On the growth of meromorphic solutions of difference equation
We estimate the order of growth of meromorphic solutions of some linear difference equations and study the relationship between the exponent of convergence of zeros and the order of growth of the entire solutions of linear difference equations.
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| author | Chen, Zong-Xuan Lan, Shuang-Ting Чень, Цзон-Сюань Лан, Шуан-Тін |
| author_facet | Chen, Zong-Xuan Lan, Shuang-Ting Чень, Цзон-Сюань Лан, Шуан-Тін |
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| description | We estimate the order of growth of meromorphic solutions of some linear difference equations and study the relationship
between the exponent of convergence of zeros and the order of growth of the entire solutions of linear difference equations. |
| first_indexed | 2026-03-24T02:15:40Z |
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UDC 517.9
Shuang-Ting Lan (Dep. Math., Guangzhou Civil Aviation College, China),
Zong-Xuan Chen (School Math. Sci., South China Normal Univ., China)
ON THE GROWTH OF MEROMORPHIC SOLUTIONS
OF DIFFERENCE EQUATION*
ПОРЯДОК РОСТУ МЕРОМОРФНИХ РОЗВ’ЯЗКIВ РIЗНИЦЕВОГО РIВНЯННЯ
We estimate the order of growth of meromorphic solutions of some linear difference equations and study the relationship
between the exponent of convergence of zeros and the order of growth of the entire solutions of linear difference equations.
Оцiнено порядок росту мероморфних розв’язкiв деяких лiнiйних рiзницевих рiвнянь та вивчено спiввiдношення
мiж показниками збiжностi нулiв та порядком зростання цiлих розв’язкiв лiнiйних рiзницевих рiвнянь.
1. Introduction and results. In this paper, we use the basic notions of Nevanlinna’s theory (see
[8, 12, 13]). In addition, we use the notations \sigma (f) to denote the order of growth of the meromorphic
function f(z), and \lambda (f) to denote the exponent of convergence of zeros of f(z).
Recently, many results of complex differences and difference equations are rapidly obtained
(see [1 – 3, 5, 7, 9, 10]). Chiang and Feng [7] studied the growth of meromorphic solutions of
homogeneous linear difference equation, when there exists only one coefficient having the maximal
order, they obtained the following result.
Theorem A. Let A0(z), . . . , An(z) be entire functions such that there exists an integer l, 0 \leq
\leq l \leq n, such that
\sigma (Al) > \mathrm{m}\mathrm{a}\mathrm{x}
1\leq j\leq n
j \not =l
\{ \sigma (Aj)\} .
If f(z) is a meromorphic solution to
An(z)y(z + n) + . . .+A1(z)y(z + 1) +A0(z)y(z) = 0,
then we have \sigma (f) \geq \sigma (Al) + 1.
Laine and Yang [11] obtained that when the dominating coefficient depending on type but not
order, Theorem A still holds. The result may be stated as follows.
Theorem B. Let A0(z), . . . , An(z) be entire functions of finite order such that among those
coefficients having the maximal order \sigma = \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (Ak), 0 \leq k \leq n\} , exactly one has its type strictly
greater than the others. If f(z) \not \equiv 0 is a meromorphic solution of equation
An(z)f(z + \omega n) + . . .+A1(z)f(z + \omega 1) +A0(z)f(z) = 0, (1.1)
then \sigma (f) \geq \sigma + 1.
Laine and Yang [11] raised the following question.
Question: Whether all meromorphic solutions f(z)(\not \equiv 0) of equation (1.1) satisfy \sigma (f) \geq
\geq 1 + \mathrm{m}\mathrm{a}\mathrm{x}0\leq j\leq n \sigma (Aj), if there is no dominating coefficient.
Giving some restriction on the coefficients of difference equation, we answer this question and
obtain the following results.
* This paper was supported by the Natural Science Foundation of Guangdong Province in China (№ 2016A030310106,
2014A030313422).
c\bigcirc SHUANG-TING LAN, ZONG-XUAN CHEN, 2016
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 11 1561
1562 SHUANG-TING LAN, ZONG-XUAN CHEN
Theorem 1.1. Let cj , j = 1, . . . , n, be distinct constants, Aj(z) = Pj(z)e
hj(z) + Qj(z), j =
= 1, . . . , n, where hj(z) are polynomials with degree k \geq 1, Pj(z)(\not \equiv 0), Qj(z) are entire functions
of order less than k. Among leading coefficients of hj(z), j \in \{ 1, . . . , n\} , having the maximal
modulus, there exists a term being unequal to the others. If f(z)(\not \equiv 0) is a meromorphic solution of
equation
An(z)f(z + cn) + . . .+A1(z)f(z + c1) = 0, (1.2)
then \sigma (f) \geq k + 1.
Corollary 1.1. Let k,Aj(z), j = 1, . . . , n, be defined as in Theorem 1.1, Bi(z), i = 1, . . . ,m,
be entire functions of order less than k, and cj , j = 1, . . . , n + m, be distinct constants. If f(z)
(\not \equiv 0) is a meromorphic solution of equation
Bm(z)f(z+ cn+m)+ . . .+B1(z)f(z+ cn+1)+An(z)f(z+ cn)+ . . .+A1(z)f(z+ c1) = 0, (1.3)
then \sigma (f) \geq k + 1.
Example 1.1. The function f(z) = ez
2
satisfies difference equation
e - 2izf(z + i) + e2izf(z - i) - 2e - 1f(z) = 0.
Obviously, \sigma (f) = 2 = \mathrm{d}\mathrm{e}\mathrm{g} h1+1 = \mathrm{d}\mathrm{e}\mathrm{g} h2+1. This example shows that the equality in Corollary 1.1
can be arrived. So the estimation in Corollary 1.1 is precise.
By Theorems A, B and 1.1, we deduce the following corollary.
Corollary 1.2. Let cj , j = 1, 2, be distinct nonzero constants, hj(z), j = 1, 2, be polynomials,
and Aj(z) ( \not \equiv 0), j = 0, 1, 2, be entire functions such that
\mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (Aj), 0 \leq j \leq 2\} < \mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{d}\mathrm{e}\mathrm{g} h1, \mathrm{d}\mathrm{e}\mathrm{g} h2\} .
If f(z) ( \not \equiv 0) is a meromorphic solution of equation
A2(z)e
h2(z)f(z + c2) +A1(z)e
h1(z)f(z + c1) +A0(z)f(z) = 0, (1.4)
then \sigma (f) \geq \mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{d}\mathrm{e}\mathrm{g} h1, \mathrm{d}\mathrm{e}\mathrm{g} h2\} + 1.
Chen [6] studied complex oscillation problems of entire solutions f(z) to homogeneous and
nonhomogeneous linear difference equations respectively, and obtained some relations between \lambda (f)
and \sigma (f). These results may be stated as follows.
Theorem C. Let Aj(z), j = 1, . . . , n, be entire functions such that there is at least one Aj being
transcendental, cj , j = 1, . . . , n, be constants which are unequal to each other. Suppose that f(z)
is a finite order transcendental entire solution of the homogeneous linear difference equation (1.2)
and satisfies \sigma (f) > \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (Aj) : 1 \leq j \leq n\} + 1.
Then \lambda (f) \geq \sigma (f) - 1. Moreover, if assume n = 2, then \lambda (f) = \sigma (f).
Theorem D. Let F (z), Aj(z), j = 1, . . . , n, be entire functions such that F (z)An(z) \not \equiv 0, ck,
k = 1, . . . , n, be constants which are unequal to each other. Suppose that f(z) is a finite order entire
solution of the nonhomogeneous linear difference equation
An(z)f(z + cn) + . . .+A1(z)f(z + c1) = F (z).
If \sigma (f) > \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (F ), \sigma (Aj) : 1 \leq j \leq n\} , then \lambda (f) = \sigma (f).
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 11
ON THE GROWTH OF MEROMORPHIC SOLUTIONS OF DIFFERENCE EQUATION 1563
In the following, we continue to study the complex oscillation problems of entire solutions to
linear difference equations (1.2) and (1.4), and obtain the following results, which extend Theorems C
and D.
Theorem 1.2. Let cj , j = 1, . . . , n, be distinct constants, and Aj(z) (\not \equiv 0), j = 1, . . . , n, be
entire functions with finite order. Suppose that f(z) is a finite order entire solution of equation (1.2)
and satisfies \sigma (f) > \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (Aj) : 1 \leq j \leq n\} +1. Then f(z) assumes every finite value d infinitely
often and \lambda (f - d) = \sigma (f).
Example 1.2. The entire function f(z) = ez
2
satisfies linear difference equation
f(z + 1) - e2z+1f(z) = 0.
Obviously, A2(z) \equiv 1, A1(z) = - e2z+1. We see \sigma (f) = 2 = \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (A1), \sigma (A2)\} + 1, but
\lambda (f) = 0 < \sigma (f). This example shows that the condition in Theorem 1.2, \sigma (f) > \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (Aj) :
1 \leq j \leq n\} + 1, can not be weaken.
By Theorems D and 1.2, we have the following corollary.
Corollary 1.3. Under conditions of Theorem 1.2, for any small entire function \varphi (z) (\not \equiv 0)
satisfying \sigma (\varphi ) < \sigma (f), we have \lambda (f - \varphi ) = \sigma (f).
Corollary 1.4. Let h1(z), h2(z) be polynomials such that
h1(z) = anz
n + . . .+ a0, h2(z) = bmzm + . . .+ b0,
where anbm \not = 0, Aj(z) (\not \equiv 0), j = 0, 1, 2, be entire functions of order less than \mathrm{m}\mathrm{a}\mathrm{x}\{ n,m\} and
ck, k = 1, 2, be distinct nonzero constants such that c2an - c1bm \not = 0 while n = m. If f(z) ( \not \equiv 0)
is a finite order entire solution of (1.4), then \lambda (f) = \sigma (f) \geq \mathrm{m}\mathrm{a}\mathrm{x}\{ n,m\} + 1.
Example 1.1 shows that the condition, c2an - c1bm \not = 0 while n = m, in Corollary 1.4 can not
be weaken.
2. Proofs of theorems and corollaries. We need the following lemmas for the proof of theorems
and corollaries.
Lemma 2.1 [4]. Suppose that f(z) is a meromorphic function with \sigma (f) = \sigma < \infty , then for
any given \varepsilon > 0, there is a set E \subset (1,\infty ) that has finite linear measure or finite logarithmic
measure, such that
| f(z)| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\sigma +\varepsilon \}
for all z satisfying | z| = r \not \in [0, 1] \cup E, r \rightarrow \infty .
Lemma 2.2 [7]. Let \eta 1, \eta 2 be two arbitrary complex numbers, and let f(z) be a meromorphic
function of finite order \sigma . Let \varepsilon > 0 be given, then there exists a subset E \subset (0,\infty ) with finite
logarithmic measure such that for all z satisfying | z| = r \not \in E \cup [0, 1], we have
\mathrm{e}\mathrm{x}\mathrm{p}\{ - r\sigma - 1+\varepsilon \} \leq
\bigm| \bigm| \bigm| \bigm| f(z + \eta 1)
f(z + \eta 2)
\bigm| \bigm| \bigm| \bigm| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\sigma - 1+\varepsilon \} .
Proof of Theorem 1.1. Contrary to our assertion, we assume \sigma (f) < k + 1. Let
hj(z) = ajkz
k + h\ast j (z), (2.1)
where ajk \not = 0 are constants, h\ast j (z) are polynomials with \mathrm{d}\mathrm{e}\mathrm{g} h\ast j \leq k - 1, j = 1, . . . , n.
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 11
1564 SHUANG-TING LAN, ZONG-XUAN CHEN
Set
I =
\biggl\{
i : | aik| = \mathrm{m}\mathrm{a}\mathrm{x}
1\leq j\leq n
| ajk|
\biggr\}
, \theta j = \mathrm{a}\mathrm{r}\mathrm{g} ajk \in [0, 2\pi ), j \in I.
There exists l \in I such that alk \not = ajk, j \in I \setminus \{ l\} . By this and the definitions of I and \theta j , we see
that
| ajk| = | alk| , \theta j \not = \theta l, j \in I \setminus \{ l\} .
Choosing \theta such that
\mathrm{c}\mathrm{o}\mathrm{s}(k\theta + \theta l) = 1. (2.2)
By \theta j \not = \theta l, j \in I \setminus \{ l\} , we have
\mathrm{c}\mathrm{o}\mathrm{s}(k\theta + \theta j) < 1, j \in I \setminus \{ l\} . (2.3)
Denote
a = \mathrm{m}\mathrm{a}\mathrm{x}
1\leq j\leq n
\{ | ajk| \} , b = \mathrm{m}\mathrm{a}\mathrm{x}
j \not \in I
\{ | ajk| \} , c = \mathrm{m}\mathrm{a}\mathrm{x}
\bigl\{
b, a \mathrm{c}\mathrm{o}\mathrm{s}(k\theta + \theta j), j \in I \setminus \{ l\}
\bigr\}
< a, (2.4)
and
\sigma = \sigma (f) < k + 1, \beta = \mathrm{m}\mathrm{a}\mathrm{x}
1\leq j\leq n
\{ \sigma (Pj), \sigma (Qj)\} < k. (2.5)
Obviously,
\sigma
\biggl(
Pj
Pl
\biggr)
\leq \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (Pj), \sigma (Pl)\} \leq \beta , 1 \leq j \leq n, j \not = l,
\sigma
\biggl(
Qj
Pl
\biggr)
\leq \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (Qj), \sigma (Pl)\} \leq \beta , 1 \leq j \leq n.
By Lemma 2.1, for any given \varepsilon , 0 < 2\varepsilon < \mathrm{m}\mathrm{i}\mathrm{n}\{ 1, k+1 - \sigma , k - \beta , a - c\} , there is a set E1 \subset (1,\infty )
with finite logarithmic measure such that for all z satisfies | z| = r \not \in E1 \cup [0, 1], we obtain\bigm| \bigm| \bigm| \bigm| Pj(z)
Pl(z)
\bigm| \bigm| \bigm| \bigm| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\beta +\varepsilon \} , 1 \leq j \leq n, j \not = l;
\bigm| \bigm| \bigm| \bigm| Qj(z)
Pl(z)
\bigm| \bigm| \bigm| \bigm| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\beta +\varepsilon \} , 1 \leq j \leq n. (2.6)
It is clear that \mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast l (z)\} is of regular order \mathrm{d}\mathrm{e}\mathrm{g} h\ast l , \mathrm{e}\mathrm{x}\mathrm{p}\{ h\ast j (z)\} , 1 \leq j \leq n, j \not = l, is of
regular order \mathrm{d}\mathrm{e}\mathrm{g} h\ast j . By \mathrm{d}\mathrm{e}\mathrm{g} h\ast j \leq k - 1, 1 \leq j \leq n, then for all large z, | z| = r, we get
| \mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast l (z)\} | \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ rk - 1+\varepsilon \} ,
\bigm| \bigm| \mathrm{e}\mathrm{x}\mathrm{p}\{ h\ast j (z)\} \bigm| \bigm| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ rk - 1+\varepsilon \} , 1 \leq j \leq n, j \not = l. (2.7)
Applying Lemma 2.2 to f(z), there is a set E2 \subset (1,\infty ) with finite logarithmic measure such
that for all z satisfies | z| = r \not \in E2 \cup [0, 1], we have\bigm| \bigm| \bigm| \bigm| f(z + cj)
f(z + cl)
\bigm| \bigm| \bigm| \bigm| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\sigma - 1+\varepsilon \} , 1 \leq j \leq n, j \not = l. (2.8)
By (1.2) and (2.1), we obtain
- \mathrm{e}\mathrm{x}\mathrm{p}\{ alkzk\} =
\sum
j\in I\setminus \{ l\}
\mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast l (z)\}
f(z + cj)
f(z + cl)
\biggl(
Pj(z)
Pl(z)
\mathrm{e}\mathrm{x}\mathrm{p}\{ ajkzk\} \mathrm{e}\mathrm{x}\mathrm{p}\{ h\ast j (z)\} +
Qj(z)
Pl(z)
\biggr)
+
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 11
ON THE GROWTH OF MEROMORPHIC SOLUTIONS OF DIFFERENCE EQUATION 1565
+
\sum
j \not \in I
\mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast l (z)\}
f(z + cj)
f(z + cl)
\biggl(
Pj(z)
Pl(z)
\mathrm{e}\mathrm{x}\mathrm{p}\{ ajkzk\} \mathrm{e}\mathrm{x}\mathrm{p}\{ h\ast j (z)\} +
Qj(z)
Pl(z)
\biggr)
+
+\mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast l (z)\}
Ql(z)
Pl(z)
. (2.9)
Let z = rei\theta , where r \not \in E1 \cup E2 \cup [0, 1]. Substituting (2.2) – (2.4), (2.6) – (2.8) into (2.9), we get
\mathrm{e}\mathrm{x}\mathrm{p}\{ ark\} \leq
\sum
j\in I\setminus \{ l\}
\mathrm{e}\mathrm{x}\mathrm{p}\{ rk - 1+\varepsilon + r\sigma - 1+\varepsilon + r\beta +\varepsilon \}
\Bigl(
\mathrm{e}\mathrm{x}\mathrm{p}\{ a \mathrm{c}\mathrm{o}\mathrm{s}(k\theta + \theta j)r
k + rk - 1+\varepsilon \} + 1
\Bigr)
+
+
\sum
j \not \in I
\mathrm{e}\mathrm{x}\mathrm{p}\{ rk - 1+\varepsilon + r\sigma - 1+\varepsilon + r\beta +\varepsilon \}
\Bigl(
\mathrm{e}\mathrm{x}\mathrm{p}\{ (b+ \varepsilon )rk + rk - 1+\varepsilon \} + 1
\Bigr)
+
+\mathrm{e}\mathrm{x}\mathrm{p}\{ rk - 1+\varepsilon + r\beta +\varepsilon \} \leq
\leq n \mathrm{e}\mathrm{x}\mathrm{p}\{ (c+ \varepsilon )rk + 2rk - 1+\varepsilon + r\sigma - 1+\varepsilon + r\beta +\varepsilon \} \leq
\leq n \mathrm{e}\mathrm{x}\mathrm{p}\{ (c+ 2\varepsilon )rk\} . (2.10)
Dividing by \mathrm{e}\mathrm{x}\mathrm{p}\{ ark\} both sides of (2.10) and letting r \rightarrow \infty , we have 1 \leq 0. A contradiction.
Hence, \sigma (f) \geq k + 1.
Proof of Corollary 1.1. We assume \sigma (f) < k + 1. Using a same method as the proof of
Theorem 1.1, we also obtain (2.1) – (2.7).
By Lemma 2.1, there is a set E3 \subset (1,\infty ) with finite logarithmic measure such that for all z
satisfies | z| = r \not \in E3 \cup [0, 1], we obtain
| Bj(z)| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\beta 1+\varepsilon \} , 1 \leq j \leq m, (2.11)
where \beta 1 = \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (Bj), 1 \leq j \leq m\} < k.
Applying Lemma 2.2 to f(z), there is a set E4 \subset (1,\infty ) with finite logarithmic measure such
that for all z satisfies | z| = r \not \in E4 \cup [0, 1], we get\bigm| \bigm| \bigm| \bigm| f(z + cj)
f(z + cl)
\bigm| \bigm| \bigm| \bigm| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\sigma - 1+\varepsilon \} , 1 \leq j \leq n+m, j \not = l. (2.12)
By (1.3) and (2.1), we have
- \mathrm{e}\mathrm{x}\mathrm{p}\{ alkzk\} =
\sum
j\in I\setminus \{ l\}
\mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast l (z)\}
f(z + cj)
f(z + cl)
\biggl(
Pj(z)
Pl(z)
\mathrm{e}\mathrm{x}\mathrm{p}\{ ajkzk\} \mathrm{e}\mathrm{x}\mathrm{p}\{ h\ast j (z)\} +
Qj(z)
Pl(z)
\biggr)
+
+
\sum
j \not \in I
\mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast l (z)\}
f(z + cj)
f(z + cl)
\biggl(
Pj(z)
Pl(z)
\mathrm{e}\mathrm{x}\mathrm{p}\{ ajkzk\} \mathrm{e}\mathrm{x}\mathrm{p}\{ h\ast j (z)\} +
Qj(z)
Pl(z)
\biggr)
+
+
n+m\sum
j=n+1
Bj(z)
f(z + cj)
f(z + cl)
+ \mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast l (z)\}
Ql(z)
Pl(z)
. (2.13)
Let z = rei\theta , where r \not \in E1 \cup E2 \cup E3 \cup E4 \cup [0, 1]. Substituting (2.2) – (2.7), (2.11) and (2.12) into
(2.13), we obtain
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 11
1566 SHUANG-TING LAN, ZONG-XUAN CHEN
\mathrm{e}\mathrm{x}\mathrm{p}\{ ark\} \leq
\sum
j\in I\setminus \{ l\}
\mathrm{e}\mathrm{x}\mathrm{p}\{ rk - 1+\varepsilon + r\sigma - 1+\varepsilon + r\beta +\varepsilon \}
\Bigl(
\mathrm{e}\mathrm{x}\mathrm{p}\{ a \mathrm{c}\mathrm{o}\mathrm{s}(k\theta + \theta j)r
k + rk - 1+\varepsilon \} + 1
\Bigr)
+
+
\sum
j \not \in I
\mathrm{e}\mathrm{x}\mathrm{p}\{ rk - 1+\varepsilon + r\sigma - 1+\varepsilon + r\beta +\varepsilon \}
\Bigl(
\mathrm{e}\mathrm{x}\mathrm{p}\{ (b+ \varepsilon )rk + rk - 1+\varepsilon \} + 1
\Bigr)
+
+m \mathrm{e}\mathrm{x}\mathrm{p}\{ r\beta 1+\varepsilon + r\sigma - 1+\varepsilon \} + \mathrm{e}\mathrm{x}\mathrm{p}\{ rk - 1+\varepsilon + r\beta +\varepsilon \} \leq
\leq n \mathrm{e}\mathrm{x}\mathrm{p}\{ (c+ \varepsilon )rk + 2rk - 1+\varepsilon + r\sigma - 1+\varepsilon + r\beta +\varepsilon \} +m \mathrm{e}\mathrm{x}\mathrm{p}\{ r\beta 1+\varepsilon + r\sigma - 1+\varepsilon \} \leq
\leq n \mathrm{e}\mathrm{x}\mathrm{p}\{ (c+ 2\varepsilon )rk\} +m \mathrm{e}\mathrm{x}\mathrm{p}\{ r\beta 1+\varepsilon + r\sigma - 1+\varepsilon \} . (2.14)
Dividing by \mathrm{e}\mathrm{x}\mathrm{p}\{ ark\} both sides of (2.14) and letting r \rightarrow \infty , we have 1 \leq 0. It is a contradiction.
So, \sigma (f) \geq k + 1 holds.
Proof of Theorem 1.2. Consider the following two cases.
Case 1: d = 0.
Contrary to our assertion, suppose that \lambda (f) < \sigma (f). Then f(z) can be written as
f(z) = H(z)eh(z), (2.15)
where H(z)(\not \equiv 0) is canonical product (or polynomial) formed by zeros of f(z) such that
\lambda (H) = \sigma (H) = \lambda (f) < \sigma (f)
and
h(z) = akz
k + ak - 1z
k - 1 + . . .+ a0, (2.16)
where k \in \BbbN + satisfying k = \sigma (f) > \lambda (f), and ak(\not = 0), ak - 1, . . . , a0 are constants.
Substituting (2.15) into (1.2), we obtain
An(z)H(z + cn) \mathrm{e}\mathrm{x}\mathrm{p}\{ h(z + cn)\} + . . .+A1(z)H(z + c1) \mathrm{e}\mathrm{x}\mathrm{p}\{ h(z + c1)\} = 0,
or
An(z) \mathrm{e}\mathrm{x}\mathrm{p}\{ h(z + cn) - h(z + c1)\} H(z + cn) + . . .
. . .+A2(z) \mathrm{e}\mathrm{x}\mathrm{p}\{ h(z + c2) - h(z + c1)\} H(z + c2) +A1(z)H(z + c1) = 0. (2.17)
Since \sigma (f) > \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (Aj) : 1 \leq j \leq n\} + 1, then \mathrm{d}\mathrm{e}\mathrm{g} h(z) = k \geq 2. By (2.16), we get
h(z + cj) - h(z + c1) = kak(cj - c1)z
k - 1 + h\ast j (z), (2.18)
where h\ast j (z) are polynomials with \mathrm{d}\mathrm{e}\mathrm{g} h\ast j \leq k - 2, j = 2, . . . , n.
Set
I = \{ i : | ci - c1| = \mathrm{m}\mathrm{a}\mathrm{x}
2\leq j\leq n
| cj - c1| \} .
We consider two cases in the following.
Case 1.1. I contains exactly one term.
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 11
ON THE GROWTH OF MEROMORPHIC SOLUTIONS OF DIFFERENCE EQUATION 1567
Without loss of generality, assume I = \{ n\} . By \sigma (Aj) < \sigma (f) - 1 = k - 1, j = 1, . . . , n, and
(2.18), we have
\sigma (Aj \mathrm{e}\mathrm{x}\mathrm{p}\{ h(z + cj) - h(z + c1)\} ) = \mathrm{d}\mathrm{e}\mathrm{g}(h(z + cj) - h(z + c1)) = k - 1, j = 2, . . . , n.
By the definition of I and I = \{ n\} , we see in the equation (2.17), the type k| ak(cn - c1)| of
coefficient An \mathrm{e}\mathrm{x}\mathrm{p}\{ h(z+ cn) - h(z+ c1)\} is strictly greater than types k| ak(cj - c1)| of coefficients
Aj \mathrm{e}\mathrm{x}\mathrm{p}\{ h(z+cj) - h(z+c1)\} , j = 2, . . . , n - 1. By this and applying Theorem B to equation (2.17),
we get \sigma (H) \geq (k - 1) + 1 = k = \sigma (f). A contradiction. So, \lambda (f) = \sigma (f).
Case 1.2. I contains more than one term.
Without loss of generality, assume I = \{ s, s+ 1, . . . , n\} , 2 \leq s < n. Set
ak = | ak| ei\theta 0 , \theta j = \mathrm{a}\mathrm{r}\mathrm{g}(cj - c1), j = s, . . . , n.
From the definition of I, we deduce
| cj - c1| < | cn - c1| , j = 1, . . . , s - 1,
| cj - c1| = | cn - c1| , j = s, . . . , n.
Since cj are distinct constants, \theta j are distinct constants, too. So we may choose \theta \in [0, 2\pi ) such
that
\mathrm{c}\mathrm{o}\mathrm{s}((k - 1)\theta + \theta 0 + \theta n) = 1. (2.19)
By \theta j \not = \theta n, j = s, . . . , n - 1, and (2.19), we see
\mathrm{c}\mathrm{o}\mathrm{s}((k - 1)\theta + \theta 0 + \theta j) < 1, j = s, . . . , n - 1. (2.20)
Denote
a = | ak(cn - c1)| , \beta = \mathrm{m}\mathrm{a}\mathrm{x}
1\leq j<s
\{ | ak(cj - c1)| \} ,
b = \mathrm{m}\mathrm{a}\mathrm{x}
s\leq j\leq n - 1
\{ a \mathrm{c}\mathrm{o}\mathrm{s}((k - 1)\theta + \theta 0 + \theta j), \beta \} , \alpha = \mathrm{m}\mathrm{a}\mathrm{x}
1\leq j\leq n
\{ \sigma (Aj), \lambda (f) - 1, k - 2\} .
(2.21)
Obviously,
\beta < a, b < a, \alpha < k - 1. (2.22)
By Lemma 2.1, for any given \varepsilon , 0 < \varepsilon < \mathrm{m}\mathrm{i}\mathrm{n}\{ a - b, 1\} , there exists a set E1 \subset (1,\infty ) having
finite logarithmic measure such that for all z satisfying | z| = r \not \in [0, 1] \cup E1, we have\bigm| \bigm| \bigm| \bigm| Aj(z)
An(z)
\bigm| \bigm| \bigm| \bigm| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\alpha +\varepsilon \} , j = 1, . . . , n - 1. (2.23)
We know both \mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast n\} and \mathrm{e}\mathrm{x}\mathrm{p}\{ h\ast j - h\ast n\} are of regular order \leq k - 2 \leq \alpha . Then for large
z, | z| = r, we obtain
| \mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast n\} | \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\alpha +\varepsilon \} ,
\bigm| \bigm| \mathrm{e}\mathrm{x}\mathrm{p}\{ h\ast j - h\ast n\}
\bigm| \bigm| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\alpha +\varepsilon \} , j = 2, . . . , n - 1. (2.24)
Applying Lemma 2.2 to H(z), there exists a set E2 \subset (1,\infty ) having finite logarithmic measure
such that for all z satisfying | z| = r \not \in [0, 1] \cup E2, we get
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 11
1568 SHUANG-TING LAN, ZONG-XUAN CHEN\bigm| \bigm| \bigm| \bigm| H(z + cj)
H(z + cn)
\bigm| \bigm| \bigm| \bigm| \leq \mathrm{e}\mathrm{x}\mathrm{p}\{ r\alpha +\varepsilon \} , j = 1, . . . , n - 1. (2.25)
By (2.17), we have
- \mathrm{e}\mathrm{x}\mathrm{p}\{ kak(cn - c1)z
k - 1\} =
n - 1\sum
j=s
Aj
An
H(z + cj)
H(z + cn)
\mathrm{e}\mathrm{x}\mathrm{p}\{ h\ast j - h\ast n\} \mathrm{e}\mathrm{x}\mathrm{p}\{ kak(cj - c1)z
k - 1\} +
+
s - 1\sum
j=2
Aj
An
H(z + cj)
H(z + cn)
\mathrm{e}\mathrm{x}\mathrm{p}\{ h\ast j - h\ast n\} \mathrm{e}\mathrm{x}\mathrm{p}\{ kak(cj - c1)z
k - 1\} +
+
A1
An
H(z + c1)
H(z + cn)
\mathrm{e}\mathrm{x}\mathrm{p}\{ - h\ast n\} . (2.26)
Take z = rei\theta , where r \not \in [0, 1] \cup E1 \cup E2. Substituting (2.19) – (2.25) into (2.26), we obtain
\mathrm{e}\mathrm{x}\mathrm{p}\{ kark - 1\} \leq (n - 2) \mathrm{e}\mathrm{x}\mathrm{p}\{ 3r\alpha +\varepsilon \} \mathrm{e}\mathrm{x}\mathrm{p}\{ kbrk - 1\} + \mathrm{e}\mathrm{x}\mathrm{p}\{ 3r\alpha +\varepsilon \} \leq
\leq (n - 1) \mathrm{e}\mathrm{x}\mathrm{p}\{ kbrk - 1 + 3r\alpha +\varepsilon \} ,
thus,
1 \leq (n - 1) \mathrm{e}\mathrm{x}\mathrm{p}\{ 3r\alpha +\varepsilon + kbrk - 1 - kark - 1\} .
Letting r \rightarrow \infty , by (2.22), we get 1 \leq 0. It is impossible. Hence, \lambda (f) = \sigma (f).
Case 2: d \not = 0.
Set g(z) = f(z) - d, then
f(z) = g(z) + d (2.27)
and
\sigma (g) = \sigma (f) > \mathrm{m}\mathrm{a}\mathrm{x}\{ \sigma (Aj) : 1 \leq j \leq n\} + 1. (2.28)
Substituting (2.27) into (1.2), we obtain
An(z)g(z + cn) + . . .+A1(z)g(z + c1) = - d(An(z) + . . .+A1(z)). (2.29)
If An(z) + . . . + A1(z) \not \equiv 0, by (2.28), (2.29) and Theorem D, we have \lambda (g) = \sigma (g), that is,
\lambda (f - d) = \sigma (f).
If An(z) + . . .+A1(z) \equiv 0, then g(z) is an entire solution of difference equation
An(z)g(z + cn) + . . .+A1(z)g(z + c1) = 0.
By (2.28) and the above Case 1, we have \lambda (g) = \sigma (g), that is, \lambda (f - d) = \sigma (f).
From the above Cases 1 and 2, we see f(z) assumes every finite value d infinitely often and
\lambda (f - d) = \sigma (f).
Proof of Corollary 1.4. Without loss of generality, assume that n \geq m. By Corollary 1.2, we
know \sigma (f) \geq n + 1. If \sigma (f) > n + 1, by Theorem 1.2, \lambda (f) = \sigma (f) holds. So, we assume
\sigma (f) = n+ 1.
Suppose that \lambda (f) < \sigma (f), then f(z) can be written as
f(z) = g(z)eh(z), (2.30)
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 11
ON THE GROWTH OF MEROMORPHIC SOLUTIONS OF DIFFERENCE EQUATION 1569
where g(z)(\not \equiv 0) is canonical product (or polynomial) formed by zeros of f(z) such that
\sigma (g) = \lambda (g) = \lambda (f) < \sigma (f) = n+ 1,
and
h(z) = dn+1z
n+1 + dnz
n + . . .+ d0 (2.31)
is a polynomial, where dn+1 \not = 0, dn . . . , d0 are constants.
Substituting (2.30), (2.31) into (1.4) and dividing by eh(z), we obtain
A2(z)e
h(z+c2) - h(z)+h2(z)g(z + c2) +A1(z)e
h(z+c1) - h(z)+h1(z)g(z + c1) +A0(z)g(z) = 0. (2.32)
By (2.31), we see
h(z + c1) - h(z) + h1(z) = ((n+ 1)c1dn+1 + an)z
n + h\ast 1(z),
h(z + c2) - h(z) + h2(z) = (n+ 1)c2dn+1z
n + bmzm + h\ast 2(z),
(2.33)
where h\ast 1(z), h
\ast
2(z) are polynomials with degree no more than n - 1.
Consider the following two cases.
Case 1: n > m.
By (n+ 1)c2dn+1 \not = 0, we see
\mathrm{d}\mathrm{e}\mathrm{g}(h(z + c2) - h(z) + h2(z)) = n \geq \mathrm{d}\mathrm{e}\mathrm{g}(h(z + c1) - h(z) + h1(z)).
Combining this with (2.32) and Corollary 1.2, we have \sigma (g) \geq n + 1. A contradiction. So,
\lambda (f) = \sigma (f) = n+ 1.
Case 2: n = m.
If (n+ 1)c1dn+1 + an \not = 0, it follows from (2.33) that
\mathrm{d}\mathrm{e}\mathrm{g}(h(z + c1) - h(z) + h1(z)) = n \geq \mathrm{d}\mathrm{e}\mathrm{g}(h(z + c2) - h(z) + h2(z)).
Combining this with (2.32) and Corollary 1.2, we have \sigma (g) \geq n+ 1 = \sigma (f). A contradiction. So,
\lambda (f) = \sigma (f) = n+ 1.
If (n+ 1)c1dn+1 + an = 0, since c1 \not = 0, we get
(n+ 1)c2dn+1 + bm = - an
c1
c2 + bm =
c1bm - c2an
c1
\not = 0,
then
\mathrm{d}\mathrm{e}\mathrm{g}(h(z + c2) - h(z) + h2(z)) = n > \mathrm{d}\mathrm{e}\mathrm{g}(h(z + c1) - h(z) + h1(z)).
Together with (2.32) and Corollary 1.2, we have \sigma (g) \geq n+1. A contradiction. So, \lambda (f) = \sigma (f) =
= n+ 1.
References
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2000. – 13. – P. 889 – 905.
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1570 SHUANG-TING LAN, ZONG-XUAN CHEN
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Received 05.07.12,
after revision — 31.08.16
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 11
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| id | umjimathkievua-article-1942 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:15:40Z |
| publishDate | 2016 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/0b/42f04c9844261bc93c4fe0c1424a810b.pdf |
| spelling | umjimathkievua-article-19422019-12-05T09:32:19Z On the growth of meromorphic solutions of difference equation Порядок росту мероморфних розв’язкiв рiзницевого рiвняння Chen, Zong-Xuan Lan, Shuang-Ting Чень, Цзон-Сюань Лан, Шуан-Тін We estimate the order of growth of meromorphic solutions of some linear difference equations and study the relationship between the exponent of convergence of zeros and the order of growth of the entire solutions of linear difference equations. Оцiнено порядок росту мероморфних розв’язкiв деяких лiнiйних рiзницевих рiвнянь та вивчено спiввiдношення мiж показниками збiжностi нулiв та порядком зростання цiлих розв’язкiв лiнiйних рiзницевих рiвнянь. Institute of Mathematics, NAS of Ukraine 2016-11-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1942 Ukrains’kyi Matematychnyi Zhurnal; Vol. 68 No. 11 (2016); 1561-1570 Український математичний журнал; Том 68 № 11 (2016); 1561-1570 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1942/924 Copyright (c) 2016 Chen Zong-Xuan; Lan Shuang-Ting |
| spellingShingle | Chen, Zong-Xuan Lan, Shuang-Ting Чень, Цзон-Сюань Лан, Шуан-Тін On the growth of meromorphic solutions of difference equation |
| title | On the growth of meromorphic solutions of difference equation |
| title_alt | Порядок росту мероморфних розв’язкiв рiзницевого рiвняння |
| title_full | On the growth of meromorphic solutions of difference equation |
| title_fullStr | On the growth of meromorphic solutions of difference equation |
| title_full_unstemmed | On the growth of meromorphic solutions of difference equation |
| title_short | On the growth of meromorphic solutions of difference equation |
| title_sort | on the growth of meromorphic solutions of difference equation |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1942 |
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