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The study of solutions for several systems of PDDEs with two complex variables

  • Yi Hui Xu , Xiao Lan Liu and Hong Yan Xu EMAIL logo
Published/Copyright: August 10, 2023
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Demonstratio Mathematica
From the journal Demonstratio Mathematica Volume 56 Issue 1

Abstract

The purpose of this article is to describe the properties of the pair of solutions of several systems of Fermat-type partial differential difference equations. Our theorems exhibit the forms of finite order transcendental entire solutions for these systems, which are some extensions and improvement of the previous theorems given by Xu, Cao, Liu, etc. Furthermore, we give a series of examples to show that the existence conditions and the forms of transcendental entire solutions with finite order of such systems are precise.

Keywords: partial differential difference equation; entire solution; Nevanlinna theory
MSC 2010: 30D 35; 35M 30; 39A 45

1 Introduction

Let us first recall the classical results that the entire solutions of the functional equation

(1.1) f 2 + g 2 = 1 ,

are f = cos a ( z ) , g = sin a ( z ) was proved by Gross [1], where a ( z ) is an entire function. This simple-looking nonlinear functional equation (1.1) can be called as the Fermat-type functional equation, analogous with the equation x 2 + y 2 = z 2 in Fermat’s last theorem in number theory. As a matter of fact, we can find that the study of the Fermat-type functional equations can be tracked back to more than 60 years ago or even earlier [2,3].

In the past 30 years, there were lots of research focusing on the solutions of functional equation (1.1), readers can refer to [4–17]. For example, Khavinson [11] in 1995 proved that any entire solutions of the partial differential equations

(1.2) ∂ f ∂ z 1 2 + ∂ f ∂ z 2 2 = 1 ,

in C 2 are necessarily linear. It should be noted that equation (1.2) is called as eiconal equation. Later, Saleeby [18,19] further proved that the entire solution of equation (1.2) is of the form f ( z 1 , z 2 ) = c 1 z 1 + c 2 z 2 + η . After theirs works, Li and co-authors [20–22] further discussed a series of deformation forms of Fermat-type partial differential equations and gave a number of important and interesting results about the existence and the forms of solutions for these partial differential equations.

Theorem A

[20, Corollary 2.3] Let P ( z 1 , z 2 ) and Q ( z 1 , z 2 ) be arbitrary polynomials in C 2 . Then, f is an entire solution of the equation

(1.3) P ∂ f ( z 1 , z 2 ) ∂ z 1 2 + Q ∂ f ( z 1 , z 2 ) ∂ z 2 2 = 1 ,

if and only if f = c 1 z 1 + c 2 z 2 + c 3 is a linear function, where c j ’s are constants, and exactly one of the following holds:

  1. c 1 = 0 and Q is a constant satisfying that ( c 2 Q ) 2 = 1 ;

  2. c 2 = 0 and P is a constant satisfying that ( c 1 P ) 2 = 1 ;

  3. c 1 c 2 ≠ 0 and P and Q are both constants satisfying that ( c 1 P ) 2 + ( c 2 Q ) 2 = 1 .

In the past 40 years, the Nevanlinna theory and the difference Nevanlinna theory of Meromorphic function with several complex variables have been developed rapidly [23–29]. Especially, Korhonen [30, Theorem 3.1] in 2012 established a logarithmic difference lemma for meromorphic functions in several variables of hyper order < 2 ∕ 3 . Later, Cao and Korhonen [23] proved that the logarithmic difference lemma for meromorphic functions holds under the condition “hyper order < 1 .” By making use of the Nevanlinna theory and difference Nevanlinna theory of several complex variables [23,30], Xu and Cao [31,32] discussed the transcendental solutions of several Fermat-type partial differential difference equations. An equation is called partial differential difference equation, if the equation includes partial derivatives, shifts or differences of f , which can be called PDDE for short.

Theorem B

[31, Theorem 1.2] Let c = ( c 1 , c 2 ) ∈ C 2 . Then, any transcendental entire solution with finite order of the PDEE

(1.4) ∂ f ( z 1 , z 2 ) ∂ z 1 2 + f ( z 1 + c 1 , z 2 + c 2 ) 2 = 1 ,

has the form of f ( z 1 , z 2 ) = sin ( A z 1 + B ) , where A is a constant on C satisfying A e i A c 1 = 1 , and B is a constant on C ; as a special case, whenever c 1 = 0 , we have f ( z 1 , z 2 ) = sin ( z 1 + B ) .

In 2020, the author of this article and his colleagues [33] studied the finite order transcendental entire solutions when equation (1.4) turns to the system of Fermat-type PDDEs and obtained Theorem C.

Theorem C

[33, Theorem 1.3] Let c = ( c 1 , c 2 ) ∈ C 2 . Then, any pair of transcendental entire solutions with finite order for the system of Fermat-type PDEEs

(1.5) ∂ f 1 ( z 1 , z 2 ) ∂ z 1 2 + f 2 ( z 1 + c 1 , z 2 + c 2 ) 2 = 1 , ∂ f 2 ( z 1 , z 2 ) ∂ z 1 2 + f 1 ( z 1 + c 1 , z 2 + c 2 ) 2 = 1

have the following forms:

( f 1 ( z ) , f 2 ( z ) ) = e L ( z ) + B 1 + e − ( L ( z ) + B 1 ) 2 , A 21 e L ( z ) + B 1 + A 22 e − ( L ( z ) + B 1 ) 2 ,

where L ( z ) = a 1 z 1 + a 2 z 2 , B 1 is a constant in C , and a 1 , c , A 21 , A 22 satisfy one of the following cases:

  1. A 21 = − i , A 22 = i , and a 1 = i , L ( c ) = 2 k + 1 2 π i , or a 1 = − i , L ( c ) = 2 k − 1 2 π i ;

  2. A 21 = i , A 22 = − i , and a 1 = i , L ( c ) = 2 k − 1 2 π i , or a 1 = − i , L ( c ) = 2 k + 1 2 π i ;

  3. A 21 = 1 , A 22 = 1 , and a 1 = i , L ( c ) = 2 k π i , or a 1 = − i , L ( c ) = ( 2 k + 1 ) π i ;

  4. A 21 = − 1 , A 22 = − 1 , and a 1 = i , L ( c ) = ( 2 k + 1 ) π i , or a 1 = − i , L ( c ) = 2 k π i .

To the best of our knowledge, there are few results about the study of systems of this Fermat-type PDDE with several complex variables. Moreover, it appears that the study of such fields has not been addressed in the literature before. Based on these, we are mainly concerned with the solutions of complex Fermat-type PDDEs, and describe the existence and form of the pair of the finite order transcendental solutions of the systems of PDDEs with constant coefficients

(1.6) μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 2 + [ α f 2 ( z + c ) − β f 1 ( z ) ] 2 = 1 , μ f 2 ( z ) + λ ∂ f 2 ∂ z 1 2 + [ α f 1 ( z + c ) − β f 2 ( z ) ] 2 = 1 ,

and

(1.7) μ f 1 ( z ) + λ 1 ∂ f 1 ∂ z 1 + λ 2 ∂ f 1 ∂ z 2 2 + [ α f 2 ( z + c ) − β f 1 ( z ) ] 2 = 1 , μ f 2 ( z ) + λ 1 ∂ f 2 ∂ z 1 + λ 2 ∂ f 2 ∂ z 2 2 + [ α f 1 ( z + c ) − β f 2 ( z ) ] 2 = 1 ,

where α , β , μ , λ , λ 1 , λ 2 , c 1 , c 2 are constants in C . Obviously, equation (1.4) and system (1.5) are the special cases of systems (1.6) and (1.7). The article is organized as follows. We will introduce our main results about the existence and the forms of entire solutions for (1.6) and (1.7) in Section 2, which generalize the previous theorems given by Xu et al. [31–33]. Meantime, we give a series of examples to explain that our results about the forms of solutions of such systems are precise. The proofs of Theorems 1.6 and 1.7 are given in Sections 4 and 5, respectively.

2 Results and examples

The first main theorem is about the existence and the forms of the solutions for system (1.6).

Theorem 2.1

Let c = ( c 1 , c 2 ) ∈ C 2 , c 2 ≠ 0 , and α , β , μ , λ be nonzero constants in C . Let ( f 1 ( z 1 , z 2 ) , f 2 ( z 1 , z 2 ) ) be a pair of transcendental entire solution with finite order of system (1.6). Then, ( f 1 ( z 1 , z 2 ) , f 2 ( z 1 , z 2 ) ) must satisfy one of the following cases:

  1. f 1 ( z 1 , z 2 ) = η 1 μ − 1 μ e − μ λ z 1 + γ z 2 + D 1 , f 2 ( z 1 , z 2 ) = δ 1 μ − 1 μ e − μ λ z 1 + γ z 2 + D 2 ,

    where η 1 , δ 1 , γ , D 1 , D 2 ∈ C satisfy

    (2.1) e D 1 − D 2 = 1 , γ = μ λ c 1 + log β α + k π i c 2 , k ∈ Z ,

    and one of the following cases:

    1. δ 1 = − η 1 = ± 1 and α = − β or δ 1 = η 1 = ± 1 , α = β ;

    2. δ 1 = η 1 and η 1 2 = μ 2 μ 2 + ( α − β ) 2 , or δ 1 = − η 1 and η 1 2 = μ 2 μ 2 + ( α + β ) 2 ;

    3. μ 2 = α 2 − β 2 , δ 1 2 + δ 2 2 = 1 , η 1 2 + η 2 2 = 1 , and α δ 2 = − β η 2 ± ( η 2 2 − 1 ) ( β 2 − α 2 ) ;

  2. if a 1 ≠ ± μ λ , then

    f 1 ( z 1 , z 2 ) = 1 2 ( λ a 1 + μ ) e a 1 z 1 + a 2 z 2 + b 1 − 1 2 ( λ a 1 − μ ) e − a 1 z 1 − a 2 z 2 − b 1 − ϑ ( z 2 ) e − μ λ z 1 + γ z 2 + D 1 , f 2 ( z 1 , z 2 ) = 1 2 ( λ a 1 + μ ) e a 1 z 1 + a 2 z 2 + b 2 − 1 2 ( λ a 1 − μ ) e − a 1 z 1 − a 2 z 2 − b 2 − ϑ ( z 2 + c 2 ) e − μ λ z 1 + γ z 2 + D 2 ,

    where ϑ ( z 2 ) is a finite order period entire function with period 2 c 2 , and a 1 , a 2 , b 1 , b 2 , γ , D 1 , D 2 ∈ C satisfy (2.1),

    (2.2) e 2 ( b 1 − b 2 ) = 1 ,

    and

    a 1 = − β i ± μ 2 − α 2 λ , e 2 ( a 1 c 1 + a 2 c 2 ) = ( λ a 1 + μ + β i ) 2 − α 2 ;

    if a 1 = μ λ , then

    f 1 ( z 1 , z 2 ) = 1 4 μ e a 1 z 1 + a 2 z 2 + b 1 + z 1 2 λ e − a 1 z 1 − a 2 z 2 − b 1 − ϑ ( z 2 ) e − μ λ z 1 + γ z 2 + D 1 ,

    f 2 ( z 1 , z 2 ) = 1 4 μ e a 1 z 1 + a 2 z 2 + b 2 + z 1 2 λ e − a 1 z 1 − a 2 z 2 − b 2 − ϑ ( z 2 + c 2 ) e − μ λ z 1 + γ z 2 + D 2 ,

    where a 1 , a 2 , b 1 , b 2 , γ ∈ C satisfy (2.1), (2.2) and

    (2.3) 2 μ β i = β 2 − α 2 , β c 1 = λ i , e 2 ( a 1 c 1 + a 2 c 2 ) = 1 − 2 μ β i ;

    if a 1 = − μ λ , then

    f 1 ( z 1 , z 2 ) = z 1 2 λ e a 1 z 1 + a 2 z 2 + b 1 + 1 4 μ e − a 1 z 1 − a 2 z 2 − b 1 − ϑ ( z 2 ) e − μ λ z 1 + γ z 2 + D 1 ,

    f 2 ( z 1 , z 2 ) = z 1 2 λ e a 1 z 1 + a 2 z 2 + b 2 + 1 4 μ e − a 1 z 1 − a 2 z 2 − b 2 − ϑ ( z 2 + c 2 ) e − μ λ z 1 + γ z 2 + D 2 ,

    where a 1 , a 2 , b 1 , b 2 , γ ∈ C satisfy (2.1), (2.2) and

    (2.4) − 2 μ β i = β 2 − α 2 , β c 1 = − λ i , e − 2 ( a 1 c 1 + a 2 c 2 ) = 1 + 2 μ β i .

The following examples show the existence of transcendental entire solutions of system (1.6).

Example 2.1

Let

f 1 ( z 1 , z 2 ) = 1 2 − 1 2 e − 2 z 1 + 3 z 2 , f 2 ( z 1 , z 2 ) = − 1 2 − 1 2 e − 2 z 1 + 3 z 2 .

Thus, ( f 1 , f 2 ) is a pair of finite order transcendental entire solution of system (1.6) with λ = 1 , μ = 2 , α = 1 , β = − 1 , and ( c 1 , c 2 ) = ( π i , π i ) .

Example 2.2

Let

f 1 ( z 1 , z 2 ) = 1 − e − z 1 + z 2 , f 2 ( z 1 , z 2 ) = 1 − e − z 1 + z 2 .

Thus, ( f 1 , f 2 ) is a pair of finite order transcendental entire solution of system (1.6) with λ = μ = α = β = 1 , and ( c 1 , c 2 ) = ( π i , π i ) .

Example 2.3

Let

f 1 ( z 1 , z 2 ) = 1 3 − e − 1 2 z 1 + z 2 , f 2 ( z 1 , z 2 ) = − 1 3 − e − 1 2 z 1 + z 2 .

Thus, ( f 1 , f 2 ) is a pair of finite order transcendental entire solution of system (1.6) with λ = 2 , μ = 1 , α = 2 2 + 2 2 i , β = 2 2 − 2 2 i , and ( c 1 , c 2 ) = ( π i , π i ) .

Example 2.4

Let

f 1 ( z 1 , z 2 ) = 2 2 + i 2 e − 2 i z 1 + ( 2 − 0.5 ) z 2 , f 2 ( z 1 , z 2 ) = 1 − 2 2 i + i 2 e − 2 i z 1 + ( 2 − 0.5 ) z 2 .

Thus, ( f 1 , f 2 ) is a pair of finite order transcendental entire solution of system (1.6) with λ = 1 , μ = 2 i , α = i , β = 1 , and ( c 1 , c 2 ) = ( π , π i ) .

Example 2.5

Let

f 1 ( z 1 , z 2 ) = 1 2 2 ( 2 + 1 ) e 2 z 1 + a 2 z 2 − 1 2 2 ( 2 − 1 ) e − 2 z 1 − a 2 z 2 − sin ( π i z 2 ) e − 2 z 1 + ( 2 + 0.5 ) π i z 2 ,

f 2 ( z 1 , z 2 ) = − 1 2 2 ( 2 + 1 ) e 2 z 1 + a 2 z 2 + 1 2 2 ( 2 − 1 ) e − 2 z 1 − a 2 z 2 + sin ( π i z 2 ) e − 2 z 1 + ( 2 + 0.5 ) π i z 2 ,

where a 2 = log ( 2 + 1 ) − π 2 i . Thus, ( f 1 , f 2 ) is a pair of finite order transcendental entire solution of system (1.6) with λ = 1 , μ = 2 , α = 1 , β = i , and ( c 1 , c 2 ) = ( π i , 1 ) .

Example 2.6

Let γ = − 1 4 ln 2 + π 8 i + 1 2 i , a 2 = 1 4 ln 2 − π 8 i − 1 2 i and

f 1 ( z 1 , z 2 ) = 1 4 e i 2 z 1 + a 2 z 2 − i z 1 4 e − i 2 z 1 − a 2 z 2 − cos ( π i z 2 ) e − i 2 z 1 + γ z 2 ,

f 2 ( z 1 , z 2 ) = 1 4 e i 2 z 1 + a 2 z 2 − i z 1 4 e − i 2 z 1 − a 2 z 2 + cos ( π i z 2 ) e − i 2 z 1 + γ z 2 .

Thus, ( f 1 , f 2 ) is a transcendental entire solution of (1.6) with λ = − 2 i , μ = 1 , α = 2 5 4 e − π 8 i , β = 2 , ( c 1 , c 2 ) = ( 1 , 1 ) , and ρ ( f ) = 1 .

For α = 1 and β = 0 in system (1.6), we have

Corollary 2.1

Let c = ( c 1 , c 2 ) ∈ C 2 , c 2 ≠ 0 , and μ , λ be nonzero constants. If ( f 1 , f 2 ) are a pair of finite order transcendental entire solution of the following system:

(2.5) μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 2 + f 2 ( z + c ) 2 = 1 , μ f 2 ( z ) + λ ∂ f 2 ∂ z 1 2 + f 1 ( z + c ) 2 = 1 ,

then, ( f 1 , f 2 ) must be of the form

f 1 ( z 1 , z 2 ) = 1 2 ( λ a 1 + μ ) e a 1 z 1 + a 2 z 2 + b 1 + 1 2 ( μ − λ a 1 ) e − a 1 z 1 − a 2 z 2 − b 1 ,

f 2 ( z 1 , z 2 ) = 1 2 ( λ a 1 + μ ) e a 1 z 1 + a 2 z 2 + b 2 + 1 2 ( μ − λ a 1 ) e − a 1 z 1 − a 2 z 2 − b 2 ,

where a 1 , a 2 , b 1 , b 2 ∈ C satisfy a 1 2 = μ 2 − 1 λ 2 and

(2.6) e 2 ( a 1 c 1 + a 2 c 2 ) = ( μ + λ a 1 ) 2 = 1 ( μ − λ a 1 ) 2 , e 2 ( b 1 − b 2 ) = − 1 ,

or

(2.7) e 2 ( a 1 c 1 + a 2 c 2 ) = − ( μ + λ a 1 ) 2 = − 1 ( μ − λ a 1 ) 2 , e 2 ( b 1 − b 2 ) = 1 .

Theorem 2.2

Let c = ( c 1 , c 2 ) ∈ C 2 , α , β , μ , λ 1 , λ 2 be nonzero constants in C , s 1 ≔ λ 2 z 1 − λ 1 z 2 , and s 0 ≔ λ 1 c 2 − λ 2 c 1 ≠ 0 . Let ( f 1 , f 2 ) be a pair of the finite order transcendental entire solutions of system (1.7). Then, ( f 1 , f 2 ) must satisfy one of the following cases:

  1. f 1 ( z 1 , z 2 ) = η 1 μ − 1 μ e − μ λ 1 z 1 + γ ( λ 1 z 2 − λ 2 z 1 ) + D 1 ,

    f 2 ( z 1 , z 2 ) = δ 1 μ − 1 μ e − μ λ 1 z 1 + γ ( λ 1 z 2 − λ 2 z 1 ) + D 2 ,

    where η 1 , δ 1 , γ , D 1 , D 2 ∈ C satisfy

    (2.8) e D 1 − D 2 = 1 , γ = μ λ 1 c 1 + log β α + 2 k π i λ 1 c 2 − λ 2 c 1 , k ∈ Z ,

    and one of the following cases:

    1. δ 1 = − η 1 = ± 1 and α = − β or δ 1 = η 1 = ± 1 , α = β ;

    2. δ 1 = η 1 and η 1 2 = μ 2 μ 2 + ( α − β ) 2 or δ 1 = − η 1 and η 1 2 = μ 2 μ 2 + ( α + β ) 2 ;

    3. μ 2 = α 2 − β 2 , δ 1 2 + δ 2 2 = 1 , η 1 2 + η 2 2 = 1 , and α δ 2 = − β η 2 ± ( η 2 2 − 1 ) ( β 2 − α 2 ) ;

  2. if μ 2 ≠ ( λ 1 a 1 + λ 2 a 2 ) 2 , then

    f 1 ( z 1 , z 2 ) = 1 2 ( λ 1 a 1 + λ 2 a 2 + μ ) e a 1 z 1 + a 2 z 2 + b 1 − 1 2 ( λ 1 a 1 + λ 2 a 2 − μ ) e − a 1 z 1 − a 2 z 2 − b 1 − ϑ ( s 1 ) e − μ λ 1 z 1 + γ s 1 + D 1 , f 2 ( z 1 , z 2 ) = 1 2 ( λ 1 a 1 + λ 2 a 2 + μ ) e a 1 z 1 + a 2 z 2 + b 2 − 1 2 ( λ 1 a 1 + λ 2 a 2 − μ ) e − a 1 z 1 − a 2 z 2 − b 2 − ϑ ( s 1 + s 0 ) e − μ λ 1 z 1 + γ s 1 + D 2 ,

    where ϑ ( s 1 ) is a finite order period entire functions with period 2 s 0 , and a 1 , a 2 , b 1 , b 2 , γ , D 1 , D 2 satisfy (2.2), (2.8), and

    (2.9) ( λ 1 a 1 + λ 2 a 2 + β i ) 2 = μ 2 − α 2 , e 2 ( a 1 c 1 + a 2 c 2 ) = λ 1 a 1 + λ 2 a 2 + β i + μ λ 1 a 1 + λ 2 a 2 + β i − μ ;

    if μ = λ 1 a 1 + λ 2 a 2 , then

    f 1 ( z 1 , z 2 ) = 1 4 μ e a 1 z 1 + a 2 z 2 + b 1 + z 1 2 λ 1 e − a 1 z 1 − a 2 z 2 − b 1 − ϑ ( s 1 ) e − μ λ 1 z 1 + γ s 1 + D 1 , f 2 ( z 1 , z 2 ) = 1 4 μ e a 1 z 1 + a 2 z 2 + b 2 + z 1 2 λ 1 e − a 1 z 1 − a 2 z 2 − b 2 − ϑ ( s 1 + s 0 ) e − μ λ 1 z 1 + γ s 1 + D 2 ,

    where a 1 , a 2 , b 1 , b 2 , γ ∈ C satisfy (2.2), (2.8), and

    (2.10) 2 μ β i = β 2 − α 2 , β c 1 = λ 1 i , e 2 ( a 1 c 1 + a 2 c 2 ) = 1 − 2 μ β i ;

    if μ = − ( λ 1 a 1 + λ 2 a 2 ) , then

    f 1 ( z 1 , z 2 ) = z 1 2 λ 1 e a 1 z 1 + a 2 z 2 + b 1 + 1 4 μ e − a 1 z 1 − a 2 z 2 − b 1 − ϑ ( s 1 ) e − μ λ 1 z 1 + γ s 1 + D 1 , f 2 ( z 1 , z 2 ) = z 1 2 λ 1 e a 1 z 1 + a 2 z 2 + b 2 + 1 4 μ e − a 1 z 1 − a 2 z 2 − b 2 − ϑ ( s 1 + s 0 ) e − μ λ 1 z 1 + γ s 1 + D 2 ,

    where a 1 , a 2 , b 1 , b 2 , γ ∈ C satisfy (2.2), (2.8), and

    (2.11) − 2 μ β i = β 2 − α 2 , β c 1 = − λ 1 i , e − 2 ( a 1 c 1 + a 2 c 2 ) = 1 + 2 μ β i .

The following examples show the existence of transcendental entire solutions of system (1.7).

Example 2.7

Let

f 1 ( z 1 , z 2 ) = − 1 2 − 1 2 e − z 1 + z 2 , f 2 ( z 1 , z 2 ) = − 1 2 − 1 2 e − z 1 + z 2 .

Thus, ( f 1 , f 2 ) is a pair of finite order transcendental entire solution of system (1.7) with λ 1 = 1 , λ 2 = − 1 , μ = 2 , α = 1 , β = 1 , and ( c 1 , c 2 ) = ( κ , κ ) , κ ∈ { C } − { 0 } .

Example 2.8

Let

f 1 ( z 1 , z 2 ) = 2 4 − 1 2 e − z 1 + z 2 , f 2 ( z 1 , z 2 ) = − 2 4 − 1 2 e − z 1 + z 2 .

Thus, ( f 1 , f 2 ) is a pair of finite order transcendental entire solution of system (1.7) with λ 1 = 1 , λ 2 = − 1 , μ = 2 , α = 1 , β = 1 , and ( c 1 , c 2 ) = ( κ , κ ) , κ ∈ { C } − { 0 } .

Example 2.9

Let

f 1 ( z 1 , z 2 ) = 1 2 2 − 1 2 e − 2 2 + 1 4 z 1 + 2 2 + 1 4 z 2 ,

f 2 ( z 1 , z 2 ) = 3 2 + 1 2 2 i − 1 2 e − 2 2 + 1 4 z 1 + 2 2 + 1 4 z 2 .

Thus, ( f 1 , f 2 ) is a pair of finite order transcendental entire solution of system (1.7) with λ 1 = 1 , λ 2 = − 1 , μ = 2 , α = 1 , β = i , and ( c 1 , c 2 ) = ( π i , π i ) .

Example 2.10

Let

f 1 ( z 1 , z 2 ) = 1 2 3 e 4 3 z 1 − 5 6 z 2 + z 1 2 3 e − 4 3 z 1 + 5 6 z 2 − cos π i κ ( z 1 − z 2 ) e − 1 2 z 1 + A ( z 1 − z 2 ) , f 2 ( z 1 , z 2 ) = 1 2 3 e 4 3 z 1 − 5 6 z 2 + z 1 2 3 e − 4 3 z 1 + 5 6 z 2 − cos π i κ ( z 1 − z 2 ) e − 1 2 z 1 + A ( z 1 − z 2 ) ,

where

A = 3 2 + 1 2 i + π 3 3 i − 9 + ( 3 3 + 4 π ) i , κ = 3 9 10 − 3 + 4 π 10 i .

Thus, ( f 1 , f 2 ) is a pair of finite order transcendental entire solution of system (1.7) with λ 1 = 3 , λ 2 = 3 , μ = 3 2 , α = 1 , β = 1 2 + 3 2 i , and

( c 1 , c 2 ) = 3 2 + 3 2 i , 3 5 + 4 3 + 2 π 5 i .

For α = 1 and β = 0 in system (1.7), we have

Corollary 2.2

Let c = ( c 1 , c 2 ) ∈ C 2 , μ , λ 1 , λ 2 be nonzero constants, and λ 1 c 2 − λ 2 c 1 ≠ 0 . If ( f 1 , f 2 ) are a pair of finite order transcendental entire solution of the following system:

(2.12) μ f 1 ( z ) + λ 1 ∂ f 1 ∂ z 1 + λ 2 ∂ f 1 ∂ z 2 2 + f 2 ( z + c ) 2 = 1 , μ f 2 ( z ) + λ 1 ∂ f 2 ∂ z 1 + λ 2 ∂ f 2 ∂ z 2 2 + f 1 ( z + c ) 2 = 1 ,

then, ( f 1 , f 2 ) must be of the form

f 1 ( z 1 , z 2 ) = 1 2 ( λ 1 a 1 + λ 2 a 2 + μ ) e a 1 z 1 + a 2 z 2 + b 1 + 1 2 ( μ − λ 1 a 1 − λ 2 a 2 ) e − a 1 z 1 − a 2 z 2 − b 1 ,

f 2 ( z 1 , z 2 ) = 1 2 ( λ 1 a 1 + λ 2 a 2 + μ ) e a 1 z 1 + a 2 z 2 + b 2 + 1 2 ( μ − λ 1 a 1 − λ 2 a 2 ) e − a 1 z 1 − a 2 z 2 − b 2 ,

where a 1 , a 2 , b 1 , b 2 ∈ C satisfy ( λ 1 a 1 + λ 2 a 2 ) 2 = μ 2 − 1 and

(2.13) e 2 ( a 1 c 1 + a 2 c 2 ) = ( μ + λ 1 a 1 + λ 2 a 2 ) 2 = 1 ( μ − λ 1 a 1 − λ 2 a 2 ) 2 , e 2 ( b 1 − b 2 ) = − 1 ,

or

(2.14) e 2 ( a 1 c 1 + a 2 c 2 ) = − ( μ + λ 1 a 1 + λ 2 a 2 ) 2 = − 1 ( μ − λ 1 a 1 − λ 2 a 2 ) 2 , e 2 ( b 1 − b 2 ) = 1 .

3 Some lemmas

The following lemmas play the key role in proving our results.

Lemma 3.1

[27,28] For an entire function F on C n , F ( 0 ) ≠ 0 and put ρ ( n F ) = ρ < ∞ . Then, there exist a canonical function f F and a function g F ∈ 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 3.1

Here denote ρ ( n F ) to be the order of the counting function of zeros of F .

Lemma 3.2

[3] If g and h are entire functions on the complex plane C and g ( h ) is an entire function of finite order, then there are only two possible cases: either

  1. the internal function h is a polynomial and the external function g is of finite order; or else

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

Lemma 3.3

[34, Lemma 3.1] Let f j ( ≢ 0 ) , j = 1 , 2 , 3 , be meromorphic functions on C m such that f 1 is not constant, and f 1 + f 2 + f 3 = 1 , and such that

∑ j = 1 3 N 2 r , 1 f j + 2 N ¯ ( r , f j ) < λ T ( r , f 1 ) + O ( log + T ( r , f 1 ) ) ,

for all r outside possibly a set with finite logarithmic measure, where λ < 1 is a positive number. Then, either f 2 = 1 or f 3 = 1 .

Remark 3.2

Here N 2 ( r , 1 f ) is the counting function of the zeros of f in ∣ z ∣ ≤ r , where the simple zero is counted once, and the multiple zero is counted twice.

Lemma 3.4

Let c = ( c 1 , c 2 ) ∈ C 2 and α , β be two nonzero constants. Let

g 1 ( z ) = θ 1 ( z 2 ) e ξ z 1 + ψ 1 ( z 2 ) , g 2 ( z ) = θ 2 ( z 2 ) e ξ z 1 + ψ 2 ( z 2 ) ,

where ξ is a constant, θ j ( z 2 ) , j = 1 , 2 are finite order entire functions, and ψ j ( z 2 ) , j = 1 , 2 are polynomials in z 2 . If ( g 1 , g 2 ) is a pair of solutions of system

(3.1) α g 2 ( z + c ) − β g 1 ( z ) = 0 , α g 1 ( z + c ) − β g 2 ( z ) = 0 ,

then, ( g 1 , g 2 ) can be expressed as the form of

g 1 ( z ) = θ ( z 2 ) e ξ z 1 + γ z 2 + D 1 , g 2 ( z ) = θ ( z 2 + c 2 ) e ξ z 1 + γ z 2 + D 2 ,

where θ ( z 2 ) are finite order entire period function with the period 2 c 2 and γ is a constant such that

e D 1 − D 2 = 1 , e γ c 2 = β α e − c 1 ξ .

Proof

Substituting g 1 , g 2 into system (3.1), we have

(3.2) e ψ 2 ( z 2 + c 2 ) − ψ 1 ( z 2 ) = θ 1 ( z 2 ) θ 2 ( z 2 + c 2 ) β α e − c 1 ξ , e ψ 1 ( z 2 + c 2 ) − ψ 2 ( z 2 ) = θ 2 ( z 2 ) θ 1 ( z 2 + c 2 ) β α e − c 1 ξ ,

which implies

(3.3) e ψ j ( z 2 + 2 c 2 ) − ψ j ( z 2 ) = θ j ( z 2 ) θ j ( z 2 + 2 c 2 ) β 2 α 2 e − 2 c 1 ξ , j = 1 , 2 .

Noting that ψ j ( z 2 ) is a polynomial, we will consider two cases as follows.

Case 1. Suppose that e ψ j ( z 2 + 2 c 2 ) − ψ j ( z 2 ) is a constant. In view of (3.3), it follows that θ j ( z 2 ) θ j ( z 2 + 2 c 2 ) is a nonzero constant for j = 1 , 2 . Set

(3.4) θ j ( z 2 ) θ j ( z 2 + 2 c 2 ) = d , j = 1 , 2 .

In view of (3.3), we can deduce that

(3.5) ψ j ( z 2 ) = h z 2 + D j , j = 1 , 2 ,

where h , D j , ( j = 1 , 2 ) are constants and

(3.6) h = γ + log d 2 c 2 , γ = − c 1 ξ + log β α c 2 .

If d = 1 , then θ j ( z 2 ) ( j = 1 , 2 ) are finite order entire period functions with period 2 c 2 and h = γ = − c 1 ξ + log β α c 2 . Substituting (3.5) and (3.6) in (3.2), we have

θ 2 ( z 2 ) e D 2 = θ 1 ( z 2 + c 2 ) e D 1 .

Thus, in view of (3.5) and (3.6), it follows that

g 1 ( z ) = θ 1 ( z 2 ) e ξ z 1 + γ z 2 + D 1 ,

and

g 2 ( z ) = θ 2 ( z 2 ) e ξ z 1 + h z 2 + D 2 = θ 1 ( z 2 + c 2 ) e ξ z 1 + γ z 2 + D 1 .

If d ≠ 1 , it follows from (3.4) that

(3.7) θ j ( z 2 ) = e − log d 2 c 2 z 2 + ε j , j = 1 , 2 ,

where ε j , j = 1 , 2 are constants. Substituting (3.5)–(3.7) in (3.2), we have e D 2 − D 1 = e ε 1 − ε 2 , which implies

(3.8) e D 1 + ε 1 − ( D 2 + ε 2 ) = 1 .

For convenience, let B 1 = D 1 + ε 1 and B 2 = D 2 + ε 2 . In view of (3.5)–(3.8), it follows that

g 1 ( z ) = θ 1 ( z 2 ) e ξ z 1 + h z 2 + D 1 = e ξ z 1 + γ z 2 + B 1

and

g 2 ( z ) = θ 2 ( z 2 ) e ξ z 1 + h z 2 + D 2 = e ξ z 1 + γ z 2 + B 2 ,

where B 1 , B 2 are constants satisfying (3.8).

Case 2. Suppose that e ψ j ( z 2 + 2 c 2 ) − ψ j ( z 2 ) ( j = 1 , 2 ) are not constants. Noting that ψ j ( z 2 ) , j = 1 , 2 , are nonconstant polynomials, we can deduce from (3.3) that θ j ( z 2 ) θ j ( z 2 + 2 c 2 ) , j = 1 , 2 are finite order transcendental entire functions. Thus, there exist two functions ϱ 1 ( z 2 ) , ϱ 2 ( z 2 ) such that

(3.9) θ j ( z 2 ) = e ϱ j ( z 2 ) , j = 1 , 2 .

This leads to

(3.10) g 1 ( z ) = e ξ z 1 + μ 1 ( z 2 ) , g 2 ( z ) = e ξ z 1 + μ 2 ( z 2 ) ,

where μ j ( z 2 ) = ψ j ( z 2 ) + ϱ j ( z 2 ) . Substituting g 1 , g 2 in (3.1), we have

(3.11) e μ 2 ( z 2 + c 2 ) − μ 1 ( z 2 ) = β α e − c 1 ξ , e μ 1 ( z 2 + c 2 ) − μ 2 ( z 2 ) = β α e − c 1 ξ ,

which implies that

(3.12) μ j ( z 2 ) = γ z 2 + D j , j = 1 , 2 ,

where γ , D j , ( j = 1 , 2 ) are constants and

(3.13) γ = − c 1 ξ + log β α c 2 , e D 1 − D 2 = 1 .

In view of (3.10) and (3.12), we have

(3.14) g 1 ( z ) = e ξ z 1 + γ z 2 + D 1 , g 2 ( z ) = e ξ z 1 + γ z 2 + D 2 ,

where γ , D 1 , D 2 satisfy (3.13).

Therefore, this completes the proof of Lemma 3.4.□

4 The proof of Theorem 2.1

Proof

Let ( f 1 , f 2 ) be a pair of transcendental entire solutions of finite order for system (1.6). Thus, we will consider the following two cases.

  1. If μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 is a constant. Denote

    (4.1) μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 = η 1 .

    In view of (1.6), it follows that

    (4.2) α f 2 ( z + c ) − β f 1 ( z ) = η 2 ,

    where η 2 is a constant satisfying

    (4.3) η 1 2 + η 2 2 = 1 .

    From (4.1) and (4.2), we have

    (4.4) μ f 2 + λ ∂ f 2 ∂ z 1 = β α μ f 1 ( z − c ) + λ ∂ f 1 ( z − c ) ∂ z 1 + μ η 2 α = η 1 β α + μ η 2 α .

    This shows that μ f 2 + λ ∂ f 2 ∂ z 1 is a constant. Let

    (4.5) μ f 2 + λ ∂ f 2 ∂ z 1 = δ 1 ,

    then α f 1 ( z + c ) − β f 2 ( z ) is a constant. Denote

    (4.6) α f 1 ( z + c ) − β f 2 ( z ) = δ 2 ,

    then it follows

    (4.7) δ 1 2 + δ 2 2 = 1 , δ 1 = β α η 1 + μ α η 2 .

    Solving equations (4.1) and (4.5), we have

    (4.8) f 1 ( z 1 , z 2 ) = η 1 μ − 1 μ e − μ λ z 1 + ϕ 1 ( z 2 ) , f 2 ( z 1 , z 2 ) = δ 1 μ − 1 μ e − μ λ z 1 + ϕ 2 ( z 2 ) ,

    where ϕ 1 ( z 2 ) , ϕ 2 ( z 2 ) are entire functions in z 2 . Substituting (4.8) in (4.2) and (4.6), we have

    α δ 1 μ − 1 μ e − μ λ ( z 1 + c 1 ) + ϕ 2 ( z 2 + c 2 ) − β η 1 μ − 1 μ e − μ λ z 1 + ϕ 1 ( z 2 ) = η 2 , α η 1 μ − 1 μ e − μ λ ( z 1 + c 1 ) + ϕ 1 ( z 2 + c 2 ) − β δ 1 μ − 1 μ e − μ λ z 1 + ϕ 2 ( z 2 ) = δ 2 ,

    which implies

    (4.9) α δ 1 − β η 1 = μ η 2 , e ϕ 2 ( z 2 + c 2 ) − ϕ 1 ( z 2 ) = β α e μ λ c 1 , α η 1 − β δ 1 = μ δ 2 , e ϕ 1 ( z 2 + c 2 ) − ϕ 2 ( z 2 ) = β α e μ λ c 1 .

    Thus, it yields that

    (4.10) ϕ 1 ( z 2 ) = γ z 2 + D 1 , ϕ 2 ( z 2 ) = γ z 2 + D 2 ,

    where γ , D 1 , D 2 are constants satisfying

    (4.11) e 2 ( D 1 − D 2 ) = 1 , γ = μ λ c 1 − log α β + k π i c 2 , k ∈ Z .

    Moreover, it follows from (4.7) and (4.9) that

    (4.12) η 1 = β α δ 1 + μ α δ 2 , ( δ 2 2 − η 2 2 ) [ μ 2 − ( α 2 − β 2 ) ] α 2 − β 2 μ 2 = 0 .

    1. If δ 2 = η 2 , it follows from (4.7) and (4.9) that

      (4.13) δ 1 = ± η 1 , ( η 1 − δ 1 ) 1 + β α = 0 .

      If δ 1 = η 1 and α = β , then δ 2 = η 2 = 0 and δ 1 = η 1 = ± 1 .

      If δ 1 = η 1 and α ≠ β , then it follows from (4.12) that η 2 = α − β μ η 1 . Substituting this into (4.3), we have

      δ 1 2 = η 1 2 = μ 2 μ 2 + ( α − β ) 2 .

      If δ 1 = − η 1 , then it follows from (4.13) that α = − β . Thus, we can deduce from (4.7) or (4.12) that δ 2 = η 2 = 0 , which implies that δ 1 = − η 1 = 1 or δ 1 = − η 1 = − 1 .

    2. If δ 2 = − η 2 , it follows from (4.2), (4.7), and (4.9) that

      (4.14) δ 1 = ± η 1 , ( η 1 + δ 1 ) 1 − β α = 0 .

      If δ 1 = − η 1 and α = − β , then δ 2 = η 2 = 0 . Then, it yields that δ 1 = − η 1 = 1 or δ 1 = − η 1 = − 1 .

      If δ 1 = − η 1 and α ≠ − β , then it follows from (4.12) that η 2 = − α + β μ η 1 . Substituting this into (4.3), we have

      δ 1 2 = η 1 2 = μ 2 μ 2 + ( α + β ) 2 .

      If δ 1 = η 1 , then it follows from (4.14) that α = β . Thus, we can deduce from (4.7) or (4.12) that δ 2 = η 2 = 0 , which implies that δ 1 = η 1 = ± 1 .

    3. If μ 2 − ( α 2 − β 2 ) = 0 , it follows from μ ≠ 0 that α ≠ ± β . Then we have

      (4.15) α δ 2 = − β η 2 ± ( η 2 2 − 1 ) ( β 2 − α 2 ) .

      Therefore, from (4.8), (4.10), (4.11) and (a), (b), (c), we obtain the conclusion (i) of Theorem 2.1.

    4. If μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 is a nonconstant, then it yields that α f 2 ( z + c ) − β f 1 ( z ) , μ f 2 + λ ∂ f 2 ∂ z 1 , and α f 1 ( z + c ) − β f 2 ( z ) are all nonconstant. Otherwise, if one of these terms is a constant, we can deduce that μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 is a constant. This is a contradiction. Thus, we can rewrite (1.6) as the form

      μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 + i ( α f 2 ( z + c ) − β f 1 ( z ) ) μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 − i ( α f 2 ( z + c ) − β f 1 ( z ) ) = 1 , μ f 2 ( z ) + λ ∂ f 2 ∂ z 1 + i ( α f 1 ( z + c ) − β f 2 ( z ) ) μ f 2 ( z ) + λ ∂ f 2 ∂ z 1 − i ( α f 1 ( z + c ) − β f 2 ( z ) ) = 1 .

      Since f 1 , f 2 are entire functions, it follows that μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 + i ( α f 2 ( z + c ) − β f 1 ( z ) ) , μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 − i ( α f 2 ( z + c ) − β f 1 ( z ) ) , μ f 2 ( z ) + λ ∂ f 2 ∂ z 1 − i ( α f 1 ( z + c ) − β f 2 ( z ) ) and μ f 2 ( z ) + λ ∂ f 2 ∂ z 1 + i ( α f 1 ( z + c ) − β f 2 ( z ) ) do not exist zeros and poles. By Lemmas 3.1 and 3.2, there exist two nonconstant polynomials p ( z ) , q ( z ) in C 2 such that

      μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 + i ( α f 2 ( z + c ) − β f 1 ( z ) ) = e p ( z ) , μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 − i ( α f 2 ( z + c ) − β f 1 ( z ) ) = e − p ( z ) , μ f 2 ( z ) + λ ∂ f 2 ∂ z 1 + i ( α f 1 ( z + c ) − β f 2 ( z ) ) = e q ( z ) , μ f 2 ( z ) + λ ∂ f 2 ∂ z 1 − i ( α f 1 ( z + c ) − β f 2 ( z ) ) = e − q ( z ) .

      The above equations lead to

      (4.16) μ f 1 ( z ) + λ ∂ f 1 ∂ z 1 = 1 2 ( e p + e − p ) ,

      (4.17) α f 2 ( z + c ) − β f 1 ( z ) = 1 2 i ( e p − e − p ) ,

      (4.18) μ f 2 ( z ) + λ ∂ f 2 ∂ z 1 = 1 2 ( e q + e − q ) ,

      (4.19) α f 1 ( z + c ) − β f 2 ( z ) = 1 2 i ( e q − e − q ) .

      In view of (4.17) and (4.18), we can deduce that

      (4.20) α μ f 2 ( z + c ) + β λ ∂ f 1 ∂ z 1 = α 2 ( e q ( z + c ) + e − q ( z + c ) ) − λ 2 i ∂ p ∂ z 1 ( e p ( z ) + e − p ( z ) ) .

      In view of (4.16) and (4.17), we have

      (4.21) α μ f 2 ( z + c ) + β λ ∂ f 1 ∂ z 1 = β i + μ 2 i e p ( z ) + β i − μ 2 i e − p ( z ) .

      By combining with (4.20) and (4.21), we have

      (4.22) λ ∂ p ∂ z 1 + β i + μ α i e q ( z + c ) + p ( z ) + λ ∂ p ∂ z 1 + β i − μ α i e q ( z + c ) − p ( z ) − e 2 q ( z + c ) ≡ 1 .

      Similar to the above argument, we can deduce from (4.16), (4.18), and (4.19) that

      (4.23) λ ∂ q ∂ z 1 + β i + μ α i e p ( z + c ) + q ( z ) + λ ∂ q ∂ z 1 + β i − μ α i e p ( z + c ) − q ( z ) − e 2 p ( z + c ) ≡ 1 .

      By Lemma 3.3, we can deduce from (4.22) and (4.23) that

      λ ∂ p ∂ z 1 + β i − μ α i e q ( z + c ) − p ( z ) ≡ 1 or λ ∂ p ∂ z 1 + β i + μ α i e q ( z + c ) + p ( z ) ≡ 1 ,

      and

      λ ∂ q ∂ z 1 + β i − μ α i e p ( z + c ) − q ( z ) ≡ 1 or λ ∂ q ∂ z 1 + β i + μ α i e p ( z + c ) + q ( z ) ≡ 1 .

      Now we will consider four cases as follows.

Case 1.

(4.24) λ ∂ p ∂ z 1 + β i − μ α i e q ( z + c ) − p ( z ) ≡ 1 , λ ∂ q ∂ z 1 + β i − μ α i e p ( z + c ) − q ( z ) ≡ 1 .

In view of (4.24), it follows that q ( z + c ) − p ( z ) = d 1 and p ( z + c ) − q ( z ) = d 2 , where d 1 , d 2 are constants in C . Thus, it yields that q ( z + 2 c ) − q ( z ) = d 1 + d 2 and p ( z + 2 c ) − p ( z ) = d 1 + d 2 . Since p , q are polynomials in C 2 , it follows that p ( z ) = L ( z ) + H ( c 2 z 1 − c 1 z 2 ) + b 1 and q ( z ) = L ( z ) + H ( c 2 z 1 − c 1 z 2 ) + b 2 , where L ( z ) is a linear form of L ( z ) = a 1 z 1 + a 2 z 2 , H ( s ) is a polynomial in s ≔ c 2 z 1 − c 1 z 2 , a 1 , a 2 , b 1 , b 2 are constants. Substituting p ( z ) , q ( z ) into (4.24), we have

(4.25) λ a 1 + λ c 2 H ′ + β i − μ α i e L ( c ) + b 2 − b 1 ≡ 1 , λ a 1 + λ c 2 H ′ + β i − μ α i e L ( c ) + b 1 − b 2 ≡ 1 .

By combining with λ ≠ 0 and c 2 ≠ 0 , it follows from (4.25) that deg s H ≤ 1 . Thus, we still write p ( z ) , q ( z ) as the forms of p ( z ) = L ( z ) + b 1 and q ( z ) = L ( z ) + b 2 . In view of (4.22)–(4.24), we have

(4.26) λ a 1 + β i + μ α i e − L ( c ) + b 1 − b 2 ≡ 1 , λ a 1 + β i + μ α i e − L ( c ) + b 2 − b 1 ≡ 1 , λ a 1 + β i − μ α i e L ( c ) + b 2 − b 1 ≡ 1 , λ a 1 + β i − μ α i e L ( c ) + b 1 − b 2 ≡ 1 .

Thus, we can deduce from (4.26) that

(4.27) ( λ a 1 + β i ) 2 = μ 2 − α 2 , e 2 ( b 1 − b 2 ) = 1 , e 2 L ( c ) = − α 2 ( λ a 1 + β i − μ ) 2 = ( λ a 1 + β i + μ ) 2 − α 2 .

If a 1 ≠ ± μ λ , solving equations (4.16) and (4.18), we have

(4.28) f 1 ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + b 1 2 ( λ a 1 + μ ) − e − a 1 z 1 − a 2 z 2 − b 1 2 ( λ a 1 − μ ) − ϑ 1 ( z 2 ) e − μ λ z 1 + ϕ 1 ( z 2 ) ,

(4.29) f 2 ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + b 2 2 ( λ a 1 + μ ) − e − a 1 z 1 − a 2 z 2 − b 2 2 ( λ a 1 − μ ) − ϑ 2 ( z 2 ) e − μ λ z 1 + ϕ 2 ( z 2 ) ,

where ϑ 1 ( z 2 ) , ϑ 2 ( z 2 ) are finite order entire functions and ϕ 1 ( z 2 ) , ϕ 2 ( z 2 ) are polynomials in z 2 . Substituting (4.28) and (4.29) in (4.17) and (4.19), and combining with (4.26) and (4.27), by Lemma 3.4, we have

(4.30) f 1 ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + b 1 2 ( λ a 1 + μ ) − e − a 1 z 1 − a 2 z 2 − b 1 2 ( λ a 1 − μ ) − ϑ ( z 2 ) e − μ λ z 1 + γ z 2 + D 1 ,

(4.31) f 2 ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + b 2 2 ( λ a 1 + μ ) − e − a 1 z 1 − a 2 z 2 − b 2 2 ( λ a 1 − μ ) − ϑ ( z 2 + c 2 ) e − μ λ z 1 + γ z 2 + D 2 ,

where ϑ ( z 2 ) is a finite order period entire function with period 2 c 2 , and γ , D 1 , D 2 satisfy (4.11) and (4.27).

If a 1 = μ λ , solving equations (4.16) and (4.18), similar to the argument as in case a 1 ≠ μ λ , we have

(4.32) f 1 ( z 1 , z 2 ) = 1 4 μ e a 1 z 1 + a 2 z 2 + b 1 + z 1 2 λ e − a 1 z 1 − a 2 z 2 − b 1 − ϑ ( z 2 ) e − μ λ z 1 + γ z 2 + D 1 ,

(4.33) f 2 ( z 1 , z 2 ) = 1 4 μ e a 1 z 1 + a 2 z 2 + b 2 + z 1 2 λ e − a 1 z 1 − a 2 z 2 − b 2 − ϑ ( z 2 + c 2 ) e − μ λ z 1 + γ z 2 + D 2 ,

where ϑ ( z 2 ) is a finite order period entire function with period 2 c 2 , and γ , D 1 , D 2 satisfy (4.11) and (4.27). Substituting (4.32) and (4.33) in (4.17) and (4.19), and combining with (4.26) and (4.27), we have β c 1 = λ i .

If a 1 = − μ λ , solving equations (4.16) and (4.18), we have

(4.34) f 1 ( z 1 , z 2 ) = 1 4 μ e − a 1 z 1 − a 2 z 2 − b 1 + z 1 2 λ e a 1 z 1 + a 2 z 2 + b 1 − ϑ ( z 2 ) e − μ λ z 1 + γ z 2 + D 1 ,

(4.35) f 2 ( z 1 , z 2 ) = 1 4 μ e − a 1 z 1 − a 2 z 2 − b 2 + z 1 2 λ e a 1 z 1 + a 2 z 2 + b 2 − ϑ ( z 2 + c 2 ) e − μ λ z 1 + γ z 2 + D 2 ,

where ϑ ( z 2 ) is a finite order period entire function with period 2 c 2 , and γ , D 1 , D 2 satisfy (4.11) and (4.27). Similar to the above argument, we have β c 1 = − λ i .

Case 2.

λ ∂ p ∂ z 1 + β i − μ α i e q ( z + c ) − p ( z ) ≡ 1 , λ ∂ q ∂ z 1 + β i + μ α i e p ( z + c ) + q ( z ) ≡ 1 .

Thus, it follows that q ( z + c ) − p ( z ) = d 1 and p ( z + c ) + q ( z ) = d 2 where d 1 , d 2 are constants. Hence, we have q ( z + 2 c ) + q ( z ) = d 1 + d 2 , which is a contradiction with the assumption of q ( z ) being nonconstant polynomial in C 2 .

Case 3.

λ ∂ p ∂ z 1 + β i + μ α i e q ( z + c ) + p ( z ) ≡ 1 , λ ∂ q ∂ z 1 + β i − μ α i e p ( z + c ) − q ( z ) ≡ 1 .

Thus, it follows that q ( z + c ) + p ( z ) = d 1 and p ( z + c ) − q ( z ) = d 2 where d 1 , d 2 are constants. Hence, we have p ( z + 2 c ) + p ( z ) = d 1 + d 2 , which is a contradiction with the assumption of p ( z ) being nonconstant polynomial in C 2 .

Case 4.

(4.36) λ ∂ p ∂ z 1 + β i + μ α i e q ( z + c ) + p ( z ) ≡ 1 , λ ∂ q ∂ z 1 + β i + μ α i e p ( z + c ) + q ( z ) ≡ 1 .

In view of (4.36), it follows that q ( z + c ) + p ( z ) = d 1 and p ( z + c ) + q ( z ) = d 2 , where d 1 , d 2 are constants in C . Thus, it yields that q ( z + 2 c ) − q ( z ) = d 1 − d 2 and p ( z + 2 c ) − p ( z ) = d 1 − d 2 . Since p , q are polynomials in C 2 , it follows that p ( z ) = L ( z ) + H ( c 2 z 1 − c 1 z 2 ) + b 1 and q ( z ) = − L ( z ) − H ( c 2 z 1 − c 1 z 2 ) + b 2 , where L ( z ) is a linear form of L ( z ) = a 1 z 1 + a 2 z 2 , H ( s ) is a polynomial in s ≔ c 2 z 1 − c 1 z 2 , a 1 , a 2 , b 1 , b 2 are constants. Similar to the argument as in Case 1, we can obtain that p ( z ) = L ( z ) + b 1 = a 1 z 1 + a 2 z 2 + b 1 and q ( z ) = − L ( z ) + b 2 = − a 1 z 1 − a 2 z 2 + b 2 . In view of (4.22), (4.23), and (4.36), it follows

(4.37) λ a 1 + β i + μ α i e − L ( c ) + b 2 + b 1 ≡ 1 , − λ a 1 + β i + μ α i e L ( c ) + b 1 + b 2 ≡ 1 , λ a 1 + β i − μ α i e L ( c ) − b 2 − b 1 ≡ 1 , − λ a 1 + β i − μ α i e − L ( c ) − b 1 − b 2 ≡ 1 .

Thus, it leads to

− λ a 1 + β i + μ α i − λ a 1 + β i − μ α i = λ a 1 + β i + μ α i λ a 1 + β i − μ α i .

By combining with α ≠ 0 , β ≠ 0 , and λ ≠ 0 , we have a 1 = 0 . Then, p ( z ) = a 2 z 2 + b 1 and q ( z ) = − a 2 z 2 + b 2 . In view of (4.37), it follows

(4.38) μ 2 = α 2 − β 2 , e 2 a 2 c 2 ≡ 1 , e 2 ( b 1 + b 2 ) = − α 2 ( β i + μ ) 2 = ( β i − μ ) 2 − α 2 .

Solving equations (4.16) and (4.18), we have

(4.39) f 1 ( z 1 , z 2 ) = 1 2 μ ( e a 2 z 2 + b 1 + e − a 2 z 2 − b 1 ) − ϑ 1 ( z 2 ) e − μ λ z 1 + ϕ 1 ( z 2 ) ,

(4.40) f 2 ( z 1 , z 2 ) = 1 2 μ ( e − a 2 z 2 + b 2 + e a 2 z 2 − b 2 ) − ϑ 2 ( z 2 ) e − μ λ z 1 + ϕ 2 ( z 2 ) ,

where ϑ 1 ( z 2 ) , ϑ 2 ( z 2 ) are finite order entire functions and ϕ 1 ( z 2 ) , ϕ 2 ( z 2 ) are polynomials in z 2 . Similar to the above argument in Case 1, it follows from (4.39) and (4.40) that

f 1 ( z 1 , z 2 ) = 1 2 μ ( e a 2 z 2 + b 1 + e − a 2 z 2 − b 1 ) − ϑ ( z 2 ) e − μ λ z 1 + γ z 2 + D 1 , f 2 ( z 1 , z 2 ) = 1 2 μ ( e − a 2 z 2 + b 2 + e a 2 z 2 − b 2 ) − ϑ ( z 2 + c 2 ) e − μ λ z 1 + γ z 2 + D 2 ,

where ϑ ( z 2 ) is a finite order period entire function with period 2 c 2 , and γ , D 1 , D 2 satisfy (4.11) and (4.27). In fact, we can see that the forms of solutions are included in case that a 1 ≠ ± μ λ in Case 1.

Therefore, this completes the proof of Theorem 2.1.□

5 The proof of Theorem 2.2

Proof

Let ( f 1 , f 2 ) be a pair of transcendental entire solutions of finite order for system (1.7). Thus, we will consider the following two cases.

  1. If μ f 1 ( z ) + λ 1 ∂ f 1 ∂ z 1 + λ 2 ∂ f 1 ∂ z 2 is a constant. Denote

    (5.1) μ f 1 ( z ) + λ 1 ∂ f 1 ∂ z 1 + λ 2 ∂ f 1 ∂ z 2 = η 1 .

In view of (1.7), it follows that

(5.2) α f 2 ( z + c ) − β f 1 ( z ) = η 2 ,

where η 2 is a constant satisfying (4.3).

From (5.1) and (5.2), we have

(5.3) μ f 2 + λ 1 ∂ f 2 ∂ z 1 + λ 2 ∂ f 2 ∂ z 2 = β α μ f 1 ( z − c ) + λ 1 ∂ f 1 ( z − c ) ∂ z 1 + λ 2 ∂ f 1 ( z − c ) ∂ z 2 + μ η 2 α = η 1 β α + μ η 2 α .

This shows that μ f 2 + λ 1 ∂ f 2 ∂ z 1 + λ 2 ∂ f 2 ∂ z 2 is a constant. Let

(5.4) μ f 2 + λ 1 ∂ f 2 ∂ z 1 + λ 2 ∂ f 2 ∂ z 2 = δ 1 ,

then α f 1 ( z + c ) − β f 2 ( z ) is a constant. Denote

(5.5) α f 1 ( z + c ) − β f 2 ( z ) = δ 2 ,

then, we have (4.7). The characteristic equations of (5.1) are

d z 1 d t = λ 1 , d z 2 d t = λ 2 , d f 1 d t = η 1 − μ f 1 .

Using the initial conditions: z 1 = 0 , z 2 = s 1 , and f 1 = f 1 ( 0 , s 1 ) ≔ ψ ( s 1 ) with a parameter s , we obtain the following parametric representation for the solutions of the characteristic equations: z 1 = λ 1 t , z 2 = λ 2 t + s 1 , and

(5.6) f 1 ( z 1 , z 2 ) = η 1 μ − 1 μ e − μ λ 1 z 1 + φ 1 ( s 1 ) ,

where φ 1 ( s 1 ) is an entire function in s 1 ≔ λ 2 z 1 − λ 1 z 2 . Similarly, solving equation (5.4), we have

(5.7) f 2 ( z 1 , z 2 ) = δ 1 μ − 1 μ e − μ λ 1 z 1 + φ 2 ( s 1 ) ,

where φ 2 ( s 1 ) is an entire function in s 1 ≔ λ 2 z 1 − λ 1 z 2 . Substituting (5.6) and (5.7) in (5.2) and (5.5), we have

α δ 1 μ − 1 μ e − μ λ ( z 1 + c 1 ) + φ 2 ( s 1 + s 0 ) − β η 1 μ − 1 μ e − μ λ z 1 + φ 1 ( s 1 ) = η 2 , α η 1 μ − 1 μ e − μ λ ( z 1 + c 1 ) + φ 1 ( s 1 + s 0 ) − β δ 1 μ − 1 μ e − μ λ z 1 + φ 2 ( s 1 ) = δ 2 ,

which implies

(5.8) α δ 1 − β η 1 = μ η 2 , e φ 2 ( s 1 + s 0 ) − φ 1 ( s 1 ) = β α e μ λ c 1 , α η 1 − β δ 1 = μ δ 2 , e φ 1 ( s 1 + s 0 ) − φ 2 ( s 1 ) = β α e μ λ c 1 .

Thus, it yields that

(5.9) φ 1 ( s 1 ) = γ s 1 + D 1 , φ 2 ( s 1 ) = γ s 1 + D 2 ,

where D 1 , D 2 are constants and satisfying

(5.10) e 2 ( D 1 − D 2 ) = 1 , γ = μ λ c 1 − log α β + k π i λ 2 c 1 − λ 1 c 2 , k ∈ Z .

Moreover, from (4.7) and (5.8), we have (4.12). Thus, from (5.6), (5.7), and (5.9), by using the same argument as in the proof of Theorem 2.1 (i), we can obtain the conclusions of Theorem 2.2 (i).

  1. If μ f 1 ( z ) + λ 1 ∂ f 1 ∂ z 1 + λ 2 ∂ f 1 ∂ z 2 is a nonconstant, then it yields that α f 2 ( z + c ) − β f 1 ( z ) , μ f 2 + λ 1 ∂ f 2 ∂ z 1 + λ 2 ∂ f 2 ∂ z 2 and α f 1 ( z + c ) − β f 2 ( z ) are all nonconstant. Otherwise, if one of these terms is a constant, we can deduce that μ f 1 ( z ) + λ 1 ∂ f 1 ∂ z 1 + λ 2 ∂ f 1 ∂ z 2 is a constant. This is a contradiction. Thus, similar to the argument as in the proof of Theorem 2.1 (ii), there exists two nonconstant polynomials p ( z ) , q ( z ) in C 2 such that

    (5.11) μ f 1 ( z ) + λ 1 ∂ f 1 ∂ z 1 + λ 2 ∂ f 1 ∂ z 2 = 1 2 ( e p + e − p ) ,

    (5.12) α f 2 ( z + c ) − β f 1 ( z ) = 1 2 i ( e p − e − p ) ,

    (5.13) μ f 2 ( z ) + λ 1 ∂ f 2 ∂ z 1 + λ 2 ∂ f 2 ∂ z 2 = 1 2 ( e q + e − q ) ,

    (5.14) α f 1 ( z + c ) − β f 2 ( z ) = 1 2 i ( e q − e − q ) .

In view of (5.11) and (5.14), we have

(5.15) λ 1 ∂ p ∂ z 1 + λ 2 ∂ p ∂ z 2 + β i + μ α i e q ( z + c ) + p ( z ) + λ 1 ∂ p ∂ z 1 + λ 2 ∂ p ∂ z 2 + β i − μ α i e q ( z + c ) − p ( z ) − e 2 q ( z + c ) ≡ 1 .

(5.16) λ 1 ∂ q ∂ z 1 + λ 2 ∂ q ∂ z 2 + β i + μ α i e p ( z + c ) + q ( z ) + λ 1 ∂ q ∂ z 1 + λ 2 ∂ q ∂ z 2 + β i − μ α i e p ( z + c ) − q ( z ) − e 2 p ( z + c ) ≡ 1 .

By Lemma 3.3, we can deduce from (5.15) and (5.16) that

λ 1 ∂ p ∂ z 1 + λ 2 ∂ p ∂ z 2 + β i − μ α i e q ( z + c ) − p ( z ) ≡ 1 or λ 1 ∂ p ∂ z 1 + λ 2 ∂ p ∂ z 2 + β i + μ α i e q ( z + c ) + p ( z ) ≡ 1 ,

and

λ 1 ∂ q ∂ z 1 + λ 2 ∂ q ∂ z 2 + β i − μ α i e p ( z + c ) − q ( z ) ≡ 1 o r λ 1 ∂ q ∂ z 1 + λ 2 ∂ q ∂ z 2 + β i + μ α i e p ( z + c ) + q ( z ) ≡ 1 .

Now we will consider four cases as follows.

Case 1.

(5.17) λ 1 ∂ p ∂ z 1 + λ 2 ∂ p ∂ z 2 + β i − μ α i e q ( z + c ) − p ( z ) ≡ 1 , λ 1 ∂ q ∂ z 1 + λ 2 ∂ q ∂ z 2 + β i − μ α i e p ( z + c ) − q ( z ) ≡ 1 .

In view of (5.17), it follows that q ( z + c ) − p ( z ) = d 1 and p ( z + c ) − q ( z ) = d 2 , where d 1 , d 2 are constants in C . Thus, it yields that q ( z + 2 c ) − q ( z ) = d 1 + d 2 and p ( z + 2 c ) − p ( z ) = d 1 + d 2 . Since p , q are polynomials in C 2 , it follows that p ( z ) = L ( z ) + H ( c 2 z 1 − c 1 z 2 ) + b 1 and q ( z ) = L ( z ) + H ( c 2 z 1 − c 1 z 2 ) + b 2 , where L ( z ) is a linear form of L ( z ) = a 1 z 1 + a 2 z 2 , H ( s ) is a polynomial in s and a 1 , a 2 , b 1 , b 2 are constants. Substituting p ( z ) , q ( z ) in (5.17), we have

(5.18) λ 1 a 1 + λ 2 a 2 + ( λ 1 c 2 − λ 2 c 1 ) H ′ + β i − μ α i e L ( c ) + b 2 − b 1 ≡ 1 ,

(5.19) λ 1 a 1 + λ 2 a 2 + ( λ 1 c 2 − λ 2 c 1 ) H ′ + β i − μ α i e L ( c ) + b 1 − b 2 ≡ 1 .

By combining with λ 1 c 2 − λ 2 c 1 ≠ 0 , it follows from (5.18) and (5.19) that deg s H ≤ 1 . Thus, we still write p ( z ) , q ( z ) as the forms of p ( z ) = L ( z ) + b 1 and q ( z ) = L ( z ) + b 2 . In view of (5.15)–(5.17), we have

(5.20) λ 1 a 1 + λ 2 a 2 + β i + μ α i e − L ( c ) + b 1 − b 2 ≡ 1 , λ 1 a 1 + λ 2 a 2 + β i + μ α i e − L ( c ) + b 2 − b 1 ≡ 1 , λ 1 a 1 + λ 2 a 2 + β i − μ α i e L ( c ) + b 2 − b 1 ≡ 1 , λ 1 a 1 + λ 2 a 2 + β i − μ α i e L ( c ) + b 1 − b 2 ≡ 1 .

Thus, we can deduce from (5.20) that

(5.21) ( λ 1 a 1 + λ 2 a 2 + β i ) 2 = μ 2 − α 2 , e 2 ( b 1 − b 2 ) = 1 ,

and

(5.22) e 2 L ( c ) = − α 2 ( λ 1 a 1 + λ 2 a 2 + β i − μ ) 2 = ( λ 1 a 1 + λ 2 a 2 + β i + μ ) 2 − α 2 .

If λ 1 a 1 + λ 2 a 2 ≠ ± μ , solving equations (5.11) and (5.13), we have

(5.23) f 1 ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + b 1 2 ( λ 1 a 1 + λ 2 a 2 + μ ) − e − a 1 z 1 − a 2 z 2 − b 1 2 ( λ 1 a 1 + λ 2 a 2 − μ ) − ϑ 1 ( s 1 ) e − μ λ z 1 + ϕ 1 ( s 1 ) ,

(5.24) f 2 ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + b 2 2 ( λ 1 a 1 + λ 2 a 2 + μ ) − e − a 1 z 1 − a 2 z 2 − b 2 2 ( λ 1 a 1 + λ 2 a 2 − μ ) − ϑ 2 ( s 1 ) e − μ λ z 1 + ϕ 2 ( s 1 ) ,

where ϑ 1 ( s 1 ) , ϑ 2 ( s 1 ) are finite order entire functions and ϕ 1 ( s 1 ) , ϕ 2 ( s 1 ) are polynomials in s 1 . Substituting (5.23) and (5.24) in (5.12) and (5.14), and combining with (5.20)–(5.22), by Lemma 3.4 we have

(5.25) f 1 ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + b 1 2 ( λ 1 a 1 + λ 2 a 2 + μ ) − e − a 1 z 1 − a 2 z 2 − b 1 2 ( λ 1 a 1 + λ 2 a 2 − μ ) − ϑ ( s 1 ) e − μ λ 1 z 1 + γ s 1 + D 1 ,

(5.26) f 2 ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + b 2 2 ( λ 1 a 1 + λ 2 a 2 + μ ) − e − a 1 z 1 − a 2 z 2 − b 2 2 ( λ 1 a 1 + λ 2 a 2 − μ ) − ϑ ( s 1 + s 0 ) e − μ λ 1 z 1 + γ s 1 + D 2 ,

where ϑ ( s 1 ) is a finite order period entire function with period 2 s 0 , and a 1 , a 2 , b 1 , b 2 , γ , D 1 , D 2 satisfy (5.10), (5.21), and (5.22).

If λ 1 a 1 + λ 2 a 2 = μ , solving equations (5.11) and (5.13), similar to the argument as in case λ 1 a 1 + λ 2 a 2 ≠ μ , we have

(5.27) f 1 ( z 1 , z 2 ) = 1 4 μ e a 1 z 1 + a 2 z 2 + b 1 + z 1 2 λ 1 e − a 1 z 1 − a 2 z 2 − b 1 − ϑ ( s 1 ) e − μ λ 1 z 1 + γ s 1 + D 1 ,

(5.28) f 2 ( z 1 , z 2 ) = 1 4 μ e a 1 z 1 + a 2 z 2 + b 2 − z 1 2 λ 1 e − a 1 z 1 − a 2 z 2 − b 2 − ϑ ( s 1 + s 0 ) e − μ λ 1 z 1 + γ s 1 + D 2 ,

where ϑ ( s 1 ) is a finite order period entire function with period 2 s 0 , and γ , D 1 , D 2 satisfy (5.10), (5.21), and (5.22). Substituting (5.27) and (5.28) in (5.12) and (5.14), and combining with (5.20)–(5.22), we have β c 1 = λ 1 i .

If λ 1 a 1 + λ 2 a 2 = − μ , solving equations (5.11) and (5.13), we have

(5.29) f 1 ( z 1 , z 2 ) = 1 4 μ e − a 1 z 1 − a 2 z 2 − b 1 + z 1 2 λ 1 e a 1 z 1 + a 2 z 2 + b 1 − ϑ ( s 1 ) e − μ λ 1 z 1 + γ s 1 + D 1 ,

(5.30) f 2 ( z 1 , z 2 ) = 1 4 μ e − a 1 z 1 − a 2 z 2 − b 2 + z 1 2 λ 1 e a 1 z 1 + a 2 z 2 + b 2 − ϑ ( s 1 + s 0 ) e − μ λ 1 z 1 + γ s 1 + D 2 ,

where ϑ ( s 1 ) is a finite order period entire function with period 2 s 0 , and γ , D 1 , D 2 satisfy (5.10), (5.21), and (5.22). Similar to the above argument, we have β c 1 = − λ 1 i .

Case 2.

λ 1 ∂ p ∂ z 1 + λ 2 ∂ p ∂ z 2 + β i − μ α i e q ( z + c ) − p ( z ) ≡ 1 , λ 1 ∂ q ∂ z 1 + λ 2 ∂ q ∂ z 2 + β i + μ α i e p ( z + c ) + q ( z ) ≡ 1 .

Thus, it follows that q ( z + c ) − p ( z ) = d 1 and p ( z + c ) + q ( z ) = d 2 where d 1 , d 2 are constants. Hence, we have q ( z + 2 c ) + q ( z ) = d 1 + d 2 , which is a contradiction with the assumption of q ( z ) being nonconstant polynomial in C 2 .

Case 3.

λ 1 ∂ p ∂ z 1 + λ 2 ∂ p ∂ z 2 + β i + μ α i e q ( z + c ) + p ( z ) ≡ 1 , λ 1 ∂ q ∂ z 1 + λ 2 ∂ q ∂ z 2 + β i − μ α i e p ( z + c ) − q ( z ) ≡ 1 .

Thus, it follows that q ( z + c ) + p ( z ) = d 1 and p ( z + c ) − q ( z ) = d 2 where d 1 , d 2 are constants. Hence, we have p ( z + 2 c ) + p ( z ) = d 1 + d 2 , which is a contradiction with the assumption of p ( z ) being nonconstant polynomial in C 2 .

Case 4.

(5.31) λ 1 ∂ p ∂ z 1 + λ 2 ∂ p ∂ z 2 + β i + μ α i e q ( z + c ) + p ( z ) ≡ 1 , λ 1 ∂ q ∂ z 1 + λ 2 ∂ q ∂ z 2 + β i + μ α i e p ( z + c ) + q ( z ) ≡ 1 .

In view of (5.31), it follows that q ( z + c ) + p ( z ) = d 1 and p ( z + c ) + q ( z ) = d 2 , where d 1 , d 2 are constants in C . Thus, it yields that q ( z + 2 c ) − q ( z ) = d 1 − d 2 and p ( z + 2 c ) − p ( z ) = d 1 − d 2 . Since p , q are polynomials in C 2 , it follows that p ( z ) = L ( z ) + H ( c 2 z 1 − c 1 z 2 ) + b 1 and q ( z ) = − L ( z ) − H ( c 2 z 1 − c 1 z 2 ) + b 2 , where L ( z ) is a linear form of L ( z ) = a 1 z 1 + a 2 z 2 , H ( s ) is a polynomial in s and a 1 , a 2 , b 1 , b 2 are constants. Similar to the argument as in Case 1, we can obtain that p ( z ) = L ( z ) + b 1 = a 1 z 1 + a 2 z 2 + b 1 and q ( z ) = − L ( z ) + b 2 = − a 1 z 1 − a 2 z 2 + b 1 .

By using the same argument as in Case 4 of Theorem 2.1, we can obtain that the forms of solutions are included in case that λ 1 a 1 + λ 2 a 2 ≠ ± μ in Case 1 of Theorem 2.2.

Therefore, this completes the proof of Theorem 2.2.□

Acknowledgments

The authors are very thankful to referees for their valuable comments which improved the presentation of the paper.

  1. Funding information: This work was supported by the National Natural Science Foundation of China (12161074), the Talent Introduction Research Foundation of Suqian University (106-CK00042/028), and the Suqian Sci & Tech Program (Grant No. K202009).

  2. Author contributions: Conceptualization, H. Y. Xu; writing-original draft preparation, H. Y. Xu, Y. H. Xu and X. L. Liu; writing – review and editing, H. Y. Xu, Y. H. Xu and X. L. Liu; and funding acquisition, H. Y. Xu and Y. H. Xu.

  3. Conflict of interest: The authors declare that none of the authors have any competing interests in the manuscript.

  4. Ethical approval: The conducted research is not related to either human or animal use.

  5. Data availability statement: All data generated or analyzed during this study are included in this published article.

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Received: 2022-10-16
Revised: 2023-03-14
Accepted: 2023-05-06
Published Online: 2023-08-10

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/dema-2022-0241
Keywords for this article
partial differential difference equation; entire solution; Nevanlinna theory
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