`Abstract and Applied AnalysisVolume 2010, Article ID 170762, 16 pageshttp://dx.doi.org/10.1155/2010/170762`
Research Article

## On Subnormal Solutions of Periodic Differential Equations

1School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China
2Department of Mathematics, College of Natural Sciences, Pusan National University, Pusan 609-735, Republic of Korea

Received 14 July 2010; Revised 22 October 2010; Accepted 12 November 2010

Copyright © 2010 Zong-Xuan Chen and Kwang Ho Shon. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We investigate the existence and the form of subnormal solutions of higher-order linear periodic differential equations, and precisely estimate the growth of all solutions.

#### 1. Introduction and Results

In this paper we use standard notations from the value distribution theory (see [13]). In addition, we denote the order of growth of by and also use the notation to denote the hyperorder of , which is defined as

Consider the second-order homogeneous linear periodic differential equation where and are polynomials in , but both are not constants. It is well known that every solution of (1.2) is an entire function.

Suppose is a solution of (1.2), and if satisfies the condition then we say that is a nontrivial subnormal solution of (1.2). For convenience, we also say that is a subnormal solution of (1.2) (see [4, 5]).

It clearly follows that (i)if then ; (ii) if , then (1.3) holds.

Wittich [5] investigated the subnormal solution of (1.2), which gives the form of all subnormal solutions of (1.2) in the following theorem.

Theorem A (see [5]). If is a subnormal solution of (1.2), then must have the form where is an integer and are constants with and .

Gundersen and Steinbart [4] refined Theorem A and got the following theorem.

Theorem B (see [4]). Under the assumption of Theorem A, the following statements hold.
(i) If and , then any subnormal solution of (1.2) must have the form where is an integer and are constants with and .(ii)If and , then any subnormal solution of (1.2) must be a constant. (iii) If , then the only subnormal solution of (1.2) is .

In [6], Chen and Shon proved that supposing that where are polynomials satisfying , if , then every solution of (1.2) satisfies .

In [6], the condition “all constant terms of and are equal to zero” plays an important role in the growth of solutions of (1.2). This makes us consider that the condition may be applied to higher-order differential equations.

Gundersen and Steinbart [4] consider a subnormal solution of higher-order linear nonhomogeneous differential equation where , are polynomials in and obtain the following theorem.

Theorem C. Suppose that, in (1.8), one has and for all . Then any subnormal solution of (1.8) must have the form where and are polynomials in .

From the proof of Theorem C, we see that the condition (1.9) of Theorem C guarantees that the corresponding homogeneous differential equation of (1.8) has no nontrivial subnormal solution.

Thus, a natural question is whether or not (1.11) has a nontrivial subnormal solution if the condition (1.9) is replaced by the condition “there exists some satisfying ”.

Examples 1.1 and 1.2 show that if , (1.11) may have a nontrivial subnormal solution.

Example 1.1. The equation has a subnormal solution .

Example 1.2. The equation has a subnormal solution .

Thus, a natural question is what conditions will guarantee that (1.11) has no nontrivial subnormal solution under the condition .

In Theorem 1.3, we answer this question. We conclude that if all constant terms of are equal to zero under the conditions and , then (1.11) has no nontrivial subnormal solution, and we also prove that all solutions of (1.11) satisfy .

Examples 1.1 and 1.2 show that the condition “all constant terms of are equal to zero” cannot be deleted in Theorem 1.3.

In this paper, we firstly investigate the existence of subnormal solutions. It is an important problem in theory of periodic differential equations.

Theorem 1.4 generalizes the result of Theorem C, shows that (1.8) has at most one nontrivial subnormal solution, gives the form of subnormal solution of (1.8), and proves that all other solutions of (1.8) satisfy .

Theorem 1.6 refines Theorem C.

Our method for obtaining the proof is totally different from the method applied in [4, 5].

Theorem 1.3. Let be polynomials in such that all constant terms of are equal to zero and , that is, where are constants and are integers. Suppose that there exists satisfying Then, one has the following properties. If , then (1.11) has no nontrivial subnormal solution and every solution of (1.11) is of hyper order . If and , then any polynomials with degree are subnormal solutions of (1.11) and all other solutions of (1.11) satisfy .

Considering proof of theorems, if the set , then (1.8) (or (1.11)) becomes an equation with rational coefficients, but the equation with rational coefficients may have nonmeromorphic solution. For example, the equation has a solution . This shows that we cannot use the transformation to prove that every solution of (1.11) is of .

Theorem 1.4. Let satisfy (1.14) and (1.15). Let and be polynomials in . If , then(i) (1.8) possesses at most one nontrivial subnormal solution , and is of the form (1.10), where and are polynomials in ;(ii) all other solutions of (1.8) satisfy except the possible subnormal solution in (i).

Example 1.5 shows the existence of subnormal solution in Theorem 1.4.

Example 1.5. The equation has a subnormal solution .

Theorem 1.6. Under the assumption of Theorem C, the following statements hold.(i) Equation (1.11) has no nontrivial subnormal solution, and all solutions of (1.11) satisfy .(ii) Equation (1.8) has at most one nontrivial subnormal solution , and is of the form (1.10); all other solutions of (1.8) satisfy .

#### 2. Lemmas for the Proofs of Theorems

Lemma 2.1 (see [7, 8]). Let be meromorphic functions, and let 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.
Then .

Lemma 2.2. Let , , , and satisfy the hypotheses of Theorem 1.3.(i)If , then (1.11) has no nonzero polynomial solution.(ii) If and , then all polynomials with degree are solutions of (1.11), and any polynomial with degree is not solution of (1.11).

Proof. (i) Firstly, by , we see that all nonzero constants cannot be a solution of (1.11). Now suppose that , are constants, ) is a solution of (1.11). If , then . Substituting into (1.11) and taking , we conclude that where is some constant. Since , we see that (2.2) is a contradiction. If , then Set . If , then we can rewrite where . Thus, we conclude by (2.3) and (2.4) that Set Since is the polynomial, we see that By Lemma 2.1, (2.5)–(2.7), we conclude that Since , by (2.6) and (2.8), we see that Since , we have . This contradicts our assumption that .
(ii) Since and , clearly all polynomials with degree are solutions of (1.11). By and (i), we see that cannot be a nonzero polynomial, and hence cannot be a polynomial with degree .

Lemma 2.3 (see [9]). Let be a transcendental meromorphic function with , and let be a finite set of distinct pairs of integers that satisfy , for , 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 , one has

Lemma 2.4 (see [10]). 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 2.5 (see [11]). Let be an entire function with . Let there exists a set that has linear measure zero, such that for any ray is a constant, and is a constant independent of ). Then is a polynomial with .

Lemma 2.6 can be obtained from [12, Theorem 4] or [2, Theorem 7.3].

Lemma 2.6. Let be entire functions of finite order. If   is a solution of equation then .

Lemma 2.7 (see [13]). Let be an entire function of infinite order with the hyperorder , and let be the central index of . Then

Lemma 2.8 (see [6]). Let be an entire function of infinite order with , and let a set have finite logarithmic measure. Then there exists such that , , such that(i)if , then for any given , (ii)if and , then for any given and for any large , one has as sufficiently large

Lemma 2.9 (see [9]). Let be a transcendental meromorphic function, and let be a given constant. Then there exist a set with finite logarithmic measure and a constant that depends only on and , such that for all satisfying , one has

Remark 2.10. From the proof of Lemma 2.9 (i.e., Theorem 3 in [9]), we can see that exceptional set satisfies that if and denote all zeros and poles of , respectively, and denote sufficiently small neighborhoods of and , respectively, then Hence, if is a transcendental entire function and is a point such that it satisfies that is sufficiently large, then (2.16) holds.

Lemma 2.11 (see [4]). Consider an th-order linear differential equation of the form where each is a polynomial in and with . Suppose that is an entire subnormal solution of (2.18), that is, an entire solution of (2.18) that also satisfies (1.3). If is periodic with period , then where and are polynomials in .

#### 3. Proof of Theorem 1.3

(i) Suppose that and are the solution of (1.11). Then is an entire function. By Lemma 2.2 (i), we see that is transcendental.

Step 1. We prove that . Suppose to the contrary that . By Lemma 2.3, we know that 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
Now we take a ray , then . We assert that is bounded on the ray . If is unbounded on the ray , then by Lemma 2.4, there exists an infinite sequence of points such that, as , and By (1.11), we get that Since and (1.14), we have where is some constant. Substituting (3.1), (3.2), and (3.4) into (3.3), we get that By , we know that when , (3.5) is a contradiction. So, on the ray , where is some constant.
Now we take a ray . Then, . If is unbounded on the ray , then by Lemma 2.4, there exists an infinite sequence of points such that, as , and By (1.11), we get that Since and (3.7), for as , By (3.8) and (3.9), we get that ; this is a contradiction. So, on the ray .

By Lemma 2.5, (3.6), and (3.10), we know that is a polynomial, which contradicts the above assertion that is transcendental. Therefore .

Step 2. We prove that (1.11) has no nontrivial subnormal solution. Now suppose that (1.11) has a nontrivial subnormal solution , and we will deduce a contradiction. By the conclusion in Step 1, satisfies (1.3) and and . By Lemma 2.6, we see that . Set . By Lemma 2.9, we see that there exist a subset having finite logarithmic measure and a constant such that for all satisfying , we have
From the Wiman-Valiron theory (see [2, page 51]), there is a set having finite logarithmic measure, so we can choose satisfying and . Thus, we get where is the central index of .
By Lemma 2.8, we see that there exists a sequence such that , , , , , and if , then by (2.14), we see that for any given , and for sufficiently large , and if , then by and (2.15), we see that for any given and for any sufficiently large , as is sufficiently large,

Since may belong to , , or , we divide this proof into three cases to prove.

Case 1. Suppose that , then . If we take , then . By , we see that there is a constant such that, as , , and . By (3.11), we see that for any given satisfying , holds for . By (3.11), (3.12), and (3.15), we see that By (1.11), we get Since and (1.14), we get that where is some constant. Substituting (3.12) and (3.18) into (3.17), we deduce that for sufficiently large , From (3.13) or (3.14), we have By (3.16), (3.19), and (3.20), we get Since , we see that (3.22) is a contradiction.

Case 2. Suppose that . By and , we see that for sufficiently large , . By (1.11), we get Since and (1.14), we get that for where are constants. By (3.12), (3.20), (3.23), and (3.24), we get that Thus, we have By (3.20), we see that (3.26) is also a contradiction.

Case 3. Suppose that or . Since the proof of is the same as the proof of , we only prove the case that . Since , for any given , we see that there is an integer , as , , and
By Lemma 2.9, there exist a subset having finite logarithmic measure and a constant such that for all satisfying , we have

Now we consider the growth of on a ray . If , then . By (1.3), for any given satisfying , If is unbounded on the ray , then by Lemma 2.4, there exists a sequence such that, as , and By Remark 2.10 and , we know that satisfies (3.28). By (3.28) and (3.29), we see that for sufficiently large , By (1.11), (3.18), (3.30), and (3.31), we deduce that Since , we know that when , (3.32) is a contradiction. Hence on the ray .

If , then . We assert that is bounded on the ray . If is unbounded on the ray , then by Lemma 2.4, there exists a sequence such that, as , and Since , for fixed , we deduce that as By (1.11), (3.34), and (3.35), we deduce that as This, (3.36) is a contradiction. Hence on the ray .

By (3.33) and (3.37), we see that satisfies on the ray .

However, since is of infinite order and satisfies , we see that for any large , as is sufficiently large Since , by (3.38) and (3.39), we see that for sufficiently large Thus and for sufficiently large where is some constant. By (1.11) and (3.12), we get that By (3.20), (3.41) and (3.42), we get that By (3.13) (or (3.14)), we see that (3.43) is a contradiction. Hence (1.11) has no nontrivial subnormal solution.

Step 3. We prove that all solutions of (1.11) satisfy . If there is a solution satisfying , then satisfies (1.3), that is, is subnormal, but this contradicts the conclusion in Step 2. Hence every solution satisfies . By this and , we get . Theorem 1.3(i) is thus proved.

(ii) Since and , we clearly see that all polynomials with degree are subnormal solutions of (1.11). By (i), we see that every satisfies or . Hence or is a polynomial with degree .

#### 4. Proof of Theorem 1.4

Suppose that and are nontrivial subnormal solutions of (1.8), then is a subnormal solution of the corresponding homogeneous equation (1.11) of (1.8). This contradicts the assertion of Theorem 1.3(i). Hence (1.8) possesses at most one nontrivial subnormal solution.

Now suppose that is a nontrivial subnormal solution of (1.8), then is also nontrivial subnormal solution, so, by the above assertion. Thus, by Lemma 2.11, we see that satisfies (1.10).

By Theorem 1.3(i), we see that all solutions of the corresponding homogeneous equation (1.11) of (1.8) are of . By variation of parameters, we see that all solutions of (1.8) satisfy . If , then clearly satisfies (1.3); that is, is subnormal. Hence all other solutions of (1.8) satisfy except at most one nontrivial subnormal solution.

#### 5. Proof of Theorem 1.6

(i) By Lemma 2.6 and , we see that . By Lemma 2.9, we see that there exist a subset having finite logarithmic measure and a constant such that for all satisfying ,

Taking , by (1.11) and (5.1), we deduce that Since , by (5.2), we get that . Hence .

(ii) Using a similar method as in the proof of Theorem 1.4, we can prove (ii).

#### Acknowledgments

The authors are grateful to referees for a number of helpful suggestions to improve the paper. Z-X. Chen was supported by the National Natural Science Foundation of China (no.: 10871076). K. H. Shon was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2010-0009646).

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