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

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Abstract

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

MSC:30D35, 39A10.

1 Introduction and preliminaries

Recently, several papers (including ) 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 ()

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  weakened the condition (1.1) of Theorem A and proved the following results.

Theorem B ()

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 ()

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 ()

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 ()

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 ()

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)

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). 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).

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).

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 )

Suppose that $P(z)=(α+iβ) z n +⋯$ (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). Set $g(z)=A(z) e P ( z )$, $z=r e i θ$, $δ(P,θ)=αcosnθ−βsinnθ$. Then, for any given $ε>0$, there exists a set $H 1 ⊂[0,2π)$ that has the linear measure zero such that for any $θ∈[0,2π)∖( H 1 ∪ H 2 )$, 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 ) |
(2.1)
2. (ii)

if $δ(P,θ)<0$, then

$exp { ( 1 + ε ) δ ( P , θ ) r n } < | g ( r e i θ ) |
(2.2)

where $H 2 ={θ∈[0,2π);δ(P,θ)=0}$ 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), 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

$m ( r , f ( z + c 2 ) f ( z ) ) ≤M r σ − 1 + ε andm ( r , f ( z + c 1 ) f ( z ) ) ≤M r σ − 1 + ε ,$
(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 $r n$

$r n σ − ε ≤T( r n ,f)≤N ( r n , 1 f ) +4 r n α + ε +2M r n σ − 1 + ε .$
(2.10)

Since $ε 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) $arga≠argb$; (b) $arga=argb$ and $|a|≠|b|$; (c) $a=b$.

Subcase (a). Since $arga≠argb$ and (2.18), it is easy to see that there exists a ray $argz= θ 0$ 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 ) 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 $δ(bz, θ 0 )=cos(argb+ θ 0 )>0$, $σ(f)=α<2$ and $ε 1 < 2 − α 2$, it is easy to see that (2.22) is a contradiction. Hence, $σ(f)≥2$.

Subcase (b). By $arga=argb$ and $|a|≠|b|$, we see that $f(z)$ cannot be a rational function. By Lemma 2.1, (2.12) holds. By $arga=argb$ and (2.18), we take $θ 1 =−arga$, then $δ(az, θ 1 )=δ(bz, θ 1 )=1$ and

$Re { a r e i θ 1 } =|a|randRe { b r e i θ 1 } =|b|r.$
(2.23)

Now suppose that $|b|>|a|$. By Lemma 2.2, for any given $ε 2$ ($0< ε 2 ),

$| h 0 ( r e i θ 1 ) e b r e i θ 1 |≥exp { ( 1 − ε 2 ) | b | r } ,$
(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 }
(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). 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). 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 ) )

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 )

If $f ∗ (z)$ ($≢ f 0 (z)$) is another meromorphic solution of (1.9) satisfying $σ( f ∗ ), then

$σ ( f ∗ − f 0 )

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 ).

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.

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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).

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Correspondence to Yanxia Liu.

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The author declares that they have no competing interests.

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