Open Access

On growth of meromorphic solutions for linear difference equations with meromorphic coefficients

Advances in Difference Equations20132013:60

https://doi.org/10.1186/1687-1847-2013-60

Received: 2 December 2012

Accepted: 26 February 2013

Published: 19 March 2013

Abstract

In this paper, we consider the value distribution of meromorphic solutions for linear difference equations with meromorphic coefficients.

MSC:30D35, 39A10.

Keywords

difference equationmeromorphic coefficientgrowthzero

1 Introduction and preliminaries

Recently, several papers (including [17]) have been published regarding value distribution of meromorphic solutions of linear difference equations. We recall the following results. Chiang and Feng proved the following theorem.

Theorem A ([2])

Let P 0 ( z ) , , P n ( z ) be polynomials such that there exists an integer l, 0 l n , such that
deg ( P l ) > max 0 j n , j l { deg ( P j ) }
(1.1)
holds. Suppose f ( z ) is a meromorphic solution of the difference equation
P n ( z ) f ( z + n ) + + P 1 ( z ) f ( z + 1 ) + P 0 ( z ) f ( z ) = 0 .
(1.2)

Then we have σ ( f ) 1 .

In this paper, we use the basic notions of Nevanlinna’s theory (see [8, 9]). In addition, we use the notation σ ( f ) to denote the order of growth of the meromorphic function f ( z ) , and λ ( f ) to denote the exponent of convergence of zeros of f ( z ) .

Chen [1] weakened the condition (1.1) of Theorem A and proved the following results.

Theorem B ([1])

Let P n ( z ) , , P 0 ( z ) be polynomials such that P n P 0 0 and
deg ( P n + + P 0 ) = max { deg P j : j = 0 , , n } 1 .
(1.3)

Then every finite order meromorphic solution f ( z ) (0) of equation (1.2) satisfies σ ( f ) 1 , and f ( z ) assumes every nonzero value a C infinitely often and λ ( f a ) = σ ( f ) .

Theorem C ([1])

Let F ( z ) , P n ( z ) , , P 0 ( z ) be polynomials such that F P n P 0 0 and (1.3). Then every finite order transcendental meromorphic solution f ( z ) of the equation
P n ( z ) f ( z + n ) + + P 1 ( z ) f ( z + 1 ) + P 0 ( z ) f ( z ) = F ( z )
(1.4)

satisfies σ ( f ) 1 and λ ( f ) = σ ( f ) .

Theorem D ([1])

Let F ( z ) , P n ( z ) , , P 0 ( z ) be polynomials such that F P n P 0 0 . Suppose that f ( z ) is a meromorphic solution with infinitely many poles of (1.2) (or (1.4)). Then σ ( f ) 1 .

For the linear difference equation with transcendental coefficients
A n ( z ) f ( z + n ) + + A 1 ( z ) f ( z + 1 ) + A 0 ( z ) f ( z ) = 0 ,
(1.5)

Chiang and Feng proved the following result.

Theorem E ([2])

Let A 0 ( z ) , , A n ( z ) be entire functions such that there exists an integer l, 0 l n , such that
σ ( A l ) > max { σ ( A j ) : 0 j n , j l } .
(1.6)

If f ( z ) is a meromorphic solution of (1.5), then we have σ ( f ) σ ( A l ) + 1 .

Laine and Yang proved the following theorem.

Theorem F ([5])

Let A 0 , , A n be entire functions of finite order so that among those having the maximal order σ : = max { σ ( A k ) : 0 k n } , exactly one has its type strictly greater than the others. Then for any meromorphic solution of
A n ( z ) f ( z + C n ) + + A 1 ( z ) f ( z + C 1 ) + A 0 ( z ) f ( z ) = 0 ,
(1.7)

we have σ ( f ) σ + 1 .

Remark 1.1 If A 0 , , A n are meromorphic functions satisfying (1.6), then Theorem E does not hold. For example, the equation
y ( z + 1 ) ( e i + e i 1 e i z 1 ) y ( z ) = 0

has a solution y ( z ) = e i z 1 , which σ ( y ) = 1 < σ ( A 0 ) + 1 .

This example shows that for the linear difference equation with meromorphic coefficients, the condition (1.6) cannot guarantee that every transcendental meromorphic solution f ( z ) of (1.7) satisfies σ ( f ) σ ( A l ) + 1 .

Thus, a natural question to ask is what conditions will guarantee every transcendental meromorphic solution f ( z ) of (1.7) with meromorphic coefficients satisfies σ ( f ) σ ( A l ) + 1 .

In this note, we consider this question and prove the following results.

Theorem 1.1 Let c 1 , c 2 ( c 1 ), a be nonzero constants, h 1 ( z ) be a nonzero meromorphic function with σ ( h 1 ) < 1 , B ( z ) be a nonzero meromorphic function.

If B ( z ) satisfies any one of the following three conditions:
  1. (i)

    σ ( B ) > 1 and δ ( , B ) > 0 ;

     
  2. (ii)

    σ ( B ) < 1 ;

     
  3. (iii)

    B ( z ) = h 0 ( z ) e b z where b is a nonzero constant, h 0 ( z ) (0) is a meromorphic function with σ ( h 0 ) < 1 ,

     
then every meromorphic solution f (0) of the difference equation
f ( z + c 2 ) + h 1 ( z ) e a z f ( z + c 1 ) + B ( z ) f ( z ) = 0
(1.8)

satisfies σ ( f ) max { σ ( B ) , 1 } + 1 .

Further, if φ ( z ) (0) is a meromorphic function with
σ ( φ ) < max { σ ( B ) , 1 } + 1 ,
then
λ ( f φ ) = σ ( f ) max { σ ( B ) , 1 } + 1 .
Corollary Under conditions of Theorem  1.1, every finite order solution f ( z ) (0) of (1.8) has infinitely many fixed points, satisfies τ ( f ) = σ ( f ) , and for any nonzero constant c,
λ ( f ( z ) c ) = σ ( f ) max { σ ( B ) , 1 } + 1 .

Example 1.1

The equation
f ( z + 2 ) 1 2 e 2 z + 3 f ( z + 1 ) 1 2 e 4 z + 4 f ( z ) = 0

satisfies conditions of Theorem 1.1 and has a solution f ( z ) = e z 2 satisfying λ ( f ) = 0 and τ ( f ) = σ ( f ) = 2 . This example shows that under conditions of Theorem 1.1, a meromorphic solution of (1.8) may have no zero.

Theorem 1.2 Let h 1 ( z ) , c 1 , c 2 , a, B ( z ) satisfy conditions of Theorem  1.1, and let F ( z ) (0) be a meromorphic function with σ ( F ) < max { σ ( B ) , 1 } + 1 . Then all meromorphic solutions with finite order of the equation
f ( z + c 2 ) + h 1 ( z ) e a z f ( z + c 1 ) + B ( z ) f ( z ) = F ( z )
(1.9)
satisfy
λ ( f ) = σ ( f ) max { σ ( B ) , 1 } + 1

with at most one possible exceptional solution with σ ( f ) < max { σ ( B ) , 1 } + 1 .

Remark 1.2 Under conditions of Theorem 1.1, equation (1.8) has no rational solution. But equation (1.9) in Theorem 1.2 may have a rational solution. For example, the equation
f ( z + 2 ) + e z f ( z + 1 ) e z f ( z ) = z + 2 e z

satisfies conditions of Theorem 1.2 and has a solution f ( z ) = z . This shows that in Theorem 1.2, there exists one possible exceptional solution with σ ( f ) < max { σ ( B ) , 1 } + 1 .

2 Proof of Theorem 1.1

We need the following lemmas to prove Theorem 1.1.

Lemma 2.1 ([2, 10])

Given two distinct complex constants η 1 , η 2 , let f be a meromorphic function of finite order σ. Then, for each ε > 0 , we have
m ( r , f ( z + η 1 ) f ( z + η 2 ) ) = O ( r σ 1 + ε ) .

Lemma 2.2 (see [11])

Suppose that P ( z ) = ( α + i β ) z n + (α, β are real numbers, | α | + | β | 0 ) is a polynomial with degree n 1 , that A ( z ) (0) is an entire function with σ ( A ) < n . Set g ( z ) = A ( z ) e P ( z ) , z = r e i θ , δ ( P , θ ) = α cos n θ β sin n θ . Then, for any given ε > 0 , there exists a set H 1 [ 0 , 2 π ) that has the linear measure zero such that for any θ [ 0 , 2 π ) ( H 1 H 2 ) , there is R > 0 such that for | z | = r > R , we have that
  1. (i)
    if δ ( P , θ ) > 0 , then
    exp { ( 1 ε ) δ ( P , θ ) r n } < | g ( r e i θ ) | < exp { ( 1 + ε ) δ ( P , θ ) r n } ;
    (2.1)
     
  2. (ii)
    if δ ( P , θ ) < 0 , then
    exp { ( 1 + ε ) δ ( P , θ ) r n } < | g ( r e i θ ) | < exp { ( 1 ε ) δ ( P , θ ) r n } ,
    (2.2)
     

where H 2 = { θ [ 0 , 2 π ) ; δ ( P , θ ) = 0 } is a finite set.

Lemma 2.3 Let c 1 , c 2 ( c 1 ), a be nonzero constants, A j ( z ) ( j = 0 , 1 , 2 ), F ( z ) be nonzero meromorphic functions. Suppose that f ( z ) is a finite order meromorphic solution of the equation
A 2 ( z ) f ( z + c 2 ) + A 1 ( z ) f ( z + c 1 ) + A 0 ( z ) f ( z ) = F ( z ) .
(2.3)

If σ ( f ) > max { σ ( F ) , σ ( A j ) ( j = 0 , 1 , 2 ) } , then λ ( f ) = σ ( f ) .

Proof Suppose that σ ( f ) = σ , max { σ ( F ) , σ ( A j ) ( j = 0 , 1 , 2 ) } = α . Then σ > α . Equation (2.3) can be rewritten as the form
1 f ( z ) = F ( z ) f ( z ) ( A 2 ( z ) f ( z + c 2 ) f ( z ) + A 1 ( z ) f ( z + c 1 ) f ( z ) + A 0 ( z ) ) .
(2.4)
Thus, by (2.4), we deduce that
T ( r , f ) = T ( r , 1 f ) + O ( 1 ) = m ( r , 1 f ) + N ( r , 1 f ) + O ( 1 ) N ( r , 1 f ) + m ( r , 1 F ) + j = 0 2 m ( r , A j ) + m ( r , f ( z + c 2 ) f ( z ) ) + m ( r , f ( z + c 1 ) f ( z ) ) + O ( 1 ) .
(2.5)
For any given ε ( 0 < ε < min { 1 4 , σ α 4 } ), and for sufficiently large r, we have that
m ( r , 1 F ) T ( r , F ) r α + ε , m ( r , A j ) r α + ε ( j = 0 , 1 , 2 ) .
(2.6)
By Lemma 2.1, we obtain
m ( r , f ( z + c 2 ) f ( z ) ) M r σ 1 + ε and m ( r , f ( z + c 1 ) f ( z ) ) M r σ 1 + ε ,
(2.7)

where M (>0) is some constant.

By σ ( f ) = σ , there exists a sequence { r n } satisfying r 1 < r 2 <  , r n such that
lim n log T ( r n , f ) log r n = σ .
(2.8)
Thus, for sufficiently large r n , we have that
T ( r n , f ) r n σ ε .
(2.9)
Substituting (2.6)-(2.9) into (2.5), we obtain for sufficiently large r n
r n σ ε T ( r n , f ) N ( r n , 1 f ) + 4 r n α + ε + 2 M r n σ 1 + ε .
(2.10)
Since ε < min { 1 4 , σ α 4 } and ε is arbitrary, by (2.10), we obtain
lim ¯ n log N ( r n , 1 f ) log r n = σ .

Hence, λ ( f ) = σ ( f ) = σ . □

Proof of Theorem 1.1 Suppose that f ( z ) (0) is a meromorphic solution of equation (1.8) with σ ( f ) < .
  1. (1)
    Suppose that B ( z ) satisfies the condition (i): σ ( B ) > 1 and δ ( , B ) = δ > 0 . Thus, for sufficiently large r,
    m ( r , B ) > δ 2 T ( r , B ) .
    (2.11)
     
Clearly, σ ( f ) σ ( B ) by (1.8). By Lemma 2.1, we see that for any given ε ( 0 < ε < σ ( B ) 1 3 ),
m ( r , f ( z + c j ) f ( z ) ) = O ( r σ ( f ) 1 + ε ) ( j = 1 , 2 ) ,
(2.12)
and
m ( r , h 1 ( z ) e a z ) T ( r , h 1 ( z ) e a z ) r 1 + ε .
(2.13)
By (1.8), we have that
B ( z ) = f ( z + c 2 ) f ( z ) + h 1 ( z ) e a z f ( z + c 1 ) f ( z ) .
(2.14)
Substituting (2.11)-(2.13) into (2.14), we deduce that
δ 2 T ( r , B ) m ( r , B ) m ( r , h 1 ( z ) e a z ) + m ( r , f ( z + c 2 ) f ( z ) ) + m ( r , f ( z + c 1 ) f ( z ) ) r 1 + ε + O ( r σ ( f ) 1 + ε ) .
(2.15)
By σ ( B ) = σ , there is a sequence r j ( 1 < r 1 < r 2 <  , r j ) satisfying
T ( r j , B ) > r j σ ( B ) ε .
(2.16)
Thus, by (2.15) and (2.16), we obtain
δ 2 r j σ ( B ) ε r j 1 + ε + M r j σ ( f ) 1 + ε ,
(2.17)
where M (>0) is some constant. Combining (2.17) and ε < σ ( B ) 1 3 , it follows that
δ 2 r j σ ( B ) ε ( 1 + o ( 1 ) ) M r j σ ( f ) 1 + ε .
So that, it follows that σ ( f ) σ ( B ) + 1 = max { σ ( B ) , 1 } + 1 .
  1. (2)

    Suppose that B ( z ) satisfies the condition (ii): σ ( B ) < 1 . Using the same method as in (1), we can obtain σ ( f ) max { σ ( B ) , 1 } + 1 .

     
  2. (3)

    Suppose that B ( z ) satisfies the condition (iii): B ( z ) = h 0 ( z ) e b z , where b is a nonzero constant, h 0 ( z ) (0) is a meromorphic function with σ ( h 0 ) < 1 .

     
Now we need to prove σ ( f ) 2 . Contrary to the assertion, suppose that σ ( f ) = α < 2 . We will deduce a contradiction. Set z = r e i θ . Then
{ Re { a z } = δ ( a z , θ ) | a | r = | a | r cos ( arg a + θ ) , Re { b z } = δ ( b z , θ ) | b | r = | b | r cos ( arg b + θ ) .
(2.18)

In what follows, we divide this proof into three subcases: (a) arg a arg b ; (b) arg a = arg b and | a | | b | ; (c) a = b .

Subcase (a). Since arg a arg b and (2.18), it is easy to see that there exists a ray arg z = θ 0 such that
{ Re { a z } = δ ( a z , θ 0 ) | a | r = | a | r cos ( arg a + θ 0 ) < 0 , Re { b z } = δ ( b z , θ 0 ) | b | r = | b | r cos ( arg b + θ 0 ) > 0 .
(2.19)
By (1.8) and (2.19), we see that f ( z ) cannot be a rational function. By Lemma 2.1, (2.12) holds. By Lemma 2.2 and (2.19), it is easy to see that for any given ε 1 ( 0 < ε 1 < min { 1 2 , 2 α 2 } ) and for sufficiently large r,
| h 0 ( r e i θ 0 ) e b r e i θ 0 | exp { ( 1 ε 1 ) | b | δ ( b z , θ 0 ) r } ,
(2.20)
and
| h 1 ( r e i θ 0 ) e a r e i θ 0 | exp { ( 1 ε 1 ) | a | δ ( a z , θ 0 ) r } < 1 .
(2.21)
Thus, by (1.8), (2.12), (2.20) and (2.21), we deduce that
exp { ( 1 ε 1 ) | b | δ ( b z , θ 0 ) r } | h 0 ( r e i θ 0 ) e b r e i θ 0 | | f ( r e i θ 0 + c 2 ) f ( r e i θ 0 ) | + | h 1 ( r e i θ 0 ) e a r e i θ 0 | | f ( r e i θ 0 + c 1 ) f ( r e i θ 0 ) | 2 exp { r σ ( f ) 1 + ε 1 } .
(2.22)

By δ ( b z , θ 0 ) = cos ( arg b + θ 0 ) > 0 , σ ( f ) = α < 2 and ε 1 < 2 α 2 , it is easy to see that (2.22) is a contradiction. Hence, σ ( f ) 2 .

Subcase (b). By arg a = arg b and | a | | b | , we see that f ( z ) cannot be a rational function. By Lemma 2.1, (2.12) holds. By arg a = arg b and (2.18), we take θ 1 = arg a , then δ ( a z , θ 1 ) = δ ( b z , θ 1 ) = 1 and
Re { a r e i θ 1 } = | a | r and Re { b r e i θ 1 } = | b | r .
(2.23)
Now suppose that | b | > | a | . By Lemma 2.2, for any given ε 2 ( 0 < ε 2 < min { 2 α , | b | | a | 2 ( | b | + | a | ) } ),
| h 0 ( r e i θ 1 ) e b r e i θ 1 | exp { ( 1 ε 2 ) | b | r } ,
(2.24)
and
| h 1 ( r e i θ 1 ) e a r e i θ 1 | exp { ( 1 + ε 2 ) | a | r } .
(2.25)
Thus, by (1.8), (2.12), (2.24) and (2.25), we deduce that
exp { ( 1 ε 2 ) | b | r } | h 0 ( r e i θ 1 ) e b r e i θ 1 | | f ( r e i θ 1 + c 2 ) f ( r e i θ 1 ) | + | h 1 ( r e i θ 1 ) e a r e i θ 1 | | f ( r e i θ 1 + c 1 ) f ( r e i θ 1 ) | exp { r σ ( f ) 1 + ε 2 } + exp { ( 1 + ε 2 ) | a | r } exp { r σ ( f ) 1 + ε 2 } .
(2.26)
Since ε 2 < 2 α , we have that σ ( f ) 1 + ε 2 = α 1 + ε 2 < 1 . Combining this and (2.26), we obtain
exp { ( 1 ε 2 ) | b | r } < exp { ( 1 + ε 2 ) | a | r ( 1 + o ( 1 ) ) } ( 1 + o ( 1 ) ) .
(2.27)

By ε 2 < | b | | a | 2 ( | b | + | a | ) , we see that (2.27) is a contradiction.

Now suppose that | b | < | a | . Using the same method as above, we can also deduce a contradiction.

Hence, σ ( f ) 2 in Subcase (b).

Subcase (c). We first affirm that f ( z ) cannot be a nonzero rational function. In fact, if f ( z ) is a rational function, then e a z [ h 1 ( z ) f ( z + c 1 ) + h 0 ( z ) f ( z ) ] = f ( z + c 2 ) is a rational function. So that h 1 ( z ) f ( z + c 1 ) + h 0 ( z ) f ( z ) 0 , that is, f ( z + c 2 ) 0 , a contradiction.

By Lemma 2.1, (2.12) holds. By a = b , equation (1.8) can be rewritten as
e a z f ( z + c 2 ) + h 1 ( z ) f ( z + c 1 ) + h 0 ( z ) f ( z ) = 0 .
(2.28)
Using the same method as in the proof of (1), we can obtain σ ( f ) 2 .
  1. (4)
    Suppose that φ ( z ) (0) is a meromorphic function with σ ( φ ) < max { σ ( B ) , 1 } + 1 . Set g ( z ) = f ( z ) φ ( z ) . Substituting f ( z ) = g ( z ) + φ ( z ) into (1.8), we obtain
    g ( z + c 2 ) + h 1 ( z ) e a z g ( z + c 1 ) + B ( z ) g ( z ) = [ φ ( z + c 2 ) + h 1 ( z ) e a z φ ( z + c 1 ) + B ( z ) φ ( z ) ] .
    (2.29)
     
If φ ( z + c 2 ) + h 1 ( z ) e a z φ ( z + c 1 ) + B ( z ) φ ( z ) 0 , then φ ( z ) is a nonzero meromorphic solution of (1.8). Thus, by the proof above, we have that σ ( φ ) max { σ ( B ) , 1 } + 1 . This contradicts our condition that σ ( φ ) < max { σ ( B ) , 1 } + 1 . Hence, φ ( z + 2 ) + h 1 ( z ) e a z φ ( z + 1 ) + B ( z ) φ ( z ) 0 , and
σ ( φ ( z + c 2 ) + h 1 ( z ) e a z φ ( z + c 1 ) + B ( z ) φ ( z ) ) < max { σ ( B ) , 1 } + 1 σ ( f ) = σ ( g ) .
Applying this and Lemma 2.3 to (2.29), we deduce that
λ ( f φ ) = λ ( g ) = σ ( g ) max { σ ( B ) , 1 } + 1 .

Thus, Theorem 1.1 is proved. □

3 Proof of Theorem 1.2

Suppose that f 0 is a meromorphic solution of (1.9) with
σ ( f 0 ) < max { σ ( B ) , 1 } + 1 .
If f ( z ) ( f 0 ( z ) ) is another meromorphic solution of (1.9) satisfying σ ( f ) < max { σ ( B ) , 1 } + 1 , then
σ ( f f 0 ) < max { σ ( B ) , 1 } + 1 .

But f f 0 is a solution of the corresponding homogeneous equation (1.8) of (1.9). By Theorem 1.1, we have σ ( f f 0 ) max { σ ( B ) , 1 } + 1 , a contradiction. Hence equation (1.9) possesses at most one exceptional solution f 0 with σ ( f 0 ) < max { σ ( B ) , 1 } + 1 .

Now suppose that f is a meromorphic solution of (1.9) with
max { σ ( B ) , 1 } + 1 σ ( f ) < .
Since σ ( f ) > max { σ ( B ) , σ ( F ) , σ ( h ( z ) e a z ) } , applying Lemma 2.3 to (1.9), we obtain
λ ( f ) = σ ( f ) .

Thus, Theorem 1.2 is proved.

Declarations

Acknowledgements

The author is grateful to the referees for a number of helpful suggestions to improve the paper. This research was partly supported by the National Natural Science Foundation of China (grant no. 11171119).

Authors’ Affiliations

(1)
School of Software Engineering, South China University of Technology, Panyu, P.R. China

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