Startseite Characterizations of transcendental entire solutions of trinomial partial differential-difference equations in ℂ2#
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Characterizations of transcendental entire solutions of trinomial partial differential-difference equations in ℂ2#

  • Hong Yan Xu und Goutam Haldar EMAIL logo
Veröffentlicht/Copyright: 21. November 2024
Veröffentlichen auch Sie bei De Gruyter Brill

Abstract

This study is devoted to exploring the existence and the precise form of finite-order transcendental entire solutions of second-order trinomial partial differential-difference equations

L ( f ) 2 + 2 h L ( f ) f ( z 1 + c 1 , z 2 + c 2 ) + f ( z 1 + c 1 , z 2 + c 2 ) 2 = e g ( z 1 , z 2 )

and

L ˜ ( f ) 2 + 2 h L ˜ ( f ) ( f ( z 1 + c 1 , z 2 + c 2 ) f ( z 1 , z 2 ) ) + ( f ( z 1 + c 1 , z 2 + c 2 ) f ( z 1 , z 2 ) ) 2 = e g ( z 1 , z 2 ) ,

where L ( f ) and L ˜ ( f ) are defined in (2.1) and (2.2), respectively, and g ( z ) is a polynomial in C 2 . Our results are the extensions of some of the previous results of Liu et al. Also, we exhibit a series of examples to explain that the forms of transcendental entire solutions of finite-order in our results are precise.

MSC 2010: 30D35; 35M30; 32W50; 39A45

1 Introduction

Fermat’s last theorem [1] says that the Fermat equation x m + y m = 1 does not admit nontrivial solutions in rational numbers when m 3 and does admit nontrivial rational solutions when m = 2 . For a positive integer m , the equation

(1.1) f m + g m = 1

is known as Fermat-type equation over function fields. With the help of the Nevanlinna theory [2,3], Montel [4], Iyer [5], and Gross [6] studied the existence and form of the solutions of the functional equation (1.1) and pointed out the following:

  1. For m = 2 , the entire solutions of (1.1) are f ( z ) = cos ( ξ ( z ) ) and g ( z ) = sin ( ξ ( z ) ) , where ξ is a non-constant entire function.

  2. For m > 2 , there are no non-constant entire solutions of (1.1).

  3. For m = 2 , the meromorphic solutions of (1.1) are of the form

    f ( z ) = 2 ϕ ( z ) 1 ϕ 2 ( z ) and g ( z ) = 1 ϕ 2 ( z ) 1 + ϕ 2 ( z ) ,

    where ϕ is a non-constant meromorphic function.

  4. For m = 3 , the meromorphic solutions of (1.1) are of the form

    f ( z ) = 1 2 ( h ) 1 + ( h ) 3 and g ( z ) = ω 2 ( h ) 1 ( h ) 3 ,

    where ω 3 = 1 and satisfies ( ) 2 = 4 3 1 .

  5. For m > 3 , there are no non-constant meromorphic solutions.

In 2004, Yang and Li [7] investigated (1.1) by replacing g with f when m = 2 , and proved that the transcendental entire solution of f ( z ) 2 + f ( z ) 2 = 1 has the form f ( z ) = A e α z 2 + e α z 2 A , where A and α are non-zero complex constants.

The advent of the difference analog lemma of the logarithmic derivative [8,9] expedites the research activity to characterize the entire or meromorphic solutions of Fermat-type difference and differential-difference equations [1013]. In 2012, Liu et al. [14] proved that the transcendental entire solutions with finite-order of the Fermat-type difference equation f ( z ) 2 + f ( z + c ) 2 = 1 must satisfy f ( z ) = sin ( A z + B ) , where B is a constant and A = ( 4 k + 1 ) π 2 c , where k is an integer. In 2016, Liu and Yang [15] studied the existence and the form of solutions of some quadratic trinomial functional equations and obtained some important results as follows. The equation f ( z ) 2 + 2 α f ( z ) f ( z ) + f 2 ( z ) = 1 has no transcendental meromorphic solutions, and the finite-order transcendental entire solution of f ( z ) 2 + 2 α f ( z ) f ( z + c ) + f ( z + c ) 2 = 1 must be of order equal to one, where in both the equations, α 0 , ± 1 .

The study of several characteristics of the solutions to partial differential equations in several complex variables [1624] is an important topic. The Fermat-type equation like appears in particle mechanics; in particular, it appears as the Lagrange function in the Lagrangian functional describing the “action” of a system. As far as the author’s knowledge is concerned, Saleebly, in 1999, first started investigation about the existence and form of entire and meromorphic solutions of Fermat-type partial differential equations [25,26]. Most noticeably, Khavinson [27] proved that any entire solution of the partial differential equation f z 1 2 + f z 2 2 = 1 must be linear, i.e., f ( z 1 , z 2 ) = a z 1 + b z 2 + c , where a , b , c C , and a 2 + b 2 = 1 . Later, Li [28,29] investigated the partial differential equations with more general forms such as f z 1 2 + f z 2 2 = p and f z 1 2 + f z 2 2 = e q , where p and q are the polynomials in C 2 .

Hereinafter, we denote by z + w = ( z 1 + w 1 , z 2 + w 2 ) for any z = ( z 1 , z 2 ) , w = ( w 1 , w 2 ) C 2 .

Recently, Xu and Cao [30] extended several results from one complex variable to several complex variables. In fact, they considered two Fermat-type equations f ( z ) 2 + f ( z + c ) 2 = 1 and f ( z + c ) 2 + f z 1 2 = 1 and proved that any finite-order transcendental entire solution f : C n P 1 ( C ) has the form f ( z ) = cos ( L ( z ) + B ) , where L is a linear function of the form L ( z ) = a 1 z 1 + + a n z n on C n such that L ( c ) = π 2 2 k π ( k Z ), and B C and f ( z ) = cos ( A 1 z 1 + A 2 z 2 + Constant ) , where A 1 c 1 + A 2 c 2 = π 2 2 k π , k Z , respectively.

In 2022, Xu et al. [31] explored the precise form of entire and meromorphic solutions of the following partial differential difference equations:

(1.2) α f z 1 + β f z 2 2 + f ( z 1 + c 1 , z 2 + c 2 ) 2 = e g ( z 1 , z 2 )

and

(1.3) α f z 1 + β f z 2 2 + ( f ( z 1 + c 1 , z 2 + c 2 ) f ( z 1 , z 2 ) ) 2 = e g ( z 1 , z 2 ) ,

where g ( z ) is a polynomial in C 2 and α and β are the constants in C . For detail study, we insist the readers to go through [31].

As far as we know, it was Saleeby [32], who first considered another quadratic functional equation

(1.4) f 2 + 2 α f g + g 2 = 1 ,

and investigated the existence and form of transcendental entire and meromorphic solutions in C n . We recall the result.

Theorem A

[32] The transcendental entire solutions of (1.4) must be of the form f ( z ) = 1 2 cos ( h ( z ) ) 1 + α + sin ( h ( z ) ) 1 α and 1 2 cos ( h ( z ) ) 1 + α sin ( h ( z ) ) 1 α , where h is entire C n . The meromorphic solutions of (1.4) must be of the form f ( z ) = α 1 α 2 β ( z ) 2 ( α 1 α 2 ) β ( z ) and g ( z ) = 1 β ( z ) 2 ( α 1 α 2 ) β ( z ) , where β ( z ) is meromorphic in C n and α 1 = α + α 2 1 , α 2 = α α 2 1 , and α 2 0 , 1 .

In 2021, Li and Xu [33] considered the quadratic trinomial partial differential difference equations

(1.5) f ( z 1 + c 1 , z 2 + c 2 ) 2 + 2 α f ( z 1 + c 1 , z 2 + c 2 ) f z 1 + f z 1 2 = e g ( z 1 + c 1 , z 2 + c 2 )

and obtained the following result.

Theorem B

[33] Let α 2 0 , 1 , c 2 0 , and g be a non-constant polynomial in C 2 , and not the form of ϕ ( z 2 ) . If (1.5) admits a finite-order transcendental entire solution f, then g must be of the form g ( z 1 , z 2 ) = a 1 z 1 + a 2 z 2 + b , where a 1 ( 0 ) , a 2 , b 2 c . Furthermore, f ( z ) must satisfy one of the following cases:

  1. f ( z 1 , z 2 ) = 2 a 1 ( A 2 ξ + A 1 ξ 1 ) e 1 2 ( a 1 z 1 + a 2 z 2 + b ) ,

    where ξ ( 0 ) , a 1 , a 2 , b , c 1 , c 2 , A 1 , A 2 C satisfying

    e 1 2 ( a 1 c 1 + a 2 c 2 ) = a 1 ( A 1 ξ + A 2 ξ 1 ) 2 ( A 2 ξ + A 1 ξ 1 ) .

  2. f ( z 1 , z 2 ) = 1 2 A 2 a 11 e a 11 z 1 + a 12 z 2 + b 1 + A 1 a 21 e a 21 z 1 + a 22 z 2 + b 2 ,

    where a j ( 0 ) , b j C , ( j = 1 , 2 ) satisfy a 11 z 1 + a 12 z 2 a 21 z 1 + a 22 z 2 ,

    g ( z 1 , z 2 ) = ( a 11 + a 21 ) z 1 + ( a 12 + a 22 ) z 2 + b 1 + b 2 , and

    e a 11 c 1 + a 12 c 2 = A 2 A 1 a 11 , e a 21 c 1 + a 2 c 2 = A 1 A 2 a 21 , e a 1 c 1 + a 2 c 2 = a 11 a 21 .

2 Main results

Inspired by Theorems A and B, and utilizing difference analogs of the Nevanlinna theory of several complex variables [34,35], we investigate the existence and forms of transcendental entire solutions of the following trinomial second-order partial differential-difference equations:

(2.1) L ( f ) 2 + 2 h L ( f ) f ( z 1 + c 1 , z 2 + c 2 ) + f ( z 1 + c 1 , z 2 + c 2 ) 2 = e g ( z 1 , z 1 )

and

(2.2) L ˜ ( f ) 2 + 2 h L ˜ ( f ) f ( z 1 + c 1 , z 2 + c 2 ) + ( f ( z 1 + c 1 , z 2 + c 2 ) f ( z 1 , z 2 ) ) 2 = e g ( z 1 , z 2 ) ,

where L ˜ ( f ) = γ 2 f ( z 1 , z 2 ) z 1 2 + δ 2 f ( z 1 , z 2 ) z 2 2 + η 2 f ( z 1 , z 2 ) z 1 z 2 , and L ( f ) = L ˜ + α f ( z 1 , z 2 ) z 1 + β f ( z 1 , z 2 ) z 2 , α , β , γ , δ , and η are the constants in C , c = ( c 1 , c 2 ) C 2 , and g ( z ) is a polynomial in C 2 .

Before we state our main results, let us first set that

(2.3) A 1 1 2 1 + h 1 2 i 1 h and A 2 1 2 1 + h + 1 2 i 1 h ,

where h 0 , ± 1 , 2 is a constant in C ,

(2.4) R 1 = α + γ a 1 + 1 2 η a 2 , R 2 = β + δ a 2 + 1 2 η a 1 , R 3 = γ c 2 2 + δ c 1 2 η c 1 c 2 , R 4 = α a 1 + β a 2 + 1 2 ( γ a 1 2 + δ a 2 2 + η a 1 a 2 ) , R j 5 = α + 2 γ a j 1 + η a j 2 , R j 6 = β + 2 δ a j 2 + η a j 1 , R j 7 = α a j 1 + β a j 2 + γ a j 1 2 + δ a j 2 2 + η a j 1 a j 2 , j = 1 , 2 .

Now, we state our results as follows.

Theorem 2.1

Let c = ( c 1 , c 2 ) C 2 , h C such that h 0 , ± 1 and g ( z 1 , z 2 ) is a polynomial in C 2 . If f ( z ) be a finite-order transcendental entire solution of equation (2.1), then one of the following cases occurs.

  1. The form of the solution f is of the form f ( z 1 , z 2 ) = A e l 1 z 1 + l 2 z 2 , where A , l 1 , and l 2 are the arbitrary constants in C with α l 1 + β l 2 + γ l 1 2 + δ l 2 2 + η l 1 l 2 = 0 and

    g ( z 1 , z 2 ) = 2 L n ± A e l 1 z 1 + l 2 z 2 , where A = A e ( l 1 c 1 + l 2 c 2 ) .

  2. g must be of the form g ( z 1 , z 2 ) = L ( z 1 , z 2 ) + H ( s 1 ) + B , where L ( z 1 , z 2 ) = a 1 z 1 + a 2 z 2 , H ( s 1 ) is a polynomial in s 1 = c 2 z 1 c 1 z 2 , a 1 , a 2 , B are constants in C , and the form of the solution is

    f ( z 1 , z 2 ) = 1 2 ( A 2 ξ + A 1 ξ 1 ) e 1 2 [ L ( z 1 , z 2 ) + H ( s 1 ) L ( c ) + B ] .

    L ( z 1 , z 2 ) satisfies relation

    e 1 2 L ( c 1 , c 2 ) = A 2 ξ + A 1 ξ 1 2 ( A 1 ξ + A 2 ξ 1 ) R 4 + ( R 1 c 2 R 2 c 1 ) a 0 + 1 2 R 3 a 0 2 ,

    where a 0 is the coefficient of s 1 of the polynomial H ( s 1 ) and R j s are defined in (2.4). In particular, if R 1 c 2 R 2 c 1 0 or R 3 0 , then H ( s 1 ) becomes linear in s 1 .

  3. g ( z 1 , z 2 ) must be of the form g ( z 1 , z 2 ) = L ( z 1 , z 2 ) + H ( s 1 ) + B , where L ( z 1 , z 2 ) = L 1 ( z 1 , z 2 ) + L 2 ( z 1 , z 2 ) and H ( s 1 ) = H 1 ( s 1 ) + H 2 ( s 1 ) with L 1 ( z 1 , z 2 ) + H 1 ( s 1 ) L 2 ( z 1 , z 2 ) + H 2 ( s 1 ) and L j ( z 1 , z 2 ) = a j 1 z 1 + a j 2 z 2 , B = B 1 + B 2 ; H j ( s 1 ) is a polynomial in s 1 = c 2 z 1 c 1 z 2 for j = 1 , 2 , and B 1 , B 2 , and a j i are the constants in C , and the form of the solution is

    f ( z 1 , z 2 ) = 1 2 [ A 2 e ( L 1 ( z 1 , z 2 ) + H 1 ( s 1 ) L 1 ( c ) + B 1 ) + A 1 e ( L 2 ( z 1 , z 2 ) + H 2 ( s 1 ) L 2 ( c ) + B 2 ) ] ,

    where L 1 ( z 1 , z 2 ) and L 2 ( z 1 , z 2 ) , respectively, satisfy the relations

    e L 1 ( c ) = A 2 A 1 [ R 17 + ( R 15 c 2 R 16 c 1 ) a 0 + R 3 a 0 2 ] and e L 2 ( c ) = A 1 A 2 [ R 27 + ( R 25 c 2 R 26 c 1 ) a 00 + R 3 a 00 2 ] ,

    a 0 and a 00 , respectively, the coefficients of the linear term of the polynomials H 1 ( s 1 ) and H 2 ( s 1 ) , and R i j s are defined in (2.4). In particular, if R 15 c 2 R 16 c 1 0 or R 3 0 , then H 1 becomes linear in s 1 . Similarly, if R 25 c 2 R 26 c 1 0 or A 3 0 , then H 2 becomes linear in s 1 .

Next, we exhibit some examples in support of Theorem 2.1.

Example 2.2

Let α = β = 0 , γ = δ = 1 , η = 2 , h = 5 4 , c 1 = c 2 = π i , and g ( z ) = 2 ( z 1 + z 2 ) . Then, it can be easily seen that f ( z 1 , z 2 ) = e z 1 + z 2 is a solution of (2.1), which is the conclusion (i) of Theorem 2.1.

Example 2.3

α = β = 0 , ξ = c 1 = c 2 = γ = δ = 1 , η = 2 , and g ( z 1 , z 2 ) = z 1 + z 2 + ( z 1 z 2 ) 2 + 1 . Then, in view of conclusion (ii) of Theorem 2.1, one can easily verify that f ( z 1 , z 2 ) = A 1 + A 2 2 e [ z 1 + z 2 + ( z 1 z 2 ) 2 + 1 ] 2 is a solution of

2 f ( z 1 , z 2 ) z 1 2 + 2 f ( z 1 , z 2 ) z 2 2 + 2 2 f ( z 1 , z 2 ) z 1 z 2 2 + 2 h 2 f ( z 1 , z 2 ) z 1 2 + 2 f ( z 1 , z 2 ) z 2 2 + 2 2 f ( z 1 , z 2 ) z 1 z 2 f ( z 1 + c 1 , z 2 + c 2 ) + f ( z 1 + c 1 , z 2 + c 2 ) 2 = e [ z 1 + z 2 + ( z 1 z 2 ) 2 + 1 ] ,

where A 1 and A 2 are given by (2.3).

Example 2.4

Let α = β = 0 , γ = δ = 1 , η = 2 , L 1 ( z ) = z 1 + z 2 , L 2 ( z ) = 2 z 1 + z 2 , H 1 ( s 1 ) = H 2 ( s 2 ) = 0 , and B 1 = B 2 = 1 . Then, in view of conclusion (iii) of Theorem 2.1, it can be easily verified that f ( z 1 , z 2 ) = 1 2 [ A 1 e z 1 + 2 z 2 + 1 + A 2 e 2 z 1 + z 2 + 1 ] is a solution of

2 f ( z 1 , z 2 ) z 1 2 + 2 f ( z 1 , z 2 ) z 2 2 2 2 f ( z 1 , z 2 ) z 1 z 2 2 + 2 h 2 f ( z 1 , z 2 ) z 1 2 + 2 f ( z 1 , z 2 ) z 2 2 2 2 f ( z 1 , z 2 ) z 1 z 2 f ( z 1 + c 1 , z 2 + c 2 ) + f ( z 1 + c 1 , z 2 + c 2 ) 2 = e [ 3 z 1 + 3 z 2 + 2 ] , where  A 1  and  A 2  are given by (2.3) .

Theorem 2.5

Let c = ( c 1 , c 2 ) C 2 , γ , δ , η , h C such that γ 0 , h 0 , ± 1 and g ( z 1 , z 2 ) is a polynomial in C 2 . Let f ( z ) be a finite-order transcendental entire solution of (2.2). Then, one of the following cases must occur.

  1. f ( z 1 , z 2 ) = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) , where ϕ 1 and ϕ 2 are finite-order transcendental entire functions in C 2 satisfying

    ϕ 1 ( β 1 z 1 α 1 z 2 + β 1 c 1 α 1 c 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 + β 2 c 1 α 2 c 2 ) ϕ 1 ( β 1 z 1 α 1 z 2 ) ϕ 2 ( β 2 z 1 α 2 z 2 ) = ± e g ( z ) 2 ,

    where α 1 α 2 = γ , β 1 β 2 = δ and α 1 β 2 + β 1 α 2 = η .

  2. g ( z 1 , z 2 ) = a 1 z 1 + a 2 z 2 + H ( c 2 z 1 c 1 z 2 ) + B , where H is a polynomial in c 2 z 1 c 1 z 2 and a 1 c 1 + a 2 c 2 = 4 k π i , k Z ,

    f ( z 1 , z 2 ) = ± 1 γ 0 z 1 α 1 0 z 1 α 2 e 1 2 [ a 1 z 1 + a 2 z 2 + H ( c 2 z 1 c 1 z 2 ) + B ] d z 1 d z 1 + 1 α 2 0 z 1 α 2 G 0 α 1 z 2 β 1 z 1 α 1 d z 1 + G 1 α 2 z 2 β 2 z 1 α 2 ,

    where α 1 , α 2 , β 1 , and β 2 are defined in (i) and G 0 and G 1 are finite-order transcendental entire functions in C 2 satisfying

    1 α 2 0 z 1 α 2 G 0 α 1 z 2 β 1 z 1 α 1 + α 1 c 2 β 1 c 1 α 1 G 0 α 1 z 2 β 1 z 1 α 1 d z 1 + G 1 α 2 z 2 β 2 z 1 α 2 + α 2 c 2 β 2 c 1 α 2 G 1 α 2 z 2 β 2 z 1 α 2 = 0 .

  3. If γ c 2 2 + δ c 1 2 η c 1 c 2 , then g must be of the form g ( z 1 , z 2 ) = a 1 z 1 + a 2 z 2 + B , a 1 , a 2 , B C , and the form of the solution is

    f ( z 1 , z 2 ) = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) + 2 2 ( A 1 ξ + A 2 ξ 1 ) γ a 1 2 + δ a 2 2 + η a 1 a 2 e 1 2 [ a 1 z 1 + a 2 z 2 + B ] ,

    where ξ ( 0 ) C , γ a 1 2 + δ a 2 2 + η a 1 a 2 0 , α 1 , α 2 , β 1 , and β 2 are the same as in (i), A 1 and A 2 are defined in (2.3), ϕ 1 and ϕ 2 are finite-order transcendental entire functions in C 2 such that

    ϕ 1 ( β 1 z 1 α 1 z 2 + β 1 c 1 α 1 c 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 + β 2 c 1 α 2 c 2 ) = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) and e 1 2 [ a 1 c 1 + a 2 c 2 ] = ( A 2 ξ + A 1 ξ 1 ) ( γ a 1 2 + δ a 2 2 + η a 1 a 2 ) 4 ( A 1 ξ + A 2 ξ 1 ) + 1 .

  4. If γ c 2 2 + δ c 1 2 η c 1 c 2 , then g must be of the form g ( z 1 , z 2 ) = L 1 ( z 1 , z 2 ) + L 2 ( z 1 , z 2 ) + B 1 + B 2 , where L j ( z 1 , z 2 ) = a j 1 z 1 + a j 2 z 2 with L 1 ( z 1 , z 2 ) L 2 ( z 1 , z 2 ) , a i j , B j C and the form of the solution is

    f ( z 1 , z 2 ) = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) + A 1 e L 1 ( z 1 , z 2 ) + B 1 2 ( γ a 11 2 + δ a 12 2 + η a 11 a 12 ) + A 2 e L 2 ( z 1 , z 2 ) + B 2 2 ( γ a 21 2 + δ a 22 2 + η a 21 a 22 ) ,

    where γ a 21 2 + δ a 22 2 + η a 21 a 22 0 , γ a 11 2 + δ a 12 2 + η a 11 a 12 0 , α 1 , α 2 , β 1 , β 2 are the same as in (i), A 1 and A 2 are defined in (2.3), ϕ 1 and ϕ 2 are finite-order transcendental entire functions in C 2 such that

    ϕ 1 ( β 1 z 1 α 1 z 2 + β 1 c 1 α 1 c 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 + β 2 c 1 α 2 c 2 ) = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) and e L 1 ( c ) = A 2 A 1 ( γ a 11 2 + δ a 12 2 + η a 11 a 12 ) + 1 , e L 2 ( c ) = A 1 A 2 ( γ a 21 2 + δ a 22 2 + η a 21 a 22 ) + 1 .

The following examples show that the forms of solutions are precise.

Example 2.6

Let α 1 = α 2 = β 1 = β 2 = 1 . Choose c = ( c 1 , c 2 ) C 2 such that c 1 c 2 = 2 log 2 . Then, in view of conclusion (i) of Theorem 2.5, we can easily deduce that f ( z 1 , z 2 ) = 2 e ( z 1 + z 2 ) 2 + e z 2 2 is a solution of (2.2) with g ( z 1 , z 2 ) = z 1 z 2 .

Example 2.7

Let α 1 = α 2 = 1 , β 1 = β 2 = 0 , a 1 = a 2 = 1 , and c 1 = c 2 = 4 π i . Let g ( z 1 , z 2 ) = z 1 + z 2 , G 0 ( z 2 ) = 2 e z 2 2 and G 1 ( z 2 ) = 4 e z 2 2 . Then, in view of conclusion (ii) of Theorem 2.5, we can easily deduce that f ( z 1 , z 2 ) = 4 e ( z 1 + z 2 ) 2 is a solution of (2.2).

Example 2.8

Let ξ = α 1 = α 2 = β 1 = β 2 = 1 , c 1 = log 2 + π i , c 2 = log 2 π i , and ψ 1 ( z 1 z 2 ) = ϕ 1 ( z 1 z 2 ) + ϕ 2 ( z 1 z 2 ) = e z 1 z 2 . Then, in view of conclusion (iii) of Theorem 2.5, we can easily see that f ( z 1 , z 2 ) = e z 1 z 2 + A 1 + A 2 2 e ( z 1 + z 2 + 1 ) 2 is a solution of (2.2), where A 1 and A 2 are given by (2.3).

Example 2.9

Let α 1 = α 2 = β 1 = β 2 = 1 , ϕ 1 ( z 1 z 2 ) = e z 1 z 2 , and ϕ 2 ( z 1 z 2 ) = 0 . Choose c = ( c 1 , c 2 ) C 2 such that c 1 c 2 = 4 π i . Let L 1 ( z ) = z 1 + 2 z 2 and L 2 ( z ) = 2 z 1 + z 2 such that e L 1 ( c ) = ( 9 A 2 + A 1 ) A 1 and e L 2 ( c ) = ( 9 A 1 + A 2 ) A 2 , where A 1 and A 2 are given by (2.3). Then, we can easily verify that f ( z 1 , z 2 ) = e z 1 z 2 + 1 9 2 ( A 1 e z 1 + 2 z 2 + 5 + A 2 e 2 z 1 + z 2 + 6 ) is a solution of (2.2) with g ( z 1 , z 2 ) = 2 z 1 + 2 z 2 + 11 .

3 Proofs of the main results

Before proving the main results, we present here some necessary lemmas that will play a key role to prove the main results of this article.

Lemma 3.1

[36] Let f j 0   ( j = 1 , 2 , 3 ) be meromorphic functions on C n such that f 1 is not constant, 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 j ) + O ( log + T ( r , f j ) )

holds 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 .

Lemma 3.2

[36] Let a 0 ( z ) , a 1 ( z ) , , a n ( z )   ( n 1 ) be meromorphic functions on C m and g 0 ( z ) , g 1 ( z ) , , g n ( z ) are entire functions on C m such that g j ( z ) g k ( z ) are not constants for 0 j < k n . If j = 0 n a j ( z ) e g j ( z ) 0 , and T ( r , a j ) = o ( T ( r ) ) , where T ( r ) = min 0 j < k n T ( r , e g j g k ) for j = 0 , 1 , , n , then a j ( z ) 0 for each j = 0 , 1 , , n .

Lemma 3.3

[3739] 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.

Lemma 3.4

[40] 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.

Proof of Theorem 2.1

Let f ( z 1 , z 2 ) be a transcendental entire solution of equation (2.1). Let us make a transformation

(3.1) L ( f ) = 1 2 ( U V ) , f ( z 1 + c 1 , z 2 + c 2 ) = 1 2 ( U + V ) .

Then, equation (2.1) becomes

( 1 + h ) U 2 + ( 1 h ) V 2 = e g ( z 1 , z 2 ) ,

which can be rewritten as

(3.2) 1 + h U e g ( z 1 , z 2 ) 2 + i 1 h V e g ( z 1 , z 2 ) 2 1 + h U e g ( z 1 , z 2 ) 2 i 1 h V e g ( z 1 , z 2 ) 2 = 1 .

Since f is finite-order transcendental entire, in view of Lemmas 3.3 and 3.4, it follows from (3.2) that

1 + h U e g ( z 1 , z 2 ) 2 + i 1 h V e g ( z 1 , z 2 ) 2 = e h 1 ( z 1 , z 2 ) 1 + h U e g ( z 1 , z 2 ) 2 i 1 h V e g ( z 1 , z 2 ) 2 = e h 1 ( z 1 , z 2 ) ,

where h 1 ( z 1 , z 2 ) is a polynomial in C 2 , from which we obtain

(3.3) U = e h 1 ( z 1 , z 2 ) + e h 1 ( z 1 , z 2 ) 2 1 + h e g ( z 1 , z 2 ) 2 and V = e h 1 ( z 1 , z 2 ) e h 1 ( z 1 , z 2 ) 2 i 1 h e g ( z 1 , z 2 ) 2 .

Set

(3.4) γ 1 ( z 1 , z 2 ) = g ( z 1 , z 2 ) 2 + h 1 ( z 1 , z 2 ) and γ 2 ( z 1 , z 2 ) = g ( z 1 , z 2 ) 2 h 1 ( z 1 , z 2 ) .

Therefore, it follows from (3.3) and (3.4) that

(3.5) U = e γ 1 ( z 1 , z 2 ) + e γ 2 ( z 1 , z 2 ) 2 1 + h and V = e γ 1 ( z 1 , z 2 ) e γ 2 ( z 1 , z 2 ) 2 i 1 h .

In view of (3.1) and (3.5), we obtain that

(3.6) L ( f ) = 1 2 [ A 1 e γ 1 ( z 1 , z 2 ) + A 2 e γ 2 ( z 1 , z 2 ) ] , f ( z 1 + c 1 , z 2 + c 2 ) = 1 2 [ A 2 e γ 1 ( z 1 , z 2 ) + A 1 e γ 2 ( z 1 , z 2 ) ] .

After simple calculation, it follows from the two equations of (3.6) that

(3.7) A 2 A 1 Q 1 ( z 1 , z 2 ) e γ 1 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) + Q 2 ( z 1 , z 2 ) e γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) A 2 A 1 e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = 1 ,

where

(3.8) Q j ( z 1 , z 2 ) = α γ j z 1 + β γ j z 2 + γ γ j z 1 2 + 2 γ j z 1 2 + δ γ j z 2 2 + 2 γ j z 2 2 + η γ j z 1 γ j z 2 + 2 γ j z 1 z 2 , j = 1 , 2 .

Now, we discuss two possible cases.

Case 1. Let γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = k , a constant in C . In view of (3.4), we conclude that h 1 ( z 1 , z 2 ) is constant. Let ξ = e h 1 ( z 1 , z 2 ) C . Then, (3.6) yields that

(3.9) L ( f ) = C 1 e g ( z 1 , z 2 ) 2 , f ( z 1 + c 1 , z 2 + c 2 ) = C 2 e g ( z 1 , z 2 ) 2 ,

where C 1 = 1 2 ( A 1 ξ + A 2 ξ 1 ) , C 2 = 1 2 ( A 2 ξ + A 1 ξ 1 ) .

Note that

(3.10) C 2 0 and C 1 2 + C 2 2 = 1 2 [ ( A 1 2 + A 2 2 ) ( ξ 2 + ξ 2 ) + 4 A 1 A 2 ] .

If C 1 = 0 , in view of (3.9) and (3.10), it follows that

(3.11) α f z 1 + β f z 2 + γ 2 f z 1 2 + δ 2 f z 2 2 + η 2 f z 1 z 2 = 0 , f ( z 1 + c 1 , z 2 + c 2 ) = ± e g ( z 1 , z 2 ) 2 .

From the first equation of (3.11), we obtain that

f ( z 1 , z 2 ) = A e l 1 z 1 + l 2 z 2 ,

where A , l 1 , and l 2 are the arbitrary constants in C satisfying the relation α l 1 + β l 2 + γ l 1 2 + δ l 2 2 + η l 1 l 2 = 0 . From the second equation of (3.11), we obtain that

g ( z 1 , z 2 ) = 2 L n ± A e l 1 z 1 + l 2 z 2 ,

where A = A e l 1 c 1 + l 2 c 2 . This is conclusion (i).

If C 1 0 , then from (3.9), we obtain that

(3.12) α g z 1 + β g z 2 + γ 1 2 g z 1 2 + 2 g z 1 2 + δ 1 2 g z 2 2 + 2 g z 2 2 + η 1 2 g z 1 g z 2 + 2 g z 1 z 2 = 2 C 1 C 2 e 1 2 [ g ( z 1 + c 1 , z 2 + c 2 ) g ( z 1 , z 2 ) ] .

Since g is a polynomial in C 2 , it follows from (3.12) that g ( z 1 + c 1 , z 2 + c 2 ) g ( z 1 , z 2 ) must be constant, say ζ C .

Then, g can be written as g ( z 1 , z 2 ) = L ( z 1 , z 2 ) + H ( s 1 ) + B , where L ( z 1 , z 2 ) = a 1 z 1 + a 2 z 2 , H ( s 1 ) is a polynomial in s 1 = c 2 z 1 c 1 z 2 , a 1 , a 2 , and B are the constants in C . Hence, it follows from (3.12) that

(3.13) ( R 1 c 2 R 2 c 1 ) H + R 3 1 2 H 2 + H = 2 ( A 1 ξ + A 2 ξ 1 ) A 2 ξ + A 1 ξ 1 e 1 2 L ( c ) R 4 ,

where R s are defined in (2.4).

If R 1 c 2 R 2 c 1 = 0 = R 3 , then from (3.13), we obtain

e 1 2 L ( c ) = A 2 ξ + A 1 ξ 1 2 ( A 1 ξ + A 2 ξ 1 ) R 4 .

If R 1 c 2 R 2 c 1 0 or (here “or” is in inclusive sense) R 3 0 , then it follows from (3.13) that H must be constant, say a 0 , which is the coefficient of s 1 in the polynomial H ( s 1 ) .

Therefore, from (3.13), we obtain that

(3.14) e 1 2 L ( c ) = A 2 ξ + A 1 ξ 1 2 ( A 1 ξ + A 2 ξ 1 ) R 4 + ( R 1 c 2 R 2 c 1 ) a 0 + 1 2 R 3 a 0 2 .

Hence, in either case L ( z ) satisfies relation (3.14).

Therefore, in view of the second equation of (3.9), we obtain the form of the solution as

f ( z 1 , z 2 ) = 1 2 ( A 2 ξ + A 1 ξ 1 ) e 1 2 [ L ( z 1 , z 2 ) + H ( s 1 ) L ( c ) + B ] .

This is conclusion (ii).

Case 2 Let γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) be non-constant. Then, in view of (3.7), it follows that Q 1 ( z 1 , z 2 ) and Q 2 ( z 1 , z 2 ) both cannot be zero at the same time.

If Q 1 ( z 1 , z 2 ) 0 and Q 2 ( z 1 , z 2 ) 0 , then (3.7) yields that

Q 2 ( z 1 , z 2 ) e γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) A 2 A 1 e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = 1 .

In view of the aforementioned equation, it follows that

N r , 1 A 2 e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) + A 1 = N r , 1 Q 2 ( z 1 , z 2 ) e γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = S ( r , e γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) ) .

Also, note that

N ( r , e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) ) = S ( r , e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) ) ,

N r , 1 e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = S ( r , e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) ) .

By the second main theorem of Nevanlinna for several complex variables, we obtain

T ( r , e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) ) N ¯ ( r , e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) ) + N ¯ r , 1 e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) + N ¯ r , 1 e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) A 1 A 2 + S ( r , e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) ) S ( r , e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) ) + S ( r , e γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) ) .

This implies that γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) is constant, which is a contradiction. Similarly, we can obtain a contradiction for the case Q 1 ( z 1 , z 2 ) 0 and Q 1 ( z 1 , z 2 ) 0 . Hence, Q 1 ( z 1 , z 2 ) 0 and Q 2 ( z 1 , z 2 ) 0 .

Since γ 1 ( z 1 , z 2 ) and γ 2 ( z 1 , z 2 ) are the polynomials in C 2 and γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) is non-constant, applying Lemma 3.1 to equation (3.7), we obtain that either

A 2 A 1 Q 1 ( z ) e γ 1 ( z ) γ 1 ( z + c ) = 1 , or Q 2 ( z 1 , z 2 ) e γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = 1 .

If

(3.15) A 2 A 1 Q 1 ( z 1 , z 2 ) e γ 1 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = 1 ,

then from (3.7), it follows that

(3.16) Q 2 ( z 1 , z 2 ) e γ 2 ( z 1 , z 2 ) γ 2 ( z 1 + c 1 , z 2 + c 2 ) = A 1 A 2 .

As γ 1 ( z 1 , z 2 ) and γ 2 ( z 1 , z 2 ) are polynomials, in view of (3.15) and (3.16), we conclude that γ 1 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) and γ 2 ( z 1 , z 2 ) γ 2 ( z 1 + c 1 , z 2 + c 2 ) both are constants in C , and hence, we can obtain that γ 1 ( z 1 , z 2 ) = L 1 ( z 1 , z 2 ) + H 1 ( s 1 ) + B 1 and γ 2 ( z 1 , z 2 ) = L 2 ( z 1 , z 2 ) + H 2 ( s 1 ) + B 2 , where L j ( z 1 , z 2 ) = a j 1 z 1 + a j 2 z 2 , H j ( s 1 ) is a polynomial in s 1 = c 2 z 1 c 1 z 2 , and a j 1 , a j 2 , B 1 , and B 2 are the constants in C for j = 1 , 2 . Note that L 1 ( z 1 , z 2 ) + H 1 ( s 1 ) L 2 ( z 1 , z 2 ) + H 2 ( s 1 ) . Otherwise, γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) would become constant, a contradiction to our assumption. Hence, the form of the polynomial g is g ( z 1 , z 2 ) = L ( z 1 , z 2 ) + H ( s 1 ) + B , where L ( z 1 , z 2 ) = L 1 ( z 1 , z 2 ) + L 2 ( z 1 , z 2 ) , H ( s 1 ) = H 1 ( s 1 ) + H 2 ( s 1 ) , and B = B 1 + B 2 .

Therefore, in view of (3.15) and (3.16), we obtain that

(3.17) ( R 15 c 2 R 16 c 1 ) H 1 + R 3 ( H 1 2 + H 1 ) = A 1 A 2 e L 1 ( c ) R 17 , ( R 25 c 2 R 26 c 1 ) H 2 + R 3 ( H 2 2 + H 2 ) = A 2 A 1 e L 2 ( c ) R 27 ,

where R s are defined in (2.4).

Then, by similar arguments as in Case 1, we obtain from (3.17) that

(3.18) e L 1 ( c ) = A 2 A 1 ( R 17 + ( R 15 c 2 R 16 c 1 ) a 0 + R 3 a 0 2 ) , e L 2 ( c ) = A 1 A 2 ( R 27 + ( R 25 c 2 R 26 c 1 ) a 00 + R 3 a 00 2 ) ,

where a 0 and a 00 , respectively, the coefficients of the linear term of the polynomials H 1 ( s 1 ) and H 2 ( s 1 ) .

Therefore, in view of the second equation of (3.6), we obtain

f ( z 1 , z 2 ) = 1 2 ( A 2 e L 1 ( z 1 , z 2 ) + H 1 ( s 1 ) L 1 ( c ) + B 1 + A 1 e L 2 ( z 1 , z 2 ) + H 2 ( s 1 ) L 2 ( c ) + B 2 ) ,

where L 1 ( c ) and L 2 ( c ) can be found from (3.18).

This is conclusion (iii).

If Q 2 ( z 1 , z 2 ) e γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = 1 , then it follows from equation (3.7) that Q 1 ( z 1 , z 2 ) e γ 1 ( z 1 , z 2 ) γ 2 ( z 1 + c 1 , z 2 + c 2 ) = 1 . Since γ 1 ( z 1 , z 2 ) and γ 2 ( z 1 , z 2 ) are both polynomials in C 2 , it follows that γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = η 1 and γ 1 ( z 1 , z 2 ) γ 2 ( z 1 + c 1 , z 2 + c 2 ) = η 2 , where η 1 , η 2 C . This implies that γ 1 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = γ 2 ( z 1 , z 2 ) γ 2 ( z 1 + c 1 , z 2 + c 2 ) = η 1 + η 2 . Therefore, we can write γ 1 ( z 1 , z 2 ) = L ( z 1 , z 2 ) + H ( s 1 ) + ζ 1 and γ 2 ( z 1 , z 2 ) = L ( z 1 , z 2 ) + H ( s 1 ) + ζ 2 . But, then we obtain γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = ζ 2 ζ 1 , constants, which is a contradiction.□

Proof of Theorem 2.5

Let f ( z 1 , z 2 ) be a finite-order transcendental entire solution of equation (2.2). Let us make a transformation

(3.19) L ˜ ( f ) = 1 2 ( U V ) , f ( z + c ) f ( z ) = 1 2 ( U + V ) .

Then, by similar arguments as in Theorem 2.1, we can obtain that

(3.20) L ˜ ( f ) = 1 2 [ A 1 e γ 1 ( z 1 , z 2 ) + A 2 e γ 2 ( z 1 , z 2 ) ] f ( z 1 + c 1 , z 2 + c 2 ) f ( z 1 , z 2 ) = 1 2 [ A 2 e γ 1 ( z 1 , z 2 ) + A 1 e γ 2 ( z 1 , z 2 ) ] ,

where γ 1 and γ 2 are defined in (3.4) and A 1 and A 2 are defined in (2.3).

In view of equations in (3.20), we obtain that

(3.21) A 2 Q 1 ( z 1 , z 2 ) + A 1 A 1 e γ 1 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) + A 1 Q 2 ( z 1 , z 2 ) + A 2 A 1 e γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) A 2 A 1 e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = 1 ,

where

(3.22) Q j ( z 1 , z 2 ) = γ γ j z 1 2 + 2 γ j z 1 2 + δ γ j z 2 2 + 2 γ j z 2 2 + η γ j z 1 γ j z 2 + 2 γ j z 1 z 2 , j = 1 , 2 .

Now, we consider two possible cases.

Case 1. Let γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = k C . In view of (3.4), we conclude that h 1 ( z 1 , z 2 ) is constant. Set e h 1 = ξ C . Then, (3.20) yields that

(3.23) L ˜ ( f ) = C 1 e g ( z 1 , z 2 ) 2 , f ( z 1 + c 1 , z 2 + c 2 ) f ( z 1 , z 2 ) = C 2 e g ( z 1 , z 2 ) 2 ,

where C 1 = 1 2 ( A 1 ξ + A 2 ξ 1 ) , C 2 = 1 2 ( A 2 ξ + A 1 ξ 1 ) .

Note that

(3.24) C 1 2 + C 2 2 = 1 2 [ ( A 1 2 + A 2 2 ) ( ξ 2 + ξ 2 ) + 4 A 1 A 2 ] .

Subcase 1.1. Let C 1 = 0 . Then, in view of (3.24), we obtain that C 2 = ± 1 . Therefore, it follows from (3.23) that

(3.25) γ 2 f ( z 1 , z 2 ) z 1 2 + δ 2 f ( z 1 , z 2 ) z 2 2 + η 2 f ( z 1 , z 2 ) z 1 z 2 = 0 , f ( z 1 + c 1 , z 2 + c 2 ) f ( z 1 , z 2 ) = ± e g ( z 1 , z 2 ) 2 .

Now, in view of the results in [35, page 2178, line 21], the first equation of (3.25) can be written as

(3.26) ( α 1 D + β 1 D ) ( α 2 D + β 2 D ) f ( z ) = 0 ,

where D z 1 , D z 2 , α 1 α 2 = γ , β 1 β 2 = δ , and α 1 β 2 + α 2 β 1 = η .

Solving (3.26), we obtain that

f ( z 1 , z 2 ) = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) ,

where ϕ 1 and ϕ 2 are finite-order transcendental entire functions in C 2 .

In view of the second equation of (3.25), we obtain that

ϕ 1 ( β 1 z 1 α 1 z 2 + β 1 c 1 α 1 c 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 + β 2 c 1 α 2 c 2 ) ϕ 1 ( β 1 z 1 α 1 z 2 ) ϕ 2 ( β 2 z 1 α 2 z 2 ) = ± e 1 2 g ( z 1 , z 2 ) .

This is conclusion (i).

Subcase 1.2. Let C 2 = 0 . Then, in view of (3.24), it yields that C 1 = ± 1 .

Therefore, it follows from (3.23) that

(3.27) γ 2 f ( z 1 , z 2 ) z 1 2 + δ 2 f ( z 1 , z 2 ) z 2 2 + η 2 f ( z 1 , z 2 ) z 1 z 2 = ± e 1 2 g ( z 1 , z 2 ) , f ( z 1 + c 1 , z 2 + c 2 ) f ( z 1 , z 2 ) = 0 .

Clearly, second equation of (3.27) shows that f is a periodic function of period c . In view of the two equations in (3.27), it follows that e 1 2 ( g ( z 1 + c 1 , z 2 + c 2 ) g ( z 1 , z 2 ) ) = 1 . This implies that g ( z 1 , z 2 ) = a 1 z 1 + a 2 z 2 + H ( c 2 z 1 c 1 z 2 ) + B , where H is a polynomial in c 2 z 1 c 1 z 2 and a 1 c 1 + a 2 c 2 = 4 k π i , k Z .

Now, in view of the results in [35, page 2178, line 21], the first equation of (3.27) can be written as

(3.28) ( α 1 D + β 1 D ) ( α 2 D + β 2 D ) f ( z 1 , z 2 ) = ± e 1 2 g ( z 1 , z 2 ) ,

where D , D , α 1 , α 2 , β 1 , and β 2 are defined in (3.26).

Let ( α 2 D + β 2 D ) f ( z 1 , z 2 ) = u ( z 1 , z 2 ) . Then, in view of the results in [[31], page 2182, line 22], we obtain

u ( z 1 , z 2 ) = ± 1 α 1 0 z 1 α 1 e 1 2 g ( z 1 , z 2 ) d z 1 + G 0 α 1 z 2 β 1 z 1 α 1 ,

where G 0 is a finite-order transcendental entire function in C 2 .

Since we have assumed that ( α 2 D + β 2 D ) f ( z 1 , z 2 ) = u ( z 1 , z 2 ) , we obtain that

(3.29) α 2 f z 1 + β 2 f z 2 = ± 1 α 1 0 z 1 α 1 e 1 2 g ( z 1 , z 2 ) d z 1 + G 0 α 1 z 2 β 1 z 1 α 1 .

By similar argument, we obtain from (3.29) that

f ( z 1 , z 2 ) = ± 1 γ 0 z 1 α 1 0 z 1 α 2 e 1 2 [ a 1 z 1 + a 2 z 2 + H ( c 2 z 1 c 1 z 2 ) + B ] d z 1 d z 1 + 1 α 2 0 z 1 α 2 G 0 α 1 z 2 β 1 z 1 α 1 d z 1 + G 1 α 2 z 2 β 2 z 1 α 2 ,

where G 1 is a finite-order transcendental entire function in C 2 .

In view of the fact that a 1 c 1 + a 2 c 2 = 4 k π i , k Z , it follows from the second equation of (3.27) that

1 α 2 0 z 1 α 2 G 0 α 1 z 2 β 1 z 1 α 1 + α 1 c 2 β 1 c 1 α 1 G 0 α 1 z 2 β 1 z 1 α 1 d z 1 + G 1 α 2 z 2 β 2 z 1 α 2 + α 2 c 2 β 2 c 1 α 2 G 1 α 2 z 2 β 2 z 1 α 2 = 0 .

This is conclusion (ii).

Subcase 1.3. Let C 1 0 and C 2 0 . Then, after simple calculations, (3.23) yields that

(3.30) γ 2 g z 1 2 + 1 2 g z 1 2 + δ 2 g z 2 2 + 1 2 g z 2 2 + η 2 g z 1 z 2 + 1 2 g z 1 g z 2 = 2 C 1 C 2 e 1 2 [ g ( z 1 + c 1 , z 2 + c 2 ) g ( z 1 , z 2 ) ] 1 .

Since g is a polynomial in C 2 , in view of (3.30), we conclude that g ( z 1 + c 1 , z 2 + c 2 ) g ( z 1 , z 2 ) = ξ , ξ C . This implies that g ( z 1 , z 2 ) = L 1 ( z 1 , z 2 ) + H ( s 1 ) + B 1 , where L 1 ( z 1 , z 2 ) = a 11 z 1 + a 12 z 2 , H ( s 1 ) is a polynomial in s 1 c 2 z 1 c 1 z 2 , a 11 , a 12 , B 1 C . Hence, we obtain from (3.30) that

(3.31) γ a 11 + 1 2 η a 12 c 2 δ a 12 + 1 2 η a 11 c 1 H + ( γ c 2 2 + δ c 1 2 η c 1 c 2 ) 1 2 H 2 + H = 2 C 1 C 2 e 1 2 L 1 ( c ) 1 .

Since γ c 2 2 + δ c 1 2 η c 1 c 2 , in view of (3.31), we conclude that H is constant. This implies that H ( s 1 ) = a 0 s 1 + b 0 . Hence, g ( z ) reduces to the form

(3.32) g ( z 1 , z 2 ) = L ( z 1 , z 2 ) + B = a 1 z 1 + a 2 z 2 + B ,

where a 1 = a 11 + a 0 c 2 , a 2 = a 12 a 0 c 1 and B = B 1 + b 0 .

Therefore, in view of (3.30) and (3.32), we obtain that

(3.33) e 1 2 [ a 1 c 1 + a 2 c 2 ] = C 2 4 C 1 ( γ a 1 2 + δ a 2 2 η a 1 a 2 ) + 1 .

Now, the first equation of (3.23) can be rewritten as

(3.34) ( γ D 2 + δ D 2 + η D D ) f ( z 1 , z 2 ) = C 1 e 1 2 [ a 1 z 1 + a 2 z 2 + B ] ,

where D z 1 and D z 2 .

Therefore, complementary function of (3.34) is C.F. = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) , where α 1 , α 2 , β 1 , and β 2 are same as in (i), ϕ 1 and ϕ 2 are finite-order transcendental entire functions in C 2 , and the particular integral is

P.I. = 4 C 1 e B 2 γ a 1 2 + δ a 2 2 + η a 1 a 2 e v d v d v = 4 C 1 γ a 1 2 + δ a 2 2 + η a 1 a 2 e 1 2 [ a 1 z 1 + a 2 z 2 + B ] ,

where v = a 1 z 1 + a 2 z 2 .

Hence, from (3.23), we obtain

f ( z 1 , z 2 ) = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) + 2 2 ( A 1 ξ + A 2 ξ 1 ) γ a 1 2 + δ a 2 2 + η a 1 a 2 e 1 2 [ a 1 z 1 + a 2 z 2 + B ] .

Substituting f ( z 1 , z 2 ) into the second equation of (3.23) and combining with (3.33), we obtain that

ϕ 1 ( β 1 z 1 α 1 z 2 + β 1 c 1 α 1 c 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 + β 2 c 1 α 2 c 2 ) = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) .

This is conclusion (iii).

Case 2. Let γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) be non-constant. Then, obviously, A 2 Q 1 ( z 1 , z 2 ) + A 1 and A 1 Q 2 ( z 1 , z 2 ) + A 2 cannot be identically zero at the same time. Otherwise, in view of (3.21), it follows that e γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) is a constant, which implies that γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) is a constant. This is a contradiction to our assumption.

If A 2 Q 1 ( z 1 , z 2 ) + A 1 0 and A 1 Q 2 ( z 1 , z 2 ) + A 2 0 , then (3.21) it yields

(3.35) ( A 1 Q 2 ( z 1 , z 2 ) + A 2 ) e γ 2 ( z 1 , z 2 ) A 2 e γ 2 ( z 1 + c 1 , z 2 + c 2 ) A 1 e γ 1 ( z 1 + c 1 , z 2 + c 2 ) 0 .

In view of Lemma 3.2 and (3.35), we can easily obtain a contradiction. Similarly, we can obtain a contradiction for the case A 2 Q 1 ( z 1 , z 2 ) + A 1 0 and A 1 Q 2 ( z 1 , z 2 ) + A 2 0 . Hence, we must have A 2 Q 1 ( z 1 , z 2 ) + A 1 0 and A 1 Q 2 ( z 1 , z 2 ) + A 2 0 .

Now, in view of Lemma 3.1, we obtain from (3.21) that either

A 2 Q 1 ( z 1 , z 2 ) + A 1 A 1 e γ 1 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) 1

or

A 1 Q 2 ( z 1 , z 2 ) + A 2 A 1 e γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) 1 .

If A 1 Q 2 ( z 1 , z 2 ) + A 2 A 1 e γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) 1 , then in view of (3.21), it follows that A 2 Q 1 ( z 1 , z 2 ) + A 1 A 2 e γ 1 ( z 1 , z 2 ) γ 2 ( z 1 + c 1 , z 2 + c 2 ) 1 . Therefore, we must obtain that γ 2 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = ξ 1 and γ 1 ( z 1 , z 2 ) γ 2 ( z 1 + c 1 , z 2 + c 2 ) = ξ 2 , ξ 1 , ξ 2 C . Thus, it follows that γ 1 ( z 1 , z 2 ) γ 1 ( z 1 + 2 c 1 , z 2 + 2 c 2 ) = γ 2 ( z 1 , z 2 ) γ 2 ( z 1 + 2 c 1 , z 2 + 2 c 2 ) = ξ 1 + ξ 2 . This implies that γ 1 ( z 1 , z 2 ) = L ( z 1 , z 2 ) + H ( s 1 ) + B 1 and γ 2 ( z 1 , z 2 ) = L ( z 1 , z 2 ) + H ( s 1 ) + B 2 , where L ( z 1 , z 2 ) = a 1 z 1 + a 2 z 2 and H ( s 1 ) is a polynomial in s 1 c 2 z 1 c 1 z 2 , a 1 , a 2 , B 1 , B 2 C . Hence, we must obtain that γ 2 ( z 1 + c 1 , z 2 + c 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = B 2 B 1 , a constant in C , which is a contradiction to the assumption.

Therefore, we must have

(3.36) A 2 Q 1 ( z 1 , z 2 ) + A 1 A 1 e γ 1 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) 1 .

In view of (3.21) and (3.36), we obtain that

(3.37) A 1 Q 2 ( z 1 , z 2 ) + A 2 A 2 e γ 2 ( z 1 , z 2 ) γ 2 ( z 1 + c 1 , z 2 + c 2 ) 1 .

Since γ 1 ( z 1 , z 2 ) and γ 2 ( z 1 , z 2 ) are the polynomials in C 2 , from (3.36) and (3.37), we can conclude that γ 1 ( z 1 , z 2 ) γ 1 ( z 1 + c 1 , z 2 + c 2 ) = η 1 and γ 2 ( z 1 , z 2 ) γ 2 ( z 1 + c 1 , z 2 + c 2 ) = η 2 , η 1 , η 2 C . Thus, we have γ 1 ( z 1 , z 2 ) = L 1 ( z 1 , z 2 ) + H 1 ( s 1 ) + B 1 and γ 2 ( z 1 , z 2 ) = L 2 ( z 1 , z 2 ) + H 2 ( s 1 ) + B 2 , where L j ( z 1 , z 2 ) = a j 1 z 1 + a j 2 z 2 and H j ( s 1 ) is a polynomial in s 1 c 2 z 1 c 1 z 2 , a j 1 , a j 2 , B j C for j = 1 , 2 .

Therefore, in view of (3.22) and (3.36), we obtain that

[ ( 2 γ a 11 + η a 12 ) c 2 ( 2 δ a 12 + η a 11 ) c 1 ] H 1 + ( γ c 2 2 + δ c 1 2 η c 1 c 2 ) ( H 1 2 + H 1 ) = A 1 A 1 [ e L 1 ( c ) 1 ] ( γ a 11 2 + δ a 12 2 + η a 11 a 12 ) .

Since γ c 2 2 + δ c 1 2 η c 1 c 2 0 , we conclude that H 1 ( s 1 ) is a linear polynomial in s 1 , and thus, L 1 ( z ) + H 1 ( s 1 ) becomes linear in C . For the sake of convenience, we still denote that γ 1 ( z ) = L 1 ( z ) + B 1 . In a similar manner, from (3.22) and (3.37), we can conclude that γ 2 ( z ) = L 2 ( z ) + B 2 .

Therefore, it follows from (3.36) and (3.37) that

(3.38) e L 1 ( c ) = A 2 A 1 ( γ a 11 2 + δ a 12 2 + η a 11 a 12 ) + 1 , e L 2 ( c ) = A 1 A 2 ( γ a 21 2 + δ a 22 2 + η a 21 a 22 ) + 1 .

Now, in view of the first equation of (3.20), it follows that

(3.39) ( γ D 2 + δ D 2 + η D D ) f ( z 1 , z 2 ) = 1 2 [ A 1 e L 1 ( z 1 , z 2 ) + B 1 + A 2 e L 2 ( z 1 , z 2 ) + B 2 ] ,

where D z 1 and D z 2 .

Now, in view of the results in [[31], page 2178, line 21], the complementary function of (3.39) is ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) , where α 1 , α 2 , β 1 , and β 2 are same as in (i), ϕ 1 and ϕ 2 are finite-order transcendental entire functions in C 2 , and the particular integral is

P.I. = A 1 e L 1 ( z 1 , z 2 ) + B 1 2 ( γ a 11 2 + δ a 12 2 + η a 11 a 12 ) + A 2 e L 2 ( z 1 , z 2 ) + B 2 2 ( γ a 21 2 + δ a 22 2 + η a 21 a 22 ) .

Hence, the form of the solution of (3.39) is

(3.40) f ( z 1 , z 2 ) = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) + A 1 e L 1 ( z 1 , z 2 ) + B 1 2 ( γ a 11 2 + δ a 12 2 + η a 11 a 12 ) + A 2 e L 2 ( z 1 , z 2 ) + B 2 2 ( γ a 21 2 + δ a 22 2 + η a 21 a 22 ) .

Substituting (3.40) into the second equation of (3.20) and combining with (3.38), we obtain that

ϕ 1 ( β 1 z 1 α 1 z 2 + β 1 c 1 α 1 c 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 + β 2 c 1 α 2 c 2 ) = ϕ 1 ( β 1 z 1 α 1 z 2 ) + ϕ 2 ( β 2 z 1 α 2 z 2 ) .

This is conclusion (iv).□


# This work was completed with the support of our TE X-pert.


Acknowledgment

Authors would like to thank all the Referees for their valuable suggestions towards the clarity of the paper.

  1. Funding information: This work was supported by the National Natural Science Foundation of China (12161074) and the Foundation of Education Department of Jiangxi (GJJ212305, GJJ202303, GJJ201813, GJJ201343) of China, the Talent Introduction Research Foundation of Suqian University (106-CK00042/028) and sponsored by Qing Lan Project and Suqian Talent Xiongying Plan of Suqian.

  2. Author contributions: Both the authors have contributed equally towards the formation of the paper.

  3. Conflict of interest: The authors declare that they have no conflict of interest.

  4. Data availability statement: Data sharing is not applicable to this article as no database was generated or analyzed during the current study.

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Received: 2023-12-16
Revised: 2024-06-08
Accepted: 2024-08-06
Published Online: 2024-11-21

© 2024 the author(s), published by De Gruyter

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

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