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On the Growth of Solutions of Some Second-Order Linear Differential Equations
Journal of Inequalities and Applications volume 2011, Article number: 635604 (2011)
Abstract
We investigate the growth of solutions of , where
and
are entire functions. When
and
satisfy some conditions, we prove that every nonzero solution of the above equation has infinite order and hyper-order 1, which improve the previous results.
1. Introduction and Results
In this paper, we willassume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna's value distribution theory of meromorphic functions (e.g., see [1–3]). In addition, we will use the notation to denote the order of growth of meromorphic function
,
to denote the hyper-order of
(see [3]).
is defined to be
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ1_HTML.gif)
We consider the second-order linear differential equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ2_HTML.gif)
where and
are entire functions of finite order. It is well known that each solution of (1.2) is an entire function, and most solutions of (1.2) have infinite order.
Thus, a natural question is what conditions on and
will guarantee that every solution
of (1.2) has infinite order? Ozawa [4], Gundersen [5], Amemiya and Ozawa [6], and Langley [7] have studied the problem with
and
is complex number or polynomial. For the case that
, and
is transcendental entire function, Gundersen proved the following in [5, Theorem A].
Theorem A.
If is a transcendental entire function with order
, then every solution
of equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ3_HTML.gif)
has infinite order.
Theorem A states that when , (1.3) may have finite-order solutions. We go deep into the problem: what condition in
when
will guarantee every solution
of (1.3) has infinite order? And more precise estimation for its rate of growth is a very important aspect. Chen investigated the problem and obtain the following in [8, Theorem B and Theorem C].
Theorem B.
Let be entire functions with
be complex numbers such that
and
. then every solution
of the equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ4_HTML.gif)
has infinite order.
Theorem C.
Let be nonzero complex numbers and
be a nonconstant polynomial or
where
is nonzero polynomial, then every solution
of the equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ5_HTML.gif)
has infinite order and .
For Theorems B and C, many authors, Wang and Lü [9],Huang, Chen, and Li [10], and Cheng and Kang [11] have made some improvement. In this paper, we areconcerned with the more general problem, and obtain the following theorem that extend and improve the previous results.
Theorem 1.1.
Let be entire functions with
be complex numbers such that
(suppose that
. If
or
, then every solution
of the equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ6_HTML.gif)
has infinite order and .
2. Remarks and Lemmas for the Proof of Theorem
Lemma 2.1 (see [12]).
Let be a transcendental meromorphic function with
be a finite set of distinct pairs of integers satisfying
. And let
be a given constant. Then,
(i)there exists a set with linear measure zero, such that, if
, then there is a constant
, such that for all
satisfying
and
and for all
, one has
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ7_HTML.gif)
(ii)there exists a set with finite logarithmic measure, such that for all
satisfying
and for all
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ8_HTML.gif)
(iii)there exists a set with finite linear measure, such that for all
satisfying
and for all
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ9_HTML.gif)
Lemma 2.2 (see [8]).
Suppose that are real numbers,
is a polynomial with degree
, that
is an entire function with
. Set
. Then for any given
, there exists a set
that has the linear measure zero, such that for any
, there is
, such that for
, we have
(i)if , then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ10_HTML.gif)
(ii)if , then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ11_HTML.gif)
where is a finite set.
Using Lemma 2.2, we can prove Lemma 2.3.
Lemma 2.3.
Suppose that is a positive entire number. Let
be nonconstant polynomials, where
are complex numbers and
. Set
, then there is a set
that has linear measure zero. If
, then there exists a ray
,
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ12_HTML.gif)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ13_HTML.gif)
where is a finite set, which has linear measure zero.
Proof.
According to the values of and
, we divide our discussion into three cases.
Case 1 ().
-
(a)
If
, let
, Then there are three cases: (i)
; (ii)
; (iii)
.
-
(i)
. By
, we know that
.
Suppose that , then take
is any constant in
.
Since has linear measure zero, there exists
such that
. Thus
. By
and
that is
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ14_HTML.gif)
Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ15_HTML.gif)
When , then
, we can prove it by using similar argument action as in the above proof.
-
(ii)
, then
. Suppose that
, then
. Let
, and take
is any constant in
.
Since has a linear measure zero, there exists
such that
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ16_HTML.gif)
Therefore
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ17_HTML.gif)
Suppose that , then
. Let
, and take
is any constant in
.
Since has linear measure zero, there exists
such that
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ18_HTML.gif)
Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ19_HTML.gif)
-
(iii)
, then
. Using similar method as in proof of (ii), we know that there exists
such that
.
-
(b)
When
, we can prove it by using the same argument action as in (a).
-
(c)
When
, we just prove the case that
(when
, we can prove it by using the same reasoning).
Let , take
is any constant in
.
Since has a linear measure zero, there exists
, such that
. Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ20_HTML.gif)
When ,
, thus,
.
When ,
, thus,
.
Therefore
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ21_HTML.gif)
Case 2.
When , or
and
, using a proof similar to Case 1, we can get the conclusion.
Case 3 ( and
).
By , there are only two cases:
; or
.
If . Take
is any constant in
.
Since has linear measure zero, there exists
such that
. Using a proof similar to Case 1(c), we can prove it.
When , we can prove it by using the same reasoning
Remark 2.4.
Using the similar reasoning of Lemma 2.3, we can obtain that, in Lemma 2.3, if is replaced by
, then it has the same result.
Lemma 2.5 (see [8]).
Let be entire functions with finite order. If
is a solution of the equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ22_HTML.gif)
then .
Lemma 2.6 (see [12]).
Let be a transcendental meromorphic function, and let
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
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ23_HTML.gif)
Remark 2.7.
In Lemma 2.6, when , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ24_HTML.gif)
Lemma 2.8 (see [13]).
Suppose that and
are nondecreasing functions, such that
, where
is a set with at most finite measure, then for any constant
, there exists
such that
for all
.
3. Proof of Theorem 1.1
Suppose that is a solution of (1.6), then,
is an entire function.
First Step
We prove that . Suppose, to the contrary, that
. By Lemma 2.1, for any given
, there exists a set
of linear measure zero, such that if
, then, there is a constant
, such that for all
satisfying
and
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ25_HTML.gif)
Let .
Case 1 (, which is
).
-
(i)
Suppose that
. By Lemmas 2.2 and 2.3, for the above
, there is a ray
, such that
(where
and
are defined as in Lemma 2.3, and
is of the linear measure zero), and satisfying
(3.2)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ27_HTML.gif)
When , for sufficiently large
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ28_HTML.gif)
Hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ29_HTML.gif)
By (1.6), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ30_HTML.gif)
Since , we know that
, then
. Substituting (3.1) and (3.5) into (3.6), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ31_HTML.gif)
By , we know that (3.7) is a contradiction.
When , using a proof similar to the above, we can also get a contradiction.
-
(ii)
Suppose that
. By Lemma 2.2, for the above
, there is a ray
such that
and
. Since
, and
, then
, thus
. For sufficiently large
, we have
(3.8)
Hence,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ33_HTML.gif)
where .
Since , we see that
, then
.
Since , we know that
, then
. Substituting (3.1) and (3.9) into (3.6), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ34_HTML.gif)
Since , we know that (3.10) is a contradiction.
Case 2 (, which is
).
-
(i)
Suppose that
, then
. By Lemma 2.2, for the above
, there is a ray
such that
and
. Because
. For sufficiently large
, we have
(3.11)
Hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ36_HTML.gif)
Using the same reasoning as in Case 1(i), we can get a contradiction.
-
(ii)
Suppose that
. By Lemma 2.2, for the above
, there is a ray
such that
, then
, Since
and
, then
. Thus,
, for sufficiently large
, we get that (3.8) and (3.9) hold.
Since , then
.
Using the same reasoning as in Case 1(ii), we can get a contradiction.
Concluding the above proof, we obtain .
Second Step
We prove that .
By Lemma 2.5 and , then
.
By Lemma 2.6 and Remark 2.7, we know that there exists a set with finite logarithmic measure and a constant
, such that for all
satisfying
, we get that (2.18) holds.
For Cases 1 and 2(i) in first step, we have proved that there is a ray satisfying
, for sufficiently large
, we get that (3.5) or (3.9) or (3.12) hold, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ37_HTML.gif)
where is a constant.
Since , then
. By (2.18), (3.6), and (3.13), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ38_HTML.gif)
By , (3.14) and Lemma 2.8, we know that there exists
, when
, we have
, then
.
For Case 2(ii) in first step, we have proved that there is a ray satisfying
, for sufficiently large
, we get (3.9) hold, and we also get that
.
By (2.18), (3.6), and (3.9), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F635604/MediaObjects/13660_2010_Article_2354_Equ39_HTML.gif)
By and (3.15) and Lemma 2.8, we know that there exists
, when
, we have
, then
.
Concluding the above proof, we obtain .
Theorem 1.1 is thus proved.
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Acknowledgment
This project was supported by the National Natural Science Foundation of China (no. 10871076).
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Peng, F., Chen, ZX. On the Growth of Solutions of Some Second-Order Linear Differential Equations. J Inequal Appl 2011, 635604 (2011). https://doi.org/10.1155/2011/635604
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DOI: https://doi.org/10.1155/2011/635604