Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 27, Number 3, June 2022
Page(s) 195 - 200
DOI https://doi.org/10.1051/wujns/2022273195
Published online 24 August 2022

© Wuhan University 2022

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction and Main Results

In this paper, Nevanlinna value distribution theory is a useful tool and its standard notations as well as well-known theorems can be found in Refs.[1,2]. Let be a meromorphic function in the complex plane . The proximity function, counting function of poles and Nevanlinna characteristic function with regard to are denoted by , and , respectively. We also use to denote any small quantity of satisfying , as , possibly outside of a set of finite logarithmic measure. The order of growth of and the exponent of convergence of zeros of are defined by

A large number of studies concerning the solvability and existence for entire and meromorphic solutions of nonlinear complex differential equations, difference equations and differential-difference equations can be found in Refs.[3-12].

The equation has exactly three nonconstant entire solutions , according to Yang and Li [11]. It is worth noting that , that is, is a linear combinations of and . As a result, it is meaningful to investigate the existence of the solutions of the differential equation

(1)

where are non-zero constants, and denotes a differential in with degree .

For Eq.(1), Li [6] obtained the following theorem.

Theorem 1 [6] Let be an integer, be a differential polynomial in of degree at most , and are three non-zero constants. If is an entire solution of equation (1), then , where and are constants and .

Later, Li [13] proved by weakening the requirement on the degree of in Theorem A.

Theorem 2 [13] Let be an integer, be a differential polynomial in of degree at most , and are three non-zero constants. If is a meromorphic solution of equation (1) and , then there exist two nonzero constants and a small function of such that .

Then it is natural to ask: is there a similar result if are polynomials? The question was answered by the following theorem.

Theorem 3 [14] Let be an integer, be a differential polynomial in of degree . Let and be two nonzero polynomials, be two nonzero constants with rational. Then the differential equation

(2)

has no transcendental entire solutions.

The exponential polynomial plays a key role in the study of non-linear complex differential equations and many interesting properties have been mentioned, for example, in 2012, Wen, Heittokangas and Laine [15] investigated and classified the finite order entire solutions of the following equation

(3)

in terms of growth and zero distribution, where is an integer, , are polynomials.

Recall that an exponential polynomial of the form

(4)

where are polynomials in such that is called an exponential polynomial of degree .

In the following, we define two classes of transcendental entire functions:

is a non-constant polynomial

is a non-constant polynomial and

Theorem 4 [15] Let be an integer, let , and let be polynomials such that is not a constant and . Then we identify the finite order entire solutions of equation (3) as follows:

(i) Every solution satisfies and is of mean type;

(ii) Every solution satisfies if and only if ;

(iii) A solution belongs to if and only if . In particular, this is the case if ;

(iv) If a solution belongs to and if is any other finite order entire solution to (3), then , where ;

(v) If is an exponential polynomial solution of the form (4), then . Moreover, if , then .

Following that, Liu [16] and Chen et al [5] investigated the cases when in Eq.(3) is replaced by or , respectively, and obtained a special form of solutions of the equations.

According to the foregoing results, all equations have only one main term on the left side.

Chen et al [17] proposed the following question based on whether a similar result can be obtained when there are two main terms.

In 2021, Chen, Hu and Wang [17] considered the following differential-difference equation

(5)

and obtained the following result.

Theorem 5 [17] If is a transcendental entire solution with finite order of Eq.(5), then the following conclusions hold:

(i) If for and for , then every solution satisfies.

(ii) If and is a solution of (5), which belongs to , then

or

where .

Inspired by the above results, we consider the following types of equations:

(6)

and

(7)

Theorem 6   Let and be integers, and be nonzero constants such that . Suppose that and . Then Eq.(6) has no transcendental entire solutions.

Theorem 7   Let and be integers, be a non-vanishing polynomial and be a non-constant polynomial. Suppose that and be nonzero constants such that . If Eq.(7) admits a finite order transcendental entire function with , then the following conclusions hold:

(i) If for , then every solution satisfies .

(ii) If a solution belongs to , then

or

where .

1 Preliminary Lemmas

In order to prove our results, we introduce the following Lemmas.

Lamma 1 [18] Let be a non-constant meromorphic function and let . Then if the growth order of is finite, we have

And if the growth order of is infinite, we have

possibly outside a set of finite linear measure.

Lamma 2 [19] Let be a transcendental meromorphic function of finite order. Then

where denotes any quantity as , possible outside of a set of finite logarithmic measure.

In the study of complex differential-difference equations, the Clunie lemma[10] plays an important part. Let's recall the Clunie lemma's precise definition.

Lamma 3 (Clunie's lemma) [10] Let be a transcendental meromorphic solution of finite order of differential-difference equation

where are polynomials in and its derivatives and its shifts with small meromorphic coefficients. If the total degree of as a polynomial in and its derivatives and its shifts is , then

for all out of a possible exceptional set of finite logarithmic measure.

Lamma 4 [18] Let be meromorphic functions, and let be entire functions satisfying

(i) ;

(ii) when , then is not a constant;

(iii) when , then

where is of finite linear measure or logarithmic measure.

Then .

2 Proof of Theorem 6

Let be a transcendental entire solution of Eq.(6).

Set ,. Then we write Eq.(6) as

(8)

Differentiating (8) yields

(9)

Eliminating from (8) and (9), we get

(10)

Differentiating (10) yields

(11)

Eliminating from (10) and (11), we get

(12)

Note that

and

Substituting into (12), we have

(13)

where

(14)

Note that and is a differential-difference polynomial in and the total degree is at most 1. By Lemma 3, we obtain

If , then

(15)

which yields a contradiction.

Thus, . From (13), we get . Then has the form

(16)

where and are constants.

By integration, we have

(17)

where and are constants.

Similarly, by and (12) we can also obtain , then we have

(18)

where and are constants. From (17) and (18), we obtain

(19)

Dividing both sides of (19) by , we get

(20)

Since , and , by (20) and Lemma 4, we obtain , that is . Similarly, dividing both sides of (19) by , we can get . So , this is a contradiction.

Therefore, Eq.(7) has no transcendental entire solutions.

3 Proof of Theorem 7

Suppose that is a transcendental entire solution of finite order of Eq.(7). In what follows, we firstly prove that .

Case A Suppose that . By Lemma 1, Lemma 2 and (7), we have

(21)

then .

Note that , therefore . Denote . Rewriting (7) in the following form:

(22)

Differentiating (22) yields

(23)

where

Eliminating from (22) and (23), we obtain

(24)

Subcase A1 If , by (24) and Lemma 4, we have

which contradicts with .

Subcase A2 If , by (24), we can get

(25)

From (25) and Lemma 4, we have

(26)

Dividing with on both sides of (26), we obtain

(27)

Since , which yields a Riccati differential equation:

(28)

where .

By a simple calculation, we can get

then

where are constants.

If , then , this is a contradiction.

If , then , which contradicts with .

Case B Suppose that . Denote ,. Equation (7) can be written as:

(29)

Differentiating (29) yields

(30)

where .

Eliminating from (29) and (30), we get

(31)

where .

Note that and is a differential-difference polynomial in and the total degree is at most 1. By Lemma 3, we obtain

and .

If , since is a transcendental entire function, then

which is a contradiction.

If , from (31) yields , then

By integration, we see that there exists a constant such that

(32)

Substituting (32) into (7) yields

(33)

Since is a transcendental entire function with , according to Hadamard decomposition theorem, can be written in the form

(34)

where is the canonical product formed by zeros of and is a non-constant polynomial which satisfies

(35)

Substituting (34) into (33), we have

(36)

Combining (35) with (36), we can see that the left order of (36) is greater than 1, but the right order is 1, which is a contradiction.

Therefore, . From (7), Lemma 1 and Lemma 2, we obtain

(37)

Note that , then , that is .

The conclusion (i) is proved. Next we will prove the conclusion (ii).

If belongs to , and noting that , we define and , where , and .

Substituting and into (7) yields

(38)

Dividing both sides of (38) by , we get

(39)

We distinguish four cases below.

Case 1 Suppose that and . By Lemma 4 and (39), we obtain , which is a contradiction.

Case 2 Suppose that and .

If , by (39) and Lemma 4, we have , this is a contradiction.

If , then , substituting these into (39) yields

(40)

by Lemma 4 and (20), we have

then reduces to a nonzero constant, and .

Case 3 Suppose that and .

If , by (39) and Lemma 4, we have

this is a contradiction.

If , , substituting these into (39) yields

(41)

by Lemma 4 and (41), we have

then reduces to a nonzero constant, and .

Case 4 Suppose that and .

If and are pairwise distinct from each other. By Lemma 4 and (39), we have

this is a contradiction.

If only two of and coincide, without loss of generality, suppose that , then (39) can be written as:

(42)

From Lemma 4 and (42) yield , this is a contradiction.

If , then we write (39) as:

By Lemma 4, we have , which implies a contradiction. The proof is complete.

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