# On the Growth of Solutions of Some Second-Order Linear Differential Equations

- Feng Peng
^{1}and - Zong-Xuan Chen
^{1}Email author

**2011**:635604

https://doi.org/10.1155/2011/635604

© Feng Peng and Zong-Xuan Chen. 2011

**Received: **10 December 2010

**Accepted: **9 February 2011

**Published: **6 March 2011

## Abstract

## 1. Introduction and Results

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.

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.

has infinite order.

Theorem C.

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.

## 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

Using Lemma 2.2, we can prove Lemma 2.3.

Lemma 2.3.

where is a finite set, which has linear measure zero.

Proof.

According to the values of and , we divide our discussion into three cases.

Suppose that , then take is any constant in .

Suppose that , then . Let , and take is any constant in .

- (iii)
- (b)
- (c)

Let , take is any constant in .

Case 2.

When , or and , using a proof similar to Case 1, we can get the conclusion.

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]).

Lemma 2.6 (see [12]).

Remark 2.7.

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

By , we know that (3.7) is a contradiction.

Since , we know that (3.10) is a contradiction.

Using the same reasoning as in Case 1(ii), we can get a contradiction.

Concluding the above proof, we obtain .

Second Step

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.

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 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.

## Declarations

### Acknowledgment

This project was supported by the National Natural Science Foundation of China (no. 10871076).

## Authors’ Affiliations

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