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Entire solutions for several complex partial differential-difference equations of Fermat type in ℂ2

  • Xian Min Gui , Hong Yan Xu EMAIL logo , Wen Ju Tang and Hua Wang
Published/Copyright: December 31, 2021
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Open Mathematics
From the journal Open Mathematics Volume 19 Issue 1

Abstract

By utilizing the Nevanlinna theory of meromorphic functions in several complex variables, we mainly investigate the existence and the forms of entire solutions for the partial differential-difference equation of Fermat type

α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 m + f ( z 1 + c 1 , z 2 + c 2 ) n = 1

and

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

where m , n are positive integers and α , β , γ 1 , γ 2 are constants in C . We give some results about the forms of solutions for these equations, which are great improvements of the previous theorems given by Xu and Cao et al. Moreover, it is very satisfactory that we give the corresponding examples to explain the conclusions of our theorems in each case.

Keywords: existence; entire solution; partial differential-difference equation
MSC 2010: 35M30; 32W50; 39A45; 30D35

1 Introduction and main results

It is well known that Wiles and Taylor [1,2] in 1995 proved Fermat’s last theorem. They pointed out that the Fermat equation x m + y n = 1 does not admit nontrivial solutions in rational numbers for m = n ≥ 3 , but there exists nontrivial rational solutions for this equation for m = n = 2 . The study of the Fermat type equation can go back to Montel [3] and Gross [4], who had discussed the existence of solutions for the Fermat type functional equation f m + g n = 1 and showed that the entire solutions are f = cos a ( z ) , g = sin a ( z ) for m = n = 2 , where a ( z ) is an entire function; there are no nonconstant entire solutions for m = n > 2 .

Of late, due to the establishment of difference analogues of Nevanlinna theory for meromorphic functions (see [5,6,7]), many scholars have paid an increasing interest in studying the properties of solutions for complex difference equations, complex differential-difference equations, and obtained a number of results about the existence and the form of solutions for some equations (including [7,8,9, 10,11,12, 13,14,15, 16,17]).

Liu et al. [18] in 2012 studied the existence of solutions for some complex difference equations and obtained the following:

Theorem A

(see [18, Theorem 1.3]). The transcendental entire solution with finite order of

f ′ ( z ) 2 + f ( z + c ) 2 = 1

must satisfy f ( z ) = sin ( z ± B i ) , where B is a constant and c = 2 k π or c = ( 2 k + 1 ) π , and k is an integer.

Theorem B

(see [18, Theorem 1.5]). The transcendental entire solutions with finite order of

f ′ ( z ) 2 + [ f ( z + c ) − f ( z ) ] 2 = 1

must satisfy f ( z ) = 12 sin ( 2 z + B i ) , where c = ( 2 k + 1 ) π , k is an integer, and B is a constant.

In recent years, with the development of difference analogues of Nevanlinna theory for meromorphic functions with several complex variables, especially the difference analogue of the logarithmic derivative lemma given by Cao and Korhonen [19]. Very recently, Xu and Cao [20,21,22] investigated the existence of the entire and meromorphic solutions for some Fermat type partial differential-difference equations and obtained the following theorems.

Theorem C

(see [20, Theorem 3.2]). Let c = ( c 1 , c 2 ) ∈ C ⧹ { 0 } . Suppose that f is a nontrivial meromorphic solution of the Fermat type partial difference equations

1 f ( z 1 + c 1 , z 2 + c 2 ) m + 1 f ( z 1 , z 2 ) m = A ( z 1 , z 2 ) f ( z 1 , z 2 ) n

or

1 f ( z 1 + c 1 , z 2 + c 2 ) m + 1 f ( z 1 + c 1 , z 2 ) m + 1 f ( z 1 , z 2 + c 2 ) m = A ( z 1 , z 2 ) f ( z 1 , z 2 ) n ,

where m ∈ N , n ∈ N ∪ { 0 } , and A ( z 1 , z 2 ) is a nonzero meromorphic function on C 2 with respect to the solution f , that is, T ( r , A ) = o ( T ( r , f ) ) . If δ f ( ∞ ) > 0 , then

lim sup r → ∞ log T ( r , f ) r > 0 .

Remark 1.1

Let n = 0 and A ( z 1 , z 2 ) = 1 , then the above equations become

1 f ( z 1 + c 1 , z 2 + c 2 ) m + 1 f ( z 1 , z 2 ) m = 1

and

1 f ( z 1 + c 1 , z 2 + c 2 ) m + 1 f ( z 1 + c 1 , z 2 ) m + 1 f ( z 1 , z 2 + c 2 ) m = 1 ,

which can be called as the partial difference equations of Fermat type.

Theorem D

(see [21, Theorem 1.1]). Let c = ( c 1 , c 2 ) ∈ C 2 . Then the Fermat type partial differential-difference equation

∂ f ( z 1 , z 2 ) ∂ z 1 n + f ( z 1 + c 1 , z 2 + c 2 ) m = 1

does not have any transcendental entire solution with finite order, where m and n are two distinct positive integers.

Theorem E

(see [21, Theorem 1.2]). Let c = ( c 1 , c 2 ) ∈ C 2 . Then any transcendental entire solution with finite order of the partial differential-difference equation

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

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

Theorems D and E suggest a question naturally: What will happen about the existence and the forms of transcendental entire solutions for the Fermat type partial differential-difference equation including both ∂ f ( z 1 , z 2 ) ∂ z 1 and ∂ f ( z 1 , z 2 ) ∂ z 2 ?

In view of the above questions, this paper is concerned with the properties of the solutions for some Fermat type equations including both difference operator and two partial differential by utilizing the Nevanlinna theory and difference Nevanlinna theory of several complex variables [19,20,23]. We describe the existence and the forms of the transcendental entire solutions with finite order for the Fermat type partial differential-difference equations with more general forms, and some results are obtained to improve the previous theorems given by Xu and Cao [21] and Liu et al. [18]. Here and below, let z + w = ( z 1 + w 1 , z 2 + w 2 ) for any z = ( z 1 , z 2 ) , w = ( w 1 , w 2 ) . The main results of this paper are listed below.

Theorem 1.1

Let c = ( c 1 , c 2 ) ∈ C 2 , m , n be two distinct positive integers, and α , β be constants in C that are not equal to zero at the same time. If the Fermat type partial differential-difference equation

(1) α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 m + f ( z 1 + c 1 , z 2 + c 2 ) n = 1

satisfies one of the conditions

  1. n > m ;

  2. m > n ≥ 2 ;

then equation (1) does not have any transcendental entire solution with finite order.

Remark 1.2

For m > n = 1 , the transcendental entire solution of equation (1) with finite order can be found. Especially, let m = 2 , n = 1 . Then equation (1) becomes of the form

(2) α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 2 + f ( z 1 + c 1 , z 2 + c 2 ) = 1 .

Here, we will give the detail to get a solution of equation (2).

Set F ( z 1 , z 2 ) = α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 , differentiating (2) for z 1 , z 2 , respectively, then it follows that

(3) F ( z 1 , z 2 ) α ∂ F ( z 1 , z 2 ) ∂ z 1 + β ∂ F ( z 1 , z 2 ) ∂ z 2 = − 1 2 F ( z 1 + c 1 , z 2 + c 2 ) .

Assuming that

(4) α ∂ F ( z 1 , z 2 ) ∂ z 1 + β ∂ F ( z 1 , z 2 ) ∂ z 2 = − 1 2 ,

(5) F ( z 1 , z 2 ) = F ( z 1 + c 1 , z 2 + c 2 ) .

For equation (4), one can find a finite order transcendental entire solution

F ( z 1 , z 2 ) = e β z 1 − α z 2 − 1 4 z 1 α + z 2 β .

By combining with (5), we can deduce that c 1 = π β i and c 2 = − π α i .

In view of the assumption, it follows that

(6) α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 = F ( z 1 , z 2 ) = e β z 1 − α z 2 − 1 4 z 1 α + z 2 β .

The characteristic equations for equation (6) are

d z 1 d t = α , d z 2 d t = β , d f d t = F ( z 1 , z 2 ) .

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

f = ∫ 0 t F ( α t , β t + s ) d t + g ( s ) = ∫ 0 t e − α s − 1 4 2 t + 1 β s d t + g ( s ) = t e − α s − 1 4 t 2 − 1 4 β s t + g ( s ) ,

where g ( s ) is an entire function with finite order in s . Then, by combining with t = z 1 α and s = z 2 − β α z 1 , the solution of equation (2) is of the form

(7) f ( z 1 , z 2 ) = 1 α z 1 e β z 1 − α z 2 − 1 4 z 1 z 2 α β + g α z 2 − β z 1 α ,

where g ( s ) satisfies the following equation:

(8) g α z 2 − β z 1 α + α c 2 − β c 1 α = 1 − e β z 1 − α z 2 − 1 4 z 1 α + z 2 β 2 + z 1 + c 1 α e β z 1 − α z 2 − 1 4 ( z 1 + c 1 ) ( z 2 + c 2 ) α β .

Let

z 1 = α α − β s + 2 α c 2 α + β , z 2 = α α − β s − 2 α c 2 α + β ,

then

α z 2 − β z 1 α + 2 c 2 = s .

Substituting z 1 , z 2 into (8), and combining with c 1 = π β i and c 2 = − π α i , we conclude

(9) g ( s ) = 1 − e − α s − 1 4 ( α + β ) 2 s − 2 ( α − β ) 2 c 2 β ( α 2 − β 2 ) 2 + s α − β − ( α − β ) c 2 β ( α + β ) e − α s − 1 4 s α − β − ( α − β ) c 2 β ( α + β ) α s β ( α − β ) − ( α − β ) c 2 α + β .

Therefore, an entire solution of equation (2) is of the form

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

where g ( s ) is stated as in (9), c 1 = π β i , and c 2 = − π α i .

For m = n = 2 , we obtain the following result.

Theorem 1.2

Let c = ( c 1 , c 2 ) ∈ C 2 and α , β be constants in C that are not equal to zero at the same time. Then any transcendental entire solution with finite order for the Fermat type partial differential-difference equation

(10) α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 2 + f ( z 1 + c 1 , z 2 + c 2 ) 2 = 1

has the following form:

f ( z 1 , z 2 ) = A 1 e a 1 z 1 + a 2 z 2 + H ( z ) + A 2 e − ( a 1 z 1 + a 2 z 2 ) − H ( z ) 2 ,

where H ( z ) ≔ H ( s 1 ) is a polynomial in s 1 satisfying ( α c 2 − β c 1 ) H ′ ≡ 0 , s 1 = c 2 z 1 − c 1 z 2 , A 1 A 2 = 1 , A 1 , A 2 are constants in C , and c , a 1 , a 2 , α , β satisfy one of the following cases:

  1. α a 1 + β a 2 = i and L ( c ) = 2 k π i , where L ( c ) = a 1 c 1 + a 2 c 2 , here and below k is an integer;

  2. α a 1 + β a 2 = − i and L ( c ) = ( 2 k + 1 ) π i .

Remark 1.3

In view of Theorem 1.2, if α = 1 , β = 0 , and c 2 ≠ 0 , then we can conclude that any finite order transcendental entire solution f of equation (10) has the form

f ( z ) = e L ( z ) + B + e − ( L ( z ) + B ) 2

satisfying a 1 = i and L ( c ) = 2 k π i , or a 1 = − i and L ( c ) = ( 2 k + 1 ) π i .

The following examples show the existence of solutions for equation (10).

Example 1.1

Let a = ( a 1 , a 2 ) = ( i , i ) and B = 0 . Thus, the function

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

satisfies equation (10) with c = ( c 1 , c 2 ) = ( π , π ) , α = 2 , β = − 1 .

Example 1.2

Let a = ( a 1 , a 2 ) = ( 1 , 1 ) and B = 0 . Thus, the function

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

satisfies equation (10) with c = ( c 1 , c 2 ) = ( 2 π i , − π i ) , α = i , β = − 2 i .

Example 1.3

Let a = ( a 1 , a 2 ) = ( 3 i , − 2 i ) , H ( z ) = 4 π 2 ( z 1 − z 2 ) 2 , and B = 0 . Thus, the function

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

satisfies equation (10) with c = ( c 1 , c 2 ) = ( 2 π , 2 π ) , α = 1 , β = 1 .

Example 1.4

Let a = ( a 1 , a 2 ) = ( 2 i , − 3 i ) , H ( z ) = π 4 ( z 1 − z 2 ) 4 , and B = 0 . Thus, the function

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

satisfies equation (10) with c = ( c 1 , c 2 ) = ( − π , − π ) , α = 1 , β = 1 .

Remark 1.4

From the conclusions of Theorems A and E, there only exist finite order transcendental entire solutions with growth order ρ ( f ) = 1 . However, we can see that there exists transcendental entire solution of (10) with growth order ρ ( f ) > 1 , that is, ρ ( f ) = 2 in Example 1.3 and ρ ( f ) = 4 in Example 1.4, these results are quite different from the previous one. Hence, our results are some improvements of the previous theorems given by Xu and Cao [21], and Liu et al. [18].

Theorem 1.3

Let c = ( c 1 , c 2 ) ∈ C 2 and c 2 , α , β 1 , β 2 be nonzero constants in C . Let f be a finite order transcendental entire solution of the Fermat type partial differential-difference equation

(11) α ∂ f ( z 1 , z 2 ) ∂ z 1 2 + [ β 1 f ( z 1 + c 1 , z 2 + c 2 ) − β 2 f ( z 1 , z 2 ) ] 2 = 1 .

  1. If ∂ f ( z 1 , z 2 ) ∂ z 1 is not constant, then the solution f ( z ) has the following form:

    f ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + B − e − ( a 1 z 1 + a 2 z 2 + B ) 2 a 1 α + e η z 2 G ( z 2 ) ,

    where a 1 , a 2 , B , η are constants in C , G ( z 2 ) is a finite order entire period function with period c 2 . Moreover,

    1. if β 1 ≠ ± β 2 , then η = log β 2 − log β 1 c 2 , and either a 1 α = i ( β 1 − β 2 ) , L ( c ) = 2 k π i , or a 1 α = − i ( β 2 + β 1 ) , L ( c ) = ( 2 k + 1 ) π i , where L ( c ) = a 1 c 1 + a 2 c 2 ;

    2. if β 1 = β 2 , then η ≡ 0 , i α a 1 = 2 β 1 , and L ( c ) = ( 2 k + 1 ) π i ;

    3. if β 1 = − β 2 , then η = log ( − 1 ) c 2 , i α a 1 = − 2 β 1 , and L ( c ) = 2 k π i ;

  2. If ∂ f ( z 1 , z 2 ) ∂ z 1 is a constant, then the solution f ( z ) has the following form:

    f ( z ) = [ e η z 2 G ( z 2 ) + τ ] + D 1 z 1 + D 2 z 2 + D 0 ,

    where τ , D 0 , D 1 , D 2 are constants in C . Moreover,

    1. if β 1 = β 2 , then η = 0 , τ ∈ C , and [ β 1 ( D 1 c 1 + D 2 c 2 ) ] 2 = 1 − ( α D 1 ) 2 ;

    2. if β 1 ≠ β 2 , then η = log β 2 − log β 1 c 2 , D 1 = D 2 = 0 , and τ , β 1 , β 2 , D 0 satisfy

      [ ( β 1 − β 2 ) ( τ − D 0 ) ] 2 = β 1 2 .

Next, we give some examples to explain the conclusions of Theorem 1.3 in each case.

Example 1.5

Let a = ( a 1 , a 2 ) = ( 1 , 1 ) . Thus, the function

f ( z 1 , z 2 ) = e z 1 + z 2 − e − ( z 1 + z 2 ) 2 i + e z 2 log 2 π i ( e 2 z 2 + 1 )

satisfies equation (11) with c = ( c 1 , c 2 ) = ( π i , π i ) , α = i , β 1 = 2 , and β 2 = 1 .

Example 1.6

Let a = ( a 1 , a 2 ) = ( − 1 , 1 ) . Thus, the function

f ( z 1 , z 2 ) = e z 1 − z 2 − e − ( z 1 − z 2 ) − 2 i + e z 2 log ( − 1.5 ) 2 π i ( e z 2 − 1 )

satisfies equation (11) with c = ( c 1 , c 2 ) = ( π i , 2 π i ) , α = i , β 1 = 2 , and β 2 = − 3 .

Example 1.7

Let a = ( a 1 , a 2 ) = ( 1 , 2 ) and B , D 0 ∈ C . Thus, the function

f ( z 1 , z 2 ) = e z 1 + 2 z 2 + B − e − ( z 1 + 2 z 2 + B ) 2 i + e 2 z 2 + D 0

satisfies equation (11) with c = ( c 1 , c 2 ) = ( − π i , π i ) , α = i , β 1 = β 2 = 1 2 .

Example 1.8

Let a = ( a 1 , a 2 ) = 1 2 , π i , D 0 = 0 , and B ∈ C . Thus, the function

f ( z 1 , z 2 ) = e 1 2 z 1 + π i z 2 + B − e − 1 2 z 1 + π i z 2 + B 2 i + e π i z 2 e 2 π i z 2

satisfies equation (11) with c = ( c 1 , c 2 ) = ( 2 π i , 1 ) , α = 2 i , β 1 = − β 2 = 1 2 .

Example 1.9

Let G ( z 2 ) = e z 2 + e − z 2 2 , τ = 0 , D 0 = 1 , D 1 = 0 and D 2 = 1 2 π i . Thus, the function

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

satisfies equation (11) with c = ( c 1 , c 2 ) = ( c 1 , 2 π i ) , α ∈ C , β 1 = β 2 = 1 .

Example 1.10

Let G ( z 2 ) = e π i z 2 + e − π i z 2 + e 2 π i z 2 , τ = 0 , D 0 = 2 , D 1 = 3 4 , and D 2 = − 1 2 . Thus, the function

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

satisfies equation (11) with c = ( c 1 , c 2 ) = ( 3 , 2 ) , α = 2 , β 1 = β 2 = 2 .

Example 1.11

Let G ( z 2 ) = e i π z 2 + e − i π z 2 , c = ( c 1 , c 2 ) = ( c 1 , 2 ) , α ∈ C , β 1 = 1 , β 2 = e 2 , D 0 = 1 , and τ = 2 − e 2 e 2 − 1 . Thus, the function

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

satisfies the following equation:

α ∂ f ( z 1 , z 2 ) ∂ z 1 2 + [ f ( z 1 + c 1 , z 2 + 2 ) − e 2 f ( z 1 , z 2 ) ] 2 = 1 .

Remark 1.5

Examples 1.5–1.8 show that the forms of solutions are true for cases ( i 1 ) – ( i 3 ) in Theorem 1.3( i ), and Examples 1.9–1.11 show that the forms of solutions are true for cases ( i i 1 ) – ( i i 2 ) in Theorem 1.3( i i ).

Naturally, we should proceed to discuss the existence and form of solutions for the following Fermat type partial differential-difference equation:

(12) α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 2 + [ γ 1 f ( z 1 + c 1 , z 2 + c 2 ) − γ 2 f ( z 1 , z 2 ) ] 2 = 1 ,

where α , β , γ 1 , γ 2 are constants in C .

Theorem 1.4

Let c = ( c 1 , c 2 ) ∈ C 2 and c 1 , c 2 , α , β , γ 1 , γ 2 be nonzero constants in C such that α c 2 − β c 1 ≠ 0 . Let f be a finite order transcendental entire solution of the Fermat type partial differential-difference equation (12)

  1. If α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 is not constant, then the solution f ( z ) has the following form:

    f ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + B − e − ( a 1 z 1 + a 2 z 2 + B ) 2 ( a 1 α + a 2 β ) + e η ( α z 2 − β z 1 ) G ( α z 2 − β z 1 ) ,

    where G ( u ) is a finite order entire period function with period α c 2 − β c 1 , a 1 , a 2 , B , η are constants in C . Moreover,

    1. if γ 1 ≠ ± γ 2 , then η = log γ 2 − log γ 1 α c 2 − β c 1 , and either a 1 α + a 2 β = i ( γ 1 − γ 2 ) , L ( c ) = 2 k π i , or a 1 α + a 2 β = − i ( γ 2 + γ 1 ) , L ( c ) = ( 2 k + 1 ) π i , where L ( c ) = a 1 c 1 + a 2 c 2 ;

    2. if γ 1 = γ 2 , then η ≡ 0 , i ( a 1 α + a 2 β ) = 2 γ 1 and L ( c ) = ( 2 k + 1 ) π i ;

    3. if γ 1 = − γ 2 , then η = log ( − 1 ) α c 2 − β c 1 , i ( a 1 α + a 2 β ) = − 2 γ 1 and L ( c ) = 2 k π i ;

  2. If α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 is a constant, then the solution f ( z ) has the following form:

    f ( z 1 , z 2 ) = D 1 z 1 + D 2 ( α z 2 − β z 1 ) + [ e η ( α z 2 − β z 1 ) G ( α z 2 − β z 1 ) + τ ] ,

    where τ , D 1 , D 2 are constants in C . Moreover,

    1. if γ 1 = γ 2 , then η = 0 , τ ∈ C and

      γ 1 2 [ D 1 c 1 + D 2 ( α c 2 − β c 1 ) ] 2 = 1 − ( α D 1 ) 2 ;

    2. if γ 1 ≠ γ 2 , then η = log γ 2 − log γ 1 α c 2 − β c 1 , D 1 = D 2 = 0 , and τ , γ 1 , γ 2 satisfy

      ( γ 1 − γ 2 ) τ = ± 1 .

Remark 1.6

Obviously, we can see that equation (12) is transformed into equation (10) when γ 2 = 0 , and equation (11) for β = 0 . Hence, we only consider the case that α , β , γ 1 , γ 2 are nonzero constants in C in Theorem 1.4.

Next, we give some examples to explain the conclusions of Theorem 1.4 in each case.

Example 1.12

Let a 1 = a 2 = 1 and B ∈ C . Thus, the function

f ( z 1 , z 2 ) = e z 1 + z 2 + B − e − ( z 1 + z 2 + B ) 2 i + e log 2 2 π i ( z 2 − z 1 ) e z 1 − z 2

satisfies equation (12) with γ 1 = 2 , γ 2 = 1 , α = β = i 2 , and c 1 = − c 2 = π i .

Example 1.13

Let a 1 = a 2 = i and B ∈ C . Thus, the function

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

satisfies equation (12) with γ 1 = γ 2 = 1 , α = 1 , β = − 3 , and c 1 = 2 π , c 2 = − π .

Example 1.14

Let a 1 = a 2 = i and B ∈ C . Thus, η = i 3 and the function

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

satisfies equation (12) with γ 1 = − γ 2 = 1 , α = β = i and c 1 = 5 π i 2 , c 2 = − π i 2 .

Example 1.15

If α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 ≡ 0 , then from (12), it follows γ 1 f ( z 1 + c 1 , z 2 + c 2 ) − γ 2 f ( z 1 , z 2 ) = ± 1 . Thus, the function f ( z 1 , z 2 ) = e β z 1 − α z 2 + β z 1 − α z 2 satisfies the following equation:

α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 2 + 1 2 π i f ( z 1 + c 1 , z 2 + c 2 ) − 1 2 π i f ( z 1 , z 2 ) 2 = 1 ,

where α , β , c 1 , c 2 are constants in C satisfying β c 1 − α c 2 = 2 π i .

Example 1.16

If α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 = 1 , then from (12), it follows that γ 1 f ( z 1 + c 1 , z 2 + c 2 ) − γ 2 f ( z 1 , z 2 ) = 0 . Let γ 1 = γ 2 = 2 , D 1 = 2 , D 2 = − 1 , α = − 1 2 , β = − 1 , c 1 = π i , and c 2 = − 2 π i . Thus, the function f ( z 1 , z 2 ) = e z 1 − 1 2 z 2 + 2 z 1 − − 1 2 z 2 + z 1 satisfies the following equation:

1 2 ∂ f ( z 1 , z 2 ) ∂ z 1 + ∂ f ( z 1 , z 2 ) ∂ z 2 2 + [ 2 f ( z 1 + π i , z 2 − 2 π i ) − 2 f ( z 1 , z 2 ) ] 2 = 1 .

Example 1.17

Let γ 1 = 2 , γ 2 = 1 , τ = 1 , and α , β , c 1 , c 2 be nonzero constants in C such that α c 2 − β c 1 = 2 . Then the function

f ( z 1 , z 2 ) = e − log 2 2 ( α z 2 − β z 1 ) e π i ( α z 2 − β z 1 ) + 1

satisfies the following equation:

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

Remark 1.7

Examples 1.12–1.14 show that the forms of solutions are true for cases ( i 1 ) – ( i 3 ) in Theorem 1.4(i), and examples 1.15–1.17 show that the forms of solutions are true for cases ( i i 1 ) – ( i i 2 ) in Theorem 1.4(ii).

2 Proof of Theorem 1.1

To prove Theorem 1.1, we require the following lemmas.

Lemma 2.1

[24,25] Let f be a nonconstant meromorphic function on C n and let I = ( i 1 , … , i n ) be a multi-index with length ∣ I ∣ = ∑ j = 1 n i j . Assume that T ( r 0 , f ) ≥ e for some r 0 . Then

m r , ∂ I f f = S ( r , f )

holds for all r ≥ r 0 outside a set E ⊂ ( 0 , + ∞ ) of finite logarithmic measure ∫ E d t t < ∞ , where ∂ I f = ∂ ∣ I ∣ f ∂ z 1 i 1 ⋯ ∂ z n i n .

Lemma 2.2

[19,23] Let f be a nonconstant meromorphic function with finite order on C n such that f ( 0 ) ≠ 0 , ∞ , and let ε > 0 . Then for c ∈ C n ,

m r , f ( z ) f ( z + c ) + m r , f ( z + c ) f ( z ) = S ( r , f )

holds for all r ≥ r 0 outside a set E ⊂ ( 0 , + ∞ ) of finite logarithmic measure ∫ E d t t < ∞ .

Proof of Theorem 1.1

Suppose that f is a finite order transcendental entire solution of equation (1). Since α , β are constants and not equal to zero at the same time, then α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 and f ( z 1 + c 1 , z 2 + c 2 ) are transcendental. Here, we will consider two cases below.

Case 1

n > m . In view of Lemma 2.2, it yields that

(13) m r , f ( z 1 , z 2 ) f ( z 1 + c 1 , z 2 + c 2 ) = S ( r , f )

holds for all r > 0 outside of a possible exceptional set E ⊂ [ 1 , + ∞ ) of finite logarithmic measure ∫ E d t t < ∞ . Thus, it follows from (13) that

(14) T ( r , f ( z 1 , z 2 ) ) = m ( r , f ( z 1 , z 2 ) ) ≤ m r , f ( z 1 , z 2 ) f ( z 1 + c 1 , z 2 + c 2 ) + m ( r , f ( z 1 + c 1 , z 2 + c 2 ) ) + log 2 = m ( r , f ( z 1 + c 1 , z 2 + c 2 ) ) + S ( r , f ) = T ( r , f ( z 1 + c 1 , z 2 + c 2 ) ) + S ( r , f ) ,

for all r ∉ E . In view of (14), Lemma 2.1, and the Mokhon’ko theorem in several complex variables [26, Theorem 3.4], it yields

(15) n T ( r , f ( z 1 , z 2 ) ) ≤ n T ( r , f ( z 1 + c 1 , z 2 + c 2 ) ) + S ( r , f ) = T ( r , f ( z 1 + c 1 , z 2 + c 2 ) n ) + S ( r , f ) = T r , α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 m − 1 + S ( r , f ) = m T r , α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 + S ( r , f )

= m m r , α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 + S ( r , f ) ≤ m m r , α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 f 1 ( z 1 , z 2 ) + m ( r , f ( z 1 , z 2 ) ) + S ( r , f ) = m T ( r , f ( z 1 , z 2 ) ) + S ( r , f ) ,

for all r ∉ E . That is,

(16) ( n − m ) T ( r , f ( z 1 , z 2 ) ) ≤ S ( r , f ) , r ∉ E ,

which is a contradiction with the assumption that f is a transcendental entire function.

Case 2

m > n ≥ 2 . Then 1 m + 1 n ≤ 2 n < 1 and n > m m − 1 . In view of the Nevanlinna second fundamental theorem, Lemma 2.2, and equation (1), it follows that

(17) ( m − 1 ) T r , α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 ≤ N ¯ r , α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 + ∑ q = 1 m N ¯ r , 1 α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 − w q + S r , α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 ≤ N ¯ r , 1 α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 m − 1 + S r , α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 ≤ N ¯ r , 1 f ( z 1 + c 1 , z 2 + c 2 ) + S ( r , f ) ≤ T ( r , f ( z 1 + c 1 , z 2 + c 2 ) ) + S ( r , f 1 ) + S ( r , f ) ,

where w q is a root of w m − 1 = 0 .

On the other hand, in view of equation (1) and the Mokhon’ko theorem in several complex variables [26, Theorem 3.4], it follows

(18) n T ( r , f ( z 1 + c 1 , z 2 + c 2 ) ) = T ( r , f ( z 1 + c 1 , z 2 + c 2 ) n ) + S ( r , f ) = T r , α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 m − 1 + S ( r , f ) = m T r , α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 + S ( r , f ) .

In view of (17)–(18) and n > m m − 1 , it follows

n − m m − 1 T ( r , f ( z 1 + c 1 , z 2 + c 2 ) ) ≤ S ( r , f ) ,

which is a contradiction with the assumption that f is a transcendental entire function.

Therefore, this completes the proof of Theorem 1.1.□

3 Proof of Theorem 1.2

The following lemmas play the key role in proving Theorems 1.2–1.3.

Lemma 3.1

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

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

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

Remark 3.1

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

Lemma 3.2

[28,29] 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 3.2

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

Lemma 3.3

[30] 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.

Now, we continue to prove Theorem 1.2.

Proof

Suppose that f is a finite order transcendental entire solution of equation (10). Since α , β are constants and not equal to zero at the same time, then it follows that α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 is transcendental. Otherwise, f ( z 1 + c 1 , z 2 + c 2 ) is not transcendental, a contradiction. We first rewrite (10) as the following form:

(19) α ∂ f ( z ) ∂ z 1 + β ∂ f ( z ) ∂ z 2 + i f ( z + c ) α ∂ f ( z ) ∂ z 1 + β ∂ f ( z ) ∂ z 2 − i f ( z + c ) = 1 .

Since α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 and f ( z 1 + c 1 , z 2 + c 2 ) are transcendental, then by Lemmas 3.2 and 3.3, it follows from (19) that

(20) α ∂ f ( z ) ∂ z 1 + β ∂ f ( z ) ∂ z 2 + if ( z + c ) = e p ( z ) , α ∂ f ( z ) ∂ z 1 + β ∂ f ( z ) ∂ z 2 − if ( z + c ) = e − p ( z ) ,

where p ( z ) is a nonconstant polynomial in z . Thus, in view of (20), it yields

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

In view of (21), we have

(22) e p ( z 1 + c 1 , z 2 + c 2 ) + e − p ( z 1 + c 1 , z 2 + c 2 ) 2 = α ∂ p ( z 1 , z 2 ) ∂ z 1 + β ∂ p ( z 1 , z 2 ) ∂ z 2 e p ( z 1 , z 2 ) + e − p ( z 1 , z 2 ) 2 i .

If α ∂ p ( z 1 , z 2 ) ∂ z 1 + β ∂ p ( z 1 , z 2 ) ∂ z 2 ≡ 0 , then from (22), it follows e p ( z 1 + c 1 , z 2 + c 2 ) + e − p ( z 1 + c 1 , z 2 + c 2 ) ≡ 0 , that is, e 2 p ( z + c ) + 1 ≡ 0 , which is impossible because p ( z ) is a nonconstant polynomial.

If α ∂ p ( z 1 , z 2 ) ∂ z 1 + β ∂ p ( z 1 , z 2 ) ∂ z 2 ≢ 0 , then (22) becomes

(23) i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 e p ( z + c ) + p ( z ) + i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 e p ( z ) − p ( z + c ) − e 2 p ( z ) ≡ 1 .

Thus, by Lemma 3.1, it yields

(24) i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 e p ( z ) − p ( z + c ) ≡ 1 .

In view of (23) and (24), it follows

(25) i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 e p ( z + c ) − p ( z ) ≡ 1 .

Since p ( z ) is a nonconstant polynomial, in view of (24) and (25), we conclude that p ( z ) = L ( z ) + H ( z ) + B , where L ( z ) is a linear function as the form L ( z ) = a 1 z 1 + a 2 z 2 , H ( z ) ≔ H ( s 1 ) is a polynomial in s 1 , s 1 ≔ c 2 z 1 − c 1 z 2 , and B is a constant in C . Substituting p ( z ) into (24), (25), we have

[ α a 1 + β a 2 + ( α c 2 − β c 1 ) H ′ ] 2 = − 1 ,

which implies that ( α c 2 − β c 1 ) H ′ = ξ , ξ ∈ C . If α c 2 − β c 1 = 0 , then ξ = 0 . If α c 2 − β c 1 ≠ 0 , then H ′ is a constant, that is, H ( z ) = b 1 s 1 + b 0 = b 1 c 2 z 1 − b 1 c 1 z 2 + b 0 , where b 1 , b 0 ∈ C . Thus, p ( z ) = L ( z ) + H ( z ) + B = L 0 ( z ) + B is still a linear form, this means H ( z ) = 0 . Hence, it follows that ( α c 2 − β c 1 ) H ′ = 0 . Thus, it yields that

(26) e L ( c ) = 1 , α a 1 + β a 2 = i , or e L ( c ) = − 1 , α a 1 + β a 2 = − i .

In view of (21), we get

(27) f ( z 1 , z 2 ) = e p ( z 1 − c 1 , z 2 − c 2 ) − e − p ( z 1 − c 1 , z 2 − c 2 ) 2 i = e L ( z ) + H ( z ) + B − L ( c ) − e − L ( z ) − H ( z ) − B + L ( c ) 2 i = A 1 e L ( z ) + H ( z ) + A 2 e − L ( z ) − H ( z ) 2 ,

where A 1 , A 2 ∈ C satisfy A 1 A 2 = 1 .

Therefore, this completes the proof of Theorem 1.2.□

4 Proof of Theorem 1.3

Proof

Suppose that f is a finite order transcendental entire solution of equation (11). Next, we consider two cases as follows.

(i) Suppose that ∂ f ∂ z 1 is not constant. Since the entire solutions of the Fermat type functional equation f 2 + g 2 = 1 are f = cos a ( z ) , g = sin a ( z ) , where a ( z ) is an entire function, then ∂ f ( z 1 , z 2 ) ∂ z 1 and β 1 f ( z 1 + c 1 , z 2 + c 2 ) − β 2 f ( z 1 , z 2 ) are transcendental. Thus, equation (11) can be rewritten as the form:

α ∂ f ( z ) ∂ z 1 + i [ β 1 f ( z + c ) − β 2 f ( z ) ] α ∂ f ( z ) ∂ z 1 − i [ β 1 f ( z + c ) − β 2 f ( z ) ] = 1 .

Since f is a finite order transcendental entire function, then by Lemmas 3.2 and 3.3, it follows

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

where p ( z 1 , z 2 ) is a nonconstant polynomial in C 2 . In view of (28), it yields

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

that is,

(29) β 2 α − i ∂ p ( z 1 , z 2 ) ∂ z 1 e p ( z 1 , z 2 ) + e − p ( z 1 , z 2 ) 2 i = β 1 e p ( z 1 + c 1 , z 2 + c 2 ) + e − p ( z 1 + c 1 , z 2 + c 2 ) 2 α .

Obviously, ∂ p ( z 1 , z 2 ) ∂ z 1 ≠ − i β 2 α , otherwise, e p ( z 1 + c 1 , z 2 + c 2 ) + e − p ( z 1 + c 1 , z 2 + c 2 ) ≡ 0 , which is a contradiction with p ( z ) is a nonconstant polynomial. Hence, in view of (29), it follows

(30) β 1 β 2 − i α ∂ p ( z ) ∂ z 1 e p ( z + c ) + p ( z ) + β 1 β 2 − i α ∂ p ( z ) ∂ z 1 e p ( z ) − p ( z + c ) − e 2 p ( z ) ≡ 1 .

Thus, by Lemma 3.1, it yields

(31) β 1 β 2 − i α ∂ p ( z ) ∂ z 1 e p ( z ) − p ( z + c ) ≡ 1 .

In view of (30) and (31), it follows

(32) β 1 β 2 − i α ∂ p ( z ) ∂ z 1 e p ( z + c ) − p ( z ) ≡ 1 .

Since p ( z ) is a nonconstant polynomial, in view of (31) and (32), we conclude that p ( z ) = L ( z ) + H ( z ) + B , where L ( z ) is a linear function as the form L ( z ) = a 1 z 1 + a 2 z 2 , H ( z ) ≔ H ( s 1 ) is a polynomial in s 1 , s 1 ≔ c 2 z 1 − c 1 z 2 , and B is a constant in C . Substituting p ( z ) into (31) and (32), it follows that α c 2 H ′ = 0 , which implies that H ( z ) is a constant since α ≠ 0 , c 2 ≠ 0 . Similar to the argument as in Theorem 1.2, H ( z ) can be considered equal to zero, this does not affect the linear form of p ( z ) . Hence, we conclude that p ( z ) = L ( z ) + B . In view of (29), let

(33) f ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + B − e − ( a 1 z 1 + a 2 z 2 + B ) 2 a 1 α + g 2 ( z 2 ) ,

where a 1 ( ≠ 0 ) , a 2 , B , D 0 are constants in C and g 2 ( x ) is a finite order entire function in x .

Substituting (33) into the second equation of (28), it follows that

e L ( z ) + B − e − ( L ( z ) + B ) 2 i = β 1 e L ( z ) + B + L ( c ) − e − ( L ( z ) + B ) − L ( c ) 2 a 1 α + β 1 g 2 ( z 2 + c 2 ) − β 2 e L ( z ) + B − e − ( L ( z ) + B ) 2 a 1 α − β 2 g 2 ( z 2 ) ,

which implies

(34) e L ( z ) + B − e − ( L ( z ) + B ) 2 i = ( β 1 e L ( c ) − β 2 ) e L ( z ) + B − ( β 1 e − L ( c ) − β 2 ) e − ( L ( z ) + B ) 2 a 1 α

and

(35) g 2 ( z 2 + c 2 ) = β 1 β 2 g 2 ( z 2 ) .

( i 1 ) If β 1 ≠ ± β 2 , in view of (34), we conclude that

β 1 e L ( c ) − β 2 = − i a 1 α , β 1 e − L ( c ) − β 2 = − i a 1 α ,

this implies that

(36) a 1 α = i ( β 1 − β 2 ) , e L ( c ) = 1 , or a 1 α = − i ( β 1 + β 2 ) , e L ( c ) = − 1 .

In view of (35), it follows that

(37) g 2 ( z 2 ) = e log β 2 − log β 1 c 2 z 2 G ( z 2 ) ,

where G ( z 2 ) is a finite order entire period function with period c 2 . Hence, f ( z 1 , z 2 ) is of the form

f ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + B − e − ( a 1 z 1 + a 2 z 2 + B ) 2 a 1 α + e log β 2 − log β 1 c 2 z 2 G ( z 2 ) ,

where G ( z 2 ) is a finite order entire period function with period c 2 , a 1 , a 2 , α , β 1 , β 2 , c satisfy (36) and (37).

( i 2 ) If β 1 = β 2 , the equality (34) implies that i α a 1 = 2 β 1 and e L ( c ) = − 1 . And (35) implies that g 2 ( z 2 ) is a finite order entire period function with period c 2 .

( i 3 ) If β 1 = − β 2 , the equality (34) implies that i α a 1 = − 2 β 1 and e L ( c ) = 1 . And (35) implies that g 2 ( z 2 ) = e z 2 log ( − 1 ) c 2 G ( z 2 ) , where G ( z 2 ) is a finite order entire period function with period c 2 .

Thus, this completes the proof of Theorem 1.3 (i).

(ii) Suppose that ∂ f ∂ z 1 is constant, let

(38) ∂ f ∂ z 1 = D 1 , D 1 ∈ C .

Then in view of (11), it follows

(39) β 1 f ( z 1 + c 1 , z 2 + c 2 ) − β 2 f ( z 1 , z 2 ) ≡ κ , ( α D 1 ) 2 + κ 2 = 1 , κ ∈ C .

Since f is a transcendental entire function of finite order, and in view of (39), we can assume that f ( z 1 , z 2 ) is of the form:

(40) f ( z 1 , z 2 ) = ψ ( z 2 ) + φ ( z 1 , z 2 ) + D 0 ,

where ψ ( z 2 ) is a finite order entire function in z 2 , φ ( z 1 , z 2 ) = D 1 z 1 + φ 1 ( z 2 ) , φ 1 ( z 2 ) is a polynomial in z 2 , D 0 are constants in C .

Substituting (40) into (39), we have

(41) κ = β 1 f ( z 1 + c 1 , z 2 + c 2 ) − β 2 f 2 ( z 1 , z 2 ) = β 1 ψ ( z 2 + c 2 ) + β 1 φ ( z 1 + c 1 , z 2 + c 2 ) + β 1 D 0 − β 2 ψ ( z 2 ) − β 2 φ ( z 1 , z 2 ) − β 2 D 0 .

( i i 1 ) If β 1 = β 2 , then in view of (41), it yields that φ ( z 1 , z 2 ) is a linear function as the form φ ( z 1 , z 2 ) = D 1 z 1 + D 2 z 2 and ψ ( x ) is a solution of equation

ψ ( x + c 2 ) = ψ ( x ) + θ 1 ,

where θ 1 = κ β 1 − ( D 1 c 1 + D 2 c 2 ) . Thus, ψ ( z 2 ) is a finite order entire periodic function with period c 2 and θ 1 ≡ 0 . Moreover, by combining with (39), we have

(42) [ β 1 ( D 1 c 1 + D 2 c 2 ) ] 2 = 1 − ( α D 1 ) 2 , D 0 ∈ C .

( i i 2 ) If β 1 ≠ β 2 , then in view of (41), it yields that φ ( z 1 , z 2 ) ≡ 0 and ψ ( z 2 ) is a solution of the difference equation

g ( x + c 2 ) = β 2 β 1 g ( x ) + θ 2 ,

where θ 2 = κ − D 0 ( β 1 − β 2 ) β 1 . Thus, ψ ( z 2 ) has the form

ψ ( z 2 ) = e log β 2 − log β 1 c 2 z 2 G ( z 2 ) + τ ,

where G ( z 2 ) is a finite order entire periodic function with period c 2 and τ satisfies τ = θ 2 1 − β 2 β 1 . Combining with (39) and θ 2 = κ − D 0 ( β 1 − β 2 ) β 1 , we have

[ ( β 1 − β 2 ) ( τ + D 0 ) ] 2 = 1 .

Therefore, this completes the proof of Theorem 1.3.□

5 Proof of Theorem 1.4

Proof

Suppose that f is a finite order transcendental entire solution of equation (12). Here, two cases will be considered below.

(i) Suppose that α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 is not constant. In view of the properties of the entire solutions of the Fermat type functional equation f 2 + g 2 = 1 , then α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 and γ 1 f ( z 1 + c 1 , z 2 + c 2 ) − γ 2 f ( z 1 , z 2 ) are transcendental. Thus, equation (12) can be rewritten as the form:

α ∂ f ∂ z 1 + β ∂ f ∂ z 2 + i [ γ 1 f ( z + c ) − γ 2 f ( z ) ] α ∂ f ∂ z 1 + β ∂ f ∂ z 2 − i [ γ 1 f ( z + c ) − γ 2 f ( z ) ] = 1 .

Since f is a finite order transcendental entire function, then we have that α ∂ f ∂ z 1 + β ∂ f ∂ z 2 + i [ γ 1 f ( z + c ) − γ 2 f ( z ) ] and α ∂ f ∂ z 1 + β ∂ f ∂ z 2 + i [ γ 1 f ( z + c ) − γ 2 f ( z ) ] have no any zeros and poles. Thus, by Lemmas 3.2 and 3.3, it yields that

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

where p ( z 1 , z 2 ) is a nonconstant polynomial in C 2 . In view of (43), it yields

e p ( z ) + e − p ( z ) 2 i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 = γ 1 e p ( z + c ) + e − p ( z + c ) 2 − γ 2 e p ( z ) + e − p ( z ) 2 ,

that is,

(44) γ 2 − i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 e p ( z ) + e − p ( z ) 2 = γ 1 e p ( z + c ) + e − p ( z + c ) 2 .

Obviously, γ 2 ≠ i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 , otherwise, e p ( z 1 + c 1 , z 2 + c 2 ) + e − p ( z 1 + c 1 , z 2 + c 2 ) ≡ 0 , which is a contradiction with p ( z ) is a nonconstant polynomial. Hence, equation (44) leads to

(45) γ 1 γ 2 − i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 e p ( z + c ) + p ( z ) + γ 1 γ 2 − i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 e p ( z ) − p ( z + c ) − e 2 p ( z ) ≡ 1 .

Thus, by Lemma 3.1, it yields

(46) γ 1 γ 2 − i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 e p ( z ) − p ( z + c ) ≡ 1 .

In view of (45) and (46), it follows

(47) γ 1 γ 2 − i α ∂ p ( z ) ∂ z 1 + β ∂ p ( z ) ∂ z 2 e p ( z + c ) − p ( z ) ≡ 1 .

Since p ( z ) is a nonconstant polynomial, in view of (46) and (47), we conclude that p ( z ) = L ( z ) + H ( z ) + B , where L ( z ) is a linear function as the form L ( z ) = a 1 z 1 + a 2 z 2 , H ( z ) ≔ H ( s 1 ) is a polynomial in s 1 , B is a constant in C . In view of α c 2 − β c 1 ≠ 0 , similar to the argument as in Theorem 1.2, we can obtain that H ( z ) ≡ 0 . Thus, in view of (46) and (47), we have

(48) γ 1 e L ( c ) − γ 2 = − i ( a 1 α + a 2 β ) , γ 1 e − L ( c ) − γ 2 = − i ( a 1 α + a 2 β ) .

In view of (43), we get the following equation:

(49) α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 = e L ( z ) + B + e − L ( z ) − B 2 .

The characteristic equations for equation (49) are

d z 1 d t = α , d z 2 d t = β , d f d t = e L ( z ) + B + e − L ( z ) − B 2 .

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

f = ∫ 0 t e ( a 1 α + a 2 β ) t + a 2 s + B + e − [ ( a 1 α + a 2 β ) t + a 2 s + B ] 2 d t + g ( s ) = e a 2 s + B 2 ∫ 0 t e ( a 1 α + a 2 β ) t d t + e − ( a 2 s + B ) 2 ∫ 0 t e − ( a 1 α + a 2 β ) t d t + g ( s ) = e a 2 s + B 2 ( a 1 α + a 2 β ) e ( a 1 α + a 2 β ) t − e − ( a 2 s + B ) 2 ( a 1 α + a 2 β ) e − ( a 1 α + a 2 β ) t + g 1 ( s ) ,

where g ( s ) is an entire function with finite order in s , and

g 1 ( s ) = g ( s ) + e − ( a 2 s + B ) 2 ( a 1 α + a 2 β ) − e a 2 s + B 2 ( a 1 α + a 2 β ) .

Substituting t = z 1 α , s = z 2 − β z 1 α into f , it follows that

(50) f ( z 1 , z 2 ) = e a 1 z 2 + a 2 z 2 + B − e − ( a 1 z 2 + a 2 z 2 + B ) 2 ( a 1 α + a 2 β ) + g 2 ( α z 2 − β z 1 ) ,

where g 2 ( s ) = g 1 ( α s ) . Substituting f ( z 1 , z 2 ) into the second equation of (43), it yields

(51) e a 1 z 2 + a 2 z 2 + B − e − ( a 1 z 2 + a 2 z 2 + B ) 2 i = γ 1 e a 1 z 2 + a 2 z 2 + B + L ( c ) − e − ( a 1 z 2 + a 2 z 2 + B ) − L ( c ) 2 ( a 1 α + a 2 β ) + γ 1 g 2 ( α z 2 − β z 1 + α c 2 − β c 1 ) − γ 2 e a 1 z 2 + a 2 z 2 + B − e − ( a 1 z 2 + a 2 z 2 + B ) 2 ( a 1 α + a 2 β ) − γ 2 g 2 ( α z 2 − β z 1 ) .

This means that

(52) e L ( z ) + B − e − ( L ( z ) + B ) 2 i = ( γ 1 e L ( c ) − γ 2 ) e L ( z ) + B − ( γ 1 e − L ( c ) − γ 2 ) e − ( L ( z ) + B ) 2 ( a 1 α + a 2 β )

and

(53) γ 1 g 2 ( α z 2 − β z 1 + α c 2 − β c 1 ) = γ 2 g 2 ( α z 2 − β z 1 ) .

( i 1 ) If γ 1 ≠ ± γ 2 , in view of (48) and (52), it yields that

(54) a 1 α + a 2 β = i ( γ 1 − γ 2 ) , e L ( c ) = 1 , or a 1 α + a 2 β = − i ( γ 1 + γ 2 ) , e L ( c ) = − 1 .

In view of (53), it follows that

(55) g 2 ( u ) = e u log β 2 − log β 1 c 2 G ( u ) ,

where G ( u ) is a finite order entire period function with period α c 2 − β c 1 and u = α z 2 − β z 1 . Hence, f ( z 1 , z 2 ) is of the form

f ( z 1 , z 2 ) = e a 1 z 1 + a 2 z 2 + B − e − ( a 1 z 1 + a 2 z 2 + B ) 2 ( a 1 α + a 2 β ) + e ( α z 2 − β z 1 ) log γ 2 − log γ 1 α c 2 − β c 1 G ( α z 2 − β z 1 ) ,

where G ( u ) is a finite order entire period function with period α c 2 − β c 1 , a 1 , a 2 , α , β , γ 1 , γ 2 , c satisfy (54) and (55).

( i 2 ) If γ 1 = γ 2 , the equality (54) implies that i ( a 1 α + a 2 β ) = 2 γ 1 and e L ( c ) = − 1 . And (55) implies that g 2 ( u ) is a finite order entire period function with period α c 2 − β c 1 .

( i 3 ) If γ 1 = − γ 2 , the equality (54) implies that i ( a 1 α + a 2 β ) = − 2 γ 1 and e L ( c ) = 1 . And (55) implies that g 2 ( u ) = e u log ( − 1 ) α c 2 − β c 1 G ( u ) , where G ( u ) is a finite order entire period function with period α c 2 − β c 1 .

Thus, this completes the proof of Theorem 1.4(i).

(ii) Suppose that α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 is constant. Let

(56) α ∂ f ( z 1 , z 2 ) ∂ z 1 + β ∂ f ( z 1 , z 2 ) ∂ z 2 = D 1 ′ , D 1 ′ ∈ C .

Then in view of (12), it follows

(57) γ 1 f ( z 1 + c 1 , z 2 + c 2 ) − γ 2 f ( z 1 , z 2 ) ≡ κ , D 1 ′ 2 + κ 2 = 1 , κ ∈ C .

By using the same argument as in the proof of Theorem 1.4(i), f ( z 1 , z 2 ) is of the form:

(58) f ( z 1 , z 2 ) = D 1 ′ z 1 α + g 2 ( α z 2 − β z 1 ) ,

where g 2 ( u ) is a finite order entire function in u , and u = α z 2 − β z 1 .

Substituting (58) into (57), we have

(59) κ = γ 1 f ( z 1 + c 1 , z 2 + c 2 ) − γ 2 f 2 ( z 1 , z 2 ) = γ 1 D 1 ′ z 1 + c 1 α + γ 1 g 2 ( α z 2 − β z 1 + α c 2 − β c 1 ) − γ 2 D 1 ′ z 1 α − γ 2 g 2 ( α z 2 − β z 1 ) .

( i i 1 ) If γ 1 = γ 2 , then (59) leads to

g 2 ( α z 2 − β z 1 + α c 2 − β c 1 ) − g 2 ( α z 2 − β z 1 ) = κ γ 1 − D 1 ′ c 1 α ,

that is,

(60) g 2 ( u + α c 2 − β c 1 ) − g 2 ( u ) = κ γ 1 − D 1 ′ c 1 α ,

for u = α z 2 − β z 1 . Thus, g 2 ( u ) can be represented as the form g 2 ( u ) = G 2 ( u ) + D 2 u + τ , where G 2 ( u ) is a finite order entire period function with period α c 2 − β c 1 , and D 2 , τ are constants in C . Combining with (60), we obtain D 2 ( α c 2 − β c 1 ) = κ γ 1 − D 1 ′ c 1 α . In view of (57), it yields

γ 1 2 [ D 1 c 1 + D 2 ( α c 2 − β c 1 ) ] 2 = 1 − ( α D 1 ) 2 ,

where D 1 = D 1 ′ α . Thus, this completes the proof of Theorem 1.4 ( i i 1 ) .

( i i 2 ) If γ 1 ≠ γ 2 , then equation (59) implies that D 1 ′ = 0 and

(61) g 2 ( u + α c 2 − β c 1 ) = γ 2 γ 1 g 2 ( u ) + κ γ 1 ,

for u = α z 2 − β z 1 . In view of (61), it follows that g 2 ( u ) = e u log γ 2 − log γ 1 α c 2 − β c 1 G ( u ) + τ , where G ( u ) is a finite order entire period function with period α c 2 − β c 1 and τ satisfies

( γ 1 − γ 2 ) τ = ± 1 .

This shows that the conclusion of Theorem 1.4 ( i i 2 ) holds.

Therefore, this completes the proof of Theorem 1.4.□

Acknowledgements

The authors thank the referee(s) for reading the manuscript very carefully and making a number of valuable and kind comments that improved the presentation.

  1. Funding information: This work was supported by the National Natural Science Foundation of China (11961030, 12161074) and the Foundation of Education Department of Jiangxi (GJJ190876, GJJ202303, GJJ201813, and GJJ191042) of China.

  2. Author contributions: Conceptualization, H.Y. Xu; writing-original draft preparation, H.Y. Xu; writing-review and editing, H.Y. Xu and W.J. Tang; funding acquisition, H.Y. Xu, X.M. Gui and H. Wang.

  3. Conflict of interest: Authors state no conflict of interest.

  4. Data availability statement: This research did not report any data.

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Received: 2021-07-12
Accepted: 2021-10-11
Published Online: 2021-12-31

© 2021 Xian Min Gui et al., 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-2021-0113
Keywords for this article
existence; entire solution; partial differential-difference equation
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  94. Entire solutions for several complex partial differential-difference equations of Fermat type in ℂ2
  95. Random attractors for stochastic plate equations with memory in unbounded domains
  96. On the convergence of two-step modulus-based matrix splitting iteration method
  97. On the separation method in stochastic reconstruction problem
  98. Robust estimation for partial functional linear regression models based on FPCA and weighted composite quantile regression
  99. Structure of coincidence isometry groups
  100. Sharp function estimates and boundedness for Toeplitz-type operators associated with general fractional integral operators
  101. Oscillatory hyper-Hilbert transform on Wiener amalgam spaces
  102. Euler-type sums involving multiple harmonic sums and binomial coefficients
  103. Poly-falling factorial sequences and poly-rising factorial sequences
  104. Geometric approximations to transition densities of Jump-type Markov processes
  105. Multiple solutions for a quasilinear Choquard equation with critical nonlinearity
  106. Bifurcations and exact traveling wave solutions for the regularized Schamel equation
  107. Almost factorizable weakly type B semigroups
  108. The finite spectrum of Sturm-Liouville problems with n transmission conditions and quadratic eigenparameter-dependent boundary conditions
  109. Ground state sign-changing solutions for a class of quasilinear Schrödinger equations
  110. Epi-quasi normality
  111. Derivative and higher-order Cauchy integral formula of matrix functions
  112. Commutators of multilinear strongly singular integrals on nonhomogeneous metric measure spaces
  113. Solutions to a multi-phase model of sea ice growth
  114. Existence and simulation of positive solutions for m-point fractional differential equations with derivative terms
  115. Bernstein-Walsh type inequalities for derivatives of algebraic polynomials in quasidisks
  116. Review Article
  117. Semiprimeness of semigroup algebras
  118. Special Issue on Problems, Methods and Applications of Nonlinear Analysis (Part II)
  119. Third-order differential equations with three-point boundary conditions
  120. Fractional calculus, zeta functions and Shannon entropy
  121. Uniqueness of positive solutions for boundary value problems associated with indefinite ϕ-Laplacian-type equations
  122. Synchronization of Caputo fractional neural networks with bounded time variable delays
  123. On quasilinear elliptic problems with finite or infinite potential wells
  124. Deterministic and random approximation by the combination of algebraic polynomials and trigonometric polynomials
  125. On a fractional Schrödinger-Poisson system with strong singularity
  126. Parabolic inequalities in Orlicz spaces with data in L1
  127. Special Issue on Evolution Equations, Theory and Applications (Part II)
  128. Impulsive Caputo-Fabrizio fractional differential equations in b-metric spaces
  129. Existence of a solution of Hilfer fractional hybrid problems via new Krasnoselskii-type fixed point theorems
  130. On a nonlinear system of Riemann-Liouville fractional differential equations with semi-coupled integro-multipoint boundary conditions
  131. Blow-up results of the positive solution for a class of degenerate parabolic equations
  132. Long time decay for 3D Navier-Stokes equations in Fourier-Lei-Lin spaces
  133. On the extinction problem for a p-Laplacian equation with a nonlinear gradient source
  134. General decay rate for a viscoelastic wave equation with distributed delay and Balakrishnan-Taylor damping
  135. On hyponormality on a weighted annulus
  136. Exponential stability of Timoshenko system in thermoelasticity of second sound with a memory and distributed delay term
  137. Convergence results on Picard-Krasnoselskii hybrid iterative process in CAT(0) spaces
  138. Special Issue on Boundary Value Problems and their Applications on Biosciences and Engineering (Part I)
  139. Marangoni convection in layers of water-based nanofluids under the effect of rotation
  140. A transient analysis to the M(τ)/M(τ)/k queue with time-dependent parameters
  141. Existence of random attractors and the upper semicontinuity for small random perturbations of 2D Navier-Stokes equations with linear damping
  142. Degenerate binomial and Poisson random variables associated with degenerate Lah-Bell polynomials
  143. Special Issue on Fractional Problems with Variable-Order or Variable Exponents (Part I)
  144. On the mixed fractional quantum and Hadamard derivatives for impulsive boundary value problems
  145. The Lp dual Minkowski problem about 0 < p < 1 and q > 0
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