1. Introduction
In this paper, we use standard notations from the value
distribution theory of meromorphic functions (see [1,2,3]). We suppose that is a meromorphic function in whole complex plane .
In addition, we denote the order of growth of by , and use the notation to denote the hyper-order of ,
defined by
where is the Nevanlinna characteristic function of .
To give the precise estimate of fixed points, we denote the exponent of convergence of fixed points by , which is defined by
and the hyper-exponent of convergence of fixed points and distinct fixed
points are denoted by and and are defined by
and
respectively, where and are respectively the integrated counting function of fixed
points and distinct fixed points of . We denote the exponent of convergence of zeros (distinct zeros) of by and the hyper-exponent of convergence of zeros (distinct zeros) of by .
Consider the second-order homogeneous linear differential equation
where and are not constants polynomials in . It’s well-known that every solution of Equation (1) is entire.
Suppose is a solution of (1). If satisfies the condition
then we say that is a nontrivial subnormal solution of (1), and if satisfies the condition [
4],
then we say that is a nontrivial -subnormal solution of (1). In [
5], Wittich investigated the subnormal
solution of (1), and obtained the form of all subnormal solutions in the following theorem:
Theorem 1. [5]
If is a subnormal solution of (1), then must have the form
where is an integer and , , ,…, are constants with .
Gundersen and Steinbert [
6] refined Theorem 1 and got the following theorem:
Theorem 2. [6]
Under the assumption of Theorem 1, the following statements hold:
- (i) If and
, then any subnormal solution of (7) must have the form
where is an integer and , , …, are constants with and .
- (ii) If and , then any subnormal solution of Equation (7) must be constant.
- (iii) If , then the only subnormal solution of (7) is
Chen and Shon [
7] investigated more general equation than (7), and got the following theorem: Set
where are
integers, are complex constants such that .
Theorem 3. [7]
Let be polynomials and satisfy (2) and (3), and . Suppose that , . If , then every solution of
equation
satisfies .
Many authors investigated the growth of solutions and
the existence of subnormal solutions for some class of higher order linear
differential equations (see [4,7,8,9,10,11,12,13]). For the higher-order linear homogeneous
differential equation
where are polynomials in , Yang and Li
[
11] generalized the result of Theorem 2 to the higher
order and obtained the following results:
Set
where are integers, are complex constants, .
Theorem 4. [11]
Let be polynomials and satisfy (5). Suppose that
where . If there exists an integer
satisfying
then every solution of Equation (4) satisfies if
one of the following condition holds:
Theorem 5. [11]
Under the assumption of Theorem 4, if
, then we have every solution of Equation (4) satisfies
In particular, they also investigated the exponents of
convergence of the fixed points of solutions and their first derivatives for
a second order Equation (1) and obtained the following
theorem:
Theorem 6. [11]
Let ,…, , ,…, be
polynomials and satisfy (2) and (3), and . Suppose that ,
. If , then every solution of Equation (1) satisfy and .
Thus, it is natural to ask what will happen if we change in the coefficients of (4) into ?
In this paper, we consider the above problem to Theorems 3, 4, 5 and 6, we obtain the following results:
We set
where is an integer and are complex constants
such that , throughout the rest of this paper.
Theorem 7.
Let be polynomials and satisfy (5). Suppose that
where . If there exists an integer satisfying
then every solution of equation
satisfies and if one of the following condition holds:
Example 1.
Let be a solution of the equation
Set
We remark that and .
Obviously, the conditions of Theorem 7 are satisfied, we see that and
Remark 1.
Very recently, Li et al., [4] have investigated subnormal solutions of the Equation (7) with
where are complex
constants instead of polynomials and obtained some results concerning their
growth.
Corollary 1.
Under the assumption of Theorem 7, if , then we have every solution of Equation (4) satisfies
In particular, we also investigate the exponents of
convergence of the fixed points of solutions and their first derivatives for
a second order equation
and we obtain the following theorems:
Theorem 8.
Let ,…, , ,…, be polynomials and satisfy (2) and (3), and .
Suppose that , . If , then every solution
of Equation (8) satisfies and .
Example 2.
Let be a solution of the equation
Set
It is clear that the conditions of Theorem 8 are satisfied with , we see that and .
Remark 2.
If then the conclusions of Theorem 8 does not hold. For instance, consider the following equation
We can easily see that (9) has solution which satisfies and In this example, we have and
Theorem 9.
Let be polynomials and
satisfy (2) and (3),
and . Suppose that
If , then every solution of equation
satisfies and .
Example 3.
Let be a solution of the equation
Set
It is clear that the conditions of Theorem 9 are satisfied with , here we have and .
Remark 3.
If then the conclusions of Theorem 9 does not hold. For instance, consider the following equation
It is easy to see that (11) has solution which satisfies and In this example, we have
and
Remark 4.
Setting , and
in Theorem 7, Corollary 1, Theorem 8 and
Theorem 9, we obtain Theorem 4,
Theorem 5, Theorem 6 and Theorem 3 respectively.
2. Auxiliary Lemmas
Recall that
we set , and , , , for , throughout the rest of this
paper. Obviously, if , as , we get
Lemma 1. [3]
Let be meromorphic functions, be entire
functions, and satisfy
- (i)
- (ii) when , then is not a constant;
- (iii) when , ,
then
where is of finite linear measure or logarithmic measure. Also, .
Lemma 2.
Let , , , , and satisfy the hypotheses of Theorem 7, then Equation (7) has no constant polynomial solution.
Proof.
Suppose that is a nonconstant polynomial solution of (7), where
are complex constants.
If , then . Taking , we have
Substituting into (7) and using (13), we conclude that
where and is some constant. Since , we see that (3) is a
contradiction. Obviously, when or , we can get that the Equation (7) has nonconstant polynomial solution from the above
process.
If , then
Set . If , then we can rewrite
for , where . Thus, we conclude by (15)
that
Set
Since and are polynomials, we see that
By Lemma 1 and (16)-(18), we conclude that
Since and , so by (16) and (19), we get a contradiction.
Lemma 3. [14,15]
Let be an entire function and suppose that
is unbounded on some ray . Then, there
exists an infinite sequence of points
, where such that and
Lemma 4. [16]
Let be a transcendental meromorphic function of finite
order Let
denote a set of distinct pairs of integers satisfying
and
let be a given constant. Then, there exists a set that has linear measure zero
such that if , then there is a constant
such that for all satisfying and
and for all , we have
Lemma 5. [17]
Let be an entire function with is a constant
and is a constant independent of .
Then is a polynomial with .
Lemma 6. [16]
Let be a transcendental meromorphic function, and be a given
constant. Then, there exists a set with
finite logarithmic measure and a constant that depends only on and ,
, such that for all satisfying ,
Remark 5.
From the proof of Lemma 6 ([16, Theorem 3]), we
can see that the exceptional set equals , where denote all zeros and poles of , and
denote sufficiently small neighborhoods of . Hence, if is a transcendental entire function and is a
point that satisfies to be sufficiently large, then the
point satisfies (20).
For details, see , [9, Remark 2.10].
Lemma 7. [10,18]
Let , , be entire functions of finite order.
If is a solution of equation
then .
Lemma 8. [19]
Let be an entire function of infinite order with the hyper-order , and let be the central index of
. Then,
Lemma 9. [7]
Let be an entire function that satisfies ; or and ; or , and a
set has a finite logarithmic
measure. Then, there exists a sequence
such that , , , , and , such that
- (i) if , then for any given ,
- (ii) if and , then for any given , we have, as is sufficiently large,
- (iii) if , then for any given ,
Lemma 10. [20]
Let be a non-constant entire function, and let . There exists
a set of finite logarithmic
measure with the following property. For , the central index of
satisfies
Lemma 11. [21,22]
Let , …, , be finite order
meromorphic functions. If is a meromorphic solution of the
equation
with and , then satisfies
and .
Lemma 12. [14]
Let and be monotone non-decreasing functions such that
for all , where
is a set of finite logarithmic measure. Let
be a given constant. Then there exists a such that for all
3. Proofs of the results
Proof of Theorem 7.
Suppose that is a solution of (7), then
is an entire function. By Lemma 2, we see that is transcendental.
First step. We prove that .
Suppose, to the contrary, that , there exists a set with linear measure zero, such that if , then there exists a constant ,
such that for all satisfying and , we have
Case 1. Take a ray , then
. We assume that is
bounded on the ray . If is
unbounded on the ray , then by Lemma 3, there exists a
sequence such that as , and
By (7), we get
For , we have
and
where . Substituting
(21), (22), (24), (25) into (23), we obtain that for sufficiently
large
where is a constant. From (26), we can get a
contradiction by and , so
on the ray .
Case 2. Now, we take a ray ,
then . If is unbounded
on the ray , then by Lemma 3, there exists a sequence such that as , and
By (7), we get
For , we have
Substituting (28) and (30) into (29), we obtain that for sufficiently large
Since , when , by (31), we get , this is a contradiction. Hence
on the ray . From Lemma 5,
(27) and (32), we know that is a
polynomial, which contradicts the assertion that is transcendental.
Therefore, .
Step 2. We prove that . By Lemma 7 and , we see that .
Now, we suppose that there exists a solution satisfies . Then we have
By Lemma 6, we see that there exists a subset
having finite logarithmic measure such that for all satisfying ,
where is some constant. From the Wiman-Valiron theory, there is a
set having finite logarithmic measure, such that
we can choose a satisfying and , then we get
where is the central index of . By Lemma 9, we see
that there exists a sequence such that , , with ,
and for any sufficiently large
Case 1. Suppose . Since is a continuous
function of , by we get . Therefore, there exists a constant such that as ,
By (33), for any given ,
By (34), (35) and (37),
we have
By (7), we get
Substituting (24), (25) and (35) into (39), we get for sufficiently large ,
By (36), (38) and (40),
we get
where is a constant. From this inequality and (38), it follows that
Since and , we see that (41) is a
contradiction.
Case 2. Suppose . Since is a continuous function of , by we get such that as ,
By (7), we can write
From (6) and , we get
where is a constant. Substituting (35) and (43) into (42), we get
where is a constant. By substituting (36) into (44), we have
Since , we see (45) is also a contradiction.
Case 3. Suppose . Since , for any given ,
such that as , , and
By Lemma 6, we se that there exist a subset
having logarithmic measure such that
for all satisfying ,
Now, we consider the growth of on a ray . By the properties
of cosine function, we suppose without loss of generality that for and for .
Subcase 3.1 For a fixed , we have . Since , we get that satisfies (33). From
where , as
outside of a possible exceptional set of finite
logarithmic measure, we get that also satisfies (33). So for any given satisfying , we have
We assert that is bounded on the ray . If is unbounded on the ray , then, by
Lemma 3, there exists a sequence such that
as , and
By Remark 5 and , we know that . By (46) and (47), we have for sufficiently large ,
Substituting (24), (25), (48) and (49) into (23)
where is a constant, which yields a contradiction by and . Hence is bounded on the ray , so
on the ray .
Subcase 3.2 For a fixed , we have . Using a
reasoning similar to that in Subcase 3.1, we obtain
on the ray .
By (51) and (52), we see that on the
ray ,
But since and satisfies , we see that, for any
large , as is sufficiently large,
Since , by (53) and (54), we see that as
. Therefore, as . Thus, for sufficiently large ,
where and is a constant. By (7), (35) and (55), we get that
i.e.,
where is a constant. Substituting (36) into (56), we obtain also a contradiction. So we have
.
Proof of Corollary 1.
From Theorem 7, we get and . Let , then . Substituting it into (7), we have
Since , from Lemma 11, and we conclude and . So
and .
Proof of Theorem 8.
From Theorem 7, we get and .
- (i) Let , then . Substituting
it into (8), we have
Since , we get . From Lemma 11, we obtain and . So and .
- (ii) Differentiating both sides of (8), we get that
By (8), we have
Substituting (58) into (57), we get
Let , then , , . Substituting these
into (59), we get that
Next, we prove that . If , then
Since , we have
Suppose By taking , we have
We get
and
So, we obtain
Since , and are polynomials and , we get
So, we have
where and are some constants. It follows that
From (61), we have
where and are some constants, which is a
contradiction. So we have If \(p
0,\) { } and
are some constants.
This is a contradiction. So, we obtain Hence, if we have From Lemma 11, we get and .
Proof of Theorem 9.
Suppose that is a solution of (10). Since
then by Lemma 7, we see that
By Lemma 6, we se that there exist a subset
having logarithmic measure such that
for all satisfying ,
Taking , in (2) and (3), we obtain
that for sufficiently large
and
Substituting (63)-(65) into (10), we deduce that for all satisfying
By (66), we deduce that for all satisfying
Since , by (67) and Lemma 12, we get
From (62) and (68) we obtain and
Acknowledgments
This paper was supported by Directorate-General for Scientific Research and Technological Development(DGRSDT).
Author Contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Conflicts of Interest
The authors declare no conflict of interest.