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Yuan and Ling Advances in Diﬀerence Equations 2014, 2014:306
http://www.advancesindifferenceequations.com/content/2014/1/306
RESEARCH
Open Access
Results on the growth of meromorphic
solutions of some linear difference equations
with meromorphic coefﬁcients
Zhang Li Yuan and Qiu Ling*
*
Correspondence:
[email protected]
College of Applied Science, Beijing
University of Technology, Beijing,
100124, China
Abstract
In this paper, we investigate the growth of meromorphic solutions of some linear
diﬀerence equations. We obtain some new results on the growth of meromorphic
solutions when most coeﬃcients in such equations are meromorphic functions,
which are supplements of previous results due to Li and Chen (Adv. Diﬀer. Equ.
2012:203, 2012) and Liu and Mao (Adv. Diﬀer. Equ. 2013:133, 2013).
MSC: 30D35; 39A10
Keywords: complex diﬀerence equation; meromorphic coeﬃcients; growth
1 Introduction and main results
In this article, a meromorphic function always means meromorphic in the whole complex
plane C, and c always means a nonzero constant. We adopt the standard notations of the
Nevanlinna value distribution theory of meromorphic functions such as T(r, f ), m(r, f )
and N(r, f ) as explained in [–]. In addition, we will use notations ρ(f ) to denote the order
of growth of a meromorphic function f (z), λ(f ) to denote the exponents of convergence
of the zero sequence of a meromorphic function f (z), λ( f ) to denote the exponents of
convergence of the pole sequence of a meromorphic function f (z), and we deﬁne them as
follows:
ρ(f ) = lim sup
r→∞
λ(f ) = lim inf
r→∞
log T(r, f )
,
log r
log N(r, f )
log r
,

log N(r, f )
= lim inf
.
λ
r→∞
f
log r
Recently, meromorphic solutions of complex diﬀerence equations have become a subject of great interest from the viewpoint of Nevanlinna theory due to the apparent role
of the existence of such solutions of ﬁnite order for the integrability of discrete diﬀerence
equations.
About the growth of meromorphic solutions of some linear diﬀerence equations, some
results can be found in [–]. Laine and Yang [] considered the entire functions coefﬁcients case and got the following.
©2014 Yuan and Ling; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons
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Theorem A [] Let Aj (z) (≡ ) (j = , , . . . , n) be entire functions of ﬁnite order such that
among those coeﬃcients having the maximal order ρ := max≤j≤n ρ(Aj ), exactly one has its
type strictly greater than the others. Then, for any meromorphic solution f (z) to
An (z)f (z + n) + · · · + A (z)f (z + ) + A (z)f (z) = ,
we have ρ(f ) ≥ ρ + .
Chiang and Feng [, ] improved Theorem A as follows.
Theorem B [, ] Let Aj (z) (≡ ) (j = , , . . . , n) be entire functions such that there exists
an integer l,  ≤ l ≤ n, such that
ρ(Al ) > max ρ(Aj ).
≤j≤n,j=l
Suppose that f (z) is a meromorphic solution to
An (z)f (z + n) + · · · + A (z)f (z + ) + A (z)f (z) = ,
then we have ρ(f ) ≥ ρ(Al ) + .
Recently in [], Peng and Chen investigated the order and the hyper-order of solutions
of some second-order linear diﬀerential equations and proved the following results.
Theorem C [] Suppose that Aj (z) (≡ ) (j = , ) are entire functions and ρ(Aj ) < . Let
α , α be two distinct complex numbers such that α α =  (suppose that |α | ≤ |α |). If
arg α = π or α < –, then every solution f (z) ≡  of the equation
f + e–z f + A eα z + A eα z f = 
has inﬁnite order and ρ (f ) = .
Moreover, Xu and Zhang [] extended the above result from entire coeﬃcients to meromorphic coeﬃcients.
It is well known that f (z) = f (z + c) – f (z) is regarded as the diﬀerence counterpart
of f . Thus a natural question is: Can we change the above second-order linear diﬀerential equation to the linear diﬀerence equation? What conditions will guarantee that every
meromorphic solution will have inﬁnite order when most coeﬃcients in such equations
are meromorphic functions?
Li and Chen [] considered the following diﬀerence equation and obtained the following theorem.
Theorem D [] Let k be a positive integer, p be a nonzero real number and f (z) be a
nonconstant meromorphic solution of the diﬀerence equation
k
An (z)f (z + n) + · · · + A (z)f (z + ) + A (z)e–pz + B (z) f (z + )
k
+ A (z)epz + B (z) f (z) = ,
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where Aj (z), B (z), B (z) (≡ ) (j = , , . . . , n) are all entire functions and max{ρ(Aj ), ρ(B ),
ρ(B )} = σ < k, then we have ρ(f ) ≥ k + .
The main purpose of this paper is to investigate the growth of meromorphic solutions
of certain linear diﬀerence equations with meromorphic coeﬃcients. The remainder of
the paper studies the properties of meromorphic solutions of a nonhomogeneous linear
diﬀerence equation. In fact, we prove the following results, in which there are still some
coeﬃcients dominating in some angles.
Theorem . Let k be a positive integer. Suppose that Aj (z), B (z) (≡ ) (j = , , . . . , n) are
all entire functions and max{ρ(Aj ), ρ(B ) :  ≤ j ≤ n} = α < k. Let α , β be two distinct
complex numbers such that α β =  (suppose that |α | ≤ |β |). Let α be a strictly negative
real constant. If arg α = π or α < α , then every meromorphic solution f (z) ≡  of the
equation
k
An (z)f (z + n) + · · · + A (z)f (z + ) + A (z)eα z f (z + )
k
k
+ A (z)eα z + B (z)eβ z f (z) = 
(.)
satisﬁes ρ(f ) ≥ k + .
Theorem . Suppose that Aj (z), B (z) (≡ ) (j = , , . . . , n) are all meromorphic functions
and max{ρ(Aj ), ρ(B ) :  ≤ j ≤ n} = α < . Let α , β be two distinct complex numbers such
that α β =  (suppose that |α | ≤ |β |). Let α be a strictly negative real constant. If arg α =
π or α < α , then every meromorphic solution f (z) ≡  of the equation
An (z)f (z + n) + · · · + A (z)f (z + ) + A (z)eα z f (z + )
+ A (z)eα z + B (z)eβ z f (z) = 
(.)
satisﬁes ρ(f ) ≥ .
Theorem . Under the assumption for the coeﬃcients of (.) in Theorem ., if f (z) is a
ﬁnite order meromorphic solution to (.), then λ(f – z) = ρ(f ). What is more, either k +  ≤
ρ(f ) ≤ max{λ(f ), λ( f )} +  or ρ(f ) = k +  > max{λ(f ), λ( f )} + .
Theorem . Under the assumption for the coeﬃcients of (.) in Theorem ., if f (z) is a
ﬁnite order meromorphic solution to (.), then λ(f – z) = ρ(f ). What is more, either  ≤
ρ(f ) ≤ max{λ(f ), λ( f )} +  or ρ(f ) =  > max{λ(f ), λ( f )} + .
Liu and Mao [] considered the meromorphic solutions of the diﬀerence equation
ak (z)f (z + k) + · · · + a (z)f (z + ) + a (z)f (z) = ,
(.)
one of their results can be stated as follows.
Theorem E [] Let aj (z) = Aj (z)ePj (z) (j = , , . . . , k), where Pj (z) = αjn zn + · · · + αj are
polynomials with degree n (≥ ), Aj (z) (≡ ) are entire functions of ρ(Aj ) < n. If αjn (j =
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, , . . . , k) are distinct complex numbers, then every meromorphic solution f (≡ ) of Eq.
(.) satisﬁes ρ(f ) ≥ max≤j≤k {ρ(aj )} + .
In this paper, we extend and improve the above result from entire coeﬃcients to meromorphic coeﬃcients in the case where the polynomials Pj (z) are of degree .
Theorem . Let aj (z) = Aj (z)eαj z (j = , , . . . , n), αj are distinct complex constants, suppose
that Aj (z) (≡ ) are meromorphic functions and ρ(Aj ) = α < , then every meromorphic
solution f (≡ ) of the equation
an (z)f (z + n) + · · · + a (z)f (z + ) + a (z)f (z) = 
(.)
satisﬁes ρ(f ) ≥ .
Theorem . Let aj (z) = Aj (z)eαj z + Dj (z) (j = , , . . . , n), αj are distinct complex constants,
suppose that Aj (z) (≡ ), Dj (z) (≡ ) are meromorphic functions and ρ(Aj ) = α < , ρ(Dj ) =
β < , then every meromorphic solution f (≡ ) of Eq. (.) satisﬁes ρ(f ) ≥ .
Next we consider the properties of meromorphic solutions of the nonhomogeneous linear diﬀerence equation corresponding to (.)
an (z)f (z + n) + · · · + a (z)f (z + ) + a (z)f (z) = F(z),
(.)
where F(z) (≡ ) is a meromorphic function.
Theorem . Let aj (z) (j = , , . . . , n) satisfy the hypothesis of Theorem . or Theorem .,
and let F(z) be a meromorphic function of ρ(F) < , then at most one meromorphic solution
f of Eq. (.) satisﬁes  ≤ ρ(f ) ≤  and max{λ(f ), λ( f )} = ρ(f ), the other solutions f satisfy
ρ(f ) ≥ .
2 Some lemmas
In this section, we present some lemmas which will be needed in the sequel.
Lemma . [] Let f (z) be a meromorphic function of ﬁnite order ρ, be a positive constant, η and η be two distinct nonzero complex constants. Then
f (z + η )
m r,
f (z + η )
= O rρ–+ ,
and there exists a subset E ⊂ (, ∞) of ﬁnite logarithmic measure such that, for all z satisfying |z| = r ∈/ [, ] ∪ E , and as r → ∞,
f (z + η ) ≤ exp rρ–+ .
exp –rρ–+ ≤ f (z + η )
Lemma . [] Let f (z) be a meromorphic function of ﬁnite order ρ, then, for any given
> , there exists a set E ⊂ (, ∞) of ﬁnite logarithmic measure such that, for all z satisfying
|z| = r ∈/ [, ] ∪ E , and as r → ∞,
exp –rρ+ ≤ f (z) ≤ exp r ρ+ .
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Lemma . [] Suppose that P(z) = (α + iβ)zn + · · · (α, β are real numbers, |α| + |β| =
)
is a polynomial with degree n ≥ , A(z) (≡ ) is an entire function with ρ(A) < n. Set g(z) =
A(z)eP(z) , z = reiθ , δ(P, θ ) = α cos nθ – β sin nθ . Then, for any given > , there exists a set
E ⊂ [, π) that has linear measure zero such that for any θ ∈ [, π)\(E ∪ E ), there is
R >  such that for |z| = r > R, we have
(i) if δ(P, θ ) > , then
exp ( – )δ(P, θ )r n < g reiθ < exp ( + )δ(P, θ )rn ;
(ii) if δ(P, θ ) < , then
exp ( + )δ(P, θ )r n < g reiθ < exp ( – )δ(P, θ )rn ,
where E = {θ ∈ [, π) : δ(P, θ ) = } is a ﬁnite set.
Lemma . applies in Theorem . where A(z) (≡ ) is an entire function.
Lemma . [] Suppose that n ≥  is a positive integer. Let Pj (z) = ajn zn + · · · (j = , )
be nonconstant polynomials, where ajq (q = , . . . , n) are distinct complex numbers and
), δ(Pj , θ ) = |ajn | cos(θj + nθ), then there is
an an = . Set z = reiθ , ajn = |ajn |eiθj , θj ∈ [– π , π

π π
a set E ⊂ [– n
, n ) that has linear measure zero. If θ = θ , then there exists a ray arg z = θ ,
π π
, n )\(E ∪ E ) such that
θ ∈ [– n
δ(P , θ ) > ,
δ(P , θ ) < 
δ(P , θ ) < ,
δ(P , θ ) > ,
or
π π
, n ) : δ(Pj , θ ) = } is a ﬁnite set, which has linear measure zero.
where E = {θ ∈ [– n
π π
π π
, n )\(E ∪ E ) is replaced by θ ∈ [ n
, n )\(E ∪ E ), then we
In Lemma ., if θ ∈ [– n
have the same result.
Lemma . [] Consider g(z) = A(z)eaz , where A(z) (≡ ) is a meromorphic function with
ρ(A) = α < , a is a complex constant, a = |a|eiϕ (ϕ ∈ [, π)). Set E = {θ ∈ [, π) : cos(ϕ +
θ ) = }, then E is a ﬁnite set. Then, for any given ( < <  – α), there exists a set E ⊂
[, π) that has linear measure zero, if z = reiθ , θ ∈ [, π)\(E ∪ E ), then we have (when r
is suﬃciently large):
(i) if δ(az, θ ) > , then
exp ( – )δ(az, θ )r ≤ g reiθ ≤ exp ( + )δ(az, θ )r ;
(ii) if δ(az, θ ) < , then
exp ( + )δ(az, θ )r ≤ g reiθ ≤ exp ( – )δ(az, θ )r ,
where δ(az, θ ) = |a| cos(ϕ + θ ).
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Lemma . applies in Theorem . where A(z) (≡ ) is a meromorphic function.
Lemma . [] Let fj (z) (j = , , . . . , n, n ≥ ) be meromorphic functions and gj (z) (j =
, , . . . , n, n ≥ ) be entire functions such that
n
gj (z)
≡ ,
(i)
j= fj (z)e
(ii) gj (z) – gk (z) are not constant functions for  ≤ j < k ≤ n,
(iii) T(r, fj ) = o(T(r, egh –gk )) (r → ∞, r ∈/ E), where E is an exceptional set of ﬁnite linear
measure,  ≤ j ≤ n and  ≤ h < k ≤ n.
Then fj (z) ≡  (j = , , . . . , n).
Lemma . [] Let G(z) = kj= Bj (z)ePj (z) , where Pj (z) = αjn zn + · · · + αj are polynomials with degree n (≥ ), Bj (≡ ) are meromorphic functions of order ρ(Bj ) < n. If αjn
(j = , , . . . , k) are distinct complex numbers, then ρ(G) = n.
3 Proofs of the results
3.1 The proof of Theorem 1.1
Suppose that (.) admits a nontrivial meromorphic solution f (z) such that ρ(f ) < k + ,
 |–|α |
}, we have
then by Lemma ., for any given such that  < < min{k +  – ρ, k – α, |β
|β |+|α |
ρ(f )–+ f (z + j) ≤ exp rρ(f )–+ ,
≤
exp –r
f (z) j = , . . . , n,
(.)
for all r outside of a possible exceptional set E with ﬁnite logarithmic measure.
Applying Lemma ., we have
exp –rα+ ≤ Aj (z) ≤ exp rα+ ,
exp –rα+ ≤ B (z) ≤ exp rα+ ,
j = , . . . , n,
(.)
(.)
for all r outside of a possible exceptional set E with ﬁnite linear measure.
Therefore from (.) we can get that
k
A (z)f (z + ) A (z)eα z f (z + )
An (z)f (z + n)
k
k
+ ··· +
+
+ A (z)eα z + B (z)eβ z = ,
f (z)
f (z)
f (z)
i.e.,
A (z)eα zk + B (z)eβ zk An (z)f (z + n) A (z)f (z + ) A (z)eα zk f (z + ) .
≤
+ ··· + +
f (z)
f (z)
f (z)
(.)
)).
Setting α = |α |eiθ , β = |β |eiϕ (θ , ϕ ∈ [– π , π

Case . arg α = π , which is θ = π .
Subcase .. Assume that θ = ϕ . By Lemma ., for the above , there is a ray arg z = θ
π π
, k )\(E ∪ E ∪ E ∪ E ) (where E and E are deﬁned as in Lemma .,
such that θ ∈ [– k
E ∪ E ∪ E ∪ E is of linear measure zero) satisfying δ(α zk , θ ) > , δ(β zk , θ ) <  or
δ(α zk , θ ) < , δ(β zk , θ ) >  for a suﬃciently large r.
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Since Aj (z), B (z) (≡ ) (j = , , . . . , n) are entire functions and max{ρ(Aj ), ρ(B ) :  ≤ j ≤
n} = α < k, then when δ(α zk , θ ) > , δ(β zk , θ ) <  for a suﬃciently large r, by Lemma .,
we have
A (z)eα zk > exp ( – )δ α zk , θ rk ,
B (z)eβ zk < exp ( – )δ β zk , θ rk < .
(.)
(.)
By (.) and (.), we have
A (z)eα zk + B (z)eβ zk ≥ A (z)eα zk – B (z)eβ zk > exp ( – )δ α zk , θ rk – 
=  – o() exp ( – ε)δ α zk , θ rk .
(.)
k
π π
Since θ ∈ [– k
, k )\(E ∪ E ∪ E ∪ E ), we know that cos kθ > , then |eα z | =
–|α |rk cos kθ
< . Therefore, by (.) we obtain
e
A (z)eα zk ≤ exp rα+ .
(.)
Substituting (.), (.), (.), (.) and (.) into (.), we obtain
 – o() exp ( – )δ α zk , θ rk < n exp rα+ + rρ(f )–+ .
(.)
By δ(α zk , θ ) > , α + < k and ρ(f ) –  + < k, we know that (.) is a contradiction.
When δ(α zk , θ ) < , δ(β zk , θ ) > , using a proof similar to the above, we can get a contradiction.
Subcase .. Assume that θ = ϕ . By Lemma ., for the above , there is a ray arg z = θ
π π
, k )\(E ∪ E ∪ E ∪ E ) (where E and E are deﬁned as in Lemma .,
such that θ ∈ [– k
E ∪ E ∪ E ∪ E is of linear measure zero) satisfying δ(α zk , θ ) > .
Since |α | ≤ |β |, α = β , and θ = ϕ , then |α | < |β |, thus δ(β zk , θ ) > δ(α zk , θ ) > .
For a suﬃciently large r, by Lemma ., we get
A (z)eα zk < exp ( + )δ α zk , θ rk ,
B (z)eβ zk > exp ( – )δ β zk , θ rk .
(.)
(.)
By (.) and (.), we get
A (z)eα zk + B (z)eβ zk ≥ B (z)eβ zk – A (z)eα zk > exp ( – )δ β zk , θ rk – exp ( + )δ α zk , θ rk
(.)
= M exp ( + )δ α zk , θ rk ,
where M = exp{[( – )δ(β zk , θ ) – ( + )δ(α zk , θ )]rk } – .
 |–|α |
}, we see that ( – )δ(β zk , θ ) – ( + )δ(α zk , θ ) >
Since  < < min{k +  – ρ, k – α, |β
|β |+|α |
, then exp{[( – )δ(β zk , θ ) – ( + )δ(α zk , θ )]rk } > , M > .
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k
π π
Since θ ∈ [– k
, k )\(E ∪ E ∪ E ∪ E ), we know that cos kθ > , then |eα z | =
–|α |rk cos kθ
e
< . Therefore, by (.) we obtain
A (z)eα zk ≤ exp rα+ .
(.)
Substituting (.), (.), (.), (.) and (.) into (.), we obtain
M exp ( + )δ α zk , θ rk ≤ n exp rα+ + rρ(f )–+ .
(.)
By δ(α zk , θ ) > , α + < k and ρ(f ) –  + < k, we know that (.) is a contradiction.
Case . α < α , which is θ = π .
Subcase .. Assume that θ = ϕ , then ϕ = π . By Lemma ., for the above , there is
π π
a ray arg z = θ such that θ ∈ [– k
, k )\(E ∪ E ∪ E ∪ E ) (where E and E are deﬁned as
in Lemma ., E ∪ E ∪ E ∪ E is of linear measure zero) satisfying δ(β zk , θ ) > . Since
cos kθ > , we have δ(α zk , θ ) = |α | cos(θ + kθ ) = –|α | cos kθ < . For a suﬃciently large r,
by Lemma ., we get
A (z)eα zk < exp ( – )δ α zk , θ rk < ,
B (z)eβ zk > exp ( – )δ β zk , θ rk .
(.)
(.)
By (.) and (.), we get
A (z)eα zk + B (z)eβ zk ≥ B (z)eβ zk – A (z)eα zk > exp ( – )δ β zk , θ rk – .
(.)
Using the same reasoning as in Subcase ., we can get a contradiction.
Subcase .. Assume that θ = ϕ , then θ = ϕ = π . By Lemma ., for the above , there
π π
is a ray arg z = θ such that θ ∈ [ k
, k )\(E ∪ E ∪ E ∪ E ) (where E and E are deﬁned
as in Lemma ., E ∪ E ∪ E ∪ E is of linear measure zero), then cos kθ < , δ(α zk , θ ) =
|α | cos(θ + kθ ) = –|α | cos kθ > , δ(β zk , θ ) = |β | cos(ϕ + kθ ) = –|β | cos kθ > .
Since |α | ≤ |β |, α = β , and θ = ϕ , then |α | < |β |, thus δ(β zk , θ ) > δ(α zk , θ ) > .
For a suﬃciently large r, we get (.), (.) and (.) hold.
Using the same reasoning as in Subcase ., we can get a contradiction. Thus we have
ρ(f ) ≥ k + .
3.2 The proof of Theorem 1.2
Since Aj (z), B (z) (≡ ) (j = , , . . . , n) are meromorphic functions and max{ρ(Aj ), ρ(B ) :
 ≤ j ≤ n} = α < , then when δ(α z, θ ) > , β(a z, θ ) <  for a suﬃciently large r, by
Lemma ., we have
A (z)eα z ≥ exp ( – )δ(α z, θ )r ,
B (z)eβ z ≤ exp ( – )δ(β z, θ )r < .
(.)
(.)
Then, using a similar argument to that of Theorem ., we obtain a contradiction.
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3.3 The proof of Theorems 1.3 and 1.4
Suppose that f (z) is a nonconstant meromorphic solution of (.) such that ρ(f ) < ∞. We
ﬁrst prove that λ(f – z) = ρ(f ). Submitting f (z) = g(z) + z into (.), we get
k
An (z)g(z + n) + · · · + A (z)g(z + ) + A (z)eα z g(z + )
k
k
+ A (z)eα z + B (z)eβ z g(z) = D(z),
where
k
k
k D(z) = –z An (z) + · · · + A (z) + A (z)eα z + A (z)eα z + B (z)eβ z
k
– nAn (z) + · · · + A (z) + A (z)eα z ≡ .
Since max{ρ(Aj ), ρ(B ) :  ≤ j ≤ n} = α < k, we have ρ(D) ≤ k.
Now, for any given > , applying Lemma . and Theorem ., we can deduce that

m r,
g(z)
k
k
k
An (z)g(z + n) + · · · + A (z)g(z + ) + A (z)eα z g(z + ) + (A (z)eα z + B (z)eβ z )g(z)
= m r,
D(z)g(z)
n
n
g(z + j)
k
≤
m r,
T r, Aj (z) + T r, eα z + T r, B (z)
+
g(z)
j=
j=
k
k
+ T r, eα z + T r, eβ z + T r, D(z) + O(log r)
= O rρ–+ = S(r, g).
This implies that


= N r,
= T(r, g) + S(r, g) = T(r, f ) + S(r, f ),
N r,
f –z
g
then λ(f – z) = ρ(f ) follows.
Next, we assert that either k +  ≤ ρ(f ) ≤ max{λ(f ), λ( f )} +  or ρ(f ) = k + . If the assertion does not hold, we have max{k, λ(f ), λ( f )} +  < ρ(f ) < ∞.
Assume that z =  is a zero (or a pole) of f (z) of order m. Applying the Hadamard factorization of a meromorphic function, we write f (z) as follows:
f (z) = zm
P (z) Q(z)
e ,
P (z)
where P (z), P (z) are entire functions such that ρ(P ) = λ(f ), ρ(P ) = λ( f ) and Q(z) is a
polynomial such that deg Q(z) = q > max{k, λ(f ), λ( f )} + .
Now, we obtain from (.) that
n
j=
hj (z)eQ(z+j) = ,
(.)
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where
P (z)
k
k
,
h = A (z)eα z + B (z)eβ z zm
P (z)
k
h = A (z)eα z (z + )m
P (z + )
,
P (z + )
and
hj (z) = Aj (z)(z + j)m
P (z + j)
P (z + j)
( ≤ j ≤ n + ).
Notice that deg(Q(z + h) – Q(z + l)) = q –  > ρ(hj ) for  ≤ h < l ≤ n and  ≤ j ≤ n. Thus,
Lemma . is valid for (.), hence we get that hj (z) ≡  for j = , , . . . , n + , a contradiction to our assumption. This completes our proof.
The proof of Theorem . is similar to that of Theorem ..
3.4 The proof of Theorem 1.5
Let f (≡ ) be a meromorphic solution of (.). Suppose that ρ(f ) < , then by Lemma .,
for any given > , there exists a subset E ⊂ (, ∞) of ﬁnite logarithmic measure such that
for all z satisfying |z| = r ∈/ [, ] ∪ E, and as r suﬃciently large, we have
f (z + j) ρ(f )–+ (j = , , . . . , n, j = l).
f (z + l) ≤ exp r
(.)
Set z = reiθ , αj = |αj |eiϕj and δ(αj z, θ ) = |αj | cos(ϕj + θ ) (j = , , . . . , k). Then E = {θ ∈ [, π) :
δ(αj z, θ ) = , j = , , . . . , n} ∪ {θ ∈ [, π) : δ(αj z – αi z, θ ) = ,  ≤ i < j ≤ n} is a set of linear
measure zero.
Consider that aj (z) = Aj (z)eαj z , Aj (z) (≡ ) are meromorphic functions and ρ(Aj ) = α < ,
by Lemma ., for the above > , there exists a set Fj ∈ [, π) of linear measure zero
such that for any z = reiθ satisfying θ ∈ [, π)\(E ∪ Fj ), and as r → ∞, we have
(i) if δ(αj z, θ ) > , then
exp ( – )δ(αj z, θ )r ≤ aj reiθ ≤ exp ( + )δ(αj z, θ )r ;
(.)
(ii) if δ(αj z, θ ) < , then
exp ( + )δ(αj z, θ )r ≤ aj reiθ ≤ exp ( – )δ(αj z, θ )r .
(.)
Set E = nj= Fj , then E is a set of linear measure zero.
Since αj are distinct complex constants, then there exists only one l ∈ {, , . . . , n} such
that δ(αl z, θ ) = max{δ(αj z, θ ) : j = , , . . . , n} for any θ ∈ [, π)\(E ∪ E ). Now we take a
ray arg z = θ ∈ [, π)\(E ∪ E ) such that δ(αl z, θ ) > .
Let δ = δ(αl z, θ ), δ = max{δ(αj z, θ ) : j = , , . . . , l, j = l}, then δ > δ . We discuss the
following two cases.
Case . δ > . We rewrite (.) in the form
–al (z) =
n
j=
aj (z)
f (z + j)
f (z + l)
(j = l).
(.)
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By (.), (.) and (.), we get for z = reiθ and suﬃciently large r ∈/ [, ] ∪ E,
n
exp ( + )δ(αj z, θ )r exp rρ(f )–+
exp ( – )δ r ≤ al reiθ ≤
j=,j=l
≤ n exp ( + )δ r exp rρ(f )–+ .
(.)

,  – ρ(f )}, by (.), we get
When  <  < min{ δδ –δ
+δ
δ – δ
r ≤ n exp rρ(f )–+ .
exp

This is impossible.
Case . δ < . By (.), (.) and (.), we get for z = reiθ and suﬃciently large r ∈/
[, ] ∪ E,
n
exp ( – )δ(αj z, θ )r exp rρ(f )–+
exp ( – )δ r ≤ al reiθ ≤
j=,j=l
≤ n exp rρ(f )–+ .
This is a contradiction. Hence we get ρ(f ) ≥ .
3.5 The proof of Theorem 1.6
Considering aj (z) = Aj (z)eαj z + Dj (z) (j = , , . . . , n), Aj (z) (≡ ), Dj (z) (≡ ) are meromorphic functions and ρ(Aj ) = α < , ρ(Dj ) = β < , by Lemmas . and ., we know that for
any given > , there exists a subset E ⊂ (, ∞) of ﬁnite logarithmic measure such that,
for all z satisfying |z| = r ∈/ [, ] ∪ E, and as r → ∞, we have
(i) if δ(αj z, θ ) > , then
exp ( – )δ(αj z, θ )r – exp rβ+
< aj reiθ < exp ( + )δ(αj z, θ )r + exp r β+
<  + o() exp ( + )δ(αj z, θ )r ;
(.)
(ii) if δ(αj z, θ ) < , then
iθ aj re < exp ( – )δ(αj z, θ )r + exp r β+ <  + o() exp rβ+ .
(.)
Then, using a similar argument to that of Theorem . and Theorem . and only replacing
(.) (or (.)) by (.) (or (.)), we can prove Theorem ..
3.6 The proof of Theorem 1.7
Let f (z) ≡  be a meromorphic solution of (.). Suppose that ρ(f ) < , then by Lemma .,
we obtain ρ(F) = ρ( kj= aj (z)f (z + j)) = . This contradicts ρ(F) < , therefore we have
ρ(f ) ≥ .
Suppose that there exist two distinct meromorphic solutions f ≡ , f ≡  of Eq. (.)
such that max{ρ(f ), ρ(f )} < , then f – f is a meromorphic solution of the homogeneous
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linear diﬀerence equation corresponding to (.) and ρ(f – f ) < . By Theorem . or
Theorem ., we get a contradiction. So Eq. (.) has at most one meromorphic solution
f satisfying  ≤ ρ(f ) ≤ .
Next we prove max{λ(f ), λ( f )} = ρ(f ) in the case ρ(f ) = . Suppose that max{λ(f ),
λ( f )} < ρ(f ), then by the Weierstrass factorization, we obtain
f (z) =
P (z) Q(z)
e ,
P (z)
(.)
where P (z), P (z) are entire functions such that ρ(P ) = λ(P ) = λ(f ), ρ(P ) = λ(P ) = λ( f )
and Q(z) is a polynomial of degree .
In the case aj (z) = Aj (z)eαj z . Substituting (.) into (.), we get
n
j=
Aj (z)
P (z + j) αj z+Q(z+j)
e
= F(z).
P (z + j)
(.)
Since αj are distinct complex numbers, by Lemma ., we obtain that the order of the
left-hand side of (.) is . This contradicts ρ(F) < .
For aj (z) = Aj (z)eαj z + Dj (z), by using an argument similar to the above, we also obtain a
contradiction.
It is obvious that max{λ(f ), λ( f )} = ρ(f ) provided that  < ρ(f ) < . Therefore we have
max{λ(f ), λ( f )} = ρ(f ).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
ZLY performed and drafted manuscript. All authors read and approved the ﬁnal manuscript.
Acknowledgements
The author thanks the referee for his/her valuable suggestions to improve the present article. The research was supported
by the Beijing Natural Science Foundation (No. 1132013) and the Foundation of Beijing University of Technology
(No. 006000514313002).
Received: 9 July 2014 Accepted: 14 November 2014 Published: 03 Dec 2014
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Cite this article as: Yuan and Ling: Results on the growth of meromorphic solutions of some linear difference
equations with meromorphic coefﬁcients. Advances in Diﬀerence Equations 2014, 2014:306
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