Entire and meromorphic solutions for systems of the differential difference equations

Hong Yan Xu; Hong Li; Xin Ding

doi:10.1515/dema-2022-0161

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Entire and meromorphic solutions for systems of the differential difference equations

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Published/Copyright: October 5, 2022

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From the journal Demonstratio Mathematica Volume 55 Issue 1

Abstract

With the help of the Nevanlinna theory of meromorphic functions, the purpose of this article is to describe the existence and the forms of transcendental entire and meromorphic solutions for several systems of the quadratic trinomial functional equations:

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

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

and

f ( z + c ) 2 + 2 α f ( z + c ) g ″ ( z ) + g ″ ( z ) 2 = 1 , g ( z + c ) 2 + 2 α g ( z + c ) f ″ ( z ) + f ″ ( z ) 2 = 1 .

We obtain a series of results on the forms of the entire solutions with finite order for such systems, which are some improvements and generalizations of the previous theorems given by Gao et al. Moreover, we provide some examples to explain the existence and forms of solutions for such systems in each case.

Keywords: Nevanlinna theory; meromorphic function; differential difference equation

MSC 2010: 30D 35; 35M 30; 32W 50; 39A 45

1 Introduction

We first assume that the reader is familiar with the basic results and notations of the Nevanlinna theory and difference Nevanlinna theory with one complex variable, which can be found in [1,2,3, 4,5]. Let us recall the classical results about the Fermat-type functional equations.

Theorem A

(See [6, Theorem 1.3].) For h : C n → C entire, the solutions of

(1.1) f 2 + g 2 = 1

are characterized as follows:

the entire solutions are f = cos ( h ) , g = sin ( h ) ;
the meromorphic solutions are of the form f = 1 − β 2 1 + β 2 , g = 2 β 1 + β 2 , with β being meromorphic on C n .

After these results, many mathematicians paid considerable attention to the solutions for complex difference equations, complex differential difference equations (DDEs), and fractional differential equations, and obtained a series of results focusing on the existence and the form of solutions for such functional equations ([3,7,8, 9,10,11, 12,13,14, 15,16,17, 18,19,20]). Liu et al. [21] in 2012 studied the existence of solutions of some complex difference equations and obtained the following.

Theorem B

(See [21, Theorem 1.1].) The transcendental entire solutions with finite order of

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

must satisfy f ( z ) = sin ( A z + B ) , where B is a constant and A = ( 4 k + 1 ) π 2 c , with k an integer.

Theorem C

(See [21, Theorem 1.3].) The transcendental entire solutions 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 , k is an integer.

In 2016, Liu and Yang [22] further investigated the entire and meromorphic solutions of some Fermat-type functional equations and obtained the following.

Theorem D

(See [22, Theorem 1.6].) If α ≠ ± 1 , 0 , then equation

(1.2) f ( z ) 2 + 2 α f ( z ) f ′ ( z ) + f ′ ( z ) 2 = 1

has no transcendental meromorphic solutions.

Theorem E

(See [22, Theorem 1.4].) If α ≠ ± 1 , 0 , then the finite order transcendental entire functions of equation

(1.3) f ( z ) 2 + 2 α f ( z ) f ( z + c ) + f ( z + c ) 2 = 1

must be of order equal to one.

In 2016, Gao [23] discussed the problem on the form of solutions for a class of system of DDEs corresponding to Theorem E and obtained the following.

Theorem F

(See [23, Theorem 1.1].) Suppose that ( f 1 , f 2 ) is a pair of finite order transcendental entire solutions for the system of DDEs

(1.4) [ f 1 ′ ( z ) ] 2 + f 2 ( z + c ) 2 = 1 , [ f 2 ′ ( z ) ] 2 + f 1 ( z + c ) 2 = 1 .

Then ( f 1 , f 2 ) satisfies

( f 1 ( z ) , f 2 ( z ) ) = ( sin ( z − b i ) , sin ( z − b 1 i ) )

( f 1 ( z ) , f 2 ( z ) ) = ( sin ( z + b i ) , sin ( z + b 1 i ) ) ,

where b , b 1 are constants, and c = k π , k is a integer.

In 2019, Liu and Gao [24] investigated the entire solutions of second-order DDEs and obtained the following.

Theorem G

(See [24, Theorem 2.1].) Suppose that f is a transcendental entire solution with finite order of the complex DDE

(1.5) f ″ ( z ) 2 + f ( z + c ) 2 = Q ( z ) ,

then Q ( z ) = c 1 c 2 is a constant, and f ( z ) satisfies

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

where a , b ∈ C , and a 4 = 1 , c = log ( − i a 2 ) + 2 k π i a , k ∈ Z .

In view of the aforementioned results, some questions can naturally be raised as follows.

Question 1.1

What will happen to entire and meromorphic solutions if equations (1.2), (1.3), and (1.5) become the quadratic trinomial functional equations?

Question 1.2

What can be said about the properties of solutions when the quadratic trinomial functional equations (systems) include the function f and its second-order differential f ″ ?

2 Results and some examples

Inspired by Questions 1.1 and 1.2, this article is devoted to the description of entire and meromorphic solutions for several quadratic trinomial difference equation and DDEs.

Theorem 2.1

Let ( f , g ) be the pair of finite order transcendental entire solutions for system

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

where α 2 ≠ 1 , α , c ∈ C . Then ( f , g ) must be one of the following forms:

f ( z ) = 1 2 cos ( a z + b 1 ) 1 + α + sin ( a z + b 1 ) 1 − α , g ( z ) = 1 2 cos ( a z + b 2 ) 1 + α + sin ( a z + b 2 ) 1 − α ,
where a , b 1 , b 2 , c satisfy
e 2 i a c ≡ − α + α 2 − 1 − α − α 2 − 1 , e 2 i ( b 1 − b 2 ) ≡ 1 ;
f ( z ) = 1 2 cos ( a z + b 1 ) 1 + α + sin ( a z + b 1 ) 1 − α , g ( z ) = 1 2 cos ( a z + b 2 ) 1 + α − sin ( a z + b 2 ) 1 − α ,
where a , b 1 , b 2 , c satisfy
e 2 i a c ≡ 1 , e 2 i ( b 1 − b 2 ) ≡ 1 .

The following examples show that equation (2.1) has a transcendental entire solution with the order ≥ 1 . This gives a very significant difference with Theorems C, E, and F.

Example 2.1

Let

f ( z ) = 2 3 sin z + 4 π 3 , g ( z ) = 2 3 sin z − 2 π 3 .

Thus, we can conclude that ρ ( ( f , g ) ) = 1 and ( f , g ) is a pair of transcendental solutions for system (2.1) with c = 2 π 3 and α = 1 2 .

Example 2.2

Let

f = 1 2 cos ( z + π ) 1 + α + sin ( z + π ) 1 − α = − 1 2 cos z 1 + α + sin z 1 − α

and

g = 1 2 cos z 1 + α − sin z 1 − α .

Thus, we can conclude that ρ ( ( f , g ) ) = 1 and ( f , g ) is a pair of transcendental solutions for system (2.1) with c = π and α ∈ C − { 0 , 1 , − 1 } .

Remark 2.1

It is easy to obtain that system (2.1) can be represented as [ f ( z ) ± g ( z + c ) ] 2 = 1 and [ g ( z ) ± f ( z + c ) ] 2 = 1 if α 2 = 1 . If f ( z ) − g ( z + c ) = 1 and g ( z ) − f ( z + c ) = 1 , then we have f ( z ) = 2 + f ( z + 2 c ) and g ( z ) = 2 + g ( z + 2 c ) . Thus, we can find the transcendental entire solutions with finite order or infinite order easily when c = 1 . For example,

( f , g ) = ( e π i z − z , − e π i z − z ) , ( f , g ) = ( sin ( e π i z ) − z , − sin ( e π i z ) − z ) .

Let us discuss the meromorphic solution ( f , g ) for system (2.1). Let α 1 = − α + α 2 − 1 and α 2 = − α − α 2 − 1 . From Theorem A(b), we have that f = α 1 − α 2 β 2 ( α 1 − α 2 ) β , g = α 1 − α 2 γ 2 ( α 1 − α 2 ) γ , where β , γ are meromorphic functions in C 2 . By combining with (2.1), it follows that

(2.2) f ( z + c ) = 1 − γ 2 ( α 1 − α 2 ) γ and g ( z + c ) = 1 − β 2 ( α 1 − α 2 ) β .

By a simple calculation, we have

(2.3) [ α 2 β ( z + c ) − γ ( z ) ] [ β ( z + c ) γ ( z ) + α 1 ] ≡ 0 , [ α 2 γ ( z + c ) − β ( z ) ] [ γ ( z + c ) β ( z ) + α 1 ] ≡ 0 .

Now, two examples of meromorphic solutions for system (2.1) are listed in the following.

Example 2.3

Let

β ( z ) = 2 1 − e π i z 1 + e π i z , γ ( z ) = 2 1 + e π i z 1 − e π i z , or β ( z ) = 2 1 − sin ( e π i z ) 1 + sin ( e π i z ) , γ ( z ) = 2 1 + sin ( e π i z ) 1 − sin ( e π i z ) .

Then we have ( f , g ) = 4 − β 2 3 β , 4 − γ 2 3 γ satisfies system (2.1) with c = 1 and α = 5 4 , which implies that β ( z ) γ ( z + c ) = 2 = − α 1 and γ ( z ) β ( z + c ) = 2 = − α 1 , where α 1 = − 2 .

Example 2.4

Let

β ( z ) = 1 3 z sin ( π i z ) , γ ( z ) = 1 3 z − sin ( π i z ) , or β ( z ) = 1 3 z sin sin ( π i z ) , γ ( z ) = 1 3 z − sin sin ( π i z ) .

Then we have ( f , g ) = 9 − β 2 8 β , 9 − γ 2 8 γ satisfies system (2.1) with c = 1 and α = − 5 3 , which implies that 3 β ( z + c ) = α 2 β ( z + c ) = γ ( z ) and 3 γ ( z + c ) = α 2 γ ( z + c ) = β ( z ) , where α 2 = 3 .

If f ( z + c ) and g ( z + c ) are replaced by f ′ ( z ) and g ′ ( z ) in system (2.1), respectively, we obtain the following.

Theorem 2.2

Let α 2 ≠ 1 , 0 and α ∈ C . Then

(2.4) f ( z ) 2 + 2 α f ( z ) g ′ ( z ) + g ′ ( z ) 2 = 1 , g ( z ) 2 + 2 α g ( z ) f ′ ( z ) + f ′ ( z ) 2 = 1 ,

there does not exist any pair of transcendental entire solutions.

If f and g are replaced by f ′ ( z ) and g ′ ( z ) for system (2.1), respectively, we obtain the following.

Theorem 2.3

Let c ∈ C − { 0 } , and α 2 ≠ 1 , 0 , α ∈ C . Then any pair of finite order transcendental entire solutions for the system of DDEs

(2.5) f ( z + c ) 2 + 2 α f ( z + c ) g ′ ( z ) + g ′ ( z ) 2 = 1 , g ( z + c ) 2 + 2 α g ( z + c ) f ′ ( z ) + f ′ ( z ) 2 = 1

must satisfy one of the following forms:

(2.6) f = 1 2 cos ( a z − a c + b 1 ) 1 + α + sin ( a z − a c + b 1 ) 1 − α ,

(2.7) g = 1 2 cos ( a z − a c + b 2 ) 1 + α + sin ( a z − a c + b 2 ) 1 − α ,
where a , b 1 , b 2 , c ∈ C satisfy
a 2 ≡ 1 , e 2 i ( b 1 − b 2 ) ≡ 1 , e 2 i a c = α + α 2 − 1 − α + α 2 − 1 ;
(2.8) f = 1 2 cos ( a z − a c + b 1 ) 1 + α + sin ( a z − a c + b 1 ) 1 − α ,

(2.9) g = 1 2 cos ( a z − a c + b 2 ) 1 + α − sin ( a z − a c + b 2 ) 1 − α ,
where a , b 1 , b 2 , c ∈ C satisfy
a 2 ≡ 1 , e 2 i ( b 1 − b 2 ) ≡ 1 , e 2 i a c ≡ − 1 .

Now we give the following examples to explain that the existence of transcendental entire solution of (2.5) and the forms of such solutions are precise.

Example 2.5

Let

f = 1 2 cos z − π 6 1 2 + sin z − π 6 3 2 = 2 3 sin z + π 6

and

g = 1 2 cos z − π 6 − π 1 2 + sin z − π 6 − π 3 2 = − 2 3 sin z + π 6 .

Then it follows that ( f , g ) is a pair of transcendental entire solutions for (2.5) with c = π 6 and α = − 1 2 . Moreover, ρ ( ( f , g ) ) = 1 , e i ( b 1 − b 2 ) = − 1 , and

e 2 i a c = e π 3 i = 1 + i 3 2 = α + α 2 − 1 − α + α 2 − 1 .

Example 2.6

Let

f = 1 2 cos z − π 2 1 + α + sin z − π 2 1 − α = 1 2 sin z 1 + α − cos z 1 − α

and

g = 1 2 cos z − π 2 1 + α − sin z − π 2 1 − α = 1 2 sin z 1 + α + cos z 1 − α .

Then ( f , g ) is a pair of transcendental entire solutions for (2.5) with c = π 2 and α ∈ C − { 0 , 1 , − 1 } . Moreover, ρ ( ( f , g ) ) = 1 and

a 2 ≡ 1 , e 2 i ( b 1 − b 2 ) ≡ 1 , e 2 i a c ≡ − 1 .

Finally, we will study the existence and the form of the solutions of some second-order quadratic trinomial differential equations (system) and DDE system and obtain the following four theorems.

Theorem 2.4

Let α 2 ≠ 1 , 0 and α ∈ C . Then the differential equation

(2.10) f ( z ) 2 + 2 α f ( z ) f ″ ( z ) + f ″ ( z ) 2 = 1

has no any transcendental entire solution with finite order.

Theorem 2.5

Let α 2 ≠ 1 , 0 and α ∈ C . Then any pair of finite order transcendental entire solutions for system of the differential equations

(2.11) f ( z ) 2 + 2 α f ( z ) g ″ ( z ) + g ″ ( z ) 2 = 1 , g ( z ) 2 + 2 α g ( z ) f ″ ( z ) + f ″ ( z ) 2 = 1

must be of the forms

f ( z ) = 1 2 cos ( a z + b 1 ) 1 + α + sin ( a z + b 1 ) 1 − α , g ( z ) = 1 2 cos ( a z − b 2 ) 1 + α − sin ( a z − b 2 ) 1 − α ,

where a , b 1 , b 2 , ∈ C satisfy a 4 ≡ 1 and e 2 i ( b 1 + b 2 ) ≡ 1 .

Theorem 2.6

Let c ∈ C − { 0 } and α 2 ≠ 1 , 0 , α ∈ C . Then any finite order transcendental entire solution of the second-order DDE

(2.12) f ( z + c ) 2 + 2 α f ( z + c ) f ″ ( z ) + f ″ ( z ) 2 = 1

must be of the form

f ( z ) = 1 2 cos [ a z − a c + b ] 1 + α + sin [ a z − a c + b ] 1 − α ,

where a , b , c ∈ C satisfy

a 4 = 1 , e 2 i a c = α + α 2 − 1 α − α 2 − 1 = 2 α 2 + 2 α α 2 − 1 − 1 .

Theorem 2.7

Let c ∈ C − { 0 } and α 2 ≠ 1 , 0 , α ∈ C . Then any pair of transcendental entire solution with finite-order of the system of the second-order DDEs

(2.13) f ( z + c ) 2 + 2 α f ( z + c ) g ″ ( z ) + g ″ ( z ) 2 = 1 , g ( z + c ) 2 + 2 α g ( z + c ) f ″ ( z ) + f ″ ( z ) 2 = 1

must be of the forms

(2.6) and (2.7), where a , b 1 , b 2 , c ∈ C satisfy
a 4 ≡ 1 , e 2 i ( b 1 − b 2 ) ≡ 1 , e 2 i a c = − α − α 2 − 1 − α + α 2 − 1 ;
(2.8) and (2.9), where a , b 1 , b 2 , c ∈ C satisfy
a 4 ≡ 1 , e 2 i ( b 1 − b 2 ) ≡ 1 , e 2 i a c ≡ 1 .

The following examples show the existence of transcendental entire solution with finite order for equation (2.12) and systems (2.11) and (2.13).

Example 2.7

Let

f ( z ) = cos z − π 3 3 + sin z − π 3 = 2 3 sin z − π 6 .

Then f ( z ) is a finite-order transcendental entire solution of equations (2.12) with c = π 3 and α = 1 2 . Obviously, we have ρ ( f ) = 1 .

Example 2.8

Let

( f , g ) = 1 2 cos ( i z ) 1 + α + sin ( i z ) 1 − α , 1 2 cos ( i z ) 1 + α − sin ( i z ) 1 − α .

Then ( f , g ) is a pair of transcendental entire solutions of system (2.11) with α ∈ C − { 0 , 1 , − 1 } . Obviously, we have ρ ( ( f , g ) ) = 1 .

Example 2.9

Let

( f , g ) = − 1 2 cos z 1 + α + sin z 1 − α , 1 2 cos z 1 + α − sin z 1 − α .

Then ( f , g ) is a pair of transcendental entire solutions of system (2.11) with α ∈ C − { 0 , 1 , − 1 } . Obviously, we have ρ ( ( f , g ) ) = 1 , 1 4 ≡ 1 , and e 2 i ( π + 0 ) ≡ 1 .

Example 2.10

Let

f ( z ) = g ( z ) = 1 2 cos ( z − 2 π 3 ) 1 ∕ 2 + sin ( z − 2 π 3 ) 3 ∕ 2 = 2 3 sin z − π 3 .

Then ( f , g ) is a pair of transcendental entire solutions of equation (2.13) with α = − 1 2 and c = 2 π 3 . Obviously, we have ρ ( ( f , g ) ) = 1 , ( − i ) 4 = 1 , e 2 i ( 0 − 0 ) = 1 , and

e 2 i a c = e 4 π 3 = − 1 − i 3 2 = − α + α 2 − 1 − α − α 2 − 1 .

Example 2.11

Let a = i , b 1 = π , b 2 = 0 , and

f = 1 2 cos ( i z ) 1 + α + sin ( i z ) 1 − α , g = − 1 2 cos ( i z ) 1 + α − sin ( i z ) 1 − α .

Then ( f , g ) is a pair of transcendental entire solutions of system (2.13) with c = π i and α ∈ C − { 0 , 1 , − 1 } . Obviously, we have ρ ( ( f , g ) ) = 1 , i 4 = 1 , e 2 i i π i = 1 , and e 2 i ( π − 0 ) = 1 .

3 Proofs of Theorems 2.1–2.3

The following lemmas play the key roles in proving our theorems.

Lemma 3.1

[25]. 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

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

Lemma 3.2

[26]. Let f j ( z ) ( j = 1 , 2 , 3 ) be meromorphic functions, f 1 ( z ) being non-constant. If ∑ j = 1 3 f j ≡ 1 and

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

where λ < 1 and T ( r ) = max 1 ≤ j ≤ 3 { T ( r , f j ) } , then f 2 ( z ) ≡ 1 or f 3 ( z ) ≡ 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.

3.1 Proof of Theorem 2.1

Assume that ( f , g ) is a pair of finite order transcendental entire solutions of system (2.1), then there exist two entire functions u , v such that

f ( z ) = 1 2 ( u + v ) , g ( z + c ) = 1 2 ( u − v ) .

Thus, we can rewrite the first equation of system (2.1) as

(3.1) ( 1 + α ) u 2 + ( 1 − α ) v 2 = 1 .

By Theorem A and Lemma 3.1, there exists a nonconstant polynomial p ( z ) such that

(3.2) f ( z ) = 1 2 cos p ( z ) 1 + α + sin p ( z ) 1 − α

and

(3.3) g ( z + c ) = 1 2 cos p ( z ) 1 + α − sin p ( z ) 1 − α .

Noting that the second equation in system (2.1), there exists a nonconstant polynomial q ( z ) such that

(3.4) g ( z ) = 1 2 cos q ( z ) 1 + α + sin q ( z ) 1 − α

and

(3.5) f ( z + c ) = 1 2 cos q ( z ) 1 + α − sin q ( z ) 1 − α .

Thus, we can deduce from (3.2)–(3.5) that

(3.6) A 1 e i [ p ( z + c ) + q ( z ) ] + A 2 e i [ q ( z ) − p ( z + c ) ] − A 2 e 2 i q ( z ) = A 1 ,

(3.7) A 1 e i [ q ( z + c ) + p ( z ) ] + A 2 e i [ p ( z ) − q ( z + c ) ] − A 2 e 2 i p ( z ) = A 1 ,

hereinafter, A 1 = 1 1 + α − i 1 − α and A 2 = 1 1 + α + i 1 − α . In view of α 2 ≠ 1 , 0 , it leads to A 1 ≠ 0 and A 2 ≠ 0 . By Lemma 3.2, it follows from (3.6) and (3.7) that

A 2 e i [ q ( z ) − p ( z + c ) ] ≡ A 1 , or e i [ p ( z + c ) + q ( z ) ] ≡ 1 ,

and

A 2 e i [ p ( z ) − q ( z + c ) ] ≡ A 1 , or e i [ q ( z + c ) + p ( z ) ] ≡ 1 .

Now, we will discuss four cases as follows.

Case 1.

(3.8) A 2 e i [ q ( z ) − p ( z + c ) ] ≡ A 1 , A 2 e i [ p ( z ) − q ( z + c ) ] ≡ A 1 .

Thus, it follows that q ( z ) − p ( z + c ) ≡ η 1 and p ( z ) − q ( z + c ) ≡ η 2 , where η 1 , η 2 ∈ C . This shows that p ( z ) − p ( z + 2 c ) ≡ η 1 + η 2 and q ( z ) − q ( z + 2 c ) ≡ η 1 + η 2 . Since p , q are nonconstant polynomials, it yields that p ( z ) = a z + b 1 and q ( z ) = a z + b 2 , where a , b 1 , b 2 ∈ C . Thus, in view of (3.6)–(3.8), it yields that

A 2 e i ( − a c + b 2 − b 1 ) ≡ A 1 , A 2 e i ( − a c + b 1 − b 2 ) ≡ A 1 , A 1 e i ( a c + b 1 − b 2 ) ≡ A 2 , A 1 e i ( a c + b 2 − b 1 ) ≡ A 2 ,

which implies that

(3.9) e 2 i a c ≡ A 2 2 A 1 2 ≡ − α + α 2 − 1 − α − α 2 − 1 , e 2 i ( b 1 − b 2 ) ≡ 1 .

Substituting these into (3.2), (3.4), combining with (3.9), we can obtain the conclusion (i) of Theorem 2.1.

Case 2.

A 2 e i [ q ( z ) − p ( z + c ) ] ≡ A 1 , e i [ q ( z + c ) + p ( z ) ] ≡ 1 .

Thus, we conclude that q ( z ) − p ( z + c ) ≡ η 1 and q ( z + c ) + p ( z ) ≡ η 2 , where η 1 , η 2 are two constants in C . It means that q ( z ) + q ( z + 2 c ) ≡ η 1 + η 2 . This is a contradiction since q ( z ) is a nonconstant polynomial.

Case 3.

e i [ p ( z + c ) + q ( z ) ] ≡ 1 , A 2 e i [ p ( z ) − q ( z + c ) ] ≡ A 1 .

Thus, we conclude that p ( z + c ) + q ( z ) ≡ η 1 and p ( z ) − q ( z + c ) ≡ η 2 , where η 1 , η 2 are two constants in C . It means that p ( z ) + p ( z + 2 c ) ≡ η 1 + η 2 . We can obtain a contradiction with the assumption of p ( z ) being a nonconstant polynomial.

Case 4.

(3.10) e i [ p ( z + c ) + q ( z ) ] ≡ 1 , e i [ q ( z + c ) + p ( z ) ] ≡ 1 .

Thus, we conclude that p ( z + c ) + q ( z ) ≡ η 1 and q ( z + c ) + p ( z ) ≡ η 2 , where η 1 , η 2 are two constants in C . This shows that p ( z + 2 c ) − p ( z ) ≡ η 1 − η 2 and q ( z + 2 c ) − q ( z ) ≡ η 2 − η 1 . Noting that p , q are nonconstant polynomials, similar to the argument as in Case 1, we have that p ( z ) = a z + b 1 ′ and q ( z ) = − a z + b 2 ′ , where a , b 1 ′ , b 2 ′ ∈ C . Thus, in view of (3.6), (3.7), and (3.10), we can deduce that

e i ( a c + b 1 ′ + b 2 ′ ) ≡ 1 , e i ( − a c + b 1 ′ + b 2 ′ ) ≡ 1 , e i ( − a c − b 1 ′ − b 2 ′ ) ≡ 1 , e i ( a c − b 2 ′ − b 1 ′ ) ≡ 1 ,

which implies that

(3.11) e 2 i a c ≡ 1 , e 2 i ( b 1 ′ + b 2 ′ ) ≡ 1 .

Let b 1 = b 1 ′ and b 2 = − b 2 ′ , substituting p , q into (3.2), (3.4), and combining with (3.11), we can obtain the conclusion of Theorem 2.1 (ii).

Thus, we complete the proof of Theorem 2.1.

3.2 Proof of Theorem 2.2

Assume that ( f , g ) is a pair of transcendental entire solutions with finite order for system (2.4). Similar to the argument as in Theorem 2.1, there exist two nonconstant polynomials p ( z ) , q ( z ) such that

(3.12) f ( z ) = 1 2 cos p ( z ) 1 + α + sin p ( z ) 1 − α , g ′ ( z ) = 1 2 cos p ( z ) 1 + α − sin p ( z ) 1 − α , g ( z ) = 1 2 cos q ( z ) 1 + α + sin q ( z ) 1 − α , f ′ ( z ) = 1 2 cos q ( z ) 1 + α − sin q ( z ) 1 − α .

Thus, it follows from (3.12) that

(3.13) − sin p 1 + α + cos p 1 − α p ′ = cos q 1 + α − sin q 1 − α

and

(3.14) − sin q 1 + α + cos q 1 − α q ′ = cos p 1 + α − sin p 1 − α .

Equations (3.13) and (3.14) can be represented as

(3.15) i A 1 p ′ e i ( p + q ) − i A 2 p ′ e i ( q − p ) − A 2 e 2 i q = A 1

and

(3.16) i A 1 q ′ e i ( p + q ) − i A 2 q ′ e i ( p − q ) − A 2 e 2 i p = A 1 .

In view of α 2 ≠ 1 , it follows that A 1 ≠ 0 and A 2 ≠ 0 . Now, we claim that p ′ ≢ 0 and q ′ ≢ 0 . In fact, if p ′ ≡ 0 , then it follows from (3.15) that − A 2 e 2 i q = A 1 , which implies that q is a constant, this is a contradiction. Similarly, we can obtain a contradiction if q ′ ≡ 0 . Thus, by Lemma 3.2, it yields from (3.15) and (3.16) that

− i A 2 p ′ e i ( q − p ) ≡ A 1 , or i p ′ e i ( p + q ) ≡ 1 ,

and

− i A 2 q ′ e i ( p − q ) ≡ A 1 , or i q ′ e i ( p + q ) ≡ 1 .

Now, we will consider four cases as follows.

Case 1.

(3.17) − i A 2 p ′ e i ( q − p ) ≡ A 1 , − i A 2 q ′ e i ( p − q ) ≡ A 1 .

Thus, we have that p − q ≡ η and both p ′ and q ′ are constants in C , where η is a constant in C . Let p ′ ≡ a . Noting that p , q are nonconstant polynomials, then it leads to p = a z + b 1 , where b 1 ∈ C . By combining with p − q ≡ η , we have q = a z + b 2 , where b 2 ∈ C . Thus, in view of (3.15)–(3.17), we can deduce that

(3.18) − i A 2 a e i ( b 2 − b 1 ) ≡ A 1 , − i A 2 a e i ( b 1 − b 2 ) ≡ A 1 , i A 1 a e i ( b 1 − b 2 ) ≡ A 2 , i A 1 a e i ( b 2 − b 1 ) ≡ A 2 .

Hence, we can deduce from (3.18) that a 2 ≡ 1 and − A 2 2 ≡ A 1 2 , that is,

− 1 1 + α − i 1 − α 2 ≡ 1 1 + α + i 1 − α 2 .

This is impossible since α ≠ 0 .

Case 2.

(3.19) − i A 2 p ′ e i ( q − p ) ≡ A 1 , i q ′ e i ( p + q ) ≡ 1 .

Thus, it yields from (3.19) that q − p ≡ η 1 and p + q ≡ η 2 , where η 1 , η 2 are constants in C . This means that 2 q ≡ η 1 + η 2 is a constant. This is a contradiction.

Case 3.

i p ′ e i ( p + q ) ≡ 1 , − i A 2 q ′ e i ( p − q ) ≡ A 1 .

Similar to the argument as in Case 2 of Theorem 2.2, we obtain that p is a constant, which is a contradiction.

Case 4.

(3.20) i p ′ e i ( p + q ) ≡ 1 , i q ′ e i ( p + q ) ≡ 1 .

Thus, we have from (3.20) that p + q is a constant, and p ′ , q ′ are constants in C . Similar to the argument as in Case 1 of Theorem 2.2, we can conclude that p = a z + b 1 and q = − a z + b 2 , where b 1 , b 2 ∈ C . Substituting these into (3.20), we obtain

(3.21) i a e i ( b 1 + b 2 ) ≡ 1 , − i a e i ( b 1 + b 2 ) ≡ 1 ,

which implies that − 1 ≡ 1 . This is a contradiction.

Therefore, this completes the proof of Theorem 2.2.

3.3 Proof of Theorem 2.3

Assume that ( f , g ) is a pair of transcendental entire solutions with finite order for system (2.5). Similar to the argument as in the proof of Theorem 2.1, there exist two nonconstant polynomials p ( z ) , q ( z ) such that

(3.22) f ( z + c ) = 1 2 cos p ( z ) 1 + α + sin p ( z ) 1 − α , g ′ ( z ) = 1 2 cos p ( z ) 1 + α − sin p ( z ) 1 − α , g ( z + c ) = 1 2 cos q ( z ) 1 + α + sin q ( z ) 1 − α , f ′ ( z ) = 1 2 cos q ( z ) 1 + α − sin q ( z ) 1 − α .

So we can deduce from (3.22) that

(3.23) − sin p ( z ) 1 + α + cos p ( z ) 1 − α p ′ = cos q ( z + c ) 1 + α − sin q ( z + c ) 1 − α

and

(3.24) − sin q ( z ) 1 + α + cos q ( z ) 1 − α q ′ = cos p ( z + c ) 1 + α − sin p ( z + c ) 1 − α .

Obviously, p ′ ≢ 0 and q ′ ≢ 0 . Otherwise, it follows from (3.23) and (3.24) that tan p ( z + c ) = 1 − α 1 + α and tan q ( z + c ) = 1 − α 1 + α , which implies that p ( z ) , q ( z ) are constants. This is a contradiction with our assumptions. Thus, in view of (3.23) and (3.24) that

(3.25) i A 1 p ′ e i ( p ( z ) + q ( z + c ) ) − i A 2 p ′ e i ( q ( z + c ) − p ( z ) ) − A 2 e 2 i q ( z + c ) ≡ A 1

and

(3.26) i A 1 q ′ e i ( q ( z ) + p ( z + c ) ) − i A 2 q ′ e i ( p ( z + c ) − q ( z ) ) − A 2 e 2 i p ( z + c ) ≡ A 1 .

Noting that α 2 ≠ 1 , 0 , we have A 1 ≢ 0 and A 2 ≢ 0 . Thus, by Lemma 3.2, we can deduce from (3.25) and (3.26) that

− i A 2 p ′ e i ( q ( z + c ) − p ( z ) ) ≡ A 1 , or i p ′ e i ( p ( z ) + q ( z + c ) ) ≡ 1 ,

and

− i A 2 q ′ e i ( p ( z + c ) − q ( z ) ) ≡ A 1 , or i q ′ e i ( q ( z ) + p ( z + c ) ) ≡ 1 .

Now, we will consider four cases as follows.

Case 1.

(3.27) − i A 2 p ′ e i ( q ( z + c ) − p ( z ) ) ≡ A 1 , − i A 2 q ′ e i ( p ( z + c ) − q ( z ) ) ≡ A 1 .

Thus, it yields that q ( z + c ) − p ( z ) ≡ η 1 and p ( z + c ) − q ( z ) ≡ η 2 , where η 1 , η 2 are two constants in C , this means that p ( z ) − p ( z + 2 c ) ≡ − ( η 1 + η 2 ) and q ( z ) − q ( z + 2 c ) ≡ − ( η 1 + η 2 ) . Since p ( z ) , q ( z ) are two nonconstant polynomials, we can conclude that p ( z ) = a z + b 1 and q ( z ) = a z + b 2 , where a , b 1 , b 2 are two constants in C . Thus, in view of (3.25)–(3.27), it follows that

(3.28) − i A 2 a e i ( a c + b 2 − b 1 ) ≡ A 1 , − i A 2 a e i ( a c + b 1 − b 2 ) ≡ A 1 , i A 1 a e i ( − a c + b 1 − b 2 ) ≡ A 2 , i A 1 a e i ( − a c + b 2 − b 1 ) ≡ A 2 ,

which implies that a 2 = 1 , and

(3.29) e 2 i ( b 1 − b 2 ) ≡ 1 , e 2 i a c = − A 1 2 A 2 2 = − − α − α 2 − 1 − α + α 2 − 1 .

In view of (3.22), we have

(3.30) f ( z ) = 1 2 cos p ( z − c ) 1 + α + sin p ( z − c ) 1 − α = 1 2 cos [ a z − a c + b 1 ] 1 + α + sin [ a z − a c + b 1 ] 1 − α

and

(3.31) g ( z ) = 1 2 cos [ a z − a c + b 2 ] 1 + α + sin [ a z − a c + b 2 ] 1 − α .

Case 2.

− i A 2 p ′ e i ( q ( z + c ) − p ( z ) ) ≡ A 1 , i q ′ e i ( q ( z ) + p ( z + c ) ) ≡ 1 .

Thus, it follows that q ( z + c ) − p ( z ) ≡ η 1 and q ( z ) + p ( z + c ) ≡ η 2 , where η 1 , η 2 are constants in C . This leads to q ( z ) + q ( z + 2 c ) ≡ η 1 + η 2 . Noting that q ( z ) is a nonconstant polynomial, we obtain a contradiction.

Case 3.

i p ′ e i ( p ( z ) + q ( z + c ) ) ≡ 1 , − i A 2 q ′ e i ( p ( z + c ) − q ( z ) ) ≡ A 1 .

Thus, it follows that p ( z ) + q ( z + c ) ≡ η 1 and p ( z + c ) − q ( z ) ≡ η 2 , where η 1 , η 2 are constants in C . This leads to p ( z ) + p ( z + 2 c ) ≡ η 1 + η 2 . Noting that p ( z ) is a nonconstant polynomial, we obtain a contradiction.

Case 4.

(3.32) i p ′ e i ( p ( z ) + q ( z + c ) ) ≡ 1 , i q ′ e i ( q ( z ) + p ( z + c ) ) ≡ 1 .

Thus, it yields that p ( z ) + q ( z + c ) ≡ η 1 and q ( z ) + p ( z + c ) ≡ η 2 , where η 1 , η 2 are two constants in C , which means that p ( z + 2 c ) − p ( z ) ≡ η 2 − η 1 and q ( z + 2 c ) − q ( z ) ≡ η 1 − η 2 . Since p ( z ) , q ( z ) are two nonconstant polynomials, and noting that p ( z + 2 c ) − p ( z ) and q ( z + 2 c ) − q ( z ) are two constants, we conclude that p ( z ) = a z + b 1 ′ and q ( z ) = − a z + b 2 ′ , where a , b 1 ′ , b 2 ′ are two constants in C . In view of (3.25), (3.26), and (3.32), we have that

(3.33) i a e i ( − a c + b 2 ′ + b 1 ′ ) ≡ 1 , − i a e i ( a c + b 1 ′ + b 2 ′ ) ≡ 1 , − i a e i ( a c − b 1 ′ − b 2 ′ ) ≡ 1 , i a e i ( − a c − b 2 ′ − b 1 ′ ) ≡ 1 .

Hence, it follows that

(3.34) a 2 ≡ 1 , e 2 i ( b 1 ′ + b 2 ′ ) ≡ 1 , e 2 i a c ≡ − 1 .

Choosing b 1 = b 1 ′ and b 2 = − b 2 ′ , it follows that e 2 i ( b 1 − b 2 ) ≡ 1 . In view of (3.22), we have (3.30) and

(3.35) f 2 ( z ) = 1 2 cos [ − a z + a c + b 2 ′ ] 1 + α + sin [ − a z + a c + b 2 ′ ] 1 − α = 1 2 cos [ a z − a c + b 2 ] 1 + α − sin [ a z − a c + b 2 ] 1 − α .

Therefore, from Cases 1–4, we complete the proof of Theorem 2.3.

4 Proofs of Theorems 2.4 and 2.6

4.1 Proof of Theorem 2.4

Let f ( z ) be a finite order transcendental entire solution of equation (2.10). Similar to the argument as in the proof of Theorem 2.1, there exists a nonconstant polynomial p ( z ) satisfying

(4.1) f ( z ) = 1 2 cos p ( z ) 1 + α + sin p ( z ) 1 − α

and

(4.2) f ″ ( z ) = 1 2 cos p ( z ) 1 + α − sin p ( z ) 1 − α .

Thus, in view of (4.1) and (4.2), it follows that

(4.3) sin p ( z ) 1 − ( p ′ ) 2 1 − α − p ″ 1 + α = cos p ( z ) 1 + ( p ′ ) 2 1 + α − p ″ 1 − α .

Now, we claim that 1 − ( p ′ ) 2 1 − α − p ″ 1 + α ≢ 0 and 1 + ( p ′ ) 2 1 + α − p ″ 1 − α ≢ 0 . In fact, if 1 − ( p ′ ) 2 1 − α − p ″ 1 + α ≡ 0 and 1 + ( p ′ ) 2 1 + α − p ″ 1 − α ≢ 0 , then it follows that cos p ≡ 0 , this thus is a contradiction. If 1 − ( p ′ ) 2 1 − α − p ″ 1 + α ≢ 0 and 1 + ( p ′ ) 2 1 + α − p ″ 1 − α ≡ 0 , then it follows that sin p ≡ 0 , we thus also obtain a contradiction. Let 1 − ( p ′ ) 2 1 − α − p ″ 1 + α ≡ 0 and 1 + ( p ′ ) 2 1 + α − p ″ 1 − α ≡ 0 . If p ″ ≢ 0 , then we have ( p ′ ) 2 = α , which implies that p ″ ≡ 0 , which is a contradiction. If p ″ ≡ 0 , then we have ( p ′ ) 2 = − 1 and ( p ′ ) 2 = 1 , which is a contradiction. Hence, it follows that

(4.4) tan p ( z ) = [ 1 + ( p ′ ) 2 ] 1 − α − p ″ 1 + α [ 1 − ( p ′ ) 2 ] 1 + α − p ″ 1 − α .

Thus, we deduce from (4.4) that T ( r , tan p ) = O { T ( r , p ) + log r } , outside possibly a set of finite Lebesgue measures, using the results (see e.g. [27, p. 99], [28]) that T ( r , F z j ) = O { T ( r , F ) } for any meromorphic function F outside a set of finite Lebesgue measures and that T ( r , P ) = O { log r } for any polynomial P . But, lim r → ∞ T ( r , tan h ) T ( r , h ) + log r = + ∞ when h is a nonconstant polynomial. Therefore, h must be constant, which is a contradiction.

Therefore, this completes the proof of Theorem 2.4.

4.2 Proof of Theorem 2.6

Let f ( z ) be a finite order transcendental entire solution of equation (2.12). Similar to the argument as in the proof of Theorem 2.1, there exists a nonconstant polynomial p ( z ) satisfying (4.2) and

(4.5) f ( z + c ) = 1 2 cos p ( z ) 1 + α + sin p ( z ) 1 − α .

Thus, we have

(4.6) f ″ ( z + c ) = 1 2 ( A cos p − B sin p ) = 1 2 cos p ( z + c ) 1 + α − sin p ( z + c ) 1 − α ,

where

A = − ( p ′ ) 2 1 + α + p ″ 1 − α and B = p ″ 1 + α + ( p ′ ) 2 1 − α .

Thus, it follows from (4.6) that

(4.7) ( A + i B ) e i p + ( A − i B ) e − i p = e i p ( z + c ) 1 1 + α + i 1 − α + e − i p ( z + c ) 1 1 + α − i 1 − α .

Now, we claim that A − i B ≢ 0 and A + i B ≢ 0 . If A − i B ≡ 0 , then p ″ − i ( p ′ ) 2 ≡ 0 . If p ′ ≡ 0 , then equation (4.7) becomes e 2 i p ( z + c ) ≡ i 1 + α − 1 − α 1 − α + i 1 + α , which is impossible since p ( z ) is a nonconstant polynomial. If p ′ ≢ 0 and p ″ − i ( p ′ ) 2 ≡ 0 , then d u d z = i u 2 , where u = p ′ . Solving this equation, we have − 1 u = i z + ξ , that is, d p d z = u = − 1 i z + ξ 1 , where ξ 1 ∈ C . Thus, it follows that p = i log [ i z + ξ 1 ] + ξ 2 , where ξ 2 ∈ C . This is a contradiction that p is a nonconstant polynomial. Hence, A − i B ≢ 0 . Similarly, we can obtain that A + i B ≢ 0 . Thus, by Lemma 3.2, we have

(4.8) ( A − i B ) e i ( p ( z + c ) − p ( z ) ) = 1 1 + α − i 1 − α .

Combining with equation (4.7), it follows that

(4.9) ( A + i B ) e i ( p ( z ) − p ( z + c ) ) = 1 1 + α + i 1 − α .

Since p ( z ) is a nonconstant polynomial, and in view of (4.8) and (4.9), we conclude that p ( z ) = a z + b , where a , b ∈ C . Thus, in view of (4.8) and (4.9), it yields that

(4.10) a 2 e i a c − 1 1 + α − i 1 − α = 1 1 + α − i 1 − α

and

(4.11) a 2 e − i a c − 1 1 + α + i 1 − α = 1 1 + α + i 1 − α .

Thus, in view of (4.5), (4.10), and (4.11), we have

f ( z ) = 1 2 cos [ a z − a c + b ] 1 + α + sin [ a z − a c + b ] 1 − α

and

a 4 = 1 , e 2 i a c = α + α 2 − 1 α − α 2 − 1 = 2 α 2 + 2 α α 2 − 1 − 1 .

Therefore, this completes the proof of Theorem 2.6.

5 Proofs of Theorems 2.5 and 2.7

5.1 Proof of Theorem 2.5

Let ( f , g ) be a pair of transcendental entire solutions with finite order for system (2.11). By using the same argument as in Theorem 2.2, there exist two nonconstant polynomials p ( z ) , q ( z ) such that

(5.1) f ( z ) = 1 2 cos p ( z ) 1 + α + sin p ( z ) 1 − α , g ″ ( z ) = 1 2 cos p ( z ) 1 + α − sin p ( z ) 1 − α , g ( z ) = 1 2 cos q ( z ) 1 + α + sin q ( z ) 1 − α , f ″ ( z ) = 1 2 cos q ( z ) 1 + α − sin q ( z ) 1 − α .

Thus, it follows from (5.1) that

(5.2) − ( p ′ ) 2 1 + α + p ″ 1 − α cos p − p ″ 1 + α + ( p ′ ) 2 1 − α sin p = cos q 1 + α − sin q 1 − α

and

(5.3) − ( q ′ ) 2 1 + α + q ″ 1 − α cos q − q ″ 1 + α + ( q ′ ) 2 1 − α sin q = cos p 1 + α − sin p 1 − α .

In view of (5.2) and (5.3), we have

(5.4) Q 11 e i ( p + q ) + Q 12 e i ( q − p ) − A 2 e 2 i q = A 1 ,

(5.5) Q 21 e i ( p + q ) + Q 22 e i ( p − q ) − A 2 e 2 i p = A 1 ,

where

Q 11 = − A 1 ( p ′ ) 2 + i A 1 p ″ , Q 12 = − A 2 ( p ′ ) 2 − i A 2 p ″ ,

and

Q 21 = − A 1 ( q ′ ) 2 + i A 1 q ″ , Q 22 = − A 2 ( q ′ ) 2 − i A 2 q ″ .

Here, we will claim that Q i j ≢ 0 for i = 1 , 2 ; j = 1 , 2 . In fact, noting that A 1 ≢ 0 and A 2 ≢ 0 , if Q 11 ≡ 0 , that is, − A 1 ( p ′ ) 2 + i A 1 p ″ ≡ 0 , set p ′ = ξ , then − A 1 ξ 2 + i A 1 ξ ′ ≡ 0 . Solving this equation, we have ξ = − 1 − i z + η , where η ∈ C . This is impossible since p is a polynomial. Similarly, we can conclude that − A 2 ( p ′ ) 2 − i A 2 p ″ ≢ 0 , − A 1 ( q ′ ) 2 + i A 1 q ″ ≢ 0 , and − A 2 ( q ′ ) 2 − i A 2 q ″ ≢ 0 . Thus, by Lemma 3.2, it yields from (5.4) and (5.5) that

Q 12 e i ( q − p ) ≡ A 1 , or Q 11 e i ( p + q ) ≡ A 1 ,

and

Q 22 e i ( p − q ) ≡ A 1 , or Q 21 e i ( p + q ) ≡ A 1 .

Now, four cases are discussed as follows.

Case 1.

(5.6) Q 12 e i ( q − p ) ≡ A 1 , Q 22 e i ( p − q ) ≡ A 1 .

Thus, it follows that p − q ≡ η , where η is a constant in C , and both p ′ and q ′ are constants, and p ″ ≡ 0 and q ″ ≡ 0 . Then, we have Q 12 = − A 2 ( p ′ ) 2 and Q 22 = − A 2 ( q ′ ) 2 . Let p ′ ≡ a . Noting that p , q are nonconstant polynomials, then it leads to p = a z + b 1 , where b 1 ∈ C . In view of p − q ≡ η , we have q = a z + b 2 , where b 2 ∈ C . Thus, in view of (5.4)–(5.6), it yields that

(5.7) − A 2 a 2 e i ( b 2 − b 1 ) ≡ A 1 , − A 2 a 2 e i ( b 1 − b 2 ) ≡ A 1 , − A 1 a 2 e i ( b 1 − b 2 ) ≡ A 2 , − A 1 a 2 e i ( b 2 − b 1 ) ≡ A 2 .

Hence, it follows from (5.7) that a 4 ≡ 1 and A 2 2 ≡ A 1 2 , that is,

1 1 + α − i 1 − α 2 ≡ 1 1 + α + i 1 − α 2 .

This is impossible.

Case 2.

(5.8) Q 12 e i ( q − p ) ≡ A 1 , Q 21 e i ( p + q ) ≡ A 1 .

Thus, it yields from (5.8) that q − p ≡ η 1 and p + q ≡ η 2 , where η 1 , η 2 are constants. This means that 2 q ≡ η 1 + η 2 is a constant. This is a contradiction.

Case 3.

Q 11 e i ( p + q ) ≡ A 1 , Q 22 e i ( p − q ) ≡ A 1 .

Similar to the argument as in Case 2 of Theorem 2.5, we obtain that p is a constant, which is a contradiction.

Case 4.

(5.9) Q 11 e i ( p + q ) ≡ A 1 , Q 21 e i ( p + q ) ≡ A 1 .

Thus, it yields from (5.9) that p + q is a constant. Similar to the argument as in Case 1 of Theorem 2.5, we conclude that p = a z + b 1 and q = − a z + b 2 , where b 1 , b 2 ∈ C . Thus, we have Q 11 = − A 1 a 2 , Q 12 = − A 2 a 2 , Q 21 = − A 1 a 2 , and Q 22 = − A 2 a 2 . Substituting these into (5.4), (5.5), and (5.9), we have

(5.10) − a 2 e i ( b 1 + b 2 ) ≡ 1 , − a 2 e − i ( b 1 + b 2 ) ≡ 1 .

This leads to a 4 ≡ 1 and e 2 i ( b 1 + b 2 ) ≡ 1 . Combining with (5.1), we obtain

(5.11) f ( z ) = 1 2 cos ( a z + b 1 ) 1 + α + sin ( a z + b 1 ) 1 − α

and

(5.12) g ( z ) = 1 2 cos [ − a z + b 2 ] 1 + α + sin [ − a z + b 2 ] 1 − α = 1 2 cos [ a z − b 2 ] 1 + α − sin [ a z − b 2 ] 1 − α .

Therefore, this completes the proof of Theorem 2.5.

5.2 Proof of Theorem 2.7

By combining with the argument as in the proofs of Theorems 2.3 and 2.6, it is easy to obtain the conclusions of Theorem 2.7.

Acknowledgments

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

Funding information: This work was supported by the National Natural Science Foundation of China no. 12161074 and the Talent introduction research Foundation of Suqian University (106-CK00042/028). Hong Li was supported by the 2021 Annual Jiangxi Social Science Planning Project (21JY14), the 2019 Annual Subject of Key Research Base of Humanities in Universities of Jiangxi Province (JD19006), the 2019 Annual Research topics of basic education in Jiangxi province (SZUGSZH2019-1086), and the Opening Project of Jiangxi Key Laboratory of Numerical Simulation and Simulation Technology in China.
Author contributions: Conceptualization, H. Y. Xu; writing–original draft preparation, H. Y. Xu, H. Li, and X. Ding; writing–review and editing, H. Y. Xu and H. Li; funding acquisition, H. Y. Xu and H. Li.
Conflict of interest: The authors declare that none of the authors have any competing interests in the manuscript.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2022-04-10

Revised: 2022-07-15

Accepted: 2022-08-17

Published Online: 2022-10-05

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

Articles in the same Issue

https://doi.org/10.1515/dema-2022-0161

Keywords for this article

Nevanlinna theory; meromorphic function; differential difference equation

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