.4. Let 
$D_1\neq 0,$ 
$g(z_1,z_2)=\alpha_1z_1+\alpha_2z_2+\beta_0$ and 
$\alpha_1,\alpha_2,\beta_0\in \mathbb{C}$. If 
$u(z_1,z_2)$ is a finite-order transcendental entire solution of equation (2.3), then 
$u(z_1,z_2)$ possesses one of the following forms:
\begin{equation*}
u(z_1,z_2)=\pm\frac{2e^{g/2}}{\sqrt{[2a_1+(\alpha_1+\alpha_2)b_1]^2+[2a_2+(\alpha_1+\alpha_2)b_2]^2}}\end{equation*}or
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{1}{2D_1}\left[(b_2+ib_1)e^{-\frac{a_1-ia_2}{b_1-ib_2}z_1+\phi_1(z_2-z_1)}
+(b_2-ib_1)e^{-\frac{a_1+ia_2}{b_1+ib_2}z_1+\phi_2(z_2-z_1)}\right],
\end{aligned}
\end{equation*}and
\begin{equation*}2(a_1b_1+a_2b_2)+(\alpha_1+\alpha_2)(b_1^2+b_2^2)=0,\end{equation*} where 
$\phi_1(z_2-z_1)$ and 
$\phi_2(z_2-z_1)$ are polynomials in 
$z_2-z_1$ satisfying the condition is given by 
$\deg_t[\phi_1(t)+\phi_2(t)]=1$ with the coefficient of t in 
$\phi_1(t)+\phi_2(t)$ being α2.
The following examples show the existence of transcendental entire solutions of equations (2.2) and (2.3).
Example 2.7. Let
\begin{equation*}
u(z_1,z_2)=\frac{1}{2}e^{z_1+z_2}.\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and 
$u(z_1,z_2)$ is a transcendental entire solution of equation (2.2) with 
$a_1=1, b_1=-1,$ 
$a_2=b_2=1$ and 
$g(z)=2z_1+2z_2$.
Example 2.8. Let
\begin{equation*}
u(z_1,z_2)=\frac{1-2i}{6}e^{\frac{1-3i}{5}z_1+z_2^3+2z_2}+\frac{1+2i}{6}e^{\frac{1+3i}{5}z_1-z_2^3+2z_2}.\end{equation*} Thus, clearly, 
$\rho(u)=3,$ and u is a transcendental entire solution of equation (2.2) with 
$a_1=1, b_1=-2$, 
$a_2=b_2=1$ and 
$g(z)=\frac{2}{5}z_1+4z_2$.
Example 2.9. Let
\begin{equation*}
u(z_1,z_2)=\frac{1}{5}e^{z_1+z_2}.\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and u is a transcendental entire solution of equation (2.3) with 
$a_1=2, b_1=a_2=b_2=1$ and 
$g(z)=2z_1+2z_2$.
Example 2.10. Let
\begin{equation*}
u(z_1,z_2)=\frac{1+i}{2}e^{-\frac{3+i}{2}z_1+(z_2-z_1)^3+2(z_2-z_1)}+\frac{1-i}{2}e^{-\frac{3-i}{2}z_1-(z_2-z_1)^3+4(z_2-z_1)}.\end{equation*} Thus, clearly, 
$\rho(u)=3,$ and u is a transcendental entire solution of equation (2.3) with 
$a_1=2, b_1=a_2=b_2=1$ and 
$g(z)=-9z_1+6z_2$.
To state the result of equation (1.7), we need to introduce some notations below. For this purpose in view, we set
\begin{align*}
&k_1:=\frac{1}{\sqrt{1+\alpha}}+\frac{1}{i\sqrt{1-\alpha}},~~k_2:=\frac{1}{\sqrt{1+\alpha}}-\frac{1}{i\sqrt{1-\alpha}},\\
&h_1:=[2(a_1+a_2)+\alpha_1(b_1+b_2)+\alpha_2(c_1+c_2)]\sqrt{1+\alpha},\\
&h_2:=[2(a_1-a_2)+\alpha_1(b_1-b_2)+\alpha_2(c_1-c_2)]\sqrt{1-\alpha},\\
&\nu_1:=\alpha_1(b_1k_1-b_2k_2)+\alpha_2(c_1k_1-c_2k_2)+2(a_1k_1-a_2k_2),\\
&\nu_2:=\alpha_1(b_2k_1-b_1k_2)+\alpha_2(c_2k_1-c_1k_2)+2(a_2k_1-a_1k_2)
\end{align*}and
\begin{align*}
&B_1:=\frac{1}{D(k_1^2-k_2^2)}\left\{(b_1k_2-b_2k_1)[\alpha_1(b_1k_1-b_2k_2)+\alpha_2(c_1k_1-c_2k_2)]\right.\\
&~~~~~\left.+2k_1k_2(a_1b_1+a_2b_2)-(k_1^2+k_2^2)(a_2b_1+a_1b_2)\right\},\\
&B_2:=\frac{1}{D(k_1^2-k_2^2)}\left\{(b_2k_2-b_1k_1)[\alpha_1(b_1k_2-b_2k_1)+\alpha_2(c_1k_2-c_2k_1)]\right.\\
&~~~~~\left.-2k_1k_2(a_1b_1+a_2b_2)+(k_1^2+k_2^2)(a_2b_1+a_1b_2)\right\}.
\end{align*}Obviously, we have
\begin{equation}
k_1k_2=\frac{2}{1-\alpha^2}, ~~k_1^2+k_2^2=\frac{-4\alpha}{1-\alpha^2}
\quad \text{and} \quad k_1^2-k_2^2=\frac{4}{i\sqrt{1-\alpha^2}}.
\end{equation}Theorem 2.5. Under the hypothesis of Theorem 2.1, let 
$\alpha^2\neq 0,1$. Also let 
$u(z_1,z_2)$ be a finite-order transcendental entire solution of (1.7).
(i) If 
$h_1\neq 0$ or 
$h_2\neq0,$ then
\begin{equation*} u(z_1,z_2)=\pm\frac{2\sqrt{2}}{\sqrt{h_1^2+h_2^2}}e^{g/2}+\eta_0e^{\frac{1}{D}[(a_2c_1-a_1c_2)z_1+(a_1b_2-b_1a_2)z_2]};\end{equation*}where η 0 is a constant;
(ii) If 
$\nu_1\nu_2\neq 0,$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{\sqrt{2}(k_1^2-k_2^2)}{4}\left[\frac{1}{\nu_1}e^{Q_3(z)}
+\frac{1}{\nu_2}e^{Q_4(z)}\right]+\eta_0e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]};
\end{aligned}
\end{equation*} (iii) If 
$\nu_1=0$ and 
$\nu_2\neq 0,$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{\sqrt{2}(k_1^2-k_2^2)}{4\nu_2}e^{Q_4(z)}+
\left(\eta_1z_1+\eta_0\right)e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]};
\end{aligned}
\end{equation*} (iv) If 
$\nu_1\neq0$ and 
$\nu_2=0,$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{\sqrt{2}(k_1^2-k_2^2)}{4\nu_1}e^{Q_3(z)}
+\left(\eta_2z_1+\eta_0\right)e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]};
\end{aligned}
\end{equation*} (v) If 
$\nu_1=0$ and 
$\nu_2=0,$ that is,
\begin{equation*}\alpha_1=\frac{2(a_2c_1-a_1c_2)}{D}, \alpha_2=\frac{2(a_1b_2-a_2b_1)}{D},\end{equation*}then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\left(z_1\eta
+\eta_0\right)e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},
\end{aligned}
\end{equation*} where 
$\eta, \eta_0, \eta_1, \eta_2$ are constants such that 
$\eta_1=e^{\beta_1},$ 
$\eta_2=e^{\beta_0-\beta_2},$ and
\begin{equation*}
\begin{aligned}
&Q_3(z)=\frac{a_2k_1-a_1k_2}{b_1k_2-b_2k_1}z_1+B_1\left(z_2-\frac{c_1k_2-c_2k_1}{b_1k_2-b_2k_1}z_1\right)+\beta_1,\\
&Q_4(z)=\frac{a_1 k_1-a_2 k_2}{b_2k_2-b_1k_1}z_1+B_2\left(z_2-\frac{c_2 k_2-c_1 k_1}{b_2k_2 -b_1k_1}z_1\right)+\beta_2.
\end{aligned}
\end{equation*}Remark 2.1. By comparing with Theorems B and 2.5, the forms of solutions in Theorem 2.5 are obviously different from those in Theorem B, and cannot be included in each other. Examples 2.11 to 2.14 can further illustrate this conclusion.
The following example shows that the forms of the finite-order transcendental entire solutions of equation (1.7) are precise.
Example 2.11. Let
\begin{equation*}
u(z)=\frac{2}{\sqrt{39}}e^{\frac{1}{2}z_1+z_2}+\eta_0e^{-\frac{1}{3}(2z_1+z_2)}.
\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and u is a transcendental entire solution of equation (1.7) with 
$a_1=1, b_1=2, c_1=-1$, 
$a_2=b_2=c_2=1,$ 
$\alpha=\frac{1}{2}$ and 
$g(z)=z_1+2z_2$. Obviously, 
$\alpha_1(a_1b_2-a_2b_1)+\alpha_2(a_1c_2-a_2c_1)=3\neq 0$. This example shows that the form of solution for Case (i) in Theorem 2.5 is precise.
Example 2.12. Let 
$\eta_0\in \mathbb{C}$ and
\begin{equation*}
u(z_1,z_2)=\frac{15-9\sqrt{3}i}{156}e^{\vartheta_1(z)}
+\frac{15+9\sqrt{3}i}{156}e^{\vartheta_2(z)}+\eta_0e^{-2z_1+3z_2},\end{equation*}where
\begin{equation*}\vartheta_1(z)=\frac{3-5\sqrt{3}i}{3}z_1+(1+4\sqrt{3}i)z_2
\quad \text{and} \quad \vartheta_2(z)=\frac{3+5\sqrt{3}i}{3}z_1+(1-4\sqrt{3}i)z_2.\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and u is a transcendental entire solution of (1.7) with 
$\alpha=-\frac{1}{2},$ 
$a_1=1, b_1=2, c_1=1,$ 
$a_2=-1,$ 
$b_2=c_2=1,$ and 
$g(z)=2z_1+2z_2$. In fact, by a simple calculation, we have
\begin{equation*}\nu_1=6\sqrt{2}-i\frac{10\sqrt{6}}{3}\neq 0
\quad \text{and} \quad \nu_2=-6\sqrt{2}-i\frac{10\sqrt{6}}{3}\neq 0.\end{equation*} This example shows that the form in the case when 
$\nu_1\nu_2\neq0$ in Theorem 2.5 is precise.
Example 2.13. Let 
$\eta_0\in \mathbb{C},$ 
$e^{\beta_0}=-3\sqrt{2}$ and
\begin{equation*}
u(z_1,z_2)=(z_1+\eta_0)e^{-\frac{1}{3}(z_1+z_2)}+\frac{\sqrt{2}}{8}e^{z_1+\frac{7}{3}z_2}.\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and u is a transcendental entire solution of equation (1.7) with 
$\alpha=\frac{5}{4},$ 
$a_1=1, b_1=2, c_1=1,$ 
$a_2=-1,$ 
$b_2=-4,$ 
$c_2=1,$ and 
$g(z)=\frac{2}{3}z_1+2z_2+\beta_0$. In fact, by a simple calculation, we have
\begin{equation*}\nu_1=0\quad \text{and} \quad \nu_2=-\frac{32}{3}\neq 0.\end{equation*} This example shows that the form in the case when 
$\nu_1=0$ and 
$\nu_2\neq0$ in Theorem 2.5 is precise.
Example 2.14. Let
\begin{equation*}
u(z_1,z_2)=\left(\frac{1}{\sqrt{6}}z_1+2\right)e^{-3z_1+z_2}.\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and u is a transcendental entire solution of equation (1.7) with 
$\alpha=2,$ 
$a_1=2, b_1=1, c_1=1,$ 
$a_2=1, b_2=1,$ 
$c_2=2,$ and 
$g(z)=-6z_1+2z_2$.
From Theorem 2.5, one can easily deduce the following corollary.
Corollary 2.2. Let 
$\alpha^2\neq 0,1,$ 
$g(z_1,z_2)=\alpha_1z_1+\alpha_2z_2+\beta_0,$ 
$\alpha_1,\alpha_2,\beta_0\in \mathbb{C}$. If 
$u(z_1,z_2)$ is a finite-order transcendental entire solution of the following equation:
\begin{equation*}\left[u+u_{z_1}\right]^2+2\alpha\left(u+u_{z_1}\right)
\left(u+u_{z_2}\right)+\left[u+u_{z_2}\right]^2=e^g.\end{equation*} (i) If 
$\alpha_1^2+2\alpha\alpha_1\alpha_2+\alpha_2^2\neq0,$ then
\begin{equation*}u(z_1,z_2)=\pm\frac{e^{g/2}}{2\sqrt{\alpha_1^2
+2\alpha\alpha_1\alpha_2+\alpha_2^2}}+\eta_0e^{-(z_1+z_2)};\end{equation*} (ii) If 
$\nu_3\nu_4\neq 0,$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{\sqrt{2}}{i\sqrt{1-\alpha^2}}
\left[\frac{e^{\gamma_3(z)}}{\nu_3}
+\frac{e^{\gamma_4(z)}}{\nu_4}\right]
+\eta_0e^{-(z_1+z_2)};
\end{aligned}
\end{equation*} (iii) If 
$\nu_3=0$ and 
$\nu_4\neq 0,$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{\sqrt{2}}{i\sqrt{1-\alpha^2}}
\left[\frac{1}{\nu_4}e^{\gamma_4(z)}
+\left(\eta_1z_1+\eta_0\right)e^{-(z_1+z_2)}
\right],
\end{aligned}
\end{equation*} (iv) If 
$\nu_3\neq 0$ and 
$\nu_4=0,$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{\sqrt{2}}{i\sqrt{1-\alpha^2}}\left[\frac{1}{\nu_3}e^{\gamma_3(z)}
+\left(\eta_2z_1+\eta_0\right)e^{-(z_1+z_2)}\right],
\end{aligned}
\end{equation*} (v) If 
$\nu_3=0$ and 
$\nu_4=0,$ that is, 
$\alpha_1=\alpha_2=-2,$ then
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{\sqrt{2}}{i\sqrt{1-\alpha^2}}\left(z_1\eta
+\eta_0\right)e^{-(z_1+z_2)},
\end{aligned}
\end{equation*} where 
$\eta,\eta_0,\eta_1,\eta_2$ are constants such that 
$\eta_1=e^{\beta_1},$ 
$\eta_2=e^{\beta_0-\beta_2},$ 
$\nu_3=k_1(\alpha_1+2)-k_2(\alpha_2+2),$ 
$\nu_4=k_2(\alpha_1+2)-k_1(\alpha_2+2),$ and
\begin{equation*}
\begin{aligned}
&\gamma_3(z)=\frac{k_1-k_2}{k_2}z_1+B_3\left(z_2+\frac{k_1}{k_2}z_1\right)+\beta_1,\\
&\gamma_4(z)=\frac{k_2-k_1}{k_1}z_1+B_4\left(z_2+\frac{k_2}{k_1}z_1\right)+\beta_2
\end{aligned}
\end{equation*}and
\begin{equation*}
B_3=\frac{-(k_1-k_2)^2+k_2(\alpha_1k_1-\alpha_2k_2)}{k_1^2-k_2^2}
\quad \text{and} \quad
B_4=\frac{-(k_1-k_2)^2+k_1(\alpha_1k_2-\alpha_2k_1)}{k_2^2-k_1^2}.\end{equation*}Corresponding to Theorem 2.2, by applying an argument similar to that in Theorem 2.5, we have the following result.
Theorem 2.6. Let 
$D\neq 0,$ 
$\alpha^2\neq 0,1,$ 
$g(z_1,z_2)=\alpha_1z_1+\alpha_2z_2+\beta_0,$ and 
$\alpha_1,\alpha_2,\beta_0\in \mathbb{C}$. Also let 
$u(z_1,z_2)$ be the finite-order transcendental entire solution of the following equation:
\begin{equation}
\left(b_1u_{z_1}+c_1u_{z_2}\right)^2+2\alpha(b_1u_{z_1}+c_1u_{z_2})(b_2u_{z_1}+c_2u_{z_2})+\left(b_2u_{z_1}+c_2u_{z_2}\right)^2=e^g.
\end{equation} (i) If 
$\epsilon_1\neq 0$ or 
$\epsilon_2\neq 0,$ then
\begin{equation*} u(z_1,z_2)=\pm\frac{2\sqrt{2}}{\sqrt{(\epsilon_1)^2+(\epsilon_2)^2}}e^{g/2}+\eta_0,\end{equation*}where
\begin{equation*}
\begin{aligned}
&\epsilon_1=\sqrt{1+\alpha}[(\alpha_1b_1+\alpha_2c_1)+(\alpha_2c_2+\alpha_1b_2)],\\
&\epsilon_2=\sqrt{1-\alpha}[(\alpha_1b_1+\alpha_2c_1)-(\alpha_2c_2+\alpha_1b_2)];
\end{aligned}
\end{equation*} (ii) If 
$B_5B_6\neq 0,$ then
\begin{equation*}
\begin{aligned}
& u(z_1,z_2) \\ & \quad=
\frac{\sqrt{2}(k_1^2\!-\!k_2^2)}{4D}\left[\frac{e^{B_5\left(z_2-\frac{c_1k_2-c_2k_1}{b_1k_2-b_2k_1}z_1\right)+\beta_1}}{\alpha_1(b_1k_1-b_2k_2)+\alpha_2(c_1k_1-c_2k_2)}
-\frac{e^{B_6\left(z_2-\frac{c_2k_2-c_1k_1}{b_2k_2-b_1k_1}z_1\right)+\beta_2}}{\alpha_1(b_1k_2-b_2k_1)+\alpha_2(c_1k_2-c_2k_1)}\right]+\eta_0;
\end{aligned}
\end{equation*} (iii) If 
$B_5=0,$ then 
$B_6=\alpha_2$ and
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&-\frac{\sqrt{2}k_2e^{\alpha_1z_1+\alpha_2z_2+\beta_2}}{4(\alpha_1b_1+\alpha_2c_1)}+
\frac{\sqrt{2}e^{\beta_1}}{4D}\left[(c_2k_1-c_1k_2)z_1+(b_1k_2-b_2k_1)z_2\right]
+\eta_0;
\end{aligned}
\end{equation*} (iv) If 
$B_6=0,$ then 
$B_5=\alpha_2$ and
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{\sqrt{2}k_1e^{\alpha_1z_1+\alpha_2z_2+\beta_1}}{4(\alpha_1b_1+\alpha_2c_1)}+
\frac{\sqrt{2}e^{\beta_2}}{4D}\left[(c_2k_2-c_1k_1)z_1+(b_1k_1-b_2k_2)z_2\right]
+\eta_0,
\end{aligned}
\end{equation*} where 
$\eta_0, \beta_0, \beta_1,\beta_2$ are constants satisfying 
$e^{\beta_1+\beta_2}=e^{\beta_0}$ and
\begin{equation*}
\begin{aligned}
&B_5=\frac{b_1k_2-b_2k_1}{(k_1^2-k_2^2)D}[\alpha_1(b_1k_1-b_2k_2)+\alpha_2(c_1k_1-c_2k_2)],\\
&B_6=\frac{b_2k_2-b_1k_1}{(k_1^2-k_2^2)D}[\alpha_1(b_1k_2-b_2k_1)+\alpha_2(c_1k_2-c_2k_1)].
\end{aligned}
\end{equation*}Correspondingly, for Theorems 2.3 and 2.4, we can apply the argument as in the proof of Theorem 2.5 in order to demonstrate the following theorems.
Theorem 2.7. Let 
$D_1\neq 0,$ 
$\alpha^2\neq 0,1$ and 
$g(z_1,z_2)=\alpha_1z_1+\alpha_2z_2+\beta_0,$ and 
$\alpha_1,\alpha_2,\beta_0\in \mathbb{C}$. If the following equation:
\begin{equation}
\left[a_1u+b_1u_{z_1}\right]^2
+2\alpha\left[a_1u+b_1u_{z_1}\right]\left[a_2u+b_2u_{z_1}\right]
+\left[a_2u+b_2u_{z_1}\right]^2=e^g
\end{equation} admits the finite-order transcendental entire solution 
$u(z_1,z_2),$ then 
$u(z_1,z_2)$ possesses the following forms:
\begin{equation*}
u(z_1,z_2)=\pm\frac{2\sqrt{2}}{\sqrt{\epsilon_3^2+\epsilon_4^2}}e^{g/2}
\end{equation*}or
\begin{equation*}
\begin{aligned}
u(z_1,z_2)\!=\!&\frac{\sqrt{2}}{4D_1}\left[(b_2 k_1-b_1 k_2)e^{\frac{a_1 k_2-a_2 k_1}{b_2 k_1-b_1 k_2}z_1+\phi_1(z_2)}
\!+\!(b_2 k_2-b_1 k_1)e^{-\frac{a_2 k_2-a_1 k_1}{b_2 k_2-b_1 k_1}z_1+\phi_2(z_2)}\right],
\end{aligned}
\end{equation*}and
\begin{equation*}
\frac{a_1 k_2-a_2 k_1}{b_2 k_1-b_1 k_2}-\frac{a_2 k_2-a_1 k_1}{b_2 k_2-b_1 k_1}=\alpha_1,\end{equation*} where 
$\phi_1(z_2),\phi_2(z_2)$ are polynomials in z2 with 
$\phi_1(z_2)+\phi_2(z_2)=\alpha_2z_2+\beta_0$, and
\begin{equation*}
\epsilon_3=[2(a_1+a_2)+\alpha_1(b_1+b_2)]\sqrt{1+\alpha}, ~~~~\epsilon_4=[2(a_1-a_2)+\alpha_1(b_1-b_2)]\sqrt{1-\alpha}.\end{equation*}Theorem 2.8. Let 
$D_1\neq0$, 
$\alpha^2\neq 0,1$ 
$g(z_1,z_2)=\alpha_1z_1+\alpha_2z_2+\beta_0,$ and 
$\alpha_1,\alpha_2,\beta_0\in \mathbb{C}$. If the following equation:
\begin{equation}
\mathcal {L}_1(u)^2+2\alpha \mathcal {L}_1(u)\mathcal {L}_2(u)+\mathcal {L}_2(u)^2=e^g
\end{equation} admits the finite-order transcendental entire solution 
$u(z_1,z_2),$ where 
$\mathcal {L}_1(u),\mathcal {L}_2(u)$ are stated as in Theorem 2.4, then 
$u(z_1,z_2)$ possesses the following forms:
\begin{equation*}
u(z_1,z_2)=\pm\frac{2\sqrt{2}}{\sqrt{\epsilon_5^2+\epsilon_6^2}}e^{g/2}
\end{equation*}or
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{\sqrt{2}}{4D_1}\left[(b_2 k_1-b_1 k_2)e^{\frac{a_1 k_2-a_2 k_1}{b_2 k_1-b_1 k_2}z_1+R_1(s)}
+(b_2 k_2-b_1 k_1)e^{-\frac{a_2 k_2-a_1 k_1}{b_2 k_2-b_1 k_1}z_1+R_2(s)}\right],
\end{aligned}
\end{equation*}and
\begin{equation*}
\frac{a_1 k_2-a_2 k_1}{b_2 k_1-b_1 k_2}-\frac{a_2 k_2-a_1 k_1}{b_2 k_2-b_1 k_1}=\alpha_1+\alpha_2,\end{equation*} where 
$R_j(s)=d_js+e_j\;\; (j=1,2),$ 
$s=z_2-z_1,$ dj and ej are constants such that 
$d_1+d_2=\alpha_2$ and 
$e_1+e_2=\beta_0,$ and
\begin{equation*}
\epsilon_5 \! =\! [2(a_1+a_2)+(\alpha_1+\alpha_2)(b_1+b_2)]\sqrt{1+\alpha}
\quad \text{and} \quad \epsilon_6\!=\![2(a_1-a_2)+(\alpha_1+\alpha_2)(b_1-b_2)]\sqrt{1\!-\!\alpha}.\end{equation*}The following examples show the existence of transcendental entire solutions of equations (2.6) and (2.7).
Example 2.15. Let
\begin{equation*}
u(z_1,z_2)=\frac{1}{\sqrt{37}}e^{z_1+z_2}.\end{equation*} Thus, clearly, 
$\rho(u)=1,$ and u is a transcendental entire solution of equation (2.6) with 
$\alpha=2,$ 
$a_1=2, b_1=a_2=b_2=1$ and 
$g(z)=2z_1+2z_2$.
Example 2.16. Let
\begin{equation*}
u(z_1,z_2)=-\sqrt{2}\left(e^{-\frac{5}{3}z_1+z_2^3+2z_2}
-e^{-\frac{4}{3}z_1-z_2^3+2z_2}\right).\end{equation*} Thus, clearly, 
$\rho(u)=3,$ and u is a transcendental entire solution of equation (2.6) with 
$\alpha=\frac{5}{4},$ 
$a_1=2, b_1=a_2=b_2=1$ and 
$g(z)=-3z_1+4z_2$.
Example 2.17. Let
\begin{equation*}
u(z_1,z_2)=\frac{1}{\sqrt{13}}e^{-z_1-2z_2}.\end{equation*} Thus, clearly, 
$\rho(u)=2,$ and u is a transcendental entire solution of equation (2.7) with 
$\alpha=2,$ 
$a_1=2, b_1=a_2=b_2=1$ and 
$g(z)=-2z_1-4z_2$.
Example 2.18. Let
\begin{equation*}
u(z_1,z_2)=-\sqrt{2}\left(e^{-\frac{11}{3}z_1+(z_2-z_1)^n
+2z_2}-e^{\frac{8}{3}z_1-(z_2-z_1)^n-4z_2}\right).\end{equation*} Thus, clearly, 
$\rho(u)=n,$ and u is a transcendental entire solution of equation (2.7) with 
$\alpha=\frac{5}{4}$, 
$a_1=2, b_1=a_2=b_2=1$ and 
$g(z)=-z_1-2z_2$.
Remark 2.2. By means of Examples 2.15 and 2.18, we can see that equations (2.6) and (2.7) admit a transcendental entire solution with any integer order.
3. Proofs of Theorems 2.1 to 2.4
As always, we assume that u is a finite-order transcendental entire solution for the equations in Theorems 2.1 to 2.4.
3.1. The proof of Theorem 2.1
Proof. First of all, we can rewrite (2.1) as follows:
\begin{equation}
\left[\frac{\mathcal {P}_1(u)}{e^{g/2}}\right]^2
+\left[\frac{\mathcal {P}_2(u)}{e^{g/2}}\right]^2=1.
\end{equation}We now divide our proof into two cases below.
Case 1. Suppose that 
$\frac{\mathcal {P}_1(u)}{e^{g/2}}$ is a constant. Also let
\begin{equation*}
\frac{\mathcal {P}_1(u)}{e^{g/2}}=\theta_1\qquad (\theta_1\in \mathbb{C}).
\end{equation*}that is,
\begin{equation}
\mathcal {P}_1(u)=a_1 u+b_1u_{z_1}+c_1u_{z_2}
=\theta_1e^{g/2}\qquad (\theta_1\in \mathbb{C}).
\end{equation} Thus, 
$\frac{\mathcal {P}_2(u)}{e^{g/2}}$ is also a constant in view of (3.1). Similarly, we let
\begin{equation}
\mathcal {P}_2(u)=a_2 u+b_2u_{z_1}+c_2u_{z_2}
=\theta_2e^{g/2}\qquad (\theta_2\in \mathbb{C}).
\end{equation}Then it follows that
\begin{equation}
\theta_1^2+\theta_2^2=1.
\end{equation} By noting that 
$D=b_1c_2-b_2c_1\neq 0$, we thus find from (3.2) and (3.3) that
\begin{equation}
u_{z_1}=\frac{1}{D}\left[(c_2\theta_1-c_1\theta_2)e^{g/2}+(a_2c_1-a_1c_2)u\right]
\end{equation}and
\begin{equation}
u_{z_2}=\frac{1}{D}\left[(b_1\theta_2-b_2\theta_1)e^{g/2}+(a_1b_2-a_2b_1)u\right].
\end{equation} Now, since 
$u_{z_1z_2}=u_{z_2 z_1}$, by differentiating (3.5) and (3.6) with respect to z2 and z 1, respectively, we have
\begin{equation}
(a_1b_2-a_2b_1)u_{z_1}-(a_2c_1-a_1c_2)u_{z_2}=-\frac{1}{2}e^{g/2}[\alpha_1(b_1\theta_2-b_2\theta_1)-\alpha_2(c_2\theta_1-c_1\theta_2)].
\end{equation} Noting that 
$a_2c_1-a_1c_2\neq 0$ and 
$a_1b_2-a_2b_1\neq 0$, upon substituting from (3.5) and (3.6) into (3.7), we get
\begin{equation}
(b_1\theta_2-b_2\theta_1)\alpha_1-(c_2\theta_1-c_1\theta_2)\alpha_2=2(a_2\theta_1-a_1\theta_2),
\end{equation}which leads to
\begin{equation}
\theta_1=\frac{2a_1+\alpha_1b_1+\alpha_2c_1}{2a_2+\alpha_1b_2+\alpha_2c_2}\theta_2.
\end{equation}Thus, Eqs. (3.4) and (3.9) can readily yield
\begin{align}
& \theta_1^2=\frac{(2a_1+\alpha_1b_1+\alpha_2c_1)^2}{A^2},~~~
\theta_2^2=\frac{(2a_2+\alpha_1b_2+\alpha_2c_2)^2}{A^2},
\end{align}where
\begin{equation*}A=\sqrt{(2a_1+\alpha_1b_1+\alpha_2c_1)^2+(2a_2+\alpha_1b_2+\alpha_2c_2)^2}.\end{equation*}On the other hand, the characteristic equations of (3.7) are given by
\begin{equation*}
\frac{dz_1}{dt}=a_2b_1-a_1b_2,\end{equation*}
\begin{equation*}\frac{dz_2}{dt}=a_2c_1-a_1c_2\end{equation*}and
\begin{equation*}\frac{du}{dt}=(a_2\theta_1-a_1\theta_2)e^{g/2}.\end{equation*} Applying the initial conditions: 
$z_1=0,z_2=s$ and 
$u=u(0,s)=\phi(s)$ with a parameter s, if we solve equation (3.7) and combine the solution with (3.8), we obtain the following parametric representation for the solutions of the characteristic equations:
\begin{equation*}z_1=(a_2b_1-a_1b_2)t,\end{equation*}
\begin{equation*}z_2=(a_2c_1-a_1c_2)t+s\end{equation*}and
\begin{equation}
u(t,s)=\int_0^t(a_2\theta_1-a_1\theta_2)e^{g/2}dt+\phi(s),
\end{equation} where 
$\phi(s)$ is a finite-order entire function with
\begin{equation*}s=z_2-\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}z_1.\end{equation*} If 
$\alpha_1(a_2b_1-a_1b_2)+\alpha_2(a_2c_1-a_1c_2)=0$, it follows from (3.8) and (3.9) that
\begin{equation}
a_2\theta_1-a_1\theta_2=\frac{\alpha_1(a_2b_1-a_1b_2)
+\alpha_2(a_2c_1-a_1c_2)}{2a_2+\alpha_1b_2+\alpha_2c_2}\theta_2=0.
\end{equation} Thus, in view of (3.11) and (3.12), it follows that 
$
u(t,s)=\phi(s)$. Upon substituting it into (3.2) or (3.3), we have
\begin{equation}
a_1 \phi(s)+ \frac{a_1(b_1c_2-b_2c_1)}{a_2b_1-a_1b_2}\phi'(s)=\theta_1e^{g/2}
\quad \text{or} \quad a_2 \phi(s)+ \frac{a_2(b_1c_2-b_2c_1)}
{a_2b_1-a_1b_2}\phi'(s)=\theta_2e^{g/2}.
\end{equation}These lead to
\begin{equation}
\phi(s)=\pm\frac{2}{A}e^{g/2}+\eta_0e^{\frac{a_1b_2-a_2b_1}{D}s}
=\pm\frac{2}{A}e^{g/2}+\eta_0e^{\frac{1}{D}
\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},
\end{equation} where 
$\eta_0\in \mathbb{C}$.
If 
$\alpha_1(a_2b_1-a_1b_2)+\alpha_2(a_2c_1-a_1c_2)\neq 0$, it follows from (3.11) that
\begin{equation}
u(t,s)=\frac{2(a_2\theta_1-a_1\theta_2)}
{\alpha_1(a_2b_1-a_1b_2)+\alpha_2(a_2c_1-a_1c_2)}e^{g/2}+\phi(s).
\end{equation}By combining it with equations (3.9), (3.10) and (3.15), we obtain
\begin{equation}
u(z_1,z_2)=\pm\frac{2}{A}e^{g/2}+\phi(s),
\end{equation} where 
$\phi(s)$ is a finite-order entire function in
\begin{equation*}s=z_2-\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}z_1.\end{equation*}Substituting from (3.16) into (3.2) or (3.3), we can deduce that
\begin{equation}
\phi(s)=\eta_0e^{\frac{a_1b_2-a_2b_1}{D}s}
=\eta_0e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]}.
\end{equation}Thus, in view of (3.16) and (3.17), this completes the proof of Case (i) of Theorem 2.1.
Case 2. Suppose that 
$\frac{\mathcal {P}_1(u)}{e^{g/2}}$ is not a constant. Thus, (1.6) can be rewritten as the following form:
\begin{equation*}
\left[\frac{\mathcal {P}_1(u)}{e^{g/2}}+i\frac{\mathcal {P}_2(u)}{e^{g/2}}\right]\left[\frac{\mathcal {P}_1(u)}{e^{g/2}}-i\frac{P_2(u)}{e^{g/2}}\right]=1.\end{equation*}Noting that the assumption that u is a transcendental entire function of finite order, it follows that
\begin{equation*}\frac{\mathcal {P}_1(u)}{e^{g/2}}+i\frac{\mathcal {P}_2(u)}{e^{g/2}}
\quad \text{and} \quad \frac{\mathcal {P}_1(u)}{e^{g/2}}-i\frac{\mathcal {P}_2(u)}{e^{g/2}}\end{equation*} have no zeros and poles. Thus, by using the conclusions in [Reference Ronkin22, Reference Stoll26] and the Hadamard factorisation theorem [Reference Pólya21], there exists a non-constant polynomial p(z) in 
$\mathbb{C}^2$such that
\begin{equation*}
\frac{\mathcal {P}_1(u)}{e^{g/2}}+i\frac{\mathcal {P}_2(u)}{e^{g/2}})=e^{p(z)}
\quad \text{and} \quad \frac{\mathcal {P}_1(u)}{e^{g/2}}-i\frac{\mathcal {P}_2(u)}{e^{g/2}}=e^{-p(z)},\end{equation*}which leads to
\begin{equation}
\mathcal {P}_1(u)=\frac{e^{Q_1}+e^{Q_2}}{2}
\quad \text{and} \quad \mathcal {P}_2(u)=\frac{e^{Q_1}-e^{Q_2}}{2i},
\end{equation}where
\begin{equation}
Q_1=\frac{g}{2}+p \quad \text{and} \quad Q_2=\frac{g}{2}-p.
\end{equation}In view of (3.18), we can deduce that
\begin{align}
&u_{z_1}=\frac{1}{D}\left\{\frac{1}{2}\left[(c_2+ic_1)e^{Q_1}+(c_2-ic_1)e^{Q_2}\right]+(a_2c_1-a_1c_2)u\right\},
\end{align}
\begin{align} &u_{z_2}=\frac{1}{D}\left\{\frac{1}{2}\left[-(b_2+ib_1)e^{Q_1}+(-b_2+ib_1)e^{Q_2}\right]+(a_1b_2-a_2b_1)u\right\}.
\end{align} By virtue of 
$u_{z_1z_2}=u_{z_2 z_1}$, and together with (3.20) and (3.21), we have
\begin{equation*}
\begin{aligned}
&\frac{1}{2}\left[(c_2+ic_1)\frac{\partial Q_1}{\partial z_2}e^{Q_1}+(c_2 -ic_1)\frac{\partial Q_2}{\partial z_2}e^{Q_2}\right]+(a_2c_1-a_1c_2)u_{z_2}\\
&\qquad =\frac{1}{2}\left[-(b_2+ib_1)\frac{\partial Q_1}{\partial z_1}e^{Q_1}-(b_2- ib_1)\frac{\partial Q_2}{\partial z_1}e^{Q_2}\right]+(a_1b_2-a_2b_1)u_{z_1}.
\end{aligned}
\end{equation*}Substituting from (3.20) and (3.21) into the above equality, we obtain
\begin{align}
e^{2p}\left[(b_{2}+ib_{1})\frac{\partial Q_{1}}{\partial z_{1}}+(c_{2}+ic_{1})\frac{\partial Q_{1}}{\partial z_{2}}+(a_{2}+ia_{1})\right]
\end{align}
\begin{align*}
&=(ib_1-b_2)\frac{\partial Q_2}{\partial z_1}+(ic_1-c_2)\frac{\partial Q_2}{\partial z_2}-(a_2-ia_1).
\end{align*} The fact that p is a non-constant polynomial in 
$\mathbb{C}^2$ leads to
\begin{equation}
(b_2+ib_1)\frac{\partial Q_1}{\partial z_1}+(c_2+ic_1)\frac{\partial Q_1}{\partial z_2}+(a_2+ia_1)\equiv 0
\end{equation}and
\begin{equation}
(ib_1-b_2)\frac{\partial Q_2}{\partial z_1}+(ic_1-c_2)\frac{\partial Q_2}{\partial z_2}-(a_2-ia_1)\equiv 0.
\end{equation} Otherwise, we can get a contradiction from (3.22). Indeed, by using the results (see, for example, [Reference Vitter27, p.99], [Reference Stoll26]) or [Reference Berenstein, Chang and Li1, Lemma 3.2], we can conclude from (3.22) that 
$T(r, e^{2p}) = O\{T(r,g)+T(r, p) + \log r\}$, outside possibly a set of finite Lebesgue measures. On the other hand, we have
\begin{equation*}\lim\limits_{r\rightarrow\infty}\frac{T(r,e^{2p})}{T(r,p)+\log r}=+\infty\end{equation*}for p being not a constant. Due to the fact that g and p are polynomials, we can see that p must be a constant. This, clearly, is a contradiction.
In view of the characteristic equations of (3.23) and (3.24), we can observe that Q 1 and Q 2 are of the following forms:
\begin{equation}
Q_1(z)=-\frac{a_2+ia_1}{b_2+ib_1}z_1+\varphi_1(s_1)
\quad \text{and} \quad Q_2(z)=-\frac{a_2-ia_1}
{b_2-ib_1}z_1+\varphi_2(s_2),
\end{equation} where 
$\varphi_j(s_j)(j=1,2)$ are entire functions in sj,
\begin{equation*}s_1=z_2-\frac{c_2+ic_1}{b_2+ib_1}z_1\end{equation*}and
\begin{equation*}s_2=z_2-\frac{c_2-ic_1}{b_2-ib_1}z_1.\end{equation*} Next, we will confirm that Q 1 and Q 2 are linear forms of z 1 and z2, that is, 
$\deg_{s_j}\varphi_j\leq 1$ 
$\;(j=1,2)$. Since p and g are polynomials in 
$\mathbb{C}^2$, it follows that 
$\varphi_j(s_j)\;\;(j=1,2)$ are polynomials in sj. Set 
$\deg_{s_1}\varphi_1=m$ and 
$\deg_{s_2}\varphi_2=n$, and let
\begin{align*}
\varphi_{1}(s_{1})=&\xi_{ms_{1}^{m}}+\xi_{m-1}s_{1^{m-1}}+\cdots+\xi_0,\\
\varphi_2(s_2)=&\zeta_ns_2^n+\zeta_{n-1}s_2^{m-1}+\cdots+\zeta_0,
\end{align*} where 
$\xi_m,\cdots,\xi_0,\zeta_n,\cdots,\zeta_0$ are constants and 
$\xi_m\zeta_n\neq0$. In view of (3.19), it follows that 
$Q_1+Q_2=g$, that is,
\begin{align}
\notag \alpha_1z_1+\alpha_2z_2+\beta_0=&-\frac{a_2+ia_1}{b_2+ib_1}z_1-\frac{a_2-ia_1}{b_2-ib_1}z_1
\end{align}
\begin{align}
&+\varphi_1\left(z_2-\frac{c_2+ic_1}{b_2+ib_1}z_1\right)+\varphi_2\left(z_2-\frac{c_2-ic_1}{b_2-ib_1}z_1\right).
\end{align} If 
$m\neq n\geqq 2$, by observing the exponents of the two sides of (3.26) in z2, we have
\begin{equation*}
\deg_{z_2}(\varphi_1+\varphi_2)=\max\{m,n\} \gt 1,\end{equation*} which is a contradiction with 
$\deg_{z_2}g=1$.
If 
$m=n\geq 2$, by observing the exponents of the two sides of (3.26) in z2 again, we have
\begin{equation*}
\xi_m=-\zeta_m,~\xi_{m-1}=-\zeta_{m-1},~\cdots,~\xi_2=-\zeta_2.
\end{equation*} Thus, we can deduce that the coefficients of the term with the degree 
$m\geqq 2$ of 
$z_2^{m-1}z_1$in the right side of (3.26) must be equal to 0, that is,
\begin{equation*}
\xi_m\left(\frac{c_2-ic_1}{b_2-ib_1}-\frac{c_2+ic_1}{b_2+ib_1}\right)\equiv 0,
\end{equation*} which implies that 
$b_1c_2-b_2c_1=0$. This is a contradiction with the assumption of Theorem 2.1.
Hence, we obtain that 
$m\leqq 1$ and 
$n\leqq 1$, that is,
\begin{align}
& \varphi_1(s_1)=\xi_1s_1+\beta_1=\xi_1\left(z_2-\frac{c_2+ic_1}{b_2+ib_1}z_1\right)+\beta_1,
\end{align}
\begin{align} & \varphi_2(s_2)=\zeta_1s_2+\beta_2=\zeta_1\left(z_2-\frac{c_2-ic_1}{b_2-ib_1}z_1\right)+\beta_2.
\end{align} In view of (3.26)-(3.28), we obtain 
$\beta_1+\beta_2=\beta_0$ and a system of the equations about ζ 1 and ξ 1 as the form given by
\begin{equation}
\left\{\begin{aligned}
&\frac{c_2+ic_1}{b_2+ib_1}\xi_1+\frac{c_2-ic_1}{b_2-ib_1}\zeta_1=-\alpha_1-\frac{2(a_1b_1+a_2b_2)}{b_1^2+b_2^2},\\
&\xi_1+\zeta_1=\alpha_2.
\end{aligned}
\right.
\end{equation}Solving the system (3.29), we have
\begin{align}
&\xi_1=-\frac{i}{2D}\left[(b_1^2+b_2^2)\alpha_1+(b_2+ib_1)(c_2-ic_1)\alpha_2+2(a_1b_1+a_2b_2)\right]:=A_1,
\end{align}
\begin{align} &\zeta_1=\frac{i}{2D}\left[(b_1^2+b_2^2)\alpha_1+(b_2-ib_1)(c_2+ic_1)\alpha_2+2(a_1b_1+a_2b_2)\right]:=A_2.
\end{align}Thus, it follows from (3.25), (3.30) and (3.31) that
\begin{align}
&Q_1(z)=-\frac{a_2+ia_1}{b_2+ib_1}z_1+A_1\left(z_2-\frac{c_2+ic_1}{b_2+ib_1}z_1\right)+\beta_1,
\end{align}
\begin{align}&Q_2(z)=-\frac{a_2-ia_1}{b_2-ib_1}z_1+A_2\left(z_2-\frac{c_2-ic_1}{b_2-ib_1}z_1\right)+\beta_2.
\end{align}In view of (3.20) and (3.21), it follows that
\begin{equation}
(a_2b_1-a_1b_2)u_{z_1}+(a_2c_1-a_1c_2)u_{z_2}=\frac{1}{2}(a_2+ia_1)e^{Q_1}+\frac{1}{2}(a_2-ia_1)e^{Q_2}.
\end{equation}The characteristic equations of (3.32) are given
\begin{equation*}
\frac{dz_1}{dt}=a_2b_1-a_1b_2,\end{equation*}
\begin{equation*}\frac{dz_2}{dt}=a_2c_1-a_1c_2\end{equation*}and
\begin{equation*}\frac{du}{dt}=\frac{1}{2}(a_2+ia_1)e^{Q_1}+\frac{1}{2}(a_2-ia_1)e^{Q_2}.\end{equation*} Using the initial conditions: 
$z_1=0$ and 
$z_2=s$, and 
$u=u(0,s)=\varphi_0(s)$ with a parameter s. Thus, the parametric representation of the solutions of the characteristic equations are 
$z_1=(a_2b_1-a_1b_2)t$, 
$z_2=(a_2c_1-a_1c_2)t+s$, and
\begin{align}
u(t,s)&=\int_0^t\left[\frac{1}{2}(a_2+ia_1)e^{Q_1}
+\frac{1}{2}(a_2-ia_1)e^{Q_2}\right]dt+\varphi_0(s),
\end{align} where 
$\varphi_0(s)$ is a finite-order entire function in s. By virtue of 
$z_1=(a_2b_1-a_1b_2)t$and 
$z_2=(a_2c_1-a_1c_2)t+s$, we rewrite the following representations of (3.32) and (3.33) with respect to t and s:
\begin{align}
&Q_1(z)=\frac{a_2+ia_1}{2}\mu_1t+A_1s+\beta_1,
\end{align}
\begin{align}&Q_2(z)=\frac{a_2-ia_1}{2}\mu_2t+A_2s+\beta_2,
\end{align}where
\begin{equation*}\mu_1=(b_1+ib_2)\alpha_1+(c_1+ic_2)\alpha_2+2(a_1+ia_2),~~
\mu_2=(b_1-ib_2)\alpha_1+(c_1-ic_2)\alpha_2+2(a_1-ia_2).\end{equation*}Next, four subcases will be considered.
Subcase 2.1. 
$a_2+ia_1\neq0$ and 
$a_2-ia_1\neq 0$. If 
$\mu_1\neq 0$ and 
$\mu_2\neq 0$, in view of (3.35) to (3.37), we have
\begin{equation*}
u(t,s)=\frac{1}{\mu_1}e^{Q_1(t,s)}+\frac{1}{\mu_2}e^{Q_2(t,s)}+\varphi_0(s).\end{equation*}Noting that
\begin{equation*}t=\frac{1}{a_2b_1-a_1b_2}z_1\quad \text{and} \quad
s=z_2-\frac{a_2c_1-a_1c_2}{a_2b_1-a_1b_2}z_1,\end{equation*}we have
\begin{equation}
u(z_1,z_2)=\frac{1}{\mu_1}e^{Q_1(z)}+\frac{1}{\mu_2}e^{Q_2(z)}+\varphi_0(s),
\end{equation} where 
$Q_1(z), Q_2(z)$ are stated as in (3.32) and (3.33). Substituting from (3.38) into (3.21) or (3.22), we can deduce that
\begin{equation*}
a_1\varphi_0(s)+ \frac{a_1(b_1c_2-b_2c_1)}{a_2b_1-a_1b_2}\varphi_0'(s)=0
\quad \text{or} \quad a_2\varphi_0(s)+ \frac{a_2(b_1c_2-b_2c_1)}{a_2b_1-a_1b_2}\varphi_0'(s)=0,
\end{equation*}which implies that
\begin{equation}
\varphi_0(s)=\eta_0 e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]}
\quad (\eta_0\in \mathbb{C}).
\end{equation}From (3.38) and (3.39), we have
\begin{equation}
u(z_1,z_2)=\frac{1}{\mu_1}e^{Q_1(z)}+\frac{1}{\mu_2}e^{Q_2(z)}
+\eta_0 e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},
\end{equation} where η 0 is a constant, and 
$Q_1(z), Q_2(z)$ are stated as in (3.32) and (3.33).
If 
$\mu_1= 0$ and 
$\mu_2\neq 0$, in view of (3.35) to (3.37), we have
\begin{equation*}
u(t,s)=\frac{a_2+ia_1}{2}te^{Q_1(t,s)}+\frac{1}{\mu_2}e^{Q_2(t,s)}+\varphi_0(s).\end{equation*}Noting that
\begin{equation*}t=\frac{1}{a_2b_1-a_1b_2}z_1\quad \text{and} \quad
s=z_2-\frac{a_2c_1-a_1c_2}{a_2b_1-a_1b_2}z_1,\end{equation*}we have
\begin{equation}
u(z_1,z_2)=\frac{a_2+ia_1}{2(a_2b_1-a_1b_2)}z_1e^{Q_1(z)}+\frac{1}{\mu_2}e^{Q_2(z)}+\varphi_0(s),
\end{equation} where 
$Q_1(z)$ and 
$Q_2(z)$ are stated as in (3.32) and (3.33). Due to 
$\mu_1=0$, this leads to
\begin{equation*}(b_1-ib_2)\alpha_1+(c_1-ic_2)\alpha_2=-2(a_1-ia_2)\end{equation*}and
\begin{equation}
Q_1(z)=A_1\left(z_2-\frac{a_2c_1-a_1c_2}{a_2b_1-a_1b_2}z_1\right)+\beta_1=\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]+\beta_1.
\end{equation} Substituting from (3.41) into (3.21) or (3.22), we find that 
$\varphi_0(s)$ is of the form (3.39). And, together with (3.41) and (3.42), we have
\begin{equation}
u(z_1,z_2)=(\eta_1z_1+\eta_0) e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]}+\frac{1}{\mu_2}e^{Q_2(z)},
\end{equation} where 
$Q_2(z)$ is stated as in (3.33) and
\begin{equation*}\eta_1=\frac{a_2+ia_1}{2(a_2b_1-a_1b_2)}e^{\beta_1}(\neq 0).\end{equation*} If 
$\mu_1\neq0$ and 
$\mu_2=0$, by using the argument as in the case for 
$\mu_1=0$ and 
$\mu_2\neq0$, we can obtain
\begin{equation}
Q_2(z)=A_2\left(z_2-\frac{a_2c_1-a_1c_2}{a_2b_1-a_1b_2}z_1\right)+\beta_2=\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]+\beta_2
\end{equation}and
\begin{equation}
u(z_1,z_2)=\frac{1}{\mu_1}e^{Q_1(z)}+(\eta_2z_1+\eta_0)
e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},
\end{equation} where 
$Q_1(z)$ is stated as in (3.32) and
\begin{equation*}\eta_2=\frac{a_2-ia_1}{2(a_2b_1-a_1b_2)}e^{\beta_2}(\neq0).\end{equation*} If 
$\mu_1=0$ and 
$\mu_2=0$, that is, 
$\alpha_1=\frac{2(a_2c_1-a_1c_2)}{D}$ and 
$\alpha_2=\frac{2(a_1b_2-a_2b_1)}{D}$, by applying the argument similar to that in the case for 
$\mu_1=0$ and 
$\mu_2\neq0$, we can obtain
\begin{equation}
u(z_1,z_2)=(\eta z_1+\eta_0) e^{\frac{1}{D}
\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},
\end{equation}where
\begin{equation*}\eta=\frac{(a_2+ia_1)e^{\beta_1}+(a_2-ia_1)e^{\beta_2}}{2(a_2b_1-a_1b_2)}.\end{equation*}It should be pointed out that the form in (3.14) is included in the case η = 0 of (3.46).
Subcase 2.2. 
$a_2+ia_1=0$ and 
$a_2-ia_1\neq 0$. Then (3.35) becomes
\begin{align}
u(t,s)&=\int_0^t\frac{1}{2}(a_2-ia_1)e^{Q_2}dt+\varphi_0(s),
\end{align} where 
$\varphi_0(s)$ is a finite-order entire function in s. If 
$\mu_2\neq 0$, by using an argument as above, we find from (3.47) that
\begin{equation}
u(z_1,z_2)=\frac{1}{\mu_2}e^{Q_2(z)}+\eta_0 e^{\frac{1}{D}
\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]}.
\end{equation} If 
$\mu_2=0$, it follows from (3.47) that
\begin{equation}
u(z_1,z_2)=\eta_0 e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]}.
\end{equation} Subcase 2.3. 
$a_2+ia_1\neq0$ and 
$a_2-ia_1=0$. Similarly, if 
$\mu_1=0$, then 
$f(z_1,z_2)$ satisfies (3.49). If 
$\mu_1\neq 0$, using similar argument as above, we can find that
\begin{equation}
u(z_1,z_2)=\frac{1}{\mu_1}e^{Q_1(z)}+\eta_0 e^{\frac{1}{D}
\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]}.
\end{equation} Subcase 2.4. 
$a_2+ia_1=0$ and 
$a_2-ia_1=0$. Thus, it follows that 
$a_1=a_2=0$. This is a contradiction with the assumption of 
$a_1b_2-a_2b_1\neq 0$.
Thus, from Subcases 2.1 to 2.4, we can obtain the conclusions of Theorem 2.1.
This evidently completes the proof of Theorem 2.1.
3.2. Proofs of Theorems 2.3 and 2.4
Here, we only give the proof of Theorem 2.3 because the proofs of Theorems 2.3 and 2.4 are very similar. We first rewrite (2.2) in the following form:
\begin{equation}
\left[\frac{a_1u+b_1u_{z_1}}{e^{g/2}}\right]^2
+\left[\frac{a_2u+b_2u_{z_1}}{e^{g/2}}\right]^2=1.
\end{equation}We will now discuss the following two cases.
Case 1. If 
$\frac{a_1u+b_1u_{z_1}}{e^{g/2}}$ is a constant, we let 
$
\frac{a_1u+b_1u_{z_1}}{e^{g/2}}=\theta_1,
$ that is,
\begin{equation}
a_1u+b_1u_{z_1}=\theta_1 e^{g/2}\qquad (\theta_1\in \mathbb{C}).
\end{equation} By virtue of (2.2), it follows that 
$\frac{a_2u+b_2u_{z_1}}{e^{g/2}}$ must be constant. Thus, we get
\begin{equation}
a_2u+b_2u_{z_1}=\theta_2 e^{g/2}\qquad (\theta_2\in \mathbb{C}).
\end{equation} This implies that 
$\theta_1^2+\theta_2^2=1$. Noting that 
$D_1:=a_1b_2-a_2b_1\neq 0$, we find from (3.52) and (3.53) that
\begin{align}
&(a_2b_1-a_1b_2)u_{z_1}=(a_2\theta_1-a_1\theta_2)e^{g/2},
\end{align}
\begin{align} &(a_1b_2-a_2b_1)u=(b_2\theta_1-b_1\theta_2)e^{g/2}.
\end{align}In view of (3.54) and (3.55), we have
\begin{equation*}
(a_2\theta_1-a_1\theta_2)e^{g/2}= (a_2b_1-a_1b_2)u_{z_1}=-\frac{1}{2}g _{z_1}(b_2\eta_1-b_1\eta_2)e^{g/2},
\end{equation*}which implies that
\begin{equation}
\theta_1=\frac{2a_1+\alpha_1b_1}{2a_2+\alpha_2b_2}\theta_2.
\end{equation} Together with 
$\theta_1^2+\theta_2^2=1$, we have
\begin{equation}
\theta_1=\pm\frac{2a_1+\alpha_1b_1}{\sqrt{(2a_1+\alpha_1b_1)^2+(2a_2+\alpha_2b_2)^2}}, ~~\theta_2=\pm\frac{2a_2+\alpha_2b_2}{\sqrt{(2a_1+\alpha_1b_1)^2+(2a_2+\alpha_2b_2)^2}}.
\end{equation}In view of (3.55) and (3.57), we obtain
\begin{equation}
u=\frac{b_2\theta_1-b_1\theta_2}{a_1b_2-a_2b_1}e^{g/2}=\pm\frac{2}{\sqrt{(2a_1+\alpha_1b_1)^2+(2a_2+\alpha_2b_2)^2}}e^{g/2}.
\end{equation}This completes the proof of Theorem 2.3.
Case 2. If 
$\frac{a_1u+b_1u_{z_1}}{e^{g/2}}$ is not a constant, by using the argument as in the proof of Theorem 2.1, there exists a non-constant polynomial p(z) in 
$\mathbb{C}^2$ such that
\begin{equation}
a_1u+b_1u_{z_1}=\frac{e^{Q_1(z)}+e^{Q_2(z)}}{2}
\quad \text{and} \quad a_2u+b_2u_{z_1}=\frac{e^{Q_1(z)}-e^{Q_2(z)}}{2i},
\end{equation}where Q 1 and Q 2 satisfy (3.19). This leads to
\begin{equation}
u=\frac{1}{D_1}\left[\frac{b_2+i b_1}{2}e^{Q_1}+\frac{b_2-ib_1}{2}e^{Q_2}\right]
\end{equation}and
\begin{equation}
u_{z_1}=-\frac{1}{D_1}\left[\frac{a_2+ia_1}{2}e^{Q_1}
+\frac{a_2-ia_1}{2}e^{Q_2}\right].
\end{equation}Now, in view of (3.60) and (3.61), it follows that
\begin{equation}
\left[(b_2+ib_1)\frac{\partial Q_1}{\partial z_1}+(a_2+ia_1)\right]e^{2p}
=(-b_2+ib_1)\frac{\partial Q_2}{\partial z_1}-(a_2-ia_1).
\end{equation}If
\begin{equation*}(b_2+ib_1)\frac{\partial Q_1}{\partial z_1}+(a_2+ia_1)\not \equiv 0,\end{equation*}then the left-hand side of (3.62) is transcendental and the right-hand side of (3.62) is polynomial. This leads to a contradiction. Hence, we have
\begin{equation*}
(b_2+ib_1)\frac{\partial Q_1}{\partial z_1}+(a_2+ia_1)\equiv 0
\quad \text{and} \quad (-b_2+ib_1)\frac{\partial Q_2}
{\partial z_1}-(a_2-ia_1)\equiv 0,\end{equation*}that is,
\begin{equation}
\frac{\partial Q_1}{\partial z_1}=-\frac{a_1-ia_2}{b_1-ib_2}
\quad \text{and} \quad \frac{\partial Q_2}{\partial z_1}
=-\frac{a_1+ia_2}{b_1+ib_2}.
\end{equation}This means that
\begin{equation}
Q_1(z)= -\frac{a_1-ia_2}{b_1-ib_2}z_1+\phi_1(z_2)
\quad \text{and} \quad Q_2(z)=-\frac{a_1+ia_2}{b_1+ib_2}z_1+\phi_2(z_2),
\end{equation} where ϕ1 and 
$\phi_2(z_2)$ are polynomials in z2.
Due to 
$Q_1+Q_2=g$, we have
\begin{equation*}2(a_1b_1+a_2b_2)+\alpha_1(b_1^2+b_2^2)=0
\quad \text{and} \quad \phi_1(z_2)+\phi_2(z_2)=\alpha_2z_2+\beta_0.\end{equation*}In view of (3.60) and (3.64), we obtain
\begin{equation*}
\begin{aligned}
u(z_1,z_2)=&\frac{1}{2D_1}\left[(b_2+ib_1)e^{-\frac{a_1-ia_2}{b_1-ib_2}z_1+\phi_1(z_2)}
+(b_2-b_1i)e^{-\frac{a_1+ia_2}{b_1+ib_2}z_1+\phi_2(z_2)}\right],
\end{aligned}
\end{equation*} where 
$\phi_1(z_2),\phi_2(z_2)$ are polynomials in z2 satisfying the condition is given by
\begin{equation*}\phi_1(z_2)+\phi_2(z_2)=\alpha_2z_2+\beta_0.\end{equation*}This completes the proof of Theorem 2.3.
4. Proof of Theorem 2.5
Proof. Let u(z) be a transcendental entire solution of equation (1.7) with finite order. Suppose that
\begin{equation}
\mathcal {P}_1(u)=\frac{1}{\sqrt{2}}(U+V)
\quad \text{and} \quad \mathcal {P}_2(u)=\frac{1}{\sqrt{2}}(U-V),
\end{equation} where 
$U,V$ are entire functions in 
$\mathbb{C}^2$. Thus, equation (1.7) can be written as follows:
\begin{equation*}
(1+\alpha)U^2+(1-\alpha)V^2=e^g,
\end{equation*}that is,
\begin{equation}
\left(\frac{\sqrt{1+\alpha}U}{e^{g/2}}\right)^2+\left(\frac{\sqrt{1-\alpha}V}{e^{g/2}}\right)^2=1.
\end{equation}Now, the following two cases will be discussed.
Case 1. If 
$\frac{\sqrt{1+\alpha}U}{e^{g/2}}$ is a constant, then it follows from (4.2) that 
$\frac{\sqrt{1-\alpha}v}{e^{g/2}}$ is also a constant. Set
\begin{equation*}
\frac{\sqrt{1+\alpha}U}{e^{g/2}}=\theta_1
\quad \text{and} \quad \frac{\sqrt{1-\alpha}V}{e^{g/2}}=\theta_2.\end{equation*} This means from (4.2) that 
$\theta_1^2+\theta_2^2=1$. Thus, we can deduce from (4.1) that
\begin{equation}
\mathcal {P}_1(u)=a_1 u+b_1u_{z_1}+c_1u_{z_2}=\tau_1e^{g/2}
\quad \text{and} \quad \mathcal {P}_2(u)=a_2 u+b_2u_{z_1}+c_2u_{z_2}=\tau_2e^{g/2},
\end{equation}where
\begin{equation*}
\tau_1=\frac{1}{\sqrt{2}}\left(\frac{\theta_1}
{\sqrt{1+\alpha}}+\frac{\theta_2}{\sqrt{1-\alpha}}\right)\quad \text{and} \quad
\tau_2=\frac{1}{\sqrt{2}}\left(\frac{\theta_1}
{\sqrt{1+\alpha}}-\frac{\theta_2}{\sqrt{1-\alpha}}\right).
\end{equation*} In view of 
$b_1c_2-b_2c_1\neq 0$, we can deduce from (4.3) that
\begin{align}
&u_{z_1}=\frac{1}{D}\left[(\tau_1c_2-\tau_2c_1)e^{g/2}+(a_2c_1-a_1c_2)u\right],
\end{align}
\begin{align} &u_{z_2}=\frac{1}{D}\left[(\tau_2b_1-\tau_1b_2)e^{g/2}+(a_1b_2-a_2b_1)u\right].
\end{align} Differentiate (4.4) and (4.5) with respect to z2 and z 1, respectively. Then, by virtue of the fact that 
$u_{z_1z_2}=u_{z_2z_1}$, we have
\begin{equation}
\frac{1}{2}\alpha_2(\tau_1c_2-\tau_2c_1)e^{g/2}+(a_2c_1-a_1c_2)u_{z_2}
=\frac{1}{2}\alpha_1(\tau_2b_1-\tau_1b_2)e^{g/2}+(a_1b_2-a_2b_1)u_{z_1}.
\end{equation}Substituting from (4.4) and (4.5) into (4.6), we obtain
\begin{equation*}
\alpha_2(\tau_1c_2-\tau_2c_1)=\alpha_1(\tau_2b_1-\tau_1b_2)+2(a_1\tau_2-a_2\tau_1),\end{equation*}that is,
\begin{equation}
(\alpha_1b_2+\alpha_2c_2+2a_2)\tau_1 =(\alpha_1b_1+\alpha_2c_1+2a_1)\tau_2.
\end{equation} Together with 
$\theta_1^2+\theta_2^2=1$, and by a simple calculation, we find that
\begin{equation}
\theta_1=\pm\frac{h_1}{\sqrt{h_1^2+h_2^2}}, ~~~\theta_2=\pm\frac{h_2}{\sqrt{h_1^2+h_2^2}},
\end{equation}where
\begin{equation*}
\begin{aligned}
&h_1:=[2(a_1+a_2)+\alpha_1(b_1+b_2)+\alpha_2(c_1+c_2)]\sqrt{1+\alpha},\\
&h_2:=[2(a_1-a_2)+\alpha_1(b_1-b_2)+\alpha_2(c_1-c_2)]\sqrt{1-\alpha}.
\end{aligned}
\end{equation*}On the other hand, in view of (4.4) and (4.5), we have the following partial differential equation:
\begin{equation}
(a_1b_2-a_2b_1)u_{z_1}+(a_1c_2-a_2c_1)u_{z_2}=(a_1\tau_2-a_2\tau_1)e^{g/2}.
\end{equation}The characteristic equations of (4.9) are given by
\begin{equation*}
\frac{dz_1}{dt}=a_1b_2-a_2b_1,\end{equation*}
\begin{equation*}\frac{dz_2}{dt}=a_1c_2-a_2c_1\end{equation*}and
\begin{equation*}\frac{du}{dt}=(a_1\tau_2-a_2\tau_1)e^{g/2}.\end{equation*} Using the initial conditions: 
$z_1=0,z_2=s$, and 
$u=u(0,s)=\varphi(s)$ with a parameter s, we obtain the following parametric representation for the solutions of the characteristic equations: 
$z_1=(a_1b_2-a_2b_1)t$, 
$z_2=(a_1c_2-a_2c_1)t+s$, and
\begin{align}
u(z_1,z_2)&=u(t,s)=\int_0^t(a_1\tau_2-a_2\tau_1)e^{g/2}dt+\varphi(s),
\end{align} where 
$\varphi(s)$ is a polynomial in s. Noting that the expressions of 
$\tau_1, \tau_2$ and 
$z_1=(a_1b_2-a_2b_1)t$, 
$z_2=(a_1c_2-a_2c_1)t+s$, we have
\begin{equation}
a_1\tau_2-a_2\tau_1=\pm\frac{\sqrt{2}}{\sqrt{h_1^2+h_2^2}}[\alpha_1(a_1b_2-a_2b_1)+\alpha_2(a_1c_2-a_2c_1)]
\end{equation}and
\begin{equation}
g(z_1,z_2)=g(t,s)=[\alpha_1(a_1b_2-a_2b_1)+\alpha_2(a_1c_2-a_2c_1)]t+\alpha_2s+\beta_0.
\end{equation} If 
$\alpha_1(a_1b_2-a_2b_1)+\alpha_2(a_1c_2-a_2c_1)=0$, in view of (4.10)-(4.12), it follows that
\begin{equation*}
u(z_1,z_2)=\varphi(s)=\varphi\left(z_2-\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}z_1\right).\end{equation*}Substituting this last equation into (4.4) or (4.5), we have
\begin{equation*}
\varphi'(s)=\frac{a_1b_2-a_2b_1}{D}\varphi(s).
\end{equation*}This leads to
\begin{equation*}\varphi(s)=\pm\frac{2\sqrt{2}}{\sqrt{h_1^2+h_2^2}}e^{g/2}
+\eta_0e^{\frac{a_1b_2-a_2b_1}{D}s},\end{equation*}which means that
\begin{equation}
u(z_1,z_2)=\pm\frac{2\sqrt{2}}{\sqrt{h_1^2+h_2^2}}e^{g/2}+\eta_0e^{\frac{1}{D}[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2]}.
\end{equation} If 
$\alpha_1(a_1b_2-a_2b_1)+\alpha_2(a_1c_2-a_2c_1)\neq0$, in view of (4.10) to (4.12), it follows that
\begin{equation*}
u(z_1,z_2)=\pm\frac{2\sqrt{2}}{\sqrt{h_1^2+h_2^2}}e^{g/2}+\varphi(s).\end{equation*}Substituting this equation into (4.4) or (4.5), and using the above argument, we have
\begin{equation}
u(z_1,z_2)=\pm\frac{2\sqrt{2}}{\sqrt{h_1^2+h_2^2}}e^{g/2}+\eta_0e^{\frac{1}{D}[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2]}.
\end{equation}This completes the proof of Theorem 2.5 (i).
Case 2. If 
$\frac{\sqrt{1+\alpha}U}{e^{g/2}}$ is not a constant, then (1.7) can be written as the form given by
\begin{equation}
\left(\frac{\sqrt{1+\alpha}U}{e^{g/2}}+i\frac{\sqrt{1-\alpha}V}{e^{g/2}}\right)
\left(\frac{\sqrt{1+\alpha}U}{e^{g/2}}-i\frac{\sqrt{1-\alpha}V}{e^{g/2}}\right)=1.
\end{equation}Noting that U and V are transcendental entire functions of finite order, we observe that
\begin{equation*}\frac{\sqrt{1+\alpha}U}{e^{g/2}}+i\frac{\sqrt{1-\alpha}V}{e^{g/2}}\end{equation*}and
\begin{equation*}\frac{\sqrt{1+\alpha}U}{e^{g/2}}-i\frac{\sqrt{1-\alpha}V}{e^{g/2}}\end{equation*} have no zeros and poles. Thus, by using the conclusions in [Reference Ronkin22, Reference Stoll26] and the Hadamard factorisation theorem [Reference Pólya21], there exists a non-constant polynomial p(z) in 
$\mathbb{C}^2$such that
\begin{equation}
\frac{\sqrt{1+\alpha}U}{e^{g/2}}+i\frac{\sqrt{1-\alpha}V}{e^{g/2}}=e^{p},
~~~\frac{\sqrt{1+\alpha}U}{e^{g/2}}-i\frac{\sqrt{1-\alpha}V}{e^{g/2}}=e^{-p}.
\end{equation}In view of (4.1) and (4.16), it follows that
\begin{align}
\mathcal {P}_1(u) & = a_1u+b_1u_{z_1}+c_1u_{z_2}\nonumber \\
& =\frac{\sqrt{2}}{4}\left[\left(\frac{1}
{\sqrt{1+\alpha}}+\frac{1}{i\sqrt{1-\alpha}}\right)e^{Q_3}+
\left(\frac{1}{\sqrt{1+\alpha}}-\frac{1}
{i\sqrt{1-\alpha}}\right)e^{Q_4}\right]\nonumber \\
& =\frac{\sqrt{2}}{4}(k_1e^{Q_3}+k_2e^{Q_4}),
\end{align}
\begin{align}
\mathcal {P}_2(u)& =a_2u+b_2u_{z_1}+c_2u_{z_2}\nonumber \\
& =\frac{\sqrt{2}}{4}\left[\left(\frac{1}
{\sqrt{1+\alpha}}-\frac{1}{i\sqrt{1-\alpha}}\right)e^{Q_3}+
\left(\frac{1}{\sqrt{1+\alpha}}+\frac{1}{i\sqrt{1-\alpha}}\right)e^{Q_4}\right]\nonumber \\
& =\frac{\sqrt{2}}{4}(k_2e^{Q_3}+k_1e^{Q_4}),
\end{align} where 
$Q_3=p+\frac{g}{2},Q_4=-p+\frac{g}{2}$.
Now, noting that 
$D=b_1c_2-b_2c_1\neq 0$, and solving the set of equations (4.17) and (4.18), we can deduce that
\begin{align}
&u_{z_1}=\frac{1}{D}\left\{\frac{\sqrt{2}}{4}
\left[(c_2 k_1-c_1 k_2)e^{Q_3}+(c_2 k_2-c_1 k_1)e^{Q_4}\right]
-(a_1c_2-a_2c_1)u\right\},
\end{align}
\begin{align} &u_{z_2}=\frac{1}{D}\left\{\frac{\sqrt{2}}{4}
\left[(b_1 k_2-b_2 k_1)e^{Q_3}+(b_1 k_1-b_2 k_2)e^{Q_4}\right]
+(a_1b_2-a_2b_1)u\right\}.
\end{align} Differentiating (4.19) and (4.20) with respect to z2 and z 1, respectively, and combining with 
$u_{z_1z_2}=u_{z_2z_1}$, we have
\begin{align}
& \left[(b_{1} k_{2}-b_{2} k_{1})\frac{\partial Q_{3}}{\partial z_{1}}
+(c_1 k_2-c_2 k_1)\frac{\partial Q_3}{\partial z_2}+(a_{1} k_2-a_2 k_1)\right]e^{2p}\nonumber \\
=& (b_2 k_2-b_1 k_1)\frac{\partial Q_4}{\partial z_1}+(c_2 k_2-c_1 k_1)
\frac{\partial Q_4}{\partial z_2}+(a_2 k_2-a_1 k_1).
\end{align}If
\begin{equation*}(b_1 k_2-b_2 k_1)\frac{\partial Q_3}{\partial z_1}+(c_1 k_2-c_2 k_1)
\frac{\partial Q_3}{\partial z_2}+(a_1 k_2-a_2 k_1)\not \equiv 0,\end{equation*}in view of the assumption that p is a non-constant polynomial, one side of (4.21) is transcendental and the other side of (4.21) is polynomial. This is a contradiction. Hence, we have
\begin{equation}
(b_1 k_2-b_2 k_1)\frac{\partial Q_3}{\partial z_1}+(c_1 k_2-c_2 k_1)
\frac{\partial Q_3}{\partial z_2}+(a_1 k_2-a_2 k_1)\equiv0.
\end{equation}It thus follows from (4.21) that
\begin{equation}
(b_2 k_2-b_1 k_1)\frac{\partial Q_4}{\partial z_1}+(c_2 k_2-c_1 k_1)
\frac{\partial Q_4}{\partial z_2}+(a_2 k_2-a_1 k_1)\equiv0.
\end{equation}We start to solve equation (4.22). The characteristic equations of (4.22) are given by
\begin{equation*}
\frac{dz_1}{dt}=b_1 k_2-b_2 k_1,\end{equation*}
\begin{equation*}\frac{dz_2}{dt}=c_1 k_2-c_2 k_1\end{equation*}and
\begin{equation*}\frac{dQ_3}{dt}=a_2 k_1-a_1 k_2.\end{equation*} Using the initial conditions: 
$z_1=0,z_2=s_1$, and 
$Q_3=Q_3(0,s_1)=\psi_1(s_1)$ with a parameter s 1. Thus, we obtain the following parametric representation for the solutions of the characteristic equations: 
$z_1=(b_1 k_2-b_2 k_1)t$, 
$z_2=(c_1 k_2-c_2 k_1)t+s_1$, and
\begin{equation}
Q_3(z_1,z_2)=\frac{a_2 k_1-a_1 k_2}{b_1k_2-b_2k_1}z_1
+\psi_1(s_1)=\frac{a_2 k_1-a_1 k_2}{b_1k_2-b_2k_1}z_1
+\psi_1(z_2-\frac{c_1 k_2-c_2 k_1}{b_1k_2-b_2k_1}z_1),
\end{equation} where 
$\psi_1(s)$ is a polynomial in
\begin{equation*}s_1=z_2-\frac{c_1 k_2-c_2 k_1}{b_1k_2-b_2k_1}z_1.\end{equation*}Similarly, by solving equation (4.23), we have
\begin{equation}
Q_4(z_1,z_2)=\frac{a_1 k_1-a_2 k_2}{b_2k_2-b_1k_1}z_1
+\psi_2(s_2)=\frac{a_1 k_1-a_2 k_2}{b_2k_2-b_1k_1}z_1
+\psi_1(z_2-\frac{c_2 k_2-c_1 k_1}{b_2k_2 -b_1k_1}z_1),
\end{equation} where 
$\psi_2(s_2)$ is a polynomial in
\begin{equation*}s_2=z_2-\frac{c_1 k_1-c_2 k_2}{b_1k_1-b_2k_2}z_1.\end{equation*} Next, we will prove that Q 3 and Q 4 are linear forms of z 1 and z2, that is, 
$\deg_{s_j}\psi_j\leq 1$ 
$\;(j=1,2)$. Since p and g are polynomials in 
$\mathbb{C}^2$, it follows that 
$\psi_j(s_j)\;\; (j=1,2)$ are polynomials in sj. Assuming that 
$\deg_{s_1}\psi_1=m$ and 
$\deg_{s_2}\psi_2=n$, we let
\begin{align*}
\notag &\psi_1(s_1)=\xi_ms_1^m+\xi_{m-1}s_1^{m-1}+\cdots+\xi_0,\\
\notag
\end{align*}
\begin{align*}
&\psi_2(s_2)=\zeta_ns_2^n+\zeta_{n-1}s_2^{m-1}+\cdots+\zeta_0,
\end{align*} where 
$\xi_m,\ldots,\xi_0,\zeta_n,\ldots,\zeta_0$ are constants and 
$\xi_m\zeta_n\neq0$. Due to 
$Q_3+Q_4=g$, that is,
\begin{align}
\alpha_1z_1+\alpha_2z_2+\beta_0 & =\frac{a_2 k_1-a_1 k_2}
{b_1k_2-b_2k_1}z_1+\frac{a_1 k_1-a_2 k_2}{b_2k_2-b_1k_1}z_1\nonumber \\
&\qquad +\psi_1\left(z_2-\frac{c_1 k_2-c_2 k_1}{b_1k_2-b_2k_1}z_1\right)
+\psi_2\left(z_2-\frac{c_2 k_2-c_1 k_1}{b_2k_2 -b_1k_1}z_1\right).
\end{align} If 
$m\neq n\geq 2$, by observing the exponents of the two sides of equation (4.26) in z 2, we have
\begin{equation*}
\deg_{z_2}(\psi_1+\psi_2)=\max\{m,n\} \gt 1,\end{equation*} which is a contradiction with 
$\deg_{z_2}g=1$.
If 
$m=n\geq 2$, by observing the exponents of the two sides of equation (4.26) in z2 again, we have
\begin{equation*}
\xi_m=-\zeta_m,~\xi_{m-1}=-\zeta_{m-1},~\cdots~,\xi_2=-\zeta_2.
\end{equation*} Thus, we can deduce that the coefficients of the term with the degree 
$m\geq 2$ of 
$z_2^{m-1}z_1$ in the right side of (4.26) must be equal to 0, that is,
\begin{equation*}
\xi_m\left(\frac{c_1 k_2-c_2 k_1}{b_1k_2-b_2k_1}
-\frac{c_2 k_2-c_1 k_1}{b_2k_2 -b_1k_1}\right)\equiv 0,
\end{equation*}which implies that
\begin{equation*}(b_1c_2-b_2c_1)\frac{4}{i\sqrt{1-\alpha^2}}=0.\end{equation*}This is a contradiction with the assumption that D ≠ 0.
Hence, we find that 
$m\leqq 1$ and 
$n\leqq 1$, that is,
\begin{align}
& \psi_{1}(s_{1})=\xi_1s_{1}+\beta_{1}=\xi_{1}
\left(z_2-\frac{c_1 k_2-c_2 k_1}{b_1k_2-b_2k_1}z_1\right)+\beta_1,
\end{align}
\begin{align} & \psi_2(s_2)=\zeta_1s_2+\beta_2
=\zeta_1\left(z_2-\frac{c_2 k_2-c_1 k_1}{b_2k_2-b_1k_1}z_1\right)+\beta_2.
\end{align} In view of equations (4.26) to (4.28), we obtain 
$\beta_1+\beta_2=\beta_0$ and a system of the equations about 
$\zeta_1,\xi_1$ in the following form:
\begin{equation}
\left\{\begin{aligned}
&\frac{c_1 k_2-c_2 k_1}{b_1k_2-b_2k_1}\xi_1+\frac{c_2 k_2-c_1 k_1}{b_2k_2 -b_1k_1}\zeta_1=\frac{a_2 k_1-a_1 k_2}{b_1k_2-b_2k_1}+\frac{a_1 k_1-a_2 k_2}{b_2k_2-b_1k_1}-\alpha_1,\\
&\xi_1+\zeta_1=\alpha_2.
\end{aligned}
\right.
\end{equation}Solving the system (4.29), we have
\begin{align}
&\xi_1=\frac{1}{D(k_1^2-k_2^2)}\left\{(b_1k_2-b_2k_1)[\alpha_1(b_1k_1-b_2k_2)+\alpha_2(c_1k_1-c_2k_2)]\right.\nonumber \\
&~~~~~\left.+2k_1k_2(a_1b_1+a_2b_2)-(k_1^2+k_2^2)(a_2b_1+a_1b_2)\right\}:=B_1,
\end{align}
\begin{align*}
\notag &\zeta_1=\frac{1}{D(k_1^2-k_2^2)}\left\{(b_2k_2-b_1k_1)[\alpha_1(b_1k_2-b_2k_1)+\alpha_2(c_1k_2-c_2k_1)]\right.
\end{align*}
\begin{align}&~~~~~\left.-2k_1k_2(a_1b_1+a_2b_2)+(k_1^2+k_2^2)(a_2b_1+a_1b_2)\right\}:=B_2.
\end{align}Thus, it follows from equations (4.25), (4.30) and (4.31) that
\begin{align}
&Q_3(z)=\frac{a_2 k_1-a_1 k_2}{b_1k_2-b_2k_1}z_1+B_1\left(z_2-\frac{c_1 k_2-c_2 k_1}{b_1k_2-b_2k_1}z_1\right)+\beta_1,
\end{align}
\begin{align}&Q_4(z)=\frac{a_2 k_2-a_1 k_1}{b_1k_1-b_2k_2}z_1+B_2\left(z_2-\frac{c_1 k_1-c_2 k_2}{b_1k_1-b_2k_2}z_1\right)+\beta_2.
\end{align}In view of equations (4.17) and (4.18), we get
\begin{equation}
(a_1b_2-a_2b_1)u_{z_1}+(a_1c_2-a_2c_1)u_{z_2}=\frac{\sqrt{2}}{4}\left[(a_1k_2-a_2k_1)e^{Q_3}+(a_1k_1-a_2k_2)e^{Q_4}\right].
\end{equation}The characteristic equations of (4.34) are
\begin{equation*}
\frac{dz_1}{dt}=a_1b_2-a_2b_1,\end{equation*}
\begin{equation*}\frac{dz_2}{dt}=a_1c_2-a_2c_1\end{equation*}and
\begin{equation*}\frac{du}{dt}=\frac{\sqrt{2}}{4}\left[(a_1k_2-a_2k_1)e^{Q_3}+(a_1k_1-a_2k_2)e^{Q_4}\right].\end{equation*} We now use the initial conditions: 
$z_1=0,z_2=s$, and 
$u=u(0,s)=\varphi_1(s)$ with a parameter s. We thus obtain the following parametric representation for the solutions of the characteristic equations: 
$z_1=(a_1b_2-a_2b_1)t$, 
$z_2=(a_1c_2-a_2c_1)t+s$, and
\begin{align}
u(t,s)&=\int_0^t\frac{\sqrt{2}}{4}\left[(a_1k_2-a_2k_1)e^{Q_3}+(a_1k_1-a_2k_2)e^{Q_4}\right]dt+\varphi_1(s),
\end{align} where 
$\varphi_1(s)$ is a finite-order entire function in s. By virtue of 
$z_1=(a_2b_1-a_1b_2)t$ and 
$z_2=(a_2c_1-a_1c_2)t+s$, we rewrite the following representations of (4.32) and (4.33) with respect to t and s:
\begin{align}
&Q_3(z)=\frac{a_1k_2-a_2k_1}{k_1^2-k_2^2}\nu_1t+B_1s+\beta_1,
\end{align}
\begin{align}
&Q_4(z)=\frac{a_2k_2-a_1k_1}{k_1^2-k_2^2}\nu_2t+B_2s+\beta_2,
\end{align}where
\begin{equation*}
\begin{aligned}&\nu_1=\alpha_1(b_1k_1-b_2k_2)
+\alpha_2(c_1k_1-c_2k_2)+2(a_1k_1-a_2k_2),\\
&\nu_2=\alpha_1(b_2k_1-b_1k_2)+\alpha_2(c_2k_1-c_1k_2)
+2(a_2k_1-a_1k_2).
\end{aligned}
\end{equation*}Next, four subcases will be considered.
Subcase 2.1. 
$a_1k_2-a_2k_1\neq 0$ and 
$a_1k_1-a_2k_2\neq 0$. If 
$\nu_1\neq 0$ and 
$\nu_2\neq 0$, in view of equations (4.35) to (4.37), we have
\begin{equation*}
u(t,s)=\frac{\sqrt{2}}{4}\left[\frac{k_1^2-k_2^2}
{\nu_1}e^{Q_3(t,s)}+\frac{k_1^2-k_2^2}{\nu_2}e^{Q_4(t,s)}\right]+\varphi_1(s).\end{equation*}Noting that
\begin{equation*}t=\frac{1}{a_2b_1-a_1b_2}z_1
\quad \text{and} \quad s=z_2-\frac{a_2c_1-a_1c_2}{a_2b_1-a_1b_2}z_1,\end{equation*}we have
\begin{equation}
u(z_1,z_2)=\frac{\sqrt{2}}{4}\left[\frac{k_1^2-k_2^2}{\nu_1}e^{Q_3(z)}
+\frac{k_1^2-k_2^2}{\nu_2}e^{Q_4(z)}\right]+\varphi_1(s),
\end{equation} where 
$Q_3(z)$ and 
$Q_4(z)$ are stated as in (4.32) and (4.33). Substituting from (4.38) into (4.17) or (4.18), we can deduce that
\begin{equation*}
a_1\varphi_1(s)+ \frac{a_1(b_1c_2-b_2c_1)}{a_2b_1-a_1b_2}\varphi_1'(s)=0\end{equation*}or
\begin{equation*}a_2\varphi_1(s)+ \frac{a_2(b_1c_2-b_2c_1)}{a_2b_1-a_1b_2}\varphi_1'(s)=0,
\end{equation*}which implies that
\begin{equation}
\varphi_1(s)=\eta_0 e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1
+(a_1b_2-a_2b_1)z_2\right]},~~\eta_0\in \mathbb{C}.
\end{equation}From (4.38) and (4.39), we have
\begin{equation}
u(z_1,z_2)=\frac{\sqrt{2}(k_1^2-k_2^2)}{4}
\left[\frac{1}{\nu_1}e^{Q_3(z)}+\frac{1}{\nu_2}e^{Q_4(z)}
\right]+\eta_0 e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},
\end{equation} where η 0 is a constant, and 
$Q_3(z)$ and 
$Q_4(z)$ are given as in (4.32) and (4.33).
If 
$\nu_1= 0$ and 
$\nu_2\neq 0$, in view of equations (4.35) to (4.37), we have
\begin{equation*}
u(t,s)=\frac{\sqrt{2}}{4}\left[(a_1k_2-a_2k_1)te^{Q_3(t,s)}
+\frac{k_1^2-k_2^2}{\nu_2}e^{Q_3(t,s)}\right]+\varphi_1(s).\end{equation*}Noting that
\begin{equation*}t=\frac{1}{a_2b_1-a_1b_2}z_1\quad \text{and} \quad
s=z_2-\frac{a_2c_1-a_1c_2}{a_2b_1-a_1b_2}z_1,\end{equation*}we have
\begin{equation}
u(z_1,z_2)=\frac{\sqrt{2}}{4}\left[\frac{a_1k_2-a_2k_1}
{a_2b_1-a_1b_2}z_1e^{Q_3(z)}+\frac{k_1^2-k_2^2}
{\nu_2}e^{Q_4(z)}\right]+\varphi_1(s),
\end{equation} where 
$Q_3(z)$ and 
$Q_4(z)$ are given as in (4.32) and (4.33). Due to 
$\nu_1=0$, this leads to
\begin{equation*}\alpha_1(b_1k_1-b_2k_2)+\alpha_2(c_1k_1-c_2k_2)=-2(a_1k_1-a_2k_2)\end{equation*}and
\begin{equation}
Q_3(z)=B_1\left(z_2-\frac{a_2c_1-a_1c_2}
{a_2b_1-a_1b_2}z_1\right)+\beta_1=\frac{1}{D}
\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]+\beta_1.
\end{equation} Substituting from (4.41) into (4.21) or (4.22), we find that 
$\varphi_1(s)$ is of the form (4.39). And, by combining it with (4.41) and (4.42), we have
\begin{equation}
u(z_1,z_2)=(\eta_1z_1+\eta_0) e^{\frac{1}{D}
\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]}
+\frac{\sqrt{2}(k_1^2-k_2^2)}{4\nu_2}e^{Q_4(z)},
\end{equation} where 
$Q_4(z)$ is given as in (3.33) and
\begin{equation*}\eta_1=\frac{\sqrt{2}}{4}\frac{a_1k_2-a_2k_1}
{a_2b_1-a_1b_2}e^{\beta_1}(\neq 0).\end{equation*} If 
$\nu_1\neq0$ and 
$\nu_2=0$, by using the argument similar to that used in the case for 
$\nu_1=0$ and 
$\nu_2\neq0$, we can obtain
\begin{equation}
Q_4(z)=B_2\left(z_2-\frac{a_2c_1-a_1c_2}
{a_2b_1-a_1b_2}z_1\right)+\beta_2
=\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]+\beta_2
\end{equation}and
\begin{equation}
u(z_1,z_2)=\frac{1}{\nu_1}e^{Q_3(z)}+(\eta_2z_1+\eta_0)
e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},
\end{equation} where 
$Q_3(z)$ is given as in (4.32) and
\begin{equation*}\eta_2=\frac{\sqrt{2}(a_1k_1-a_2k_2)}{4(a_2b_1-a_1b_2)}
e^{\beta_2}(\neq 0).\end{equation*} If 
$\nu_1=0$ and 
$\nu_2=0$, that is,
\begin{equation*}\alpha_1=\frac{2(a_2c_1-a_1c_2)}{D}
\quad \text{and} \quad \alpha_2=\frac{2(a_1b_2-a_2b_1)}{D},\end{equation*} by applying the argument similar to that in the case for 
$\nu_1=0$ and 
$\nu_2\neq 0$, we can obtain
\begin{equation}
u(z_1,z_2)=(\eta z_1+\eta_0) e^{\frac{1}{D}
\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]},
\end{equation}where
\begin{equation*}\eta=\frac{\sqrt{2}[(a_1k_2-a_2k_1)e^{\beta_1}
+(a_1k_1-a_2k_2)e^{\beta_2}}{4(a_2b_1-a_1b_2)}.\end{equation*} Subcase 2.2. 
$a_1k_2-a_2k_1=0$ and 
$a_1k_1-a_2k_2\neq 0$. Then (4.35) becomes
\begin{align}
u(t,s)&=\int_0^t\frac{\sqrt{2}}{4}(a_1k_1-a_2k_2)e^{Q_4}dt+\varphi_1(s),
\end{align} where 
$\varphi_1(s)$ is a finite-order entire function in s. If 
$\nu_2\neq 0$, by using the above argument, we find from (4.47) that
\begin{equation}
u(z_1,z_2)=\frac{\sqrt{2}}{4}\frac{k_1^2-k_2^2}{\nu_2}
e^{Q_4(z)}+\eta_0 e^{\frac{1}{D}\left[(a_2c_1-a_1c_2)z_1
+(a_1b_2-a_2b_1)z_2\right]}.
\end{equation} If 
$\nu_2=0$, it follows from (3.47) that
\begin{equation}
u(z_1,z_2)=\eta_0 e^{\frac{1}{D}
\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]}.
\end{equation} Subcase 2.3. 
$a_1k_2-a_2k_1\neq0$ and 
$a_1k_1-a_2k_2=0$. Similarly, if 
$\nu_1=0$, then u satisfies (4.49). If 
$\nu_1\neq 0$, similar to the above argument, we find that
\begin{equation}
u(z_1,z_2)=\frac{\sqrt{2}}{4}\frac{k_1^2-k_2^2}{\nu_1}
e^{Q_3(z)}+\eta_0 e^{\frac{1}{D}
\left[(a_2c_1-a_1c_2)z_1+(a_1b_2-a_2b_1)z_2\right]}.
\end{equation} Subcase 2.4. 
$a_1k_2-a_2k_1=0$ and 
$a_1k_1-a_2k_2=0$. Then it follows that 
$a_1=a_2=0$. This leads to a contradiction with 
$a_1b_2-a_2b_1=0$.
Thus, from Subcases 2.1-2.4, we are led to the conclusions of Theorem 2.5.
This completes the proof of Theorem 2.5.
Acknowledgements
The authors are very thankful to the referees for their valuable comments which improved the presentation of the paper.
Funding Statement
This work was supported by the National Natural Science Foundation of China (12161074), by the Foundation of the Education Department of Jiangxi (GJJ211333, GJJ202303, GJJ201813, GJJ201343) of China, by the Talent Introduction Research Foundation of Suqian University (106-CK00042/028), by the Suqian Science and Technology Program Grant No. M202206), by the Qing Lan Project of the Jiangsu Province, and by the Suqian Talent Xiongying Plan of Suqian.
Competing interests
The authors declare that they have no competing interests in the manuscript.
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