Properties of meromorphic solutions of first-order differential-difference equations

Lihao Wu; Baoqin Chen; Sheng Li

doi:10.1515/math-2023-0147

Article Open Access

Properties of meromorphic solutions of first-order differential-difference equations

Lihao Wu , Baoqin Chen and Sheng Li

Published/Copyright: December 1, 2023

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$Open Mathematics$

From the journal Open Mathematics Volume 21 Issue 1

Abstract

For the first-order differential-difference equations of the form

A ( z ) f ( z + 1 ) + B ( z ) f ′ ( z ) + C ( z ) f ( z ) = F ( z ) ,

where A ( z ) , B ( z ) , C ( z ) , and F ( z ) are polynomials, the existence, growth, zeros, poles, and fixed points of their nonconstant meromorphic solutions are investigated. It is shown that all nonconstant meromorphic solutions are transcendental when deg B ( z ) < deg { A ( z ) + C ( z ) } + 1 and all transcendental solutions are of order at least 1. For the finite-order transcendental solution f ( z ) , the relationship between ρ ( f ) and max { λ ( f ) , λ ( 1 ∕ f ) } is discussed. Some examples for sharpness of our results are provided.

Keywords: differential-difference equation; growth and zeros; meromorphic solution

MSC 2010: 30D35; 39A06

1 Introduction

A meromorphic function f ( z ) is always understood to be nonconstant and meromorphic in the whole complex plane C in this article. Concerning the value distribution of meromorphic functions, we assume that the reader is familiar with the basic Nevanlinna value distribution theory and its standard notations, such as m ( r , f ) , N ( r , f ) , T ( r , f ) , and S ( r , f ) , see, e.g., [1,2]. In particular, for a meromorphic function f ( z ) , the notations of order, the exponent of convergence of a-points (zeros) of f , the exponent of convergence of poles of f , and the exponent of convergence of fixed points of f are as follows:

ρ ( f ) ≔ limsup r → ∞ log + T ( r , f ) log r , λ ( f = a ) ≔ limsup r → ∞ log N r , 1 f − a log r , λ ( 1 ∕ f ) ≔ limsup r → ∞ log N ( r , f ) log r , τ ( f ) ≔ limsup r → ∞ log N r , 1 f − z log r ,

respectively, and denote λ ( f ) ≔ λ ( f = 0 ) .

As is well known, Nevanlinna value distribution theory of meromorphic functions have been extensively applied to investigate the growth, value distribution, and solvability of meromorphic solutions of linear and nonlinear differential equations [1,3–5]. Meromorphic solutions of complex difference equations have been a subject of great interest since 2000 (see, e.g., [6–13]) due to the application of classical Nevanlinna theory in difference by Ablowitz et al. [14], and number of fundamental results on difference analogues of Nevanlinna value distribution have been obtained (see, e.g., [8–12,15]). We first recall some relative results.

Theorem A

[8,9] Let A 0 ( z ) , A 1 ( z ) , … , A n ( z ) be polynomials such that there exists an integer l , 0 ≤ l ≤ n , such that

deg ( A l ) > max 0 ≤ j ≤ n , j ≠ l { deg ( A j ) } .

Suppose that f ( z ) is a meromorphic solution to

A n ( z ) f ( z + n ) + ⋯ + A 1 ( z ) f ( z + 1 ) + A 0 ( z ) f ( z ) = 0 ,

then we have ρ ( f ) ≥ 1 .

Theorem B

[6] Let A 2 ( z ) ≢ 0 , A 1 ( z ) , and F ( z ) be polynomials and c 2 and c 1 ( ≠ c 2 ) be constants. Suppose that f ( z ) is a transcendental meromorphic solution of difference equation

A 2 ( z ) f ( z + c 2 ) + A 1 ( z ) f ( z + c 1 ) = F ( z ) .

Then, ρ ( f ) ≥ 1 .

In particular, if f ( z ) is an entire function of finite-order, then we have two cases as follows:

if F ( z ) ≢ 0 , then λ ( f ) = ρ ( f ) ;
if F ( z ) ≡ 0 and ρ ( f ) > 1 , then λ ( f ) ≥ ρ ( f ) − 1 ; if F ( z ) ≡ 0 and ρ ( f ) = 1 , then λ ( f ) = 1 or f ( z ) has only finitely many zeros.

Theorem C

[13] Let A 0 ( z ) , A 1 ( z ) , … , A n ( z ) , and B ( z ) be polynomials such that

A 0 ( z ) A n ( z ) ≢ 0 , deg ∑ deg A j = d A j = d ,

where d = max 0 ≤ j ≤ n { deg A j } . If f ( z ) is a transcendental meromorphic solution of

∑ j = 0 n A j ( z ) f ( z + j ) = B ( z ) ,

then ρ ( f ) ≥ 1 . Moreover, if f ( z ) is of finite-order, then 1 ≤ ρ ( f ) ≤ 1 + max { λ ( f ) , λ ( 1 ∕ f ) } .

As continuation of Theorems A, B, and C, we consider the first-order differential-difference equations of form

(1.1) A ( z ) f ( z + 1 ) + B ( z ) f ′ ( z ) + C ( z ) f ( z ) = F ( z ) ,

where A ( z ) ( ≢ 0 ) , B ( z ) ( ≢ 0 ) , C ( z ) , and F ( z ) are polynomials.

We will consider the existences of rational solutions of equation (1.1) in Section 2, and the growth, zeros, poles, and fixed points of transcendental meromorphic solutions of them in Section 3.

2 Existences of rational solutions

For the homogeneous case, we obtain the following result:

Theorem 2.1

Let A ( z ) ( ≢ 0 ) , B ( z ) ( ≢ 0 ) , and C ( z ) be polynomials. If

(2.1) deg B ( z ) < deg { A ( z ) + C ( z ) } + 1 ,

then the first-order differential-difference equation

(2.2) A ( z ) f ( z + 1 ) + B ( z ) f ′ ( z ) + C ( z ) f ( z ) = 0

has no nonconstant rational solutions. If

(2.3) deg B ( z ) = deg { A ( z ) + C ( z ) } + 1 ,

then all nonconstant rational solutions of equation (2.2) satisfy that

lim z → ∞ f ( z ) = a ∈ { 0 , ∞ } .

(2.4) deg B ( z ) > deg { A ( z ) + C ( z ) } + 1 ,

then all nonconstant rational solutions of equation (2.2) satisfy that

lim z → ∞ f ( z ) = a ∉ { 0 , ∞ } .

Example 2.1

Here, we give three examples for Theorem 2.1.

The equation
f ( z + 1 ) + ( − z 2 − z − 1 ) f ′ ( z ) + z f ( z ) = 0
has a solution f ( z ) = z such that lim z → ∞ f ( z ) = ∞ , where deg B ( z ) = deg { A ( z ) + C ( z ) } + 1 .
The equation
( z + 1 ) f ( z + 1 ) + z f ′ ( z ) − ( z − 1 ) f ( z ) = 0
has a solution f ( z ) = 1 ∕ z such that lim z → ∞ f ( z ) = 0 , where deg B ( z ) < deg { A ( z ) + C ( z ) } + 1 .
The equation
( z + 1 ) f ( z + 1 ) − z 3 f ′ ( z ) − 2 z f ( z ) = 0
has a solution f ( z ) = 1 + 1 ∕ z such that lim z → ∞ f ( z ) = 1 ∉ { 0 , ∞ } , where deg B ( z ) > deg { A ( z ) + C ( z ) } + 1 .

For the nonhomogeneous case, we should point out that equation (1.1) always has rational solutions no matter what the relationship between deg B ( z ) and ( deg { A ( z ) + C ( z ) } + 1 ) is (see examples as follows).

Example 2.2

The equation
f ( z + 1 ) + ( z 3 − z 2 − z − 1 ) f ′ ( z ) + z f ( z ) = z 3
has a solution f ( z ) = z , where deg B > deg { A ( z ) + C ( z ) } + 1 .
The equation
f ( z + 1 ) − z f ′ ( z ) − f ( z ) = 1 − z
has a solution f ( z ) = z , where deg B ( z ) = deg { A ( z ) + C ( z ) } + 1 .
The equation
( z + 1 ) f ( z + 1 ) + z f ′ ( z ) + f ( z ) = 1
has a solution f ( z ) = 1 z , where deg B ( z ) < deg { A ( z ) + C ( z ) } + 1 .

To prove Theorem 2.1, we give a simple conclusion first.

Lemma 2.1

Suppose that R ( z ) is a nonconstant rational function, then we have

R ( z + 1 ) = R ( z ) ( 1 + o ( 1 ) ) , a n d z R ′ ( z ) = O ( R ( z ) ) , a s z → ∞ .

More precisely, denote

lim z → ∞ R ( z ) = a ,

then

z R ′ ( z ) = K R ( z ) ( 1 + o ( 1 ) ) , a s z → ∞ ,

if a ∈ { 0 , ∞ } , where K ( ≠ 0 ) is a constant, and

z R ′ ( z ) = o ( R ( z ) ) , a s z → ∞ ,

if a ∉ { 0 , ∞ } .

Proof

If R ( z ) is a nonconstant polynomial, we can easily prove our conclusions. In the following, suppose that R ( z ) is a non-polynomial rational function of the form

R ( z ) = P 1 ( z ) + Q ( z ) P 2 ( z ) ,

where P 1 ( z ) , P 2 ( z ) , and Q ( z ) are polynomials such that

P 2 ( z ) = a n z n + a n − 1 z n − 1 + … + a 1 z + a 0 ,

Q ( z ) = b m z m + b m − 1 z m − 1 + … + b 1 z + b 0 ,

and a n , … , a 0 , b n , … , b 0 are constants such that a n b m ≠ 0 and m < n .

Case 1: a = 0 . That is, P 1 ( z ) ≡ 0 . As

R ( z + 1 ) = b m ( z + 1 ) m + b m − 1 ( z + 1 ) m − 1 + … + b 1 ( z + 1 ) + b 0 a n ( z + 1 ) n + a n − 1 ( z + 1 ) n − 1 + … + a 1 ( z + 1 ) + a 0 = b m z m + d m − 1 z m − 1 + … + d 1 z + d 0 a n z n + c n − 1 z n − 1 + … + c 1 z + c 0 ,

where c n − 1 , … , c 0 , d n − 1 , … , d 0 are constants, we immediately obtain that

lim z → ∞ R ( z + 1 ) R ( z ) = 1 .

With a simple calculation, we can obtain that

z R ′ ( z ) R ( z ) = z P 2 ( z ) Q ′ ( z ) − P 2 ′ ( z ) Q ( z ) P 2 2 ( z ) P 2 ( z ) Q ( z ) = z Q ′ ( z ) Q ( z ) − z P 2 ′ ( z ) P 2 ( z ) = m b m z m + … + b 1 z b m z m + … + b 1 z + b 0 − n a n z n + … + a 1 z a n z n + … + a 1 z + a 0 ,

and hence

lim z → ∞ z R ′ ( z ) R ( z ) = m − n .

Case 2: a = ∞ . That is, P 1 ( z ) is a nonconstant polynomial such that deg P 1 ( z ) = d ≥ 1 . Then, it is not difficult to find that

R ( z + 1 ) = P 1 ( z + 1 ) ( 1 + o ( 1 ) ) = P 1 ( z ) ( 1 + o ( 1 ) ) = R ( z ) ( 1 + o ( 1 ) ) , as z → ∞ ,

and

z R ′ ( z ) = z P 1 ′ ( z ) + z Q ( z ) P 2 ( z ) ′ = d P 1 ( z ) ( 1 + o ( 1 ) ) = d R ( z ) ( 1 + o ( 1 ) ) , as z → ∞ .

Case 3: a ≠ 0 , ∞ . That is, P 1 ( z ) ≡ a . Then,

R ( z + 1 ) = a + o ( 1 ) = R ( z ) + o ( 1 ) = R ( z ) ( 1 + o ( 1 ) ) , as z → ∞ .

With a similar calculation as in Case 1, we can deduce that

lim z → ∞ z R ′ ( z ) R ( z ) − a = lim z → ∞ z Q ′ ( z ) Q ( z ) − z P 2 ′ ( z ) P 2 ( z ) = m − n .

This indicates that

□ z R ′ ( z ) = ( m − n ) [ R ( z ) − a ] ( 1 + o ( 1 ) ) = o ( 1 ) = o ( R ( z ) ) , as z → ∞ .

Proof of Theorem 2.1

Let us finish our proof case by case.

Case 1: Equation (2.1) holds. Suppose that equation (2.2) has a nonconstant rational solution f ( z ) . Then, from Lemma 2.1, we have

(2.5) f ( z + 1 ) = f ( z ) ( 1 + o ( 1 ) ) , and z f ′ ( z ) = O ( f ( z ) ) , as z → ∞ .

From equations (2.2) and (2.5), we obtain

∣ z [ A ( z ) + C ( z ) ] f ( z ) ( 1 + o ( 1 ) ) ∣ = ∣ z [ A ( z ) f ( z + 1 ) + C ( z ) f ( z ) ] ∣ = ∣ − B ( z ) z f ′ ( z ) ∣ ≤ ∣ B ( z ) f ( z ) ( 1 + o ( 1 ) ) ∣ , as z → ∞ .

Thus, we obtain

∣ z [ A ( z ) + C ( z ) ] ∣ ≤ ∣ B ( z ) ( 1 + o ( 1 ) ) ∣ , as z → ∞ ,

which contradicts to equation (2.1). Therefore, we prove that equation (2.2) has no nonconstant rational solutions in this case.

Case 2: Equation (2.3) holds. Suppose that equation (2.2) has a nonconstant rational solution f ( z ) such that

lim z → ∞ f ( z ) = a ∉ { 0 , ∞ } .

Then, from Lemma 2.1, we have

(2.6) f ( z + 1 ) = f ( z ) ( 1 + o ( 1 ) ) , and z f ′ ( z ) = o ( f ( z ) ) , as z → ∞ .

Similarly, we obtain from equations (2.2) and (2.6) that

∣ z [ A ( z ) + C ( z ) ] ∣ = ∣ o ( B ( z ) ) ∣ , as z → ∞ ,

which contradicts to (2.3). Therefore, we prove that a ∈ { 0 , ∞ } in this case.

Case 3: Equation (2.4) holds. Suppose that equation (2.2) has a nonconstant rational solution f ( z ) such that

lim z → ∞ f ( z ) = a ∈ { 0 , ∞ } .

Then, from Lemma 2.1, we have

(2.7) f ( z + 1 ) = f ( z ) ( 1 + o ( 1 ) ) , and z f ′ ( z ) = K f ( z ) ( 1 + o ( 1 ) ) ,

as z → ∞ , where K ( ≠ 0 ) is a constant. Now we obtain from equations (2.2) and (2.7) that

∣ z [ A ( z ) + C ( z ) ] ∣ = ∣ K B ( z ) ( 1 + o ( 1 ) ) ∣ , as z → ∞ ,

which contradicts to equation (2.4). Therefore, we prove that a ∉ { 0 , ∞ } in this case.□

3 Growth and value distribution of transcendental meromorphic solution

We consider the general case first and prove Theorem 3.1.

Theorem 3.1

Let A ( z ) ≢ 0 , B ( z ) , C ( z ) , and F ( z ) ≢ 0 be polynomials, and let f ( z ) be a transcendental meromorphic solution of equation (1.1), then ρ ( f ) ≥ 1 .

In particular, if f ( z ) is of finite-order, then we have two cases as follows:

if F ( z ) ≢ 0 , then λ ( f ) = ρ ( f ) ;
if F ( z ) ≡ 0 , then 1 ≤ ρ ( f ) ≤ 1 + max { λ ( f ) , λ ( 1 ∕ f ) } , and ρ ( f ) = max { λ ( f ) , λ ( 1 ∕ f ) } = 1 when ρ ( f ) = 1 .

To prove Theorem 3.1, we need some lemmas.

Lemma 3.1

[8] Let f ( z ) be a meromorphic function of finite-order ρ , ε > 0 , and let c 1 and c 2 be two distinct nonzero complex constants. Then,

m r , f ( z + η 1 ) f ( z + η 2 ) = O ( r ρ − 1 + ε ) ,

and there exists a subset E ⊂ ( 1 , + ∞ ) of finite logarithmic measure, such that for all z satisfying ∣ z ∣ = r ∉ [ 0 , 1 ] ∪ E , where r is sufficiently large,

exp { − r ρ − 1 + ε } ≤ f ( z + η 1 ) f ( z + η 2 ) ≤ exp { r ρ − 1 + ε } .

Remark 3.1

From Lemma 3.1 (see also Lemma 3.3 in [16]), we have

f ( z + η 1 ) f ( z + η 2 ) → 1 ,

as ∣ z ∣ = r ∉ E ∪ [ 0 , 1 ] , r → ∞ .

Let ν ( r , f ) be the central index of the Taylor expansion of f ( z ) , and we have

Lemma 3.2

[1] Let f ( z ) be a transcendental entire function. Let 0 < δ < 1 4 and z be such that ∣ z ∣ = r and that

∣ f ( z ) ∣ > M ( r , f ) ν ( r , f ) − 1 4 + δ

holds. Then, there exists a set E ⊂ ( 1 , + ∞ ) of finite logarithmic measure, i.e., l m E = ∫ E d t ∕ t < + ∞ , such that

f ( m ) ( z ) f ( z ) = ( 1 + o ( 1 ) ) ν ( r , f ) z m

holds for all m ≥ 0 and all r ∉ E .

Lemma 3.3

[17] Suppose that f ( z ) is a transcendental entire function with finite-order ρ ( f ) = ρ < ∞ , and a set E ⊂ [ 1 , + ∞ ) has a finite logarithmic measure. Then, there exist a sequence of points: { z k = r k e i θ k } satisfying ∣ f ( z k ) ∣ = M ( r k , f ) , θ k ∈ [ 0 , 2 π ) , and lim k → ∞ θ k = θ 0 ∈ [ 0 , 2 π ) r k ∉ E , r k → ∞ , such that for any given ε > 0 , as r k is sufficiently large, we have

r k ρ − ε < ν ( r k , f ) < r k ρ + ε .

Since each transcendental entire function f ( z ) satisfies,

lim r → ∞ ν ( r , f ) = + ∞ ,

and we can obtain from Lemmas 3.2 and 3.3 that

Lemma 3.4

Suppose that f ( z ) is a transcendental entire function with finite-order ρ ( f ) = ρ < ∞ , and a set E ⊂ [ 1 , + ∞ ) has a finite logarithmic measure. Then, a sequence of points: { z k = r k e i θ k } satisfying ∣ f ( z k ) ∣ = M ( r k , f ) , θ k ∈ [ 0 , 2 π ) , lim k → ∞ θ k = θ 0 ∈ [ 0 , 2 π ) , and r k ∉ E , r k → ∞ , such that for any given ε > 0 , as r k is sufficiently large, we have

1 2 r k m ( ρ − 1 − ε ) < f ( m ) ( z k ) f ( z k ) < 2 r k m ( ρ − 1 + ε ) .

Proof of Theorem 3.1

Let f ( z ) be a transcendental meromorphic solution of equation (2.2). If B ( z ) ≡ 0 , then, from Theorem B, there is nothing to prove. For the case B ( z ) ≢ 0 , we will discuss in three steps.

Step 1: We prove that ρ ( f ) ≥ 1 . Otherwise, we have ρ ( f ) < 1 , and hence f ( z ) has infinitely many poles or infinitely many zeros.

Step 1.1: We show that f ( z ) has finitely many poles. Suppose that f ( z ) has infinitely many poles. Set D = { z : ∣ Re z ∣ ≤ M , ∣ Im z ∣ ≤ M } , where M > 0 is some constant, such that all zeros of B ( z ) are in D . Then, there exists at least one of the following four regions:

D 1 = { z : Re z > M } , D 2 = { z : Re z < − M } ,

D 3 = { z : Im z > M } , D 4 = { z : Im z < − M } ,

say D 1 , such that in D 1 , f ( z ) has infinitely many poles and B ( z ) has no zeros.

Let z 0 ∈ D 1 be a pole of f ( z ) with multiplicity k 0 , then z 0 ∈ D 1 is a pole of f ′ ( z ) with multiplicity ( k 0 + 1 ) . From equation (2.2), we find that z 0 + 1 must be a pole of f ( z ) with multiplicity ( k 0 + 1 ) . By induction, we can prove that z 0 + k ( k = 1 , 2 , … ), are all in D 1 , and are poles of f ( z ) . Thus, λ ( 1 ∕ f ) ≥ 1 , and hence ρ ( f ) ≥ 1 . This contradicts to our assumption ρ ( f ) < 1 .

Step 1.2: We continue our proof by assuming that f ( z ) has no poles, without loss of generality. By Remark 3.1, there exists a set E ⊂ [ 1 , + ∞ ) , which has a finite logarithmic measure such that

(3.1) f ( z + 1 ) = f ( z ) ( 1 + o ( 1 ) ) ,

as ∣ z ∣ = r → ∞ and r ∈ [ 1 , + ∞ ) \ E .

From Lemma 3.4, for the exceptional set E above and ε 0 = min ρ 2 , 1 − ρ 2 > 0 , there exists a sequence of points: { z k = r k e i θ k } satisfying ∣ f ( z k ) ∣ = M ( r k , f ) , θ k ∈ [ 0 , 2 π ) , lim k → ∞ θ k = θ 0 ∈ [ 0 , 2 π ) , and r k ∉ E , r k → ∞ , such that for any given ε > 0 , as r k is sufficiently large, we have

1 2 r k ρ ∕ 2 − 1 ≤ 1 2 r k ρ − 1 − ε 0 < f ′ ( z k ) f ( z k ) < 2 r k ρ − 1 + ε 0 ≤ 2 r k ( 1 + ρ ) ∕ 2 − 1 .

Thus,

(3.2) 1 2 r k ρ ∕ 2 − 1 ∣ f ( z k ) ∣ < ∣ f ′ ( z k ) ∣ < 2 r k ( 1 + ρ ) ∕ 2 − 1 ∣ f ( z k ) ∣ .

From equations (3.1) and (3.2), we obtain that

(3.3) A ( z k ) f ( z k + 1 ) + C ( z k ) f ( z k ) = [ A ( z k ) + C ( z k ) ] f ( z k ) ( 1 + o ( 1 ) ) ,

and

(3.4) 1 2 r k ρ ∕ 2 − 1 ∣ B ( z k ) f ( z k ) ∣ < ∣ B ( z k ) f ′ ( z k ) ∣ < 2 r k ( 1 + ρ ) ∕ 2 − 1 ∣ B ( z k ) f ( z k ) ∣ ,

as ∣ z k ∣ = r k → ∞ and r k ∈ [ 1 , + ∞ ) \ E , such that ∣ f ( z k ) ∣ = M ( r k , f ) .

As f ( z ) is a transcendental entire function and F ( z ) is a polynomial, we have

(3.5) F ( z k ) = o ( f ( z k ) ) ,

as ∣ z k ∣ = r k → ∞ and r k ∈ [ 1 , + ∞ ) \ E , such that ∣ f ( z k ) ∣ = M ( r k , f ) .

Denote

d 1 = deg [ A ( z ) + C ( z ) ] , d 2 = deg B ( z ) ,

and we discuss two cases as follows:

Case 1: d 2 ≥ d 1 + 1 . By equations (3.1) and (3.3)–(3.5), we obtain that

K 1 r k d 1 ∣ f ( z k ) ∣ = ∣ [ A ( z k ) + C ( z k ) ] f ( z k ) ( 1 + o ( 1 ) ) ∣ = ∣ [ A ( z k ) f ( z k + 1 ) + C ( z k ) f ( z k ) ] ∣ = ∣ B ( z k ) f ′ ( z k ) − F ( z k ) ∣ > 1 2 K 2 r k d 2 + ρ ∕ 2 − 1 ∣ f ( z k ) ∣ − ∣ F ( z k ) ∣ = 1 2 r k d 1 + ρ ∕ 2 ∣ f ( z k ) ∣ ( 1 + o ( 1 ) ) ,

where K 1 , K 2 > 0 are constants. This leads to a contradiction that

K 1 r k d 1 > 1 2 ( 1 + o ( 1 ) ) K 2 r k d 1 + ρ ∕ 2 ,

as z k → ∞ .

Case 2: d 2 < d 1 + 1 . Now, d 2 ≤ d 1 . By equations (3.1) and (3.3)–(3.5), we obtain that

K 1 r k d 1 ∣ f ( z k ) ∣ = ∣ [ A ( z k ) + C ( z k ) ] f ( z k ) ( 1 + o ( 1 ) ) ∣ = ∣ [ A ( z k ) f ( z k + 1 ) + C ( z k ) f ( z k ) ] ∣ = ∣ B ( z k ) f ′ ( z k ) − F ( z k ) ∣ < 2 K 3 r k d 2 + ( 1 − ρ ) ∕ 2 − 1 ∣ f ( z k ) ∣ + ∣ F ( z k ) ∣ ≤ 2 K 3 r k d 1 + ( 1 − ρ ) ∕ 2 − 1 ∣ f ( z k ) ∣ ( 1 + o ( 1 ) ) ,

where K 3 > 0 is a constant. This leads to another contradiction that

K 1 r k d 1 < 2 K 3 r k d 1 + ( 1 − ρ ) ∕ 2 − 1 ( 1 + o ( 1 ) ) ,

as z k → ∞ .

To sum up, we have proved that ρ ( f ) ≥ 1 .

Step 2: We will prove conclusion (i) provided that f ( z ) is of finite-order and F ( z ) ≢ 0 .

From Lemma 3.1, the lemma on the logarithmic derivative, and equation (3.1), we obtain

m r , 1 f = m r , − 1 F ( z ) A ( z ) f ( z + 1 ) f ( z ) + B ( z ) f ′ ( z ) f ( z ) + C ( z ) ≤ m r , f ( z + 1 ) f ( z ) + m r , f ′ ( z ) f ( z ) + S ( r , f ) = S ( r , f ) ,

which means that

N r , 1 f = T ( r , f ) + S ( r , f ) .

Hence, λ ( f ) = ρ ( f ) .

Step 3: We will prove conclusion (ii) provided that f ( z ) is of finite-order and F ( z ) ≡ 0 .

Suppose that 1 + λ f < ρ ( f ) = ρ < ∞ , where λ f = max { λ ( f ) , λ ( 1 ∕ f ) } . Assume that z = 0 is a zero (or pole) of f ( z ) of multiplicity d . By Hadamard’s factorization theorem for meromorphic function (see [2], Theorem 2.7), we can write f ( z ) as follows:

(3.6) f ( z ) = z d p ( z ) q ( z ) e g ( z ) ,

where p ( z ) and q ( z ) are entire functions such that ρ ( p ) = λ ( p ) = λ ( f ) , ρ ( q ) = λ ( q ) = λ ( 1 ∕ f ) , and g ( z ) is a polynomial such that deg g ( z ) = m . Since ρ ( f ) > 1 + λ f , we see that m = ρ ( f ) > 1 + max { ρ ( p ) , ρ ( q ) } .

Submitting equation (3.6) into equation (2.2), we obtain that

e g ( z + 1 ) − g ( z ) = − 1 A ( z ) h ( z + 1 ) [ B ( z ) h ′ ( z ) g ′ ( z ) − C ( z ) h ( z ) ] ,

where

h ( z ) = z d p ( z ) q ( z ) .

This yields that

m − 1 = deg { g ( z + 1 ) − g ( z ) } = ρ ( e g ( z + 1 ) − g ( z ) ) = ρ − 1 A ( z ) h ( z + 1 ) [ B ( z ) h ′ ( z ) g ′ ( z ) − C ( z ) h ( z ) ] < m − 1 ,

a contradiction.

These two contradictions and Step 1 ensure that 1 ≤ ρ ( f ) ≤ 1 + λ f .

If ρ ( f ) = 1 , we claim that λ f = 1 . Otherwise, 0 ≤ λ f < 1 . Then, g ( z ) = a z + b in equation (3.6), where a ( ≠ 0 ) and b are constants, and ρ ( h ) = ρ ( z d p ∕ q ) < 1 . Thus, equation (3.1) is of the form

e a A ( z ) h ( z + 1 ) + B ( z ) h ′ ( z ) + C ( z ) = 0 .

With a similar argument to that for Step 1, we can prove that ρ ( h ) ≥ 1 . This is impossible. The proof of Theorem 3.1 is completed.□

From Theorems 2.1 and 3.1, we obtain the following Corollary 3.1.

Corollary 3.1

If f ( z ) is a nonconstant meromorphic solution of equation (2.1), where A ( z ) ( ≢ 0 ) , B ( z ) ( ≢ 0 ) , and C ( z ) are polynomials such that deg B ( z ) < deg { A ( z ) + C ( z ) } + 1 , then ρ ( f ) ≥ 1 .

Moreover, if f ( z ) is of finite-order, then 1 ≤ ρ ( f ) ≤ 1 + max { λ ( f ) , λ ( 1 ∕ f ) } , and ρ ( f ) = max { λ ( f ) , λ ( 1 ∕ f ) } = 1 , when ρ ( f ) = 1 .

Considering fixed points of transcendental meromorphic solutions of equation (1.1), we obtain the following results.

Theorem 3.2

Let f ( z ) be a finite-order transcendental meromorphic solution of equation (1.1). If

A ( z ) ( z + 1 ) + B ( z ) + z C ( z ) ≢ F ( z ) ,

then f ( z ) must have infinitely many fixed points such that τ ( f ) = ρ ( f ) ≥ 1 .

Corollary 3.2

If f ( z ) is a nonconstant meromorphic solution of equation (2.1), where A ( z ) ( ≢ 0 ) , B ( z ) ( ≢ 0 ) , and C ( z ) are polynomials such that deg B ( z ) < deg { A ( z ) + C ( z ) } + 1 , then f ( z ) must have infinitely many fixed points such that τ ( f ) = ρ ( f ) ≥ 1 .

Remark 3.2

By Theorems 2.1 and 3.1, we can prove Corollary 3.2 easily with the idea in the proof of Theorem 3.2. Thus, we omit all those details.

Proof Theorem 3.2

Let f ( z ) be a transcendental meromorphic solution of the equation (1.1), then from Theorem 3.1, we see that ρ ( f ) ≥ 1 . Set g ( z ) = f ( z ) − z , then obviously, T ( r , f ) = T ( r , g ) , and we can rewrite equation (3.1) as follows:

(3.7) A ( z ) g ( z + 1 ) + B ( z ) g ′ ( z ) + C ( z ) g ( z ) = D ( z ) ,

where D ( z ) = F ( z ) − [ A ( z ) ( z + 1 ) + B ( z ) + z C ( z ) ] ≢ 0 .

From Lemma 3.1, the lemma on the logarithmic derivative, and equation (3.1), we obtain

m r , 1 g = m r , − 1 D ( z ) A ( z ) g ( z + 1 ) g ( z ) + B ( z ) g ′ ( z ) g ( z ) + C ( z ) ≤ m r , g ( z + 1 ) g ( z ) + m r , g ′ ( z ) g ( z ) + S ( r , g ) = S ( r , g ) ,

which means that

N r , 1 g = T ( r , g ) + S ( r , g ) .

Hence, τ ( f ) = λ ( g ) = ρ ( g ) = ρ ( f ) .□

lish_ls@gdou.edu.cn

# Lihao Wu and Baoqin Chen contributed equally to this work.

Acknowledgements

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

Funding information: This work is supported by the National Natural Science Fund of China (No. 12101138), and the Open Foundation of the Guangdong Provincial Key Laboratory of Electronic Information Products Reliability Technology.
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Conflict of interest: The authors declare that they have no competing interests.
Data availability statement: No data were used to support this study.

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Received: 2023-07-31

Revised: 2023-10-19

Accepted: 2023-10-20

Published Online: 2023-12-01

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

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https://doi.org/10.1515/math-2023-0147

Keywords for this article

differential-difference equation; growth and zeros; meromorphic solution

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