# Subnormal Solutions of Second-Order Nonhomogeneous Linear Differential Equations with Periodic Coefficients

- Zhi-Bo Huang
^{1}, - Zong-Xuan Chen
^{1}Email author and - Qian Li
^{2}

**2009**:416273

https://doi.org/10.1155/2009/416273

© Zhi-Bo Huang et al. 2009

**Received: **8 February 2009

**Accepted: **24 May 2009

**Published: **29 June 2009

## Abstract

## 1. Introduction

We use the standard notations from Nevanlinna theory in this paper (see [1–3]).

The study of the properties of the solutions of a linear differential equation with periodic coefficients is one of the difficult aspects in the complex oscillation theory of differential equations. However, it is also one of the important aspects since it relates to many special functions. Some important researches were done by different authors; see, for instance, [4–9].

where and are polynomials in and are not both constants. It is well known that every solution of (1.1) is an entire function.

- H.
Wittich has given the general forms of all subnormal solutions of (1.1) that are shown in the following theorem.

Theorem 1 A (see [9]).

- G.
G. Gundersen and E. M. Steinbart refined Theorem A and obtained the exact forms of subnormal solutions of (1.1) as follows.

Theorem 1 B (see [6]).

In addition to the statement of Theorem A, the following statements hold with regard to the subnormal solutions of (1.1).

where is an integer and are constants with

(ii)If and , then any subnormal solution of (1.1) must be a constant.

(iii)If , then the only subnormal solution of (1.1) is .

where , and are polynomials in such that are not both constants. They found the exact forms of all subnormal solutions of (1.5), that is, what is mentioned in [6, Theorem ?2.2, Theorem ?2.3 and Theorem ?2.4].

where , and are polynomials in such that , , , and are not all constants?

where , and are polynomials in and are not all constants.

In this paper, we obtain the forms of subnormal solutions of nonhomogeneous linear differential equation (1.6) when . We have the following theorem.

Theorem 1.1.

Suppose that is a subnormal solution of (1.6), where , , and are polynomials in such that and are not all constants.

where is a constant, and are polynomials in .

where is a constant, and are constants that may or may not be equal to zero, may be equal to zero or may be a polynomial in , , and are polynomials in with .

## 2. Lemmas for the Proof

In order to prove Theorem 1.1, we need some lemmas.

Lemma 2.1 (see [7]).

Suppose that is a subnormal solution of (1.7), where , , and are polynomials in and are not all constants.

(i)If and then any subnormal solution must be a constant.

where is a polynomial in with .

Lemma 2.2 (see [10]).

Let be a transcendental meromorphic function, let be a given real constant, and let . Then there exists a constant such that the following two statements hold (where ).

(ii)There exists a set that has finite logarithmic measure such that (2.2) holds for all satisfying .

Lemma 2.3 (see [6]).

Lemma 2.4 (see [8]).

where is a constant and and are polynomials in .

As an application of Lemma 2.4, one has the following lemma.

Lemma 2.5.

where is a constant and and are polynomials in .

Proof.

Since is entire and is linearly dependent with , can be written as (see [11, page 382]), where is a constant and is analytic on . Then we have the representation from Lemma 2.4.

Lemma 2.6.

where and are polynomials in with

Proof.

So and are polynomials in and , respectively, and by (2.13), but and have the exact representations that depend on the relations of , and . If , then (2.14) is of the form (2.11), and (2.12) gives (2.10). If , then we repeat the above process finite times until we obtain (2.10) and (2.11). This completes the proof of Lemma 2.6.

## 3. Proof of Theorem

In this section, we will prove Theorem 1.1.

- (i)Suppose that is a subnormal solution of (1.6) with and . If is a polynomial solution of (1.6), then must be a constant, which is of the form (1.8). Thus we suppose that is transcendental. It follows from Lemma 2.2(i) that there exists a set that has linear measure zero such that if , then there is a constant such that for all satisfying and , we have

where is a constant and . It also follows from Lemma 2.2(ii) that there exists a set that has finite logarithmic measure such that (3.1) holds for all satisfying .

From above, we have that (3.1) holds on the set

It follows from (3.4)–(3.6) and that (3.6) yields as on the set This is a contradiction.

where the integral of is defined on the simple contour , extending from a point to a point in the complex domain.

as in the angular domain . Together with (3.8) and (3.11), is bounded in the angular domain .

If , it follows from Lemma 2.5 that must have the form (1.8).

will be a subnormal solution of (1.7). Since we suppose that , we will discuss the following two cases.

Case 1.

where is a constant, and are polynomials in . Thus, has the form of (1.8).

Case 2.

where is a polynomial in with .

- (ii)

Then is a subnormal solution of (1.7). Now if , this shows that and has the form of (1.9) by Lemma 2.5. Thus, we suppose that in the following.

If it follows from the proof of Case 1 of Theorem 1.1(i) that has the form of (1.9).

where is a polynomial in with .

where is an integer and are constants with

Now, we will discuss the following two cases.

Case A.

where and are polynomials in with

It follows from (1.6), (3.25), (3.29) and (3.30) that

where and are constants, , and are polynomials in with . This is the form of (1.9).

Case B.

where and are constants, and, are polynomials in with . Set . Then is a polynomial in by the hypotheses of in (3.36). This is the form of (1.9). We have proved Theorem 1.1(ii) when

Now we suppose that . By Lemma 2.6, there exists a polynomial in , satisfies (2.10) and (2.11).

Since and since we have proved Theorem 1.1 holds in the cases when holds, we can apply this result to (2.11).

where is a constant, and are polynomials in . This is a form of (1.9).

where and are polynomials in with , and are constants that may or may not be equal to zero. By (2.10) and (3.40), we obtain that has the form of (1.9). Theorem 1.1(ii) is completed.

Now, we give some examples to show that Theorem 1.1 is correct.

Example 3.1.

This is an example of Theorem 1.1(i).

Example 3.2.

This is an example of Theorem 1.1(ii) with and .

Example 3.3.

## Declarations

### Acknowledgements

The authors are very grateful to the referee for his (her) many valuable comments and suggestions which greatly improved the presentation of this paper. The project was supposed by the National Natural Science Foundation of China (no. 10871076), and also partly supposed by the School of Mathematical Sciences Foundation of SCNU, China.

## Authors’ Affiliations

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