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On the Growth of Solutions of Some SecondOrder Linear Differential Equations
Journal of Inequalities and Applications volumeÂ 2011, ArticleÂ number:Â 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 hyperorder 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 [1â€“3]). In addition, we will use the notation to denote the order of growth of meromorphic function , to denote the hyperorder of (see [3]). is defined to be
We consider the secondorder linear differential equation
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
has infinite order.
Theorem A states that when , (1.3) may have finiteorder 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
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
has infinite order and .
For Theorems B and C, many authors, Wang and LÃ¼ [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
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
(ii)there exists a set with finite logarithmic measure, such that for all satisfying and for all , we have
(iii)there exists a set with finite linear measure, such that for all satisfying and for all , we have
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
(ii)if , then
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
or
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
Therefore,
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 ,
Therefore
Suppose that , then . Let , and take is any constant in .
Since has linear measure zero, there exists such that ,
Therefore,

(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
When , , thus, .
When , , thus, .
Therefore
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
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 ,
Remark 2.7.
In Lemma 2.6, when , we have
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
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
When , for sufficiently large , we have
Hence
By (1.6), we obtain
Since , we know that , then . Substituting (3.1) and (3.5) into (3.6), we get
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,
where .
Since , we see that , then .
Since , we know that , then . Substituting (3.1) and (3.9) into (3.6), we obtain
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
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,
where is a constant.
Since , then . By (2.18), (3.6), and (3.13), we obtain
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
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 SecondOrder 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