# The Fixed Points of Solutions of Someq-Difference Equations

Authors: Xiu-Min Zheng; Hong-Yan Xu; Jun-Feng Xu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

The purpose of this paper is to investigate the fixed points of solutionsf
(
z
) of some q-difference equations and obtain some results about the exponents of convergence of fixed points of f
(
z
) and f
(
q
j
z
)
(
j
∈
N
+
), q-differences Δ
q
f
(
z
)
=
f
(
q
z
)
-
f
(
z
), and q-divided differences Δ
q
f
(
z
)
/
f
(
z
).


---

## Body

## 1. Introduction and Main Results

Throughout this paper, we will assume that the readers are familiar with basic notations such asm
(
r
,
f
), N
(
r
,
f
), and T
(
r
,
f
) of Nevanlinna theory (see Hayman [1], Yang [2], and Yang and Yi [3]). We use ρ
(
f
), λ
(
f
), and λ
(
1
/
f
) to denote the order, the exponent of convergence of zeros, and the exponent of convergence of poles of f
(
z
), respectively, and we also use the notation τ
(
f
) to denote the exponent of convergence of fixed points of f
(
z
), which is defined as
(1)
τ
(
f
)
=
limsup
⁡
r
→
∞
log
⁡
N
(
r
,
1
/
(
f
(
z
)
-
z
)
)
log
⁡
r
,
and S
(
r
,
f
) to denote any quantity satisfying S
(
r
,
f
)
=
o
(
T
(
r
,
f
)
) for all r on a set F of logarithmic density 1, where the logarithmic density of a set F is defined by
(2)
lim
⁡
r
→
∞
1
log
⁡
r
∫
[
1
,
r
]
∩
F
‍
1
t
d
t
.
Throughout this paper, the set F of logarithmic density 1 will be not necessarily the same at each occurrence.Recently, a number of papers (including [4–9]) focused on complex difference equations, system of complex difference equations, and difference analogues of Nevanlinna theory. Correspondingly, there are many papers focusing on the q-difference (or q-shift difference) equations, such as  [10–16].In 2013, Zhang [17] investigated the growth of meromorphic solutions of some complex q-difference equations and the exponents of convergence of fixed points and zeros of transcendental meromorphic solutions of the second order q-difference equation and obtained the following theorem.Theorem 1 (see [17]).
Suppose thatf
(
z
) is a transcendental meromorphic solution of the equation
(3)
f
(
q
2
z
)
+
γ
1
f
(
q
z
)
=
α
0
+
α
1
f
(
z
)
+
α
2
f
2
(
z
)
β
0
+
β
1
f
(
z
)
+
β
2
f
2
(
z
)
,
where |
q
|
<
1, coefficients γ
1, α
0, α
1, α
2, β
0, β
1, and β
2 are constants, and at least one of α
2
,
β
2 is nonzero. Then, ρ
(
f
)
=
0 and (i) f
(
z
) has infinitely many fixed points, and (ii) f
(
z
) has infinitely many zeros, whenever α
0
≠
0.Our first result of this paper is about the exponents of convergence of fixed points and zeros of transcendental meromorphic solutions of the higher orderq-difference equation as follows.Theorem 2.
Suppose thatf
(
z
) is a transcendental meromorphic solution of the equation
(4)
f
(
q
n
z
)
+
∑
t
=
1
n
-
1
‍
γ
t
f
(
q
t
z
)
=
∑
j
=
0
n
α
j
f
j
(
z
)
∑
j
=
0
n
β
j
f
j
(
z
)
,
where q
∈
C, |
q
|
<
1, coefficients γ
t (t
=
1
,
…
,
n
-
1), α
j, β
j, (j
=
0
,
…
,
n), are constants, and at least one of α
n
,
β
n is nonzero. Then, ρ
(
f
)
=
0 and (i) f
(
z
) has infinitely many fixed points, and (ii) f
(
z
) has infinitely many zeros, whenever α
0
≠
0.From Theorem2, it is a natural question to ask, What will happen if the right-hand side of (4) is a rational function in both arguments?Regarding the above question, we will investigate the exponents of convergence of fixed points of meromorphic solutions of theq-difference equation
(5)
f
(
q
z
)
=
R
(
z
)
f
(
z
)
Q
(
z
)
+
P
(
z
)
f
(
z
)
,
where P
(
z
), Q
(
z
), and R
(
z
) are nonzero polynomials, q
∈
C, and |
q
|
≠
0,1. Similar to [18, Page 99], we can call (5) a q-Pielou logistic equation, which is a special form of nonautonomous Schröder equations.Theorem 3.
LetP
(
z
), Q
(
z
), and R
(
z
) be nonzero polynomials such that
(6)
deg
⁡
P
(
z
)
≥
max
⁡
{
deg
⁡
R
(
z
)
,
deg
⁡
Q
(
z
)
,
1
}
.
Set Δ
q
f
(
z
)
=
f
(
q
z
)
-
f
(
z
), where q
∈
C and |
q
|
≠
0,1. Then every transcendental meromorphic solution f
(
z
) of (5) satisfies the following statements:(i)
f
(
q
j
z
) has infinitely many fixed points and τ
(
f
(
q
j
z
)
)
=
ρ
(
f
), (
j
=
0,1
,
2
,
…
);
(ii)
ifR
(
z
)
-
(
z
+
1
)
Q
(
z
)
≢
0, then Δ
q
f
(
z
)
/
f
(
z
) has infinitely many fixed points and τ
(
Δ
q
f
/
f
)
=
ρ
(
f
).We also study fixed points of transcendental meromorphic solutions of the followingq-difference equations:
(7)
a
n
(
z
)
f
(
q
n
z
)
+
⋯
+
a
1
(
z
)
f
(
q
z
)
+
a
0
(
z
)
f
(
z
)
=
0
,
(8)
a
n
(
z
)
f
(
q
n
z
)
+
⋯
+
a
1
(
z
)
f
(
q
z
)
+
a
0
(
z
)
f
(
z
)
=
F
(
z
)
,
where 0
<
|
q
|
<
1, a
j
(
z
) (j
=
0,1
,
…
,
n), and F
(
z
) are polynomials and a
n
(
z
)
a
0
(
z
)
≢
0, and obtain the following results.Theorem 4.
Letq
∈
C, 0
<
|
q
|
<
1, let a
j
(
z
) (j
=
0,1
,
…
,
n) be polynomials, and let a
n
(
z
)
a
0
(
z
)
≢
0. If a
0
(
z
)
,
a
1
(
z
)
,
…
,
a
n
(
z
) satisfy one of the following conditions:(i)
there exists an integers (0
≤
s
≤
n) such that
(9)
deg
⁡
a
s
(
z
)
>
max
⁡
{
deg
⁡
a
j
(
z
)
,
j
=
0,1
,
…
,
n
,
j
≠
s
}
;
(ii)
(10)
q
n
a
n
(
z
)
+
⋯
+
q
a
1
(
z
)
+
a
0
(
z
)
≢
0
,then every transcendental meromorphic solution f
(
z
) of (7) satisfies that f
(
q
j
z
) has infinitely many fixed points and τ
(
f
(
q
j
z
)
)
=
ρ
(
f
) for j
∈
N.By using the same argument as that in Theorem4, we can easily obtain the following theorem.Theorem 5.
Letq
∈
C, 0
<
|
q
|
<
1, a
j
(
z
) (j
=
0,1
,
…
,
n), and F
(
z
) be polynomials and let a
n
(
z
)
a
0
(
z
)
≢
0. If a
0
(
z
)
,
a
1
(
z
)
,
…
,
a
n
(
z
)
,
F
(
z
) satisfy one of the following conditions:(i)
a
0
(
z
)
,
a
1
(
z
)
,
…
,
a
n
(
z
) and F
(
z
) contain just one term of maximal total degree;
(ii)
(11)
q
n
a
n
(
z
)
+
⋯
+
q
a
1
(
z
)
+
a
0
(
z
)
-
F
(
z
)
≢
0
,then every transcendental meromorphic solution f
(
z
) of (8) satisfies that f
(
q
j
z
) has infinitely many fixed points and τ
(
f
(
q
j
z
)
)
=
ρ
(
f
) for j
∈
N.


## 2. Some Lemmas

The following result is a difference counterpart to the standard result due to A. A. Mohon’ko and V. D. Mohon’ko [19].Lemma 6 (see [20], Theorem 2.2).
Letf
(
z
) be a nonconstant zero-order meromorphic solution of P
(
z
,
f
)
=
0, where P
(
z
,
f
) is a q-difference polynomial in f
(
z
). If P
(
z
,
a
)
≢
0 for a slowly moving target a
(
z
), then
(12)
m
(
r
,
1
f
-
a
)
=
S
(
r
,
f
)
,
on a set of logarithmic density 1.Lemma 7 (see [21, 22]).
Leta
j
(
z
), j
=
0,1
,
…
,
n, and Q
(
z
) be rational functions, and let a
0
(
z
)
≢
0, a
n
(
z
)
≡
1, and q (0
<
|
q
|
<
1). Then(i)
all meromorphic solutions of the equation(13)
∑
j
=
0
n
‍
a
j
(
z
)
f
(
q
j
z
)
=
Q
(
z
)
satisfy T
(
r
,
f
)
=
O
(
(
log
⁡
r
)
2
);
(ii)
all transcendental meromorphic solutions of (13) satisfy (
log
⁡
r
)
2
=
O
(
T
(
r
,
f
)
).Lemma 8 (see [17], Theorem 2).
Suppose thatf
(
z
) is a nonconstant meromorphic solution of the equation
(14)
∑
j
=
1
n
‍
γ
j
(
z
)
f
(
q
j
z
)
=
R
(
z
,
f
(
z
)
)
=
∑
i
=
0
s
α
i
(
z
)
f
i
(
z
)
∑
i
=
0
t
β
i
(
z
)
f
i
(
z
)
,
where q (0
<
|
q
|
<
1) is a complex number, α
j
(
z
) (j
=
0,1
,
…
,
s), α
s
(
z
)
≢
0, β
j
(
z
) (j
=
0,1
,
…
,
t), β
t
(
z
)
≢
0, γ
n
(
z
)
≡
1, and γ
j
(
z
) (j
=
0,1
,
…
,
n) are small functions of f
(
z
), and R
(
z
,
f
) is irreducible in f
(
z
). Then, d
=
max
⁡
{
s
,
t
}
≤
n and ρ
(
f
)
≤
(
log
⁡
n
-
log
⁡
d
)
/
-
log
⁡
|
q
|.Lemma 9 (see [21, page 249] or [23, Theorem 1.1]).
Letf
(
z
) be a transcendental meromorphic function of zero-order and let q be a nonzero complex constant. Then
(15)
T
(
r
,
f
(
q
z
)
)
=
T
(
r
,
f
)
+
S
(
r
,
f
)
on a set of logarithmic density 1.


## 3. Proof of Theorem2

Suppose thatf
(
z
) is a transcendental meromorphic solution of (4). From the assumptions of Theorem 2, it follows from Lemma 8 that ρ
(
f
)
≤
0
=
(
log
⁡
n
-
log
⁡
n
)
/
-
log
⁡
|
q
|. Thus, ρ
(
f
)
=
0. Clearly, we have λ
(
f
)
=
τ
(
f
)
=
ρ
(
f
)
=
0.(i) Firstly, we prove thatf
(
z
) has infinitely many fixed points. Set g
(
z
)
=
f
(
z
)
-
z. Then g
(
z
) is transcendental, T
(
r
,
g
)
=
T
(
r
,
f
)
+
O
(
log
⁡
r
), and S
(
r
,
f
)
=
S
(
r
,
g
). So, g
(
z
) is of zero-order. Then substituting f
(
z
)
=
g
(
z
)
+
z into (4), we get that
(16)
g
(
q
n
z
)
+
∑
t
=
1
n
-
1
‍
γ
t
g
(
q
t
z
)
+
q
n
z
+
∑
t
=
1
n
-
1
‍
γ
t
q
t
z
=
∑
j
=
0
n
α
j
(
g
(
z
)
+
z
)
j
∑
j
=
0
n
β
j
(
g
(
z
)
+
z
)
j
.
Set A
(
z
)
=
g
(
q
n
z
)
+
∑
t
=
1
n
-
1
γ
t
g
(
q
t
z
)
+
q
n
z
+
∑
t
=
1
n
-
1
γ
t
q
t
z and
(17)
P
1
(
z
,
g
(
z
)
)
:
=
A
(
z
)
∑
j
=
0
n
‍
β
j
(
g
(
z
)
+
z
)
j
-
∑
j
=
0
n
‍
α
j
(
g
(
z
)
+
z
)
j
.
It follows from (17) that
(18)
P
1
(
z
,
0
)
=
(
q
n
+
∑
t
=
1
n
-
1
‍
γ
t
q
t
)
∑
j
=
0
n
‍
β
j
z
j
+
1
-
∑
j
=
0
n
‍
α
j
z
j
=
(
q
n
+
∑
t
=
1
n
-
1
‍
γ
t
q
t
)
β
n
z
n
+
1
+
∑
j
=
0
n
-
1
‍
[
(
q
n
+
∑
t
=
1
n
-
1
‍
γ
t
q
t
)
β
j
-
α
j
+
1
]
z
j
+
1
-
α
0
.Suppose thatP
1
(
z
,
0
)
≡
0. If q
n
+
∑
t
=
1
n
-
1
γ
t
q
t
=
0, then it follows from (18) that α
0
=
α
1
=
⋯
=
α
n
=
0. Thus, the right-hand side of (4) is 0, which is in contradiction with the assumption of Theorem 2. If q
n
+
∑
t
=
1
n
-
1
γ
t
q
t
≠
0, it follows from (18) that β
n
=
α
0
=
0 and
(19)
α
j
+
1
β
j
=
q
n
+
∑
t
=
1
n
-
1
‍
γ
t
q
t
,
j
=
0,1
,
…
,
n
-
1
.
Thus, we have from (4) and (19) that
(20)
f
(
q
n
z
)
+
∑
t
=
1
n
-
1
‍
γ
t
f
(
q
t
z
)
=
(
q
n
+
∑
t
=
1
n
-
1
‍
γ
t
q
t
)
f
(
z
)
,
which is in contradiction with the assumption of Theorem 2. Hence, we have P
1
(
z
,
0
)
≢
0. By Lemma 6, we get that
(21)
m
(
r
,
1
g
)
=
S
(
r
,
g
)
=
S
(
r
,
f
)
on a set of logarithmic density 1. Thus, it follows from (21) that
(22)
N
(
r
,
1
f
-
z
)
=
N
(
r
,
1
g
)
=
T
(
r
,
f
)
+
S
(
r
,
f
)
on a set of logarithmic density 1. Since f
(
z
) is a transcendental meromorphic solution of (4), then it follows from (22) that f
(
z
) has infinitely many fixed points.(ii) From (4), we have
(23)
P
2
(
z
,
f
(
z
)
)
∶
=
[
f
(
q
n
z
)
+
∑
t
=
1
n
-
1
‍
γ
t
f
(
q
t
z
)
]
∑
j
=
0
n
‍
β
j
f
j
(
z
)
-
∑
j
=
0
n
‍
α
j
f
j
(
z
)
.
Since α
0
≠
0 and from (23), we derive that
(24)
P
2
(
z
,
0
)
=
α
0
≢
0
.
Thus, it follows from Lemma 6 that
(25)
m
(
r
,
1
f
)
=
S
(
r
,
f
)
on a set of logarithmic density 1; that is,
(26)
N
(
r
,
1
f
)
=
T
(
r
,
f
)
+
S
(
r
,
f
)
on a set of logarithmic density 1. Since f
(
z
) is a transcendental solution of (4), then it follows from (26) that f
(
z
) has infinitely many zeros.Thus, this completes the proof of Theorem2.


## 4. Proof of Theorem3

Suppose thatf
(
z
) is a transcendental meromorphic solution of (5). Since q
∈
C, |
q
|
≠
0,1, and P
(
z
), Q
(
z
), and R
(
z
) are polynomials, it follows from Lemma 8 and [11] that f
(
z
) is of zero-order.(i) We first prove thatf
(
z
) has infinitely many fixed points and τ
(
f
)
=
ρ
(
f
). Set g
(
z
)
=
f
(
z
)
-
z. Then g
(
z
) is transcendental, T
(
r
,
g
)
=
T
(
r
,
f
)
+
O
(
log
⁡
r
), and S
(
r
,
g
)
=
S
(
r
,
f
). Then it follows that g
(
z
) is of zero-order. Set
(27)
P
3
(
z
,
f
(
z
)
)
∶
=
P
(
z
)
f
(
z
)
f
(
q
z
)
+
f
(
q
z
)
Q
(
z
)
-
R
(
z
)
f
(
z
)
≡
0
.
Then substituting f
(
z
)
=
g
(
z
)
+
z into (27), we have
(28)
P
4
(
z
,
g
(
z
)
)
=
P
(
z
)
(
g
(
z
)
+
z
)
(
g
(
q
z
)
+
q
z
)
+
Q
(
z
)
(
g
(
q
z
)
+
q
z
)
-
R
(
z
)
(
g
(
z
)
+
z
)
=
0
.
It follows from (28) that
(29)
P
4
(
z
,
0
)
=
q
z
2
P
(
z
)
+
q
z
Q
(
z
)
-
z
R
(
z
)
.
Thus, we derive by (6) and (29) that P
4
(
z
,
0
)
≢
0. Thus, by Lemma 6 and P
4
(
z
,
0
)
≢
0, we have
(30)
m
(
r
,
1
g
)
=
S
(
r
,
g
)
=
S
(
r
,
f
)
on a set of logarithmic density 1; that is,
(31)
N
(
r
,
1
f
-
z
)
=
N
(
r
,
1
g
)
=
T
(
r
,
f
)
+
S
(
r
,
f
)
on a set of logarithmic density 1.Sincef
(
z
) is a transcendental meromorphic solution of (5), then it follows from (31) that f
(
z
) has infinitely many fixed points.Next, we prove thatf
(
q
z
) has infinitely many fixed points and τ
(
f
(
q
z
)
)
=
ρ
(
f
). From (5), we have
(32)
f
(
q
z
)
-
z
=
(
R
(
z
)
-
z
P
(
z
)
)
f
(
z
)
-
z
Q
(
z
)
Q
(
z
)
+
P
(
z
)
f
(
z
)
=
(
R
(
z
)
-
z
P
(
z
)
)
[
f
(
z
)
-
z
Q
(
z
)
/
(
R
(
z
)
-
z
P
(
z
)
)
]
Q
(
z
)
+
P
(
z
)
f
(
z
)
.
By (6), we have R
(
z
)
-
z
P
(
z
)
≢
0. Since f
(
z
) is transcendental and P
(
z
), Q
(
z
), and R
(
z
) are polynomials, we have by (32) the fact that f
(
z
)
-
z
Q
(
z
)
/
(
R
(
z
)
-
z
P
(
z
)
) and Q
(
z
)
+
P
(
z
)
f
(
z
) have the same poles, except possibly finitely many poles. Moreover, we can get that (
R
(
z
)
-
z
P
(
z
)
)
f
(
z
)
-
z
Q
(
z
) and Q
(
z
)
+
P
(
z
)
f
(
z
) have at most finitely many common zeros. In fact, suppose that z
0 is a common zero of (
R
(
z
)
-
z
P
(
z
)
)
f
(
z
)
-
z
Q
(
z
) and Q
(
z
)
+
P
(
z
)
f
(
z
). Then (
R
(
z
0
)
-
z
0
P
(
z
0
)
)
f
(
z
0
)
-
z
0
Q
(
z
0
)
=
0; that is, f
(
z
0
)
=
z
0
Q
(
z
0
)
/
(
R
(
z
0
)
-
z
0
P
(
z
0
)
). Substituting it into Q
(
z
0
)
+
P
(
z
0
)
f
(
z
0
), we have
(33)
z
0
Q
(
z
0
)
R
(
z
0
)
-
z
0
P
(
z
0
)
P
(
z
0
)
+
Q
(
z
0
)
=
R
(
z
0
)
Q
(
z
0
)
R
(
z
0
)
-
z
0
P
(
z
0
)
=
0
.
Thus, this shows that z
0 must be the zeros of R
(
z
)
Q
(
z
)
/
(
R
(
z
)
-
z
P
(
z
)
). Since P
(
z
), Q
(
z
), and R
(
z
) are polynomials, then R
(
z
)
Q
(
z
)
/
(
R
(
z
)
-
z
P
(
z
)
) has only finitely many zeros. So, f
(
z
)
-
z
Q
(
z
)
/
(
R
(
z
)
-
z
P
(
z
)
) and Q
(
z
)
+
P
(
z
)
f
(
z
) have at most finitely many common zeros. Then it follows from (32) that
(34)
τ
(
f
(
q
z
)
)
=
λ
(
f
(
q
z
)
-
z
)
=
λ
(
f
(
z
)
-
z
Q
(
z
)
R
(
z
)
-
z
P
(
z
)
)
.From (27), we have
(35)
P
3
(
z
,
z
Q
(
z
)
R
(
z
)
-
z
P
(
z
)
)
=
P
(
z
)
z
Q
(
z
)
R
(
z
)
-
z
P
(
z
)
q
z
Q
(
q
z
)
R
(
q
z
)
-
q
z
P
(
q
z
)
+
q
z
Q
(
q
z
)
R
(
q
z
)
-
q
z
P
(
q
z
)
Q
(
z
)
-
R
(
z
)
z
Q
(
z
)
R
(
z
)
-
z
P
(
z
)
=
(
R
(
q
z
)
q
z
2
P
(
q
z
)
Q
(
z
)
R
(
z
)
+
q
z
Q
(
q
z
)
Q
(
z
)
R
(
z
)
h
-
z
Q
(
z
)
R
(
z
)
R
(
q
z
)

q
z
2
P
(
q
z
)
Q
(
z
)
R
(
z
)
)
×
(
(
R
(
z
)
-
z
P
(
z
)
)
(
R
(
q
z
)
-
q
z
P
(
q
z
)
)
)
-
1
.
Since deg
⁡
P
(
z
)
≥
max
⁡
{
deg
⁡
R
(
z
)
,
deg
⁡
Q
(
z
)
} and deg
⁡
P
(
q
z
)
=
deg
⁡
P
(
z
), then we have deg
⁡
{
q
z
2
P
(
q
z
)
Q
(
z
)
R
(
z
)
+
q
z
Q
(
q
z
)
Q
(
z
)
R
(
z
)
-
z
Q
(
z
)
R
(
z
)
R
(
q
z
)
}
≥
1. Thus, it follows from (35) that P
3
(
z
,
z
Q
(
z
)
/
(
R
(
z
)
-
z
P
(
z
)
)
)
≢
0. Since f
(
z
) is transcendental function of zero-order and z
Q
(
z
)
/
(
R
(
z
)
-
z
P
(
z
)
) is a rational function, then we have by Lemma 6 the fact that
(36)
m
(
r
,
1
f
(
z
)
-
z
Q
(
z
)
/
(
R
(
z
)
-
z
P
(
z
)
)
)
=
S
(
r
,
f
)
on a set of logarithmic density 1; that is,
(37)
N
(
r
,
1
f
(
z
)
-
z
Q
(
z
)
/
(
R
(
z
)
-
z
P
(
z
)
)
)
=
T
(
r
,
f
)
+
S
(
r
,
f
)
on a set of logarithmic density 1. Since f
(
z
) is transcendental, we can derive from (34) and (37) that f
(
q
z
) has infinitely many fixed points and τ
(
f
(
q
z
)
)
=
ρ
(
f
).Now, we prove thatf
(
q
2
z
) has infinitely many fixed points and τ
(
f
(
q
2
z
)
)
=
ρ
(
f
). From (5), we have
(38)
f
1
(
q
z
)
=
R
(
q
z
)
f
1
(
z
)
Q
(
q
z
)
+
P
(
q
z
)
f
1
(
z
)
,
where f
1
(
z
)
=
f
(
q
z
). By Lemma 9, we have ρ
(
f
1
)
=
ρ
(
f
)
=
0. Obviously, deg
⁡
P
(
q
z
)
=
deg
⁡
P
(
z
)
≥
1, deg
⁡
R
(
q
z
)
=
deg
⁡
R
(
z
), and deg
⁡
Q
(
q
z
)
=
deg
⁡
Q
(
z
). Thus, by using the same argument as in the proof of τ
(
f
(
q
z
)
)
=
ρ
(
f
), we can prove that f
1
(
q
z
)
=
f
(
q
2
z
) has infinitely many fixed points and τ
(
f
(
q
2
z
)
)
=
τ
(
f
1
(
q
z
)
)
=
ρ
(
f
1
)
=
ρ
(
f
).Thus, by using the same method as above, we can obtain thatf
(
q
j
z
) has infinitely many fixed points and τ
(
f
(
q
j
z
)
)
=
ρ
(
f
) for j
=
0,1
,
….(ii) Now, we prove thatΔ
q
f
(
z
)
/
f
(
z
) has infinitely many fixed points and
(39)
τ
(
Δ
q
f
f
)
=
ρ
(
f
)
.
By (5) and from R
(
z
)
-
(
z
+
1
)
Q
(
z
)
≢
0, we have
(40)
Δ
q
f
(
z
)
f
(
z
)
-
z
=
f
(
q
z
)
-
f
(
z
)
f
(
z
)
-
z
=
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
-
(
z
+
1
)
P
(
z
)
f
(
z
)
Q
(
z
)
+
P
(
z
)
f
(
z
)
=
-
(
z
+
1
)
P
(
z
)
×
(
f
(
z
)
-
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
(
z
+
1
)
P
(
z
)
)
×
(
Q
(
z
)
+
P
(
z
)
f
(
z
)
)
-
1
.
Since R
(
z
)
-
(
z
+
1
)
Q
(
z
)
≢
0, f
(
z
) is transcendental, and P
(
z
), Q
(
z
), and R
(
z
) are polynomials, we have by (40) the fact that f
(
z
)
-
(
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
)
/
(
z
+
1
)
P
(
z
) and Q
(
z
)
+
P
(
z
)
f
(
z
) have the same poles, except possibly finitely many poles. Moreover, by using the same argument as in (i), we can get that R
(
z
)
-
(
z
+
1
)
Q
(
z
)
-
(
z
+
1
)
P
(
z
)
f
(
z
) and Q
(
z
)
+
P
(
z
)
f
(
z
) have at most finitely many common zeros. Then it follows from (40) that
(41)
τ
(
Δ
q
f
f
)
=
λ
(
Δ
q
f
f
-
z
)
=
λ
(
f
(
z
)
-
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
(
z
+
1
)
P
(
z
)
)
.
From (27), we have
(42)
P
3
(
z
,
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
(
z
+
1
)
P
(
z
)
)
=
P
(
z
)
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
(
z
+
1
)
P
(
z
)
R
(
q
z
)
-
(
q
z
+
1
)
Q
(
q
z
)
(
q
z
+
1
)
P
(
q
z
)
+
R
(
q
z
)
-
(
q
z
+
1
)
Q
(
q
z
)
(
q
z
+
1
)
P
(
q
z
)
Q
(
z
)
-
R
(
z
)
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
(
z
+
1
)
P
(
z
)
∶
=
B
(
z
)
(
q
z
+
1
)
(
z
+
1
)
P
(
q
z
)
P
(
z
)
,
where
(43)
B
(
z
)
=
(
q
z
+
1
)
(
z
+
1
)
P
(
q
z
)
Q
(
z
)
R
(
z
)
-
(
q
z
+
1
)
P
(
z
)
Q
(
q
z
)
R
(
z
)
-
(
q
z
+
1
)
P
(
q
z
)
R
2
(
z
)
+
P
(
z
)
R
(
z
)
R
(
q
z
)
.
Since P
(
z
), Q
(
z
), and R
(
z
) are polynomials satisfying (6), then it follows from (43) that B
(
z
) is a polynomial of degree t
≥
1. Thus, from (42) we have P
3
(
z
,
(
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
)
/
(
z
+
1
)
P
(
z
)
)
≢
0. Since f
(
z
) is transcendental function of zero-order and (
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
)
/
(
z
+
1
)
P
(
z
) is a rational function, then we have by Lemma 6 the fact that
(44)
m
(
r
,
1
f
(
z
)
-
(
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
)
/
(
z
+
1
)
P
(
z
)
)
=
S
(
r
,
f
)
on a set of logarithmic density 1; that is,
(45)
N
(
r
,
1
f
(
z
)
-
(
R
(
z
)
-
(
z
+
1
)
Q
(
z
)
)
/
(
z
+
1
)
P
(
z
)
)
=
T
(
r
,
f
)
+
S
(
r
,
f
)
on a set of logarithmic density 1. Since f
(
z
) is transcendental, we can derive from (41) and (45) that Δ
q
f
(
z
)
/
f
(
z
) has infinitely many fixed points and τ
(
Δ
q
f
/
f
)
=
ρ
(
f
).Thus, this completes the proof of Theorem3.


## 5. Proof of Theorem4

Suppose thatf
(
z
) is a transcendental meromorphic solution of (7). Since q
∈
C, 0
<
|
q
|
<
1, and a
j
(
z
), j
=
0,1
,
…
,
n, are polynomials, by Lemma 7, we see that f
(
z
) is of zero-order. Set
(46)
P
5
(
z
,
f
(
z
)
)
∶
=
a
n
(
z
)
f
(
q
n
z
)
+
⋯
+
a
1
(
z
)
f
(
q
z
)
+
a
0
(
z
)
f
(
z
)
=
0
.
Thus, it follows from (46) that
(47)
P
5
(
z
,
z
)
=
a
n
(
z
)
q
n
z
+
⋯
+
a
1
(
z
)
q
z
+
a
0
(
z
)
z
=
z
[
q
n
a
n
(
z
)
+
⋯
+
q
a
1
(
z
)
+
a
0
(
z
)
]
.(i) Suppose thata
0
(
z
)
,
…
,
a
n
(
z
) satisfy condition (9). Then it follows that P
5
(
z
,
z
)
≢
0. Since f
(
z
) is a transcendental solution of zero-order, then it follows from Lemma 6 that
(48)
m
(
r
,
1
f
-
z
)
=
S
(
r
,
f
)
on a set of logarithmic density 1. So,
(49)
N
(
r
,
1
f
-
z
)
=
T
(
r
,
f
)
+
S
(
r
,
f
)
on a set of logarithmic density 1. Thus, it follows that f
(
z
) has infinitely many fixed points and τ
(
f
)
=
ρ
(
f
).Now, we prove thatf
(
q
z
) has infinitely many fixed points and τ
(
f
(
q
z
)
)
=
ρ
(
f
). By (7), we derive
(50)
a
n
(
q
z
)
f
1
(
q
n
z
)
+
⋯
+
a
1
(
q
z
)
f
1
(
q
z
)
+
a
0
(
q
z
)
f
1
(
z
)
=
0
,
where f
1
(
z
)
=
f
(
q
z
). Since f
(
z
) is a transcendental meromorphic function of zero-order, then we have by Lemma 9 the fact that f
1
(
z
) is a transcendental and ρ
(
f
1
)
=
ρ
(
f
). By deg
⁡
a
j
(
q
z
)
=
deg
⁡
a
j
(
z
), j
=
0,1
,
…
,
n, and (9), we have
(51)
deg
⁡
a
s
(
q
z
)
=
deg
⁡
a
s
(
z
)
>
max
⁡
{
a
j
(
q
z
)
,
j
=
0,1
,
…
,
n
,
j
≠
s
}
.
Thus, by the above proof of τ
(
f
)
=
ρ
(
f
), we see that f
1
(
z
)
=
f
(
q
z
) has infinitely many fixed points and τ
(
f
1
)
=
τ
(
f
(
q
z
)
)
=
ρ
(
f
1
)
=
ρ
(
f
). Continuing to use the same method as the above, we can prove that f
(
q
j
z
) has infinitely many fixed points and τ
(
f
(
q
j
z
)
)
=
ρ
(
f
) for j
=
0,1
,
….(ii) Suppose thata
0
(
z
)
,
…
,
a
n
(
z
) satisfy the condition (10).By using the same argument as the one above, we can prove thatf
(
z
) has infinitely many fixed points and τ
(
f
)
=
ρ
(
f
) easily.Now, we prove thatf
(
q
z
) has infinitely many fixed points and τ
(
f
(
q
z
)
)
=
ρ
(
f
). Set
(52)
P
6
(
z
,
f
1
(
z
)
)
∶
=
a
n
(
q
z
)
f
1
(
q
n
z
)
+
⋯
+
a
1
(
q
z
)
f
1
(
q
z
)
+
a
0
(
q
z
)
f
1
(
z
)
=
0
.
Thus, it follows from (10) that
(53)
P
6
(
z
,
z
)
=
z
[
q
n
a
n
(
q
z
)
+
⋯
+
q
a
1
(
q
z
)
+
a
0
(
q
z
)
]
≢
0
.
In fact, if P
6
(
z
,
z
)
≡
0, replacing z by z
/
q into (53), we have
(54)
P
6
(
z
q
,
z
q
)
=
z
q
[
q
n
a
n
(
z
)
+
⋯
+
q
a
1
(
z
)
+
a
0
(
z
)
]
≡
0
,
which is in contradiction with the condition (10). Since f
1
(
z
)
=
f
(
q
z
) and f
(
z
) is transcendental meromorphic of zero-order, then it follows from (53) and Lemma 6 that f
1
(
z
)
=
f
(
q
z
) has infinitely many fixed points and τ
(
f
1
)
=
τ
(
f
(
q
z
)
)
=
ρ
(
f
1
)
=
ρ
(
f
). Continuing to use the same method as the one above, we can prove that f
(
q
j
z
) has infinitely many fixed points and τ
(
f
(
q
j
z
)
)
=
ρ
(
f
) for j
=
0,1
,
….Thus, this completes the proof of Theorem4.



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*Source: 262570-2014-10-14.xml*