Home Study on solutions of the systems of complex product-type PDEs with more general forms in ℂ2
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Study on solutions of the systems of complex product-type PDEs with more general forms in ℂ2

  • Hong Li , Zhao Sheng Luo EMAIL logo and Hong Yan Xu EMAIL logo
Published/Copyright: November 16, 2024
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Open Mathematics
From the journal Open Mathematics Volume 22 Issue 1

Abstract

With the help of the Nevanlinna theory and the Hadamard factorization theory of meromorphic functions, we mainly give a description of the existence and the forms of the transcendental entire solutions of several product-type complex partial differential equations systems

( a 1 f z 1 + a 2 f z 2 ) ( a 3 g z 1 + a 4 g z 2 ) = e h 1 , ( a 1 g z 1 + a 2 g z 2 ) ( a 3 f z 1 + a 4 f z 2 ) = e h 2 ,

and

( a 1 f z 1 + a 2 g z 1 ) ( a 3 f z 2 + a 4 g z 2 ) = e h 1 , ( a 1 f z 2 + a 2 g z 2 ) ( a 3 f z 1 + a 4 g z 1 ) = e h 2 ,

where f z t = ∂ f t ∂ z t , g z t = ∂ g t ∂ z t , t = 1 , 2 ; a j ∈ C , j = 1 , 2 , 3 , 4 , and h 1 and h 2 are the polynomials in C 2 . More importantly, we list some examples to explain that our results about the forms of solutions of such systems are precise to some extent.

Keywords: Nevanlinna theory; entire function; system of functional equations; several complex variables
MSC 2010: 30D35; 35M30

1 Introduction

The purpose of this article is to study the transcendental entire solutions of two systems of the complex partial differential equations (PDEs) in C 2 . A large amount of literature indicates that the study of PDEs has penetrated into various fields of real life (can be found in Courant and Hilbert [1] and Garabedian [2]). For example, the Laplace equation, wave equation, heat conduction equation, and Helmholtz equation play very important roles in mathematical physics, engineering, especially in fields such as electromagnetic, fluid dynamics, and heat conduction. In general, it is not easy to solve the exact solution of a PDE. But, for the first-order nonlinear PDE

(1.1) u x 2 + u y 2 = 1 ,

it is said as the eikonal equation in real variable case, which is concerned with the problems of geometrical optics and wave propagation, where u x = ∂ u ⁄ ∂ x and u y = ∂ u ⁄ ∂ y . The classic conclusion that entire solution of eikonal equation (1.1) in C 2 must be a linear function of x , y , i . e . , u = a x + b y + c and a 2 + b 2 = 1 , c ∈ C , was proved by Khavinson [3, Theorem 1], Saleeby [4, Theorem 1], and Li [5, Corollary 2.2]. Then, their work inspired several interest results such as Khavinson [6], Khavinson and Lundberg [7], Li [8,9], Hemmati [10], Saleeby [11–13], Chen and Han [14,15] and Lü and Li [16], Xu and his colleagues [17–19] on complex analytic solutions of equation (1.1) and its variants in C 2 including u x 2 + u y 2 = p , u 2 + u x 2 + u y 2 = 1 , u t − C u m u x = q , ∏ j = 1 3 ( a j 1 u z 1 + a j 2 u z 2 + a j 3 u z 3 ) = 1 , etc, where p and q are the polynomials in C 2 . In 2005, Li [8] further discussed a nonlinear PDE ( u z 1 ) 2 + ( u z 2 ) 2 = e g , which is more general than equation (1.1). In fact, this equation in the real variable case always appears 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 e g .

Theorem A

[8, Theorem 2.1] Let g be a polynomial in C 2 . Then, u is an entire solution of the PDE

(1.2) ( u z 1 ) 2 + ( u z 2 ) 2 = e g ,

in C 2 if and only if

  1. u = f ( c 1 z 1 + c 2 z 2 ) or

  2. u = ϕ 1 ( z 1 + i z 2 ) + ϕ 2 ( z 1 − i z 2 ) ,

  3. where f is an entire function in C satisfying

    f ′ ( c 1 z 1 + c 2 z 2 ) = ± e g ⁄ 2 ,

    c 1 and c 2 are two constants satisfying c 1 2 + c 2 2 = 1 , and ϕ 1 and ϕ 2 are entire functions in C satisfying

    ϕ 1 ′ ( z 1 + i z 2 ) ϕ 2 ′ ( z 1 − i z 2 ) = 1 4 e g .

    Moreover, the forms of f , ϕ 1 , and ϕ 2 can be found in [8].

Khavinson [3] or Li [20] remarked, under the linear transformation x = z 1 + i z 2 and y = z 1 − i z 2 , equation (1.2) can be written as the form

(1.3) U x U y = P ,

where U ( x , y ) = u ( z 1 , z 2 ) , and P ( x , y ) = e g . By a simple calculation, one can obtain

(1.4) A ( U x x U y y − U x y 2 ) + B U x y + C = 0 ,

where A = U x U y , B = U x P y + U y P x , and C = P x P y . Obviously, equation (1.4) can be seen as a non-degenerate Monge-Ampère equation, which is widely used in differential geometry, variational methods, optimization problems, and transmission problems. There is an enormous literature dedicated to the study of Monge-Ampère equations. Inspired by this remark in Khavinson [3] and Li [20], Lü [21] studied the solutions of some first-order nonlinear PDEs with product-type and obtained

Theorem B

[21, Theorem 1] Let g be a polynomial in C 2 , and let m be a non-negative integer. Then, u is an entire solution of the PDE u x u y = x m e g in C 2 if and only if the following assertions hold:

  1. u = ϕ 1 ( x ) + ϕ 2 ( x ) , where ϕ 1 ′ ( x ) = x m e α ( x ) and ϕ 2 ′ ( y ) = e β ( y ) satisfying α ( x ) + β ( y ) = g ( x , y ) ;

  2. u = F ( y + A x m + 1 ) , where A is a non-zero constant and ( m + 1 ) A F ′ 2 ( y + A x m + 1 ) = e g ;

  3. u = ( x k + 1 ⁄ ( k + 1 ) ) e a y + b + C , where ( a ⁄ ( k + 1 ) ) e 2 ( a y + b ) = e g , m = 2 k + 1 and a ( ≠ 0 ) , b , C are constants.

In 2022, Chen and Han [14] further gave the description of entire solutions for a series of product-type PDEs and obtained

Theorem C

[14, Theorem 1.7] Assume that g ( z , w ) is a polynomial in C 2 , p ( z ) ≢ 0 is a polynomial in C , and l ≥ 0 and m , n ≥ 1 are integers. u ( z , w ) in C 2 is an entire solution to the first-order nonlinear PDE

(1.5) ( u l u z ) m ( u l u w ) n = p ( z ) e g ( z , w ) ,

if and only if one of the following situations occurs:

(1)  l = 0 and u ( z , w ) = c 1 − n m ∫ p 1 m ( z ) e α ( z ) d z + c 1 ∫ e β ( w ) w + c 0 for some constants c 0 , c 1 , where α ( z ) and β ( w ) are the polynomials in C such that m α ( z ) + n β ( w ) = g ( z , w ) ; in particular, if g ( z , w ) = 0 and p ( z ) = K for a constant K ( ≠ 0 ) , then u ( z , w ) is linear.

(2.1)  l = 0 and u ( z , w ) = c 1 e − κ ∫ e f ( s ) d s + c 0 for some constants c 0 , c 1 , and κ , where f ( s ) is an entire function of s ∈ C with s = s ( z , w ) ≔ c 1 − 1 − n m e κ ∫ p 1 m ( z ) d z + w and

g ( z , w ) = ( m + n ) f c 1 − 1 − n m e κ ∫ p 1 m ( z ) d z + w − n κ .

(2.2)  l ≥ 1 and u ( z , w ) = ( ( l + 1 ) c 1 c 2 e − κ ) 1 l + 1 e f ( s ) l + 1 for an additional constant c 2 , provided f ( s ) is linear in s ∈ C , with all other notations appearing exactly the same as in (2.1).

(3.1)  l = 0 and u ( z , w ) = c 3 − 1 e − κ r ( z ) e a 0 ( z ) + c 3 w + c 0 for some constants c 0 , c 3 , and κ , where a 0 ( z ) is a polynomial of z ∈ C satisfying a 0 ( z ) = g ( z , w ) + n κ m + n − c 3 w and r ( z ) is a polynomial of z ∈ C (together with a 0 ( z ) ) satisfying p ( z ) = c 3 − m e − m κ ( r ′ ( z ) + r ( z ) a 0 ′ ( z ) ) m r n ( z ) .

(3.2)  l ≥ 1 and u ( z , w ) = ( ( l + 1 ) c 3 − 1 e − κ r ( z ) ) 1 l + 1 e a 0 ( z ) + c 3 w l + 1 , provided r ( z ) = r ˜ k ( l + 1 ) ( z ) for an integer k ≥ 1 and a polynomial r ˜ ( z ) , with all other notations the same as in (3.1).

Recently, by making use of the difference analogs of Nevanlinna theory in C n [22,23], many scholars paid considerable attention to describing the properties on the solutions of some complex differential difference equations (can be found in [24–28]). In particular, Xu et al. [23] and [29–31] dealt with the existence of the solutions for some Fermat-type complex PDE and obtained that the transcendental entire solution with finite-order of ∂ f ( z 1 , z 2 ) ∂ z 1 2 + f ( z 1 , z 2 ) 2 = 1 , has the form of f ( z 1 , z 2 ) = sin ( A z 1 + g ( z 2 ) ) , where A is a constant on C satisfying A 2 = 1 , and g ( z 2 ) is a polynomial in one variable z 2 (see [30, Corollary 1.4]). Later, Xu et al. [32] in 2020 extended their results when the equation is turned into the system of complex PDEs and obtained

Theorem D

[32, Corollary 1.1] Any pair of transcendental entire solutions with finite-order for the system of Fermat-type PDEs

(1.6) ∂ f 1 ( z 1 , z 2 ) ∂ z 1 2 + f 2 ( z 1 , z 2 ) 2 = 1 , ∂ f 2 ( z 1 , z 2 ) ∂ z 1 2 + f 1 ( z 1 , z 2 ) 2 = 1

have the following forms:

( f 1 ( z ) , f 2 ( z ) ) = e L ( z ) + B 1 + e − ( L ( z ) + B 1 ) 2 , η e L ( z ) + B 1 + 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 and η satisfy (i) a 1 = i and η = 1 ; (ii) a 1 = − i and η = − 1 .

In 2023, Xu et al. [33] further extended the aforementioned result from the system to the systems of the quadratic trinomial PDEs and obtained

Theorem E

[33, Theorem 2.2] Let D ≔ a d − b c ≠ 0 and ( u , v ) be a pair of transcendental entire solutions with finite-order for system

(1.7) ( a u z 1 + b u z 2 ) ( c v z 1 + d v z 2 ) = e h , ( a v z 1 + b v z 2 ) ( c u z 1 + d u z 2 ) = e h .

Then, ( u , v ) is one of the following forms:

  1. If h ( z ) = ψ 1 ( z 1 ) + ψ 2 ( z 1 ) − ξ and a − c A = 0 , then

    ( u , v ) = ( F 1 ( z 1 ) ⁄ a , G 1 ( z 1 ) ⁄ a ) ,

    where A = e ξ , F 1 ′ ( t ) = e ψ 1 ( t ) and G 1 ′ ( t ) = e ψ 2 ( t ) , ψ j ( t ) , j = 1 , 2 are polynomials in t .

  2. If h ( z ) = ϕ 1 ( z 2 ) + ϕ 2 ( z 2 ) − ξ and b − d A = 0 , then

    ( u , v ) = ( F 2 ( z 2 ) ⁄ b , G 2 ( z 2 ) ⁄ b ) ,

    where F 2 ′ ( t ) = e ϕ 1 ( t ) and G 2 ′ ( t ) = e ϕ 2 ( t ) , ϕ j ( t ) , j = 1 , 2 are polynomials in t.

  3. If h ( z ) = φ 1 z 2 − b − d A a − c A z 1 + φ 2 z 2 − b − d A a − c A z 1 + ξ and ( b − d A ) ( a − c A ) ≠ 0 , then

    ( u , v ) = a A − 1 − c D F 3 z 2 − b − d A a − c A z 1 , a A − 1 − c D G 3 z 2 − b − d A a − c A z 1 ,

    where F 3 ′ ( t ) = e φ 1 ( t ) , G 3 ′ ( t ) = e φ 2 ( t ) , φ j ( t ) , j = 1 , 2 are nonconstant polynomials in C .

  4. If h ( z ) = ψ 3 z 2 − d c z 1 + ϕ 3 z 2 − b a z 1 = ψ 4 z 2 − d c z 1 + ϕ 4 z 2 − b a z 1 , then

    ( u , v ) = a F 4 z 2 − b a z 1 − c F 5 z 2 − d c z 1 D , a G 4 z 2 − b a z 1 − c G 5 z 2 − d c z 1 D ,

    where F 4 ′ ( t ) = e ϕ 4 ( t 1 ) , F 5 ′ ( t ) = e ψ 3 ( t 2 ) , G 4 ′ ( t 1 ) = e ϕ 3 ( t 1 ) , G 5 ′ ( t 2 ) = e ψ 4 ( t 2 ) , ϕ j ( t ) , ψ j ( t ) , j = 3 , 4 are nonconstant polynomials in C .

By observing the forms of system (1.7) in Theorem E, one can suggest the following question

Question 1.1

What had happened about the solutions if the polynomial h in system (1.7) is replaced by two polynomials h 1 and h 2 in C 2 such that e h 1 ≠ e h 2 ?

2 Results

The main aim of this article is to give a positive answer to Question 1.1. More precisely, we will provide the description of the transcendental entire solutions for the system of the first-order nonlinear PDEs with product-type

(2.1) ( a 1 f z 1 + a 2 f z 2 ) ( a 3 g z 1 + a 4 g z 2 ) = e h 1 , ( a 1 g z 1 + a 2 g z 2 ) ( a 3 f z 1 + a 4 f z 2 ) = e h 2 ,

and

(2.2) ( a 1 f z 1 + a 2 g z 1 ) ( a 3 f z 2 + a 4 g z 2 ) = e h 1 , ( a 1 f z 2 + a 2 g z 2 ) ( a 3 f z 1 + a 4 g z 1 ) = e h 2 ,

in C 2 , where f z t = ∂ f ∂ z t , g z t = ∂ g ∂ z t , t = 1 , 2 , a j ∈ C , j = 1 , 2 , 3 , 4 , and h 1 and h 2 are polynomials in C 2 . It is note

worthy, as far as we know, that the subject of this article seems to be discussed not much in the field of the nonlinear PDEs. We obtain

Theorem 2.1

Let D ≔ a 1 a 4 − a 2 a 3 ≠ 0 , a j ∈ C , h 1 , h 2 be two polynomials in C 2 satisfying e h 1 ≠ e h 2 . Let ( f , g ) be a pair of the finite-order transcendental entire solutions of system (2.1).

(i) If h 1 = φ 1 ( z 1 ) + ψ 1 ( a 2 z 1 − a 1 z 2 ) , h 2 = φ 1 ( z 1 ) + φ 2 ( a 4 z 1 − a 3 z 2 ) − ξ , and e ξ = a 1 a 3 , then

( f , g ) = F 1 ( z 1 ) a 1 , G 1 ( a 4 z 1 − a 3 z 2 ) − G 2 ( a 2 z 1 − a 1 z 2 ) D ,

where F 1 ′ ( t ) = e φ 1 ( t ) , φ 1 ( t ) is a polynomial in t .

If h 1 = φ 3 ( z 2 ) + ψ 1 ( a 2 z 1 − a 1 z 2 ) , h 2 = φ 3 ( z 2 ) + φ 2 ( a 4 z 1 − a 3 z 2 ) − ξ , and e ξ = a 2 a 4 , then

( f , g ) = F 2 ( z 2 ) a 2 , G 1 ( a 4 z 1 − a 3 z 2 ) − G 2 ( a 2 z 1 − a 1 z 2 ) D ,

where F 2 ′ ( t ) = e φ 3 ( t ) , φ 3 ( t ) is a polynomial in t.

If h 1 = φ 4 z 2 − a 2 − A a 4 a 1 − A a 3 z 1 + ψ 1 ( a 2 z 1 − a 1 z 2 ) , h 2 = φ 4 z 2 − a 2 − A a 4 a 1 − A a 3 z 1 + φ 2 ( a 4 z 1 − a 3 z 2 ) − ξ , and A = e ξ , then

( f , g ) = A − 1 a 1 − a 3 D F 3 z 2 − a 2 − A a 4 a 1 − A a 3 z 1 , G 1 ( a 4 z 1 − a 3 z 2 ) − G 2 ( a 2 z 1 − a 1 z 2 ) D ,

where F 3 ′ ( t ) = e φ 4 ( t ) , φ 4 ( t ) is a polynomial in t, G 1 ′ ( t ) = e φ 2 ( t ) , and G 2 ′ ( t ) = e ψ 1 ( t ) , φ 2 ( t ) , ψ 1 ( t ) are polynomials in t.

(ii) If h 1 = ψ 3 ( z 1 ) + φ 5 ( a 4 z 1 − a 3 z 2 ) , h 2 = ψ 3 ( z 1 ) + ψ 2 ( a 2 z 1 − a 1 z 2 ) − ξ , and e ξ = a 3 a 1 , then

( f , g ) = F 4 ( a 4 z 1 − a 3 z 2 ) − F 5 ( a 2 z 1 − a 1 z 2 ) D , G 3 ( z 1 ) a 3 ,

where G 3 ′ ( t ) = e ψ 3 ( t ) , ψ 3 ( t ) is a polynomial in t.

If h 1 = ψ 4 ( z 2 ) + φ 5 ( a 4 z 1 − a 3 z 2 ) , h 2 = ψ 4 ( z 2 ) + ψ 2 ( a 2 z 1 − a 1 z 2 ) − ξ , and e ξ = a 4 a 2 , then

( f , g ) = F 4 ( a 4 z 1 − a 3 z 2 ) − F 5 ( a 2 z 1 − a 1 z 2 ) D , G 4 ( z 2 ) a 4 ,

where G 4 ′ ( t ) = e ψ 4 ( t ) , ψ 4 ( t ) is a polynomial in t.

If h 1 = ψ 5 ( z 2 − B a 1 − a 3 B a 2 − a 4 z 1 ) + φ 5 ( a 4 z 1 − a 3 z 2 ) , h 2 = ψ 5 ( z 2 − B a 1 − a 3 B a 2 − a 4 z 1 ) + ψ 2 ( a 2 z 1 − a 1 z 2 ) − ξ , and ( B a 1 − a 3 ) ( B a 2 − a 4 ) ≠ 0 , where e ξ = B , then

( f , g ) = F 4 ( a 4 z 1 − a 3 z 2 ) − F 5 ( a 2 z 1 − a 1 z 2 ) D , B − 1 a 4 − a 3 D G 5 z 2 − B a 1 − a 3 B a 2 − a 4 z 1 ,

where G 5 ′ ( t ) = e ψ 5 ( t ) , ψ 5 ( t ) is a polynomial in t , F 4 ′ ( t ) = e φ 5 ( t ) , F 5 ′ ( t ) = e ψ 2 ( t ) , φ 5 ( t ) , ψ 2 ( t ) are polynomials in t.

(iii) If h 1 = φ 6 ( a 4 z 1 − a 3 z 2 ) + ψ 6 ( a 2 z 1 − a 1 z 2 ) , h 2 = φ 7 ( a 4 z 1 − a 3 z 2 ) + ψ 7 ( a 2 z 1 − a 1 z 2 ) , and both of φ 6 ( a 4 z 1 − a 3 z 2 ) − ψ 7 ( a 2 z 1 − a 1 z 2 ) and ψ 6 ( a 2 z 1 − a 1 z 2 ) − φ 7 ( a 4 z 1 − a 3 z 2 ) are not constants, then

( f , g ) = F 6 ( a 4 z 1 − a 3 z 2 ) − F 7 ( a 2 z 1 − a 1 z 2 ) D , G 7 ( a 4 z 1 − a 3 z 2 ) − G 6 ( a 2 z 1 − a 1 z 2 ) D ,

where F 6 ′ ( t ) = e φ 6 ( t ) , G 6 ′ ( t ) = e ψ 6 ( t ) , F 7 ′ ( t ) = e ψ 7 ( t ) , G 7 ′ ( t ) = e φ 7 ( t ) , φ 6 , φ 7 , ψ 6 , ψ 7 are nonconstant polynomials in C .

(iv) If h 1 − h 2 = ξ 1 + ξ 2 , h 1 = φ 8 ( z 1 ) + ψ 8 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 , and e ξ 1 = a 1 a 3 , then

( f , g ) = 1 a 1 F 8 ( z 1 ) , a 1 − a 3 e − ξ 2 D G 8 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 ,

where F 8 ′ ( t ) = e φ 8 ( t ) , G 8 ′ ( t ) = e ψ 8 ( t ) , φ 8 ( t ) , ψ 8 ( t ) are nonconstant polynomials in C .

If h 1 − h 2 = ξ 1 + ξ 2 , h 1 = ψ 9 ( z 1 ) + φ 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 and e ξ 2 = a 3 a 1 , then

( f , g ) = a 1 e − ξ 1 − a 3 D F 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 , 1 a 3 G 9 ( z 1 ) ,

where F 9 ′ ( t ) = e φ 9 ( t ) , G 9 ′ ( t ) = e ψ 9 ( t ) , φ 9 ( t ) , ψ 9 ( t ) are nonconstant polynomials in C .

If h 1 − h 2 = ξ 1 + ξ 2 , and

h 1 = φ 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 + ψ 10 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 ,

then

( f , g ) = a 1 e − ξ 1 − a 3 D F 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 , a 1 − a 3 e − ξ 2 D G 10 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 ,

where F 9 ′ ( t ) = e φ 9 ( t ) , G 10 ′ ( t ) = e ψ 10 ( t ) , φ 9 ( t ) , ψ 10 ( t ) are nonconstant polynomials in C .

The following examples show that the forms of solutions of system (2.1) are precise.

Example 2.1

Let a 1 = 1 , a 2 = − 1 , a 3 = 2 , and a 4 = 1 in Theorem 2.1, we then have the following system:

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

Thus,

  1. ( f , g ) = ∫ e z 1 3 d z 1 , 1 3 ∫ e ( z 1 − 2 z 2 ) 5 d z 1 − 1 3 ∫ e 2 ( − z 1 − z 2 ) 3 d z 1

    is a pair of finite-order transcendental entire solutions of (2.3) with

    h 1 = z 1 3 + 2 ( − z 1 − z 2 ) 3 , h 2 = z 1 3 + ( z 1 − 2 z 2 ) 5 + log 2 ;

  2. ( f , g ) = − ∫ e z 2 2 d z 1 , 1 3 ∫ e ( z 1 − 2 z 2 ) 5 d z 1 − 1 3 ∫ e 2 ( − z 1 − z 2 ) 3 d z 1

    is a pair of finite-order transcendental entire solutions of (2.3) with

    h 1 = z 2 2 + 2 ( − z 1 − z 2 ) 3 , h 2 = z 2 2 + ( z 1 − 2 z 2 ) 5 − π i ;

  3. ( f , g ) = − 3 5 ∫ e ( z 2 − 2 3 z 1 ) 2 d z 1 , 1 3 ∫ e ( z 1 − 2 z 2 ) 5 d z 1 − 1 3 ∫ e 2 ( − z 1 − z 2 ) 3 d z 1

    is a pair of finite-order transcendental entire solutions of (2.3) with

    h 1 = z 2 − 2 3 z 1 2 + 2 ( − z 1 − z 2 ) 3 and h 2 = z 2 − 2 3 z 1 2 + ( z 1 − 2 z 2 ) 5 − log 5 .

    This corresponds to conclusion (i) in Theorem 2.1. Moreover, one can give the corresponding example for the conclusion (ii) in Theorem 2.1.

Remark 2.1

We can see that the forms of solutions (1), (2), and (3) in Example 2.1 do not include any case of Theorem E. Thus, this shows that the conclusions of Theorem 2.1 are improvement of Theorem E to some extent.

Example 2.2

Let

( f , g ) = ∫ e z 1 − z 2 d z 1 − ∫ e ( z 1 − 2 z 2 ) 4 d z 1 , ∫ e ( z 1 − z 2 ) 3 d z 1 − ∫ e ( z 1 − 2 z 2 ) 2 d z 1 .

Thus, ( f , g ) is a pair of transcendental entire solution of system (2.1) with a 1 = 2 , a 2 = a 3 = a 4 = 1 , and

h 1 = ( z 1 − z 2 ) + ( z 1 − 2 z 2 ) 2 , h 2 = ( z 1 − z 2 ) 3 + ( z 1 − 2 z 2 ) 4 .

This corresponds to conclusion (iii) in Theorem 2.1.

Example 2.3

Let a 1 = 3 , a 2 = 2 , a 3 = 1 , a 4 = 1 in Theorem 2.1, we then have the following system:

(2.4) ( 3 f z 1 + 2 f z 2 ) ( g z 1 + g z 2 ) = e h 1 , ( 3 g z 1 + 2 g z 2 ) ( f z 1 + f z 2 ) = e h 2 .

Thus,

  1. ( f , g ) = 1 3 ∫ e z 1 3 + z 1 2 d z 1 , 5 2 ∫ e ( z 2 − 3 5 z 2 ) 2 d z 2

    is a pair of finite-order transcendental entire solutions of (2.4) with

    h 1 = z 1 3 + z 2 + z 2 − 3 5 z 1 2 , h 2 = z 1 3 + z 2 + z 2 − 3 5 z 1 2 − log 6 ;

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

    is a pair of finite-order transcendental entire solutions of (2.4) with

    h 1 = z 1 2 + 2 z 1 + z 2 − 1 2 z 1 3 , h 2 = z 1 2 + 2 z 1 + z 2 − 1 2 z 1 3 + log 3 + 2 π i ;

  3. ( f , g ) = 2 ∫ e z 2 − 1 2 z 1 2 d z 1 , 11 4 ∫ e z 2 − 7 11 z 1 3 d z 2

    is a pair of finite-order transcendental entire solutions of (2.4) with

    h 1 = z 2 − 1 2 z 1 2 + z 2 − 7 11 z 1 3 , h 2 = z 2 − 1 2 z 1 2 + z 2 − 7 11 z 1 3 + log 4 + 2 π i .

    This corresponds to conclusion (iv) in Theorem 2.1.

Similar to the argument as in the proof of Theorem 2.1, one can obtain the following corollary easily.

Corollary 2.1

Let h 1 and h 2 be two nonconstant polynomials in C 2 such that e h 1 ≠ e h 2 , and ( f , g ) be the pair of finite-order transcendental entire solutions for system

(2.5) f z 1 g z 2 = e h 1 , g z 1 f z 2 = e h 2 .

  1. If h 1 = φ 1 ( z 1 ) + φ 2 ( z 2 ) , h 2 = φ 3 ( z 1 ) + φ 4 ( z 2 ) , then we have

    ( f , g ) = ( F 1 ( z 1 ) + F 2 ( z 2 ) , G 1 ( z 1 ) + G 2 ( z 2 ) ) ,

    where F 1 ′ = e φ 1 , F 2 ′ = e φ 2 , G 1 ′ = e φ 3 , G 2 ′ = e φ 4 .

  2. If h 1 = φ 5 ( z 1 ) + φ 7 ( z 1 + A z 2 ) , h 2 = φ 6 ( z 2 ) + φ 7 ( z 1 + A z 2 ) − ξ , and A = e ξ , then we have

    ( f , g ) = ( F 3 ( z 1 ) + F 4 ( z 2 ) , A − 1 G 3 ( z 1 + A z 2 ) ) .

    If h 1 = φ 8 ( z 2 ) + φ 10 ( z 1 + A − 1 z 2 ) , h 2 = φ 9 ( z 1 ) + φ 10 ( z 1 + A − 1 z 2 ) − ξ , and A = e ξ , then we have

    ( f , g ) = ( F 5 ( z 1 + A − 1 z 2 ) , G 4 ( z 1 ) + G 5 ( z 2 ) ) ,

    where F 3 ′ ( x ) = e φ 5 ( x ) , F 4 ′ ( x ) = e φ 6 ( x ) , F 5 ′ ( t ) = e φ 10 ( t ) , G 3 ′ ( x ) = e φ 7 ( x ) , G 4 ′ ( z 1 ) = e φ 9 ( z 1 ) , G 5 ′ ( z 2 ) = e φ 8 ( z 2 ) .

  3. If h 1 = φ 11 ( z 1 + B 1 − 1 z 2 ) + φ 12 ( z 1 + B 2 z 2 ) , h 2 = φ 11 ( z 1 + B 1 − 1 z 2 ) + φ 12 ( z 1 + B 2 z 2 ) − ξ 1 − ξ 2 , and B 1 = e ξ 1 , B 2 = e ξ 2 , then we have

    ( f , g ) = ( F 6 ( z 1 + B 1 − 1 z 2 ) , G 6 ( z 1 + B 2 z 2 ) ) ,

    where F 6 ′ ( x ) = e φ 11 ( x ) , G 6 ′ ( x ) = e φ 12 ( x ) .

For system (2.2), we obtain

Theorem 2.2

Let D ≔ a 1 a 4 − a 2 a 3 ≠ 0 , a j ∈ C , h 1 , and h 2 be two polynomials in C 2 satisfying e h 1 ≠ e h 2 . Let ( f , g ) be a pair of the finite-order transcendental entire solutions of system (2.2).

  1. If h 1 = φ 1 ( z 1 ) + φ 3 ( z 1 + A z 2 ) + ξ , h 2 = φ 2 ( z 2 ) + φ 3 ( z 1 + A z 2 ) , where A = e ξ , ξ ∈ C , then

    f = 1 D [ a 4 ( F 1 ( z 1 ) + F 2 ( z 2 ) ) − a 2 F 3 ( z 1 + A z 2 ) ] , g = 1 D [ a 1 G 3 ( z 1 + A z 2 ) − a 3 ( G 1 ( z 1 ) + G 2 ( z 2 ) ) ] ,

    if h 1 = φ 4 ( z 2 ) + φ 6 ( z 1 + A − 1 z 2 ) , h 2 = φ 5 ( z 1 ) + φ 6 ( z 1 + A − 1 z 2 ) − ξ , then

    f = 1 D [ a 4 F 6 ( z 1 + A − 1 z 2 ) − a 2 ( F 5 ( z 1 ) + F 4 ( z 2 ) ) ] , g = 1 D [ a 1 G 5 ( z 1 ) + a 1 G 4 ( z 2 ) − a 3 ( G 6 ( z 1 + A − 1 z 2 ) ) ] .

  2. If h 1 = φ 7 ( z 1 ) + φ 9 ( z 2 ) , h 2 = φ 8 ( z 2 ) + φ 10 ( z 1 ) , then

    f = 1 D [ a 4 ( F 7 ( z 1 ) + F 8 ( z 2 ) ) − a 2 ( F 9 ( z 2 ) + F 10 ( z 1 ) ) ] , g = 1 D [ a 1 ( G 10 ( z 1 ) + G 9 ( z 2 ) ) − a 3 ( G 7 ( z 1 ) + G 8 ( z 2 ) ) ] .

  3. If h 1 = φ 11 ( z 1 + A − 1 z 2 ) + φ 12 ( z 1 + B z 2 ) + ξ 2 , h 2 = φ 11 ( z 1 + A − 1 z 2 ) + φ 12 ( z 1 + B z 2 ) − ξ 1 , ξ 1 , ξ 2 ∈ C , and A 1 = e ξ 1 , A 2 = e ξ 2 , then

    f = 1 D [ a 4 F 11 ( z 1 + A 1 − 1 z 2 ) − a 2 F 12 ( z 1 + A 2 z 2 ) ] , g = 1 D [ a 1 G 12 ( z 1 + A 2 z 2 ) − a 3 G 11 ( z 1 + A 1 − 1 z 2 ) ] ,

    where φ j ( t ) , j = 1 , … , 12 are polynomials in t , F j , G j , j = 1 , … , 12 are finite-order entire functions satisfying F j ′ ( t ) = G j ′ ( t ) = e φ j ( t ) , j = 1 , … , 12 .

The following example shows that every case in the forms of solutions in Theorem 2.2 can be occurred.

Example 2.4

Let a 1 = 1 , a 2 = 2 , a 3 = − 1 , a 4 = − 1 in Theorem 2.2, we then have the following system:

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

  1. Let

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

    Thus, ( f , g ) is a pair of finite-order transcendental entire solutions of (2.4) with

    h 1 = z 1 2 + 2 z 1 + ( z 1 + 2 z 2 ) 2 + log 2 , h 2 = z 2 2 + ( z 1 + 2 z 2 ) 2 .

  2. Let

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

    Thus, ( f , g ) is a pair of finite-order transcendental entire solutions of (2.4) with

    h 1 = z 2 2 + ( z 1 + 1 2 z 2 ) 3 , h 2 = z 1 3 + ( z 1 + 1 2 z 2 ) 3 − log 2 .

    Cases (1) and (2) correspond to conclusion (i) of Theorem 2.2.

  3. Let

    f = − ∫ e z 1 3 + 2 z 1 d z 1 − ∫ e 3 z 2 d z 2 − 2 ∫ e z 2 + z 2 2 d z 2 − 2 ∫ e z 1 + z 1 2 d z 1 , g = ∫ e z 1 + z 1 2 d z 1 + ∫ e z 2 2 + z 2 d z 2 + ∫ e z 1 3 + 2 z 1 d z 1 + ∫ e 3 z 2 d z 2 .

    Thus, ( f , g ) is a pair of finite-order transcendental entire solutions of (2.4) with

    h 1 = z 1 3 + 2 z 1 + z 2 + z 2 2 , h 2 = 3 z 2 + z 1 + z 1 2 .

    This corresponds to conclusion (ii) in Theorem 2.2.

  4. Let

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

    Thus, ( f , g ) is a pair of finite-order transcendental entire solutions of (2.4) with

    h 1 = ( z 1 + 3 z 2 ) 2 + z 1 + 2 z 2 + log 2 , h 2 = ( z 1 + 3 z 2 ) 2 + z 1 + 2 z 2 + log 3 .

    This corresponds to conclusion (iii) in Theorem 2.2.

3 Proofs of Theorem 2.1

We first start the following lemmas, which play key roles in proving our main theorems.

Lemma 3.1

[34,35] 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, ρ ( n F ) denotes the order of the counting function of zeros of F .

Lemma 3.2

[36] If g and h are the entire functions 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

[37, Theorem 1.106] Suppose that a 0 ( z ) , a 1 ( z ) , … , a n ( z ) ( n ≥ 1 ) are meromorphic functions on C m and g 0 ( z ) , g 1 ( z ) , … , g n ( z ) are the entire functions on C m such that g j ( z ) − g k ( z ) are not constants for 0 ≤ j < k ≤ n . If the following conditions

∑ j = 0 n a j ( z ) e g j ( z ) ≡ 0

and

‖ T ( r , a j ) = o ( T ( r ) ) , j = 0 , 1 , … , n

hold, where T ( r ) = min 0 ≤ j < k ≤ n T ( r , e g j − g k ) , then a j ( z ) ≡ 0 ( j = 0 , 1 , 2 , … , n ) .

3.1 Proof of Theorem 2.1

Assume that ( f , g ) is a pair of transcendental entire solutions of system (2.1). Noting that f , g , h 1 , and h 2 are entire functions, and in view of the form of system (2.1), we can deduce that a 1 f z 1 + a 2 f z 2 , a 3 g z 1 + a 4 g z 2 , a 1 g z 1 + a 2 g z 2 , a 3 f z 1 + a 4 f z 2 have no zeros and poles. Otherwise, we will obtain a contradiction with the assumption of f and g being entire functions. Thus, by Lemmas 3.1 and 3.2, there exist four polynomials α ˜ , α ˜ ˜ , β ˜ , and β ˜ ˜ in C 2 satisfying

(3.1) a 1 f z 1 + a 2 f z 2 = e α ˜ , a 3 g z 1 + a 4 g z 2 = e β ˜ ,

and

(3.2) a 1 g z 1 + a 2 g z 2 = e α ˜ ˜ , a 3 f z 1 + a 4 f z 2 = e β ˜ ˜ ,

which implies that

(3.3) α ˜ + β ˜ = h 1 and α ˜ ˜ + β ˜ ˜ = h 2 .

Due to D = a 1 a 4 − a 2 a 3 ≠ 0 , and by solving the aforementioned systems, we can obtain

(3.4) f z 1 = 1 D ( a 4 e α ˜ − a 2 e β ˜ ˜ ) , f z 2 = 1 D ( a 1 e β ˜ ˜ − a 3 e α ˜ ) ,

and

(3.5) g z 1 = 1 D ( a 4 e α ˜ ˜ − a 2 e β ˜ ) , g z 2 = 1 D ( a 1 e β ˜ − a 3 e α ˜ ˜ ) .

In view of the fact that f z 1 z 2 = f z 2 z 1 and g z 1 z 2 = g z 2 z 1 , it yields from (3.4) and (3.5) that

a 4 α ˜ z 2 e α ˜ − a 2 β ˜ ˜ z 2 e β ˜ ˜ = a 1 β ˜ ˜ z 1 e β ˜ ˜ − a 3 α ˜ z 1 e α ˜ , a 4 α ˜ ˜ z 2 e α ˜ ˜ − a 2 β ˜ z 2 e β ˜ = a 1 β ˜ z 1 e β ˜ − a 3 α ˜ ˜ z 1 e α ˜ ˜ ,

i.e.,

(3.6) ( a 4 α ˜ z 2 + a 3 α ˜ z 1 ) e α ˜ = ( a 1 β ˜ ˜ z 1 + a 2 β ˜ ˜ z 2 ) e β ˜ ˜

and

(3.7) ( a 4 α ˜ ˜ z 2 + a 3 α ˜ ˜ z 1 ) e α ˜ ˜ = ( a 1 β ˜ z 1 + a 2 β ˜ z 2 ) e β ˜ .

Now, four cases will be discussed in the following.

Case 1. If α ˜ − β ˜ ˜ = ξ 1 , ξ 1 ∈ C and β ˜ − α ˜ ˜ is a nonconstant polynomial, then it follows that

(3.8) α ˜ z 1 = β ˜ ˜ z 1 and α ˜ z 2 = β ˜ ˜ z 2 .

Thus, we can deduce from (3.6) and (3.8) that

(3.9) ( a 1 − a 3 e ξ 1 ) α ˜ z 1 + ( a 2 − a 4 e ξ 1 ) α ˜ z 2 ≡ 0 .

If a 4 α ˜ ˜ z 2 + a 3 α ˜ ˜ z 1 ≠ 0 , we can rewrite (3.7) as the following form:

e α ˜ ˜ − β ˜ = a 1 β ˜ z 1 + a 2 β ˜ z 2 a 3 α ˜ ˜ z 1 + a 4 α ˜ ˜ z 2 .

Indeed, using the results (see, for example, [38, p. 99], [35]) or [39, Lemma 3.2], we can conclude from the aforementioned equality that T ( r , e α ˜ ˜ − β ˜ ) = O { log r } , outside possibly a set of finite Lebesgue measure. On the other hand, we have

lim r → ∞ T ( r , e α ˜ ˜ − β ˜ ) log r = + ∞ ,

since α ˜ ˜ − β ˜ is not a constant. This, clearly, is

a 1 β ˜ z 1 + a 2 β ˜ z 2 ≡ 0 and a 3 α ˜ ˜ z 1 + a 4 α ˜ ˜ z 2 ≡ 0 .

By solving the aforementioned equations, we have

(3.10) β ˜ = ψ 1 ( a 2 z 1 − a 1 z 2 ) and α ˜ ˜ = φ 2 ( a 4 z 1 − a 3 z 2 ) ,

where ψ 1 ( x ) and φ 2 ( x ) are the polynomials in x . By (3.5) and (3.10), we have

(3.11) g = G 1 ( a 4 z 1 − a 3 z 2 ) − G 2 ( a 2 z 1 − a 1 z 2 ) D ,

where G 1 ( x ) and G 2 ( x ) are the finite-order transcendental entire functions satisfying

G 1 ′ ( x ) = e φ 2 ( x ) and G 2 ′ ( x ) = e ψ 1 ( x ) .

Due to (3.9), it obviously follows that a 1 − a 3 e ξ 1 = 0 and a 2 − a 4 e ξ 1 = 0 cannot hold simultaneously. Otherwise, we can deduce that a 1 a 4 − a 2 a 3 = a 2 a 3 e ξ − a 2 a 3 e ξ = 0 ; this is a contradiction with D ≠ 0 .

If a 1 − a 3 e ξ 1 = 0 and a 2 − a 4 e ξ 1 ≠ 0 , then it follows from (3.9) that α ˜ z 2 ≡ 0 and β ˜ ˜ z 2 ≡ 0 , which means that α ˜ = φ 1 ( z 1 ) , where φ 1 ( z 1 ) is a polynomial in z 1 . By combining with (3.3) and α ˜ − β ˜ ˜ = ξ 1 , we have β ˜ ˜ = φ 1 ( z 1 ) − ξ 1 . Substituting these into (3.4), we have

(3.12) f z 1 = 1 D ( a 4 − a 2 e − ξ 1 ) e α ˜ = 1 a 1 e φ 1 ( z 1 ) and f z 2 = 1 D ( a 1 e − ξ 1 − a 3 ) e α ˜ ≡ 0 ,

which lead to

(3.13) f = F 1 ( z 1 ) a 1 ,

where F 1 ′ ( z 1 ) = e φ 1 ( z 1 ) . Noting that (3.3), (3.10), and α ˜ = φ 1 ( z 1 ) , we have

(3.14) h 1 = φ 1 ( z 1 ) + ψ 1 ( a 2 z 1 − a 1 z 2 ) , and h 2 = φ 1 ( z 1 ) + φ 2 ( a 4 z 1 − a 3 z 2 ) − ξ 1 .

If a 1 − a 3 e ξ 1 ≠ 0 and a 2 − a 4 e ξ 1 = 0 , using the same argument as in above, we have

(3.15) f = F 2 ( z 2 ) a 2 , g = G 1 ( a 4 z 1 − a 3 z 2 ) − G 2 ( a 2 z 1 − a 1 z 2 ) D ,

where F 2 ′ ( z 2 ) = e φ 3 ( z 2 ) , φ 3 ( z 2 ) is a polynomial in z 2 , G 1 , and G 2 are stated as in (3.11) and

(3.16) h 1 = φ 3 ( z 2 ) + ψ 1 ( a 2 z 1 − a 1 z 2 ) and h 2 = φ 3 ( z 2 ) + φ 2 ( a 4 z 1 − a 3 z 2 ) − ξ 1 .

If ( a 1 − a 3 e ξ 1 ) ( a 2 − a 4 e ξ 1 ) ≠ 0 , then it follows from (3.9) and (3.10) that

(3.17) α ˜ z 1 + a 2 − A a 4 a 1 − A a 3 α ˜ z 2 ≡ 0 ,

where e ξ 1 = A . The characteristic equations of (3.16) are

d z 1 d t = 1 , d z 2 d t = a 2 − A a 4 a 1 − A a 3 , and d α ˜ d t = 0 .

We can conclude that α ˜ = φ 4 z 2 − a 2 − A a 4 a 1 − A a 3 z 1 , where φ 4 ( x ) is a polynomial in x = z 2 − a 2 − A a 4 a 1 − A a 3 z 1 . By combining with α ˜ − β ˜ ˜ = ξ 1 , we have β ˜ ˜ = φ 4 ( z 2 − a 2 − A a 4 a 1 − A a 3 z 1 ) − ξ 1 . Substituting these into (3.4) and (3.5), we have

f z 1 = a 4 − A − 1 a 2 D e φ 4 ( z 2 − a 2 − A a 4 a 1 − A a 3 z 1 ) , and f z 2 = A − 1 a 1 − a 3 D e φ 4 ( z 2 − a 2 − A a 4 a 1 − A a 3 z 1 ) .

Thus, by a simple calculation, we can deduce that

(3.18) f = A − 1 a 1 − a 3 D F 3 z 2 − a 2 − A a 4 a 1 − A a 3 z 1 and g = G 1 ( a 4 z 1 − a 3 z 2 ) − G 2 ( a 2 z 1 − a 1 z 2 ) D ,

where F 3 ′ ( t ) = e φ 4 ( t ) , G 1 , and G 2 are stated as in (3.11), and

h 1 = φ 4 z 2 − a 2 − A a 4 a 1 − A a 3 z 1 + ψ 1 ( a 2 z 1 − a 1 z 2 ) and h 2 = φ 4 z 2 − a 2 − A a 4 a 1 − A a 3 z 1 + φ 2 ( a 4 z 1 − a 3 z 2 ) − ξ 1 .

Thus, this completes the proof of Theorem 2.1 (i).

Case 2. If β ˜ − α ˜ ˜ = ξ 2 and α ˜ − β ˜ ˜ is a nonconstant polynomial, then it follows that

(3.19) α ˜ ˜ z 1 = β ˜ z 1 and α ˜ ˜ z 2 = β ˜ z 2 .

Thus, we can deduce from (3.6) and (3.7) that

(3.20) ( a 1 e ξ 2 − a 3 ) β ˜ z 1 + ( a 2 e ξ 2 − a 4 ) β ˜ z 2 ≡ 0 .

Using the same argument as in Case 1, we have

(3.21) a 1 β ˜ ˜ z 1 + a 2 β ˜ ˜ z 2 = 0 and a 3 α ˜ z 1 + a 4 α ˜ z 2 = 0 ,

which implies that

(3.22) α ˜ = φ 5 ( a 4 z 1 − a 3 z 2 ) and β ˜ ˜ = ψ 2 ( a 2 z 1 − a 1 z 2 ) ,

where φ 5 ( x ) and ψ 2 ( x ) are the polynomials in x . In view of (3.4) and (3.22), we can deduce that

(3.23) f = F 4 ( a 4 z 1 − a 3 z 2 ) − F 5 ( a 2 z 1 − a 1 z 2 ) D ,

where F 4 ′ ( x ) = e φ 5 ( x ) and F 5 ′ ( x ) = e ψ 2 ( x ) .

If a 1 e ξ 2 − a 3 = 0 and a 2 e ξ 2 − a 4 ≠ 0 , similar to the argument as in Case 1 for (3.20), we have

(3.24) g = G 3 ( z 1 ) a 3 ,

where G 3 ′ ( z 1 ) = e ψ 3 ( z 1 ) , ψ 3 ( z 1 ) is a polynomial in z 1 , and

h 1 = φ 5 ( a 4 z 1 − a 3 z 2 ) + ψ 3 ( z 1 ) and h 2 = ψ 3 ( z 1 ) + ψ 2 ( a 2 z 1 − a 1 z 2 ) − ξ 2 .

If a 1 e ξ 2 − a 3 ≠ 0 and a 2 e ξ 2 − a 4 = 0 , similar to the argument as in Case 1 for (3.20), we have

(3.25) g = G 4 ( z 2 ) a 4 ,

where G 4 ′ ( z 2 ) = e ψ 4 ( z 2 ) , ψ 4 ( z 2 ) is a polynomial in z 2 , and

h 1 = φ 5 ( a 4 z 1 − a 3 z 2 ) + ψ 4 ( z 2 ) and h 2 = ψ 4 ( z 2 ) + ψ 2 ( a 2 z 1 − a 1 z 2 ) − ξ 2 .

If a 1 e ξ 2 − a 3 ≠ 0 and a 2 e ξ 2 − a 4 ≠ 0 , similar to the argument as in Case 1 for (3.20), we have

(3.26) g = B − 1 a 4 − a 2 D G 5 z 2 − a 1 B − a 3 a 2 B − a 4 z 1 ,

where B = e ξ 2 , G 5 ′ ( x ) = e ψ 5 ( x ) , ψ 5 ( x ) is a polynomial in x = z 2 − a 1 B − a 3 a 2 B − a 4 z 1 , and

h 1 = φ 5 ( a 4 z 1 − a 3 z 2 ) + ψ 5 z 2 − a 1 B − a 3 a 2 B − a 4 z 1 and h 2 = ψ 2 ( a 2 z 1 − a 1 z 2 ) + ψ 5 z 2 − a 1 B − a 3 a 2 B − a 4 z 1 − ξ 2 .

Thus, from (3.22)–(3.26), we complete the proof of Theorem 2.1 (ii).

Case 3. Both of α ˜ − β ˜ ˜ and β ˜ − α ˜ ˜ are not constants, by the results in [38, p. 99], [35] or [39, Lemma 3.2], it thus follows from (3.6) and (3.7) that

a 3 α ˜ z 1 + a 4 α ˜ z 2 ≡ 0 , a 1 β ˜ ˜ z 1 + a 2 β ˜ ˜ z 2 ≡ 0 ,

and

a 3 α ˜ ˜ z 1 + a 4 α ˜ ˜ z 2 ≡ 0 , a 1 β ˜ z 1 + a 2 β ˜ z 2 ≡ 0 .

By solving the aforementioned equations, we have

(3.27) α ˜ = φ 6 ( a 4 z 1 − a 3 z 2 ) , α ˜ ˜ = φ 7 ( a 4 z 1 − a 3 z 2 ) ,

and

(3.28) β ˜ = ψ 6 ( a 2 z 1 − a 1 z 2 ) , β ˜ ˜ = ψ 7 ( a 2 z 1 − a 1 z 2 ) ,

where φ 6 ( x ) , φ 7 ( x ) , ψ 6 ( x ) , ψ 7 ( x ) are the polynomials in x . Thus, we can deduce from (3.4), (3.5), (3.27), and (3.28) that

(3.29) f = F 6 ( a 4 z 1 − a 3 z 2 ) − F 7 ( a 2 z 1 − a 1 z 2 ) D and g = G 7 ( a 4 z 1 − a 3 z 2 ) − G 6 ( a 2 z 1 − a 1 z 2 ) D ,

where F 6 , F 7 , G 6 , and G 7 are entire functions satisfying

F 6 ′ ( t 2 ) = e φ 6 ( t 2 ) , F 7 ′ ( t 1 ) = e ψ 7 ( t 1 ) , G 6 ′ ( t 1 ) = e ψ 6 ( t 1 ) , and G 7 ′ ( t 2 ) = e φ 7 ( t 2 ) .

These lead to

(3.30) h 1 = φ 6 ( a 4 z 1 − a 3 z 2 ) + ψ 6 ( a 2 z 1 − a 1 z 2 ) and h 2 = φ 7 ( a 4 z 1 − a 3 z 2 ) + ψ 7 ( a 2 z 1 − a 1 z 2 ) .

This completes the proof of Theorem 2.1 (iii).

Case 4. Both α ˜ − β ˜ ˜ and β ˜ − α ˜ ˜ are constants, let us assume that α ˜ − β ˜ ˜ = ξ 1 and β ˜ − α ˜ ˜ = ξ 2 , where ξ 1 , ξ 2 ∈ C . Thus, it follows from (3.3) that

(3.31) α ˜ z 1 = β ˜ ˜ z 1 , α ˜ z 2 = β ˜ ˜ z 2 , β ˜ z 1 = α ˜ ˜ z 1 , β ˜ z 2 = α ˜ ˜ z 2 ,

and

(3.32) h 1 − h 2 = ξ 1 + ξ 2 .

Substituting (3.31) into (3.6) and (3.7), we have

(3.33) ( a 1 − a 3 e ξ 1 ) α ˜ z 1 + ( a 2 − a 4 e ξ 1 ) α ˜ z 2 ≡ 0 ,

and

(3.34) ( a 1 e ξ 2 − a 3 ) β ˜ z 1 + ( a 2 e ξ 2 − a 4 ) β ˜ z 2 ≡ 0 .

If a 1 − a 3 e ξ 1 = 0 , noting that D = a 1 a 4 − a 2 a 3 ≠ 0 , we have a 2 − a 4 e ξ 1 ≠ 0 . It follows that α ˜ z 2 ≡ 0 , i.e.,

(3.35) α ˜ = φ 8 ( z 1 ) ,

where φ 8 ( z 1 ) is a polynomial in z 1 . In view of (3.32) and e h 1 ≠ e h 2 , it follows that a 1 e ξ 2 − a 3 ≠ 0 . Thus, by solving equation (3.34), we have

(3.36) β ˜ = ψ 8 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 ,

where ψ 8 ( x ) is a polynomial in x , especially, β ˜ = ψ 8 ( z 2 ) if a 2 e ξ 2 − a 4 = 0 . In view of (3.3), (3.5), (3.35), and (3.36), we have

(3.37) f = 1 a 1 F 8 ( z 1 )

and

g = 1 D ( a 1 − a 3 e − ξ 2 ) G 8 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 ,

where F 8 ′ ( z 1 ) = e φ 8 ( z 1 ) , G 8 ′ ( x ) = e ψ 8 ( x ) . Noting that a 1 − a 3 e ξ 1 = 0 and e h 1 − h 2 = e ξ 1 + ξ 2 , we can deduce that

(3.38) g = ( 1 − e h 2 − h 1 ) a 1 D G 8 z 2 − a 2 a 3 e h 1 − h 2 − a 1 a 4 a 1 a 3 ( e h 1 − h 2 − 1 ) z 1 ,

especially, g = 1 a 4 G 8 ( z 2 ) if a 2 e ξ 2 − a 4 = 0 . Thus, we have

h 1 = φ 8 ( z 1 ) + ψ 8 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 and h 2 = φ 8 ( z 1 ) + ψ 8 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 − ξ 1 − ξ 2 .

If a 1 − a 3 e ξ 1 ≠ 0 , by solving equation (3.33), we have

(3.39) α ˜ = φ 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 ,

where φ 9 ( x ) is a polynomial in x , especially, α ˜ = φ 9 ( z 2 ) if a 2 − a 4 e ξ 1 = 0 . If a 1 e ξ 2 − a 3 = 0 , it thus follows from (3.34) that β ˜ = ψ 9 ( z 1 ) , where ψ 9 ( z 1 ) is a polynomial in z 1 . Noting that α ˜ − β ˜ ˜ = ξ 1 and β ˜ − α ˜ ˜ = ξ 2 , we have

(3.40) β ˜ ˜ = φ 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 − ξ 1 , α ˜ ˜ = ψ 9 ( z 1 ) − ξ 2 .

Thus, we can deduce from (3.4), (3.5), and (3.40) that

(3.41) f = 1 D ( a 1 e − ξ 1 − a 3 ) F 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 , g = 1 a 3 G 9 ( z 1 ) ,

where F 9 , G 9 are the finite-order transcendental entire functions satisfying F 9 ′ ( x ) = e φ 9 ( x ) , G 9 ′ ( z 1 ) = e ψ 9 ( z 1 ) . Thus, we have

h 1 = φ 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 + ψ 9 ( z 1 ) and h 2 = φ 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 + ψ 9 ( z 1 ) − ξ 1 − ξ 2 .

If a 1 e ξ 2 − a 3 ≠ 0 , it follows from (3.34) that

(3.42) β ˜ = ψ 10 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 ,

where ψ 10 ( x ) is a polynomial in x , especially, β ˜ = ψ 10 ( z 2 ) if a 2 e ξ 2 − a 4 = 0 . By combining with α ˜ − β ˜ ˜ = ξ 1 and β ˜ − α ˜ ˜ = ξ 2 , we can deduce from (3.5) that

(3.43) g = a 1 − a 3 e − ξ 2 D G 10 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 ,

where G 10 is a finite-order transcendental entire function satisfying G 10 ′ ( x ) = e ψ 10 ( x ) . Thus, we have

h 1 = φ 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 + ψ 10 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 ,

h 2 = φ 9 z 2 − a 2 − a 4 e ξ 1 a 1 − a 3 e ξ 1 z 1 + ψ 10 z 2 − a 2 e ξ 2 − a 4 a 1 e ξ 2 − a 3 z 1 − ξ 1 − ξ 2 .

Hence, we prove the conclusions of Theorem 2.1 (iv).

Therefore, from Cases 1–4, we complete the proof of Theorem 2.1.

4 Proof of Theorem 2.2

Assume that ( f , g ) is a pair of finite-order transcendental entire solutions of system (2.2). Since h 1 and h 2 are nonconstant polynomials in C 2 , in view of Lemmas 3.1 and 3.2, there exist four polynomials α ˜ , α ˜ ˜ , β ˜ , and β ˜ ˜ in C 2 such that

(4.1) a 1 f z 1 + a 2 g z 2 = e α ˜ , a 3 f z 2 + a 4 g z 1 = e β ˜ ,

(4.2) a 1 f z 2 + a 2 g z 1 = e α ˜ ˜ , a 3 f z 1 + a 4 g z 2 = e β ˜ ˜ ,

and

(4.3) α ˜ + β ˜ = h 1 , α ˜ ˜ + β ˜ ˜ = h 2 .

Thus, we can deduce from (4.1) and (4.2) that

(4.4) f z 1 = 1 D ( a 4 e α ˜ − a 2 e β ˜ ˜ ) , f z 2 = 1 D ( a 4 e α ˜ ˜ − a 2 e β ˜ )

and

(4.5) g z 1 = 1 D ( a 1 e β ˜ ˜ − a 3 e α ˜ ) , g z 2 = 1 D ( a 1 e β ˜ − a 3 e α ˜ ˜ ) .

Noting that f z 1 z 2 = f z 2 z 1 and g z 1 z 2 = g z 2 z 1 , in view if (4.4) and (4.5), we have

(4.6) a 4 α ˜ z 2 e α ˜ + a 2 β ˜ z 1 e β ˜ − a 2 β ˜ ˜ z 2 e β ˜ ˜ − a 4 α ˜ ˜ z 1 e α ˜ ˜ ≡ 0

and

(4.7) a 3 α ˜ z 2 e α ˜ + a 1 β ˜ z 1 e β ˜ − a 1 β ˜ ˜ z 2 e β ˜ ˜ − a 3 α ˜ ˜ z 1 e α ˜ ˜ ≡ 0 .

Here, we will discuss five cases as follows.

Case 1. Assume that only one of α ˜ z 2 , α ˜ ˜ z 1 , β ˜ z 1 , and β ˜ ˜ z 2 is equal to 0. Let us assume that α ˜ z 2 ≡ 0 and α ˜ ˜ z 1 β ˜ z 1 β ˜ ˜ z 2 ≠ 0 . Thus, (4.6) and (4.7) can be written as

(4.8) a 2 ( β ˜ z 1 e β ˜ − β ˜ ˜ z 2 e β ˜ ˜ ) = a 4 α ˜ ˜ z 1 e α ˜ ˜

and

(4.9) a 1 ( β ˜ z 1 e β ˜ − β ˜ ˜ z 2 e β ˜ ˜ ) = a 3 α ˜ ˜ z 1 e α ˜ ˜ .

Noting that a j ∈ C − { 0 } , j = 1 , 2 , 3 , 4 , we have from (4.8) and (4.9) that

α ˜ ˜ z 1 e α ˜ ˜ ( a 1 a 4 − a 2 a 3 ) = 0 ,

which is a contradiction with the assumptions of D ≠ 0 and α ˜ ˜ z 1 ≠ 0 .

Case 2. Assume that only two terms of α ˜ z 2 , α ˜ ˜ z 1 , β ˜ z 1 , β ˜ ˜ z 2 is equal to 0. If α ˜ z 2 = β ˜ z 1 = 0 and α ˜ ˜ z 1 β ˜ ˜ z 2 ≠ 0 , it thus follows from (4.6) and (4.7) that

− a 2 β ˜ ˜ z 2 e β ˜ ˜ = a 4 α ˜ ˜ z 1 e α ˜ ˜ , − a 1 β ˜ ˜ z 2 e β ˜ ˜ = a 3 α ˜ ˜ z 1 e α ˜ ˜ .

These lead to a 1 a 2 = a 3 a 4 , i.e., a 1 a 4 − a 2 a 3 = 0 , which is a contradiction.

If α ˜ z 2 = β ˜ ˜ z 2 = 0 and α ˜ ˜ z 1 β ˜ z 1 ≠ 0 , it thus follows from (4.6) and (4.7) that

a 2 β ˜ z 1 e β ˜ = a 4 α ˜ ˜ z 1 e α ˜ ˜ , a 1 β ˜ z 1 e β ˜ = a 3 α ˜ ˜ z 1 e α ˜ ˜ ,

which leads to a 1 a 4 − a 2 a 3 = 0 , which is a contradiction.

If β ˜ z 1 = α ˜ ˜ z 1 = 0 and α ˜ z 2 β ˜ ˜ z 2 ≠ 0 , it thus follows from (4.6) and (4.7) that

a 4 α ˜ z 2 e α ˜ = a 2 β ˜ ˜ z 2 e β ˜ ˜ , a 3 α ˜ z 2 e α ˜ = a 1 β ˜ ˜ z 2 e β ˜ ˜ ,

which leads to a 1 a 4 − a 2 a 3 = 0 , which is a contradiction.

If β ˜ ˜ z 2 = α ˜ ˜ z 1 = 0 and α ˜ z 2 β ˜ z 1 ≠ 0 , it thus follows from (4.6) and (4.7) that

a 4 α ˜ z 2 e α ˜ = − a 2 β ˜ z 1 e β ˜ , a 3 α ˜ z 2 e α ˜ = − a 1 β ˜ z 1 e β ˜ ,

which leads to a 1 a 4 − a 2 a 3 = 0 , which is a contradiction.

If α ˜ z 2 = α ˜ ˜ z 1 = 0 and β ˜ z 1 β ˜ ˜ z 2 ≠ 0 , it follows that

(4.10) α ˜ = φ 1 ( z 1 ) , α ˜ ˜ = φ 2 ( z 2 ) ,

where φ 1 ( x ) , φ 2 ( x ) are two polynomials in x , and in view of (4.6) and (4.7), we have

(4.11) β ˜ z 1 e β ˜ = β ˜ ˜ z 2 e β ˜ ˜ .

This means that β ˜ − β ˜ ˜ = ξ , where ξ ∈ C . Otherwise, we can obtain a contradiction using the results (see, for example, [38, p. 99], [35]) or [39, Lemma 3.2]. So, it yields that β ˜ z 1 = β ˜ ˜ z 1 , β ˜ z 2 = β ˜ ˜ z 2 and

(4.12) A β ˜ ˜ z 1 − β ˜ ˜ z 2 = 0 ,

where A = e ξ . Thus, by solving equation (4.12), we have β ˜ ˜ = φ 3 ( z 1 + A z 2 ) = β ˜ − ξ , where φ 3 ( x ) is a polynomial in x . This yields that

(4.13) h 1 = α ˜ + β ˜ = φ 1 ( z 1 ) + φ 3 ( z 1 + A z 2 ) + ξ and h 2 = φ 2 ( z 2 ) + φ 3 ( z 1 + A z 2 ) .

By solving the first equation of (4.4), we have

f = 1 D [ a 4 F 1 ( z 1 ) − a 2 F 3 ( z 1 + A z 2 ) ] + Φ ( z 2 ) ,

where Φ ( z 2 ) is a finite-order entire function in z 2 and F 1 ′ ( z 1 ) = e φ 1 ( z 1 ) and F 3 ′ ( t ) = e φ 3 ( t ) . Substituting this into the second equation of (4.4), we have Φ ′ ( z 2 ) = a 4 D e φ 2 ( z 2 ) , which leads to Φ ( z 2 ) = F 2 ( z 2 ) , where F 2 ′ ( z 2 ) = e φ 2 ( z 2 ) . Thus, we have

(4.14) f = a 4 [ F 1 ( z 1 ) + F 2 ( z 2 ) ] − a 2 F 3 ( z 1 + A z 2 ) D ,

where F 1 ′ ( z 1 ) = e φ 1 ( z 1 ) , F 2 ′ ( z 2 ) = e φ 2 ( z 2 ) , F 3 ′ ( x ) = e φ 3 ( x ) , φ 1 ( z 1 ) , φ 2 ( z 2 ) , φ 3 ( x ) are the polynomials in z 1 , z 2 , x = z 1 + A z 2 , respectively. Similarly, we can obtain

(4.15) g = a 1 G 3 ( z 1 + A z 2 ) − a 3 [ G 1 ( z 1 ) + G 2 ( z 2 ) ] D ,

where G 1 ′ ( z 1 ) = e φ 1 ( z 1 ) , G 2 ′ ( z 2 ) = e φ 2 ( z 2 ) , G 3 ′ ( t ) = e φ 3 ( t ) , φ 1 ( z 1 ) , φ 2 ( z 2 ) , φ 3 ( t ) are the polynomials in z 1 , z 2 , t = z 1 + A z 2 , respectively.

If β ˜ z 1 = β ˜ ˜ z 2 = 0 and α ˜ z 2 α ˜ ˜ z 1 ≠ 0 , it follows that

(4.16) β ˜ = φ 4 ( z 2 ) , and β ˜ ˜ = φ 5 ( z 1 ) ,

where φ 4 ( x ) , φ 5 ( x ) are the polynomials in x , and in view of (4.6) and (4.7), we have

(4.17) α ˜ z 2 e α ˜ = α ˜ ˜ z 1 e α ˜ ˜ .

This means that α ˜ − α ˜ ˜ = ξ , where ξ ∈ C . Otherwise, we can obtain a contradiction using the results (see, for example, [38, p. 99], [35]) or [39, Lemma 3.2]. So, it yields that α ˜ z 1 = α ˜ ˜ z 1 , α ˜ z 2 = α ˜ ˜ z 2 and

(4.18) α ˜ z 1 − A α ˜ z 2 = 0 ,

where A = e ξ . Thus, by solving equation (4.18), we have α ˜ = φ 6 ( z 1 + A − 1 z 2 ) = α ˜ ˜ + ξ , where φ 6 ( x ) is a polynomial in x . Using the same argument as in the case α ˜ z 2 = α ˜ ˜ z 1 = 0 and β ˜ z 1 β ˜ ˜ z 2 ≠ 0 , we have

(4.19) f = a 4 F 6 ( z 1 + A − 1 z 2 ) − a 2 [ F 5 ( z 1 ) + F 4 ( z 2 ) ] D , g = a 1 [ G 5 ( z 1 ) + G 4 ( z 2 ) ] − a 3 G 6 ( z 1 + A − 1 z 2 ) D ,

where A = e ξ , and

F 4 ′ ( z 2 ) = G 4 ′ ( z 2 ) = e φ 4 ( z 2 ) , F 5 ′ ( z 1 ) = G 5 ′ ( z 1 ) = e φ 5 ( z 1 ) , F 6 ′ ( x ) = G 6 ′ ( x ) = e φ 6 ( x ) ,

φ 4 ( z 2 ) , φ 5 ( z 1 ) , φ 6 ( x ) are the polynomials in z 2 , z 1 , x = z 1 + A − 1 z 2 , respectively, satisfying

h 1 = φ 4 ( z 2 ) + φ 6 ( z 1 + A − 1 z 2 ) and h 2 = φ 5 ( z 1 ) + φ 6 ( z 1 + A − 1 z 2 ) − ξ .

Case 3. Assume that only three terms of α ˜ z 2 , α ˜ ˜ z 1 , β ˜ z 1 , and β ˜ ˜ z 2 are equal to 0. Thus, from (4.6) and (4.7), we can obtain a contradiction with the assumption of a j ∈ C − { 0 } , j = 1 , 2 , 3 , 4 .

Case 4. Assume that all of α ˜ z 2 , α ˜ ˜ z 1 , β ˜ z 1 , and β ˜ ˜ z 2 are equal to 0. Thus, it follows that

(4.20) α ˜ = φ 7 ( z 1 ) , α ˜ ˜ = φ 8 ( z 2 ) , β ˜ = φ 9 ( z 2 ) , and β ˜ ˜ = φ 10 ( z 1 ) ,

where φ 7 ( z 1 ) , φ 8 ( z 2 ) , φ 9 ( z 2 ) , and φ 10 ( z 1 ) are the polynomials. In view of (4.4), (4.5), and (4.20), we have

(4.21) f = 1 D [ a 4 ( F 7 ( z 1 ) + F 8 ( z 2 ) ) − a 2 ( F 9 ( z 2 ) + F 10 ( z 1 ) ) ] ,

and

(4.22) g = 1 D [ a 1 ( G 10 ( z 1 ) + G 9 ( z 2 ) ) − a 3 ( G 7 ( z 1 ) + G 8 ( z 2 ) ) ] ,

where F j , G j , j = 7 , … , 10 are finite-order entire functions satisfying

F j ′ ( x ) = G j ′ ( x ) = e φ j ( x ) , j = 7 , 8 , 9 , 10 ,

and

h 1 = α ˜ + β ˜ = φ 7 ( z 1 ) + φ 9 ( z 2 ) , h 2 = α ˜ ˜ + β ˜ ˜ = φ 8 ( z 2 ) + φ 10 ( z 1 ) .

Thus, this is the conclusion (ii) of Theorem 2.2.

Case 5. Assume that all of α ˜ z 2 , α ˜ ˜ z 1 , β ˜ z 1 , and β ˜ ˜ z 2 are not equal to 0. Here, we will divide into four subcases as follows.

Subcase 5.1. Suppose that all of the differences of any two terms of α ˜ , α ˜ ˜ , β ˜ , and β ˜ ˜ are not constants, by Lemma 3.3, we can deduce that all of α ˜ z 2 , α ˜ ˜ z 1 , β ˜ z 1 , and β ˜ ˜ z 2 are equal to 0, which is a contradiction with the begin assumptions in Case 5.

Subcase 5.2. Suppose that only one of the differences of any two terms of α ˜ , α ˜ ˜ , β ˜ , and β ˜ ˜ is a constant; let us assume that α ˜ − α ˜ ˜ = ξ , where ξ ∈ C . Thus, we can rewrite (4.6) as the following form:

(4.23) a 4 α ˜ z 2 e ξ + a 2 β ˜ z 1 e β ˜ − α ˜ ˜ − a 2 β ˜ ˜ z 2 e β ˜ ˜ − α ˜ ˜ − a 4 α ˜ ˜ z 1 ≡ 0 .

If a 4 α ˜ z 2 e ξ − a 4 α ˜ ˜ z 1 = 0 , it follows that β ˜ z 1 e β ˜ − α ˜ ˜ − β ˜ ˜ z 2 e β ˜ ˜ − α ˜ ˜ = 0 . Noting that β ˜ z 1 β ˜ ˜ z 2 ≠ 0 , we have

e β ˜ − β ˜ ˜ = β ˜ ˜ z 2 β ˜ z 1 ,

which is a contradiction since β ˜ − β ˜ ˜ which is not a constant.

If a 4 α ˜ z 2 e ξ − a 4 α ˜ ˜ z 1 ≠ 0 , and noting that β ˜ − α ˜ ˜ , β ˜ ˜ − α ˜ ˜ , and β ˜ − β ˜ ˜ are not constants, by applying the second basic theorem for β ˜ z 1 e β ˜ − α ˜ ˜ , we have

T ( r , β ˜ z 1 e β ˜ − α ˜ ˜ ) < N ( r , β ˜ z 1 e β ˜ − α ˜ ˜ ) + N r , 1 β ˜ z 1 e β ˜ − α ˜ ˜ + N r , 1 β ˜ z 1 e β ˜ − α ˜ ˜ − a 4 ( α ˜ z 2 e ξ − α ˜ ˜ z 1 ) ⁄ a 2 + S ( r , β ˜ z 1 e β ˜ − α ˜ ˜ ) < O ( log r ) + N r , 1 β ˜ ˜ z 2 e β ˜ ˜ − α ˜ ˜ + S ( r , β ˜ z 1 e β ˜ − α ˜ ˜ ) < O ( log r ) + S ( r , β ˜ z 1 e β ˜ − α ˜ ˜ ) ,

which is a contradiction with β ˜ − α ˜ ˜ being a nonconstant polynomial in C 2 .

Subcase 5.3. Suppose that only two of the differences of any two terms of α ˜ , α ˜ ˜ , β ˜ , and β ˜ ˜ are constants; we thus only discuss three cases such as (1) α ˜ − β ˜ = ξ 1 and α ˜ ˜ − β ˜ ˜ = ξ 2 ; (2) α ˜ − β ˜ ˜ = ξ 1 and β ˜ − α ˜ ˜ = ξ 2 ; (3) α ˜ − α ˜ ˜ = ξ 1 and β ˜ − β ˜ ˜ = ξ 2 , where ξ 1 , ξ 2 ∈ C . We can easily obtain contradictions from other cases. For example, let α ˜ − β ˜ = ξ 1 and α ˜ − β ˜ ˜ = ξ 2 , then it yields β ˜ − β ˜ ˜ = ξ 2 − ξ 1 . Thus, we obtain that three of the differences of any two terms of α ˜ , α ˜ ˜ , β ˜ , and β ˜ ˜ are constants. This is a contradiction with the assumptions of Subcase 5.3.

  1. α ˜ − β ˜ = ξ 1 and α ˜ ˜ − β ˜ ˜ = ξ 2 . Thus, we have

    (4.24) α ˜ z 1 = β ˜ z 1 , α ˜ z 2 = β ˜ z 2 , α ˜ ˜ z 1 = β ˜ ˜ z 1 , and α ˜ ˜ z 2 = β ˜ ˜ z 2 .

    Substituting (4.24) into (4.6) and (4.7), we have

    ( a 4 β ˜ z 2 e ξ 1 + a 2 β ˜ z 1 ) e β ˜ = ( a 2 α ˜ ˜ z 2 e ξ 2 + a 4 α ˜ ˜ z 1 ) e α ˜ ˜

    and

    ( a 3 β ˜ z 2 e ξ 1 + a 1 β ˜ z 1 ) e β ˜ = ( a 3 α ˜ ˜ z 2 e ξ 2 + a 1 α ˜ ˜ z 1 ) e α ˜ ˜ .

    Since β ˜ − α ˜ ˜ is not a constant, it follows that

    (4.25) a 4 β ˜ z 2 e ξ 1 + a 2 β ˜ z 1 ≡ 0 , a 2 α ˜ ˜ z 2 e ξ 2 + a 4 α ˜ ˜ z 1 ≡ 0

    and

    (4.26) a 3 β ˜ z 2 e ξ 1 + a 1 β ˜ z 1 ≡ 0 , a 3 α ˜ ˜ z 2 e ξ 2 + a 1 α ˜ ˜ z 1 ≡ 0 .

    Noting that β ˜ z 1 β ˜ z 2 α ˜ ˜ z 1 α ˜ ˜ z 2 ≠ 0 , we have from (4.25) and (4.26) that a 4 a 3 = a 2 a 1 , which is a contradiction with D ≠ 0 .

  2. α ˜ − β ˜ ˜ = ξ 1 and β ˜ − α ˜ ˜ = ξ 2 . Thus, we have

    (4.27) α ˜ z 1 = β ˜ ˜ z 1 , α ˜ z 2 = β ˜ ˜ z 2 , α ˜ ˜ z 1 = β ˜ z 1 , and α ˜ ˜ z 2 = β ˜ z 2 .

    Substituting (4.27) into (4.6) and (4.7), we have

    ( a 4 α ˜ z 2 e ξ 1 − a 2 β ˜ ˜ z 2 ) e β ˜ ˜ = ( a 4 α ˜ ˜ z 1 − a 2 β ˜ z 1 e ξ 2 ) e α ˜ ˜

    and

    ( a 3 α ˜ z 2 e ξ 1 − a 1 β ˜ ˜ z 2 ) e β ˜ ˜ = ( a 3 α ˜ ˜ z 1 − a 1 β ˜ z 1 e ξ 2 ) e α ˜ ˜ .

    Since β ˜ ˜ − α ˜ ˜ is not a constant and β ˜ z 1 β ˜ z 2 α ˜ ˜ z 1 α ˜ ˜ z 2 ≠ 0 , we can deduce that

    a 4 e ξ 1 = a 2 and a 4 = a 2 e ξ 2 ,

    which implies e ξ 1 + ξ 2 ≡ 1 . Noting that h 1 − h 2 = ξ 1 + ξ 2 , we have e h 1 = e h 2 . This is a contradiction with the assumption of Theorem 2.2.

  3. α ˜ − α ˜ ˜ = ξ 1 and β ˜ − β ˜ ˜ = ξ 2 . Thus, we have

    (4.28) α ˜ z 1 = α ˜ ˜ z 1 , α ˜ z 2 = α ˜ ˜ z 2 , β ˜ z 1 = β ˜ ˜ z 1 , β ˜ z 2 = β ˜ ˜ z 2 .

Substituting (4.28) into (4.6) and (4.7), we have

a 4 ( α ˜ z 2 e ξ 1 − α ˜ ˜ z 1 ) e α ˜ ˜ = a 2 ( β ˜ ˜ z 2 − β ˜ z 1 e ξ 2 ) e β ˜ ˜

and

a 3 ( α ˜ z 2 e ξ 1 − α ˜ ˜ z 1 ) e α ˜ ˜ = a 1 ( β ˜ ˜ z 2 − β ˜ z 1 e ξ 2 ) e β ˜ ˜ .

Since α ˜ ˜ − β ˜ is not a constant, using the results (see, for example, [38, p. 99], [35]) or [39, Lemma 3.2], we can deduce that

α ˜ z 1 − α ˜ z 2 e ξ 1 ≡ 0 and β ˜ ˜ z 1 − β ˜ ˜ z 2 e − ξ 2 ≡ 0 .

By solving the aforementioned equations, we have

(4.29) α ˜ = φ 11 ( z 1 + B 1 − 1 z 2 ) and β ˜ ˜ = φ 12 ( z 1 + B 2 z 2 ) ,

where φ 11 ( x ) , φ 12 ( x ) are two polynomials in x , and B 1 = e ξ 1 , B 2 = e ξ 2 . In view of (4.4) and (4.5), we have

(4.30) f = 1 D [ a 4 F 11 ( z 1 + B 1 − 1 z 2 ) − a 2 F 12 ( z 1 + B 2 z 2 ) ] , g = 1 D [ a 1 G 12 ( z 1 + B 2 z 2 ) − a 3 G 11 ( z 1 + B 1 − 1 z 2 ) ] ,

where F 11 , F 12 , G 11 , and G 12 are finite-order entire functions satisfying

F 11 ′ ( x ) = G 11 ′ ( x ) = e φ 11 ( x ) and F 12 ′ ( x ) = G 12 ′ ( x ) = e φ 12 ( x ) .

Here, we have

h 1 = φ 11 ( z 1 + B 1 − 1 z 2 ) + φ 12 ( z 1 + B 2 z 2 ) + ξ 2 and h 2 = φ 11 ( z 1 + B 1 − 1 z 2 ) + φ 12 ( z 1 + B 2 z 2 ) − ξ 1 .

Therefore, this is the conclusion (iii) of Theorem 2.2.

Subcase 5.4. Suppose that three of the differences of any two terms of α ˜ , α ˜ ˜ , β ˜ , and β ˜ ˜ are constants; here, we only prove the case

(4.31) α ˜ − α ˜ ˜ = ξ 1 , β ˜ − α ˜ ˜ = ξ 2 , and β ˜ ˜ − α ˜ ˜ = ξ 3 ,

where ξ 1 , ξ 2 , ξ 3 ∈ C . Thus, it yields that all of the differences of two terms of α ˜ , α ˜ ˜ , β ˜ , and β ˜ ˜ are constants. In fact, by combining with (4.6) or (4.7), we can prove that all of the differences of two terms of α ˜ , α ˜ ˜ , β ˜ , and β ˜ ˜ also are constants for the other cases. Set ξ 0 = ξ 1 − ξ 2 . Noting that α ˜ + β ˜ = h 1 and α ˜ ˜ + β ˜ ˜ = h 2 , we thus have from (4.31) that

(4.32) α ˜ − β ˜ = ξ 0 , β ˜ = h 1 − ξ 0 2 , α ˜ ˜ = h 2 − ξ 3 2 , and h 1 − h 2 = ξ 1 + ξ 2 − ξ 3 .

In view of (4.6), (4.7), and (4.31), we have

(4.33) a 4 α ˜ z 2 e ξ 1 + a 2 β ˜ z 1 e ξ 2 − a 2 β ˜ ˜ z 2 e ξ 3 − a 4 α ˜ ˜ z 1 = 0

and

(4.34) a 3 α ˜ z 2 e ξ 1 + a 1 β ˜ z 1 e ξ 2 − a 1 β ˜ ˜ z 2 e ξ 3 − a 3 α ˜ ˜ z 1 = 0 .

Noting that α ˜ z 1 = α ˜ ˜ z 1 = β ˜ z 1 = β ˜ ˜ z 1 and α ˜ z 2 = α ˜ ˜ z 2 = β ˜ z 2 = β ˜ ˜ z 2 , we can deduce from (4.33) and (4.34) that

( a 4 e ξ 1 − a 2 e ξ 3 ) α ˜ z 2 = ( a 4 − a 2 e ξ 2 ) α ˜ z 1

and

( a 3 e ξ 1 − a 1 e ξ 3 ) α ˜ z 2 = ( a 3 − a 1 e ξ 2 ) α ˜ z 1 .

In view of α ˜ z 1 α ˜ ˜ z 2 ≠ 0 , it follows that

(4.35) ( a 1 a 4 − a 2 a 3 ) e ξ 3 ( 1 − e ξ 1 + ξ 2 − ξ 3 ) = 0 ,

Noting that D = a 1 a 4 − a 2 a 3 ≠ 0 , (4.35) can lead to e ξ 1 + ξ 2 − ξ 3 ≡ 1 . Since h 1 − h 2 = ξ 1 + ξ 2 − ξ 3 , we can obtain a contradiction with the assumption of e h 1 ≠ e h 2 .

Therefore, from Case 1 to Case 5, we complete the proof of Theorem 2.2.

Acknowledgements

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 Science and Technology Research Project of Jiangxi Provincial Department of Education (GJJ2201238), the Project of Jiangxi Province Education Science Planning Project in China (No. 23YB176), the Research Foundation of Suqian University (106-CK00042/028), and sponsored by Qing Lan Project and Suqian Talent Xiongying Plan of Suqian.

  2. Author contributions: HL and HYX completed the main part of this article. HL, ZSL, and HYX corrected the main theorems. All authors gave final approval for publication.

  3. Conflict of interest: The authors state no conflict of interests.

  4. Data availability statement: No data were used to support this study.

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Received: 2024-01-02
Revised: 2024-09-23
Accepted: 2024-09-30
Published Online: 2024-11-16

© 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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https://doi.org/10.1515/math-2024-0086
Keywords for this article
Nevanlinna theory; entire function; system of functional equations; several complex variables
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Articles in the same Issue

  1. Special Issue on Contemporary Developments in Graph Topological Indices
  2. On the maximum atom-bond sum-connectivity index of graphs
  3. Upper bounds for the global cyclicity index
  4. Zagreb connection indices on polyomino chains and random polyomino chains
  5. On the multiplicative sum Zagreb index of molecular graphs
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  7. Special Issue on Convex Analysis and Applications - Part I
  8. Weighted Hermite-Hadamard-type inequalities without any symmetry condition on the weight function
  9. Scattering threshold for the focusing energy-critical generalized Hartree equation
  10. (p, q)-Compactness in spaces of holomorphic mappings
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  69. Small values and functional laws of the iterated logarithm for operator fractional Brownian motion
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  80. A quasi-reversibility method for solving nonhomogeneous sideways heat equation
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  86. Power vector inequalities for operator pairs in Hilbert spaces and their applications
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  92. Vector optimization problems with weakened convex and weakened affine constraints in linear topological spaces
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  94. Brouwer's conjecture for the sum of the k largest Laplacian eigenvalues of some graphs
  95. High-order finite-difference ghost-point methods for elliptic problems in domains with curved boundaries
  96. Riemannian invariants for warped product submanifolds in Q ε m × R and their applications
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  103. Dynamics in a predator-prey model with predation-driven Allee effect and memory effect
  104. A note on orthogonal decomposition of 𝔰𝔩n over commutative rings
  105. On the δ-chromatic numbers of the Cartesian products of graphs
  106. Binomial convolution sum of divisor functions associated with Dirichlet character modulo 8
  107. Commutator of fractional integral with Lipschitz functions related to Schrödinger operator on local generalized mixed Morrey spaces
  108. System of degenerate parabolic p-Laplacian
  109. Stochastic stability and instability of rumor model
  110. Certain properties and characterizations of a novel family of bivariate 2D-q Hermite polynomials
  111. Stability of an additive-quadratic functional equation in modular spaces
  112. Monotonicity, convexity, and Maclaurin series expansion of Qi's normalized remainder of Maclaurin series expansion with relation to cosine
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  114. On the existence of tripartite graphs and n-partite graphs
  115. Classifying pentavalent symmetric graphs of order 12pq
  116. Almost periodic functions on time scales and their properties
  117. Some results on uniqueness and higher order difference equations
  118. Coding of hypersurfaces in Euclidean spaces by a constant vector
  119. Cycle integrals and rational period functions for Γ0+(2) and Γ0+(3)
  120. Degrees of (L, M)-fuzzy bornologies
  121. A matrix approach to determine optimal predictors in a constrained linear mixed model
  122. On ideals of affine semigroups and affine semigroups with maximal embedding dimension
  123. Solutions of linear control systems on Lie groups
  124. A uniqueness result for the fractional Schrödinger-Poisson system with strong singularity
  125. On prime spaces of neutrosophic extended triplet groups
  126. On a generalized Krasnoselskii fixed point theorem
  127. On the relation between one-sided duoness and commutators
  128. Non-homogeneous BVPs for second-order symmetric Hamiltonian systems
  129. Erratum
  130. Erratum to “Infinitesimals via Cauchy sequences: Refining the classical equivalence”
  131. Corrigendum
  132. Corrigendum to “Matrix stretching”
  133. Corrigendum to “A comprehensive review of the recent numerical methods for solving FPDEs”
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