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`Abstract and Applied AnalysisVolume 2013, Article ID 626898, 7 pageshttp://dx.doi.org/10.1155/2013/626898`
Research Article

## The Hyperorder of Solutions of Second-Order Linear Differential Equations

School of Mathematics and Statistics, Anyang Normal University, Anyang 455000, China

Received 15 June 2013; Accepted 29 July 2013

Copyright © 2013 Guowei Zhang. 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 prove that the hyperorder of every nontrivial solution of homogenous linear differential equations of type and nonhomogeneous equation of type is one, where are entire functions of order less than one, improving the previous results of Chen, Wang, and Laine.

#### 1. Introduction

We assume that the reader is familiar with the usual notations and the basic results of the Nevanlinna theory (see [14]). We also use basic notions and the results of the Wiman-Valiron theory; see [5]. Let be a nonconstant meromorphic function in the complex plane. We remark that , respectively, will be used to denote the order, respectively, the hyperorder, of . In particular, the hyperorder is defined as see [1, 2, 4]. For a set , let , respectively, , denote the linear measure, respectively, the logarithmic measure of . Moreover, the upper logarithmic density and the lower logarithmic density of are defined by Observe that may have a different meaning at different occurrences in what follows.

We now recall some previous results concerning linear differential equations of type where are entire functions of order less than one, and are complex constants.

Chen proved the following theorem; see [6].

Theorem A. Let , be entire functions of order less than one, and the complex constants satisfy and . Then every nontrivial solution of (3) is of infinite order.

Wang and Laine investigated the nonhomogeneous equation (4) and got the following; see [7].

Theorem B. Suppose that , , are entire functions of order less than one, and the complex constants satisfy and . Then every nontrivial solution of (4) is of infinite order.

Theorem C. Suppose that , , , , are entire functions of order less than one, and the complex constants satisfy and . Then every nontrivial solution of equation is of infinite order.

In this paper, we investigate the hyperorder of the nontrivial solutions of (3), (4), and (5) and obtain the following theorems.

Theorem 1. Suppose that , ,  are entire functions of order less than one, and the complex constants satisfy and . Then the hyperorder of every nontrivial solution of (4) is one.

Corollary 2. Let , be entire functions of order less than one, and the complex constants satisfy and . Then the hyperorder of every nontrivial solution of (3) is one.

Theorem 3. Suppose that , , , , are entire functions of order less than one, and the complex constants satisfy and . Then the hyperorder of every nontrivial solution of (5) is one.

#### 2. Lemmas

Lemma 4 (see [5]). Let be an entire function of infinite order and let be the central index of , then the hyperorder

Lemma 5 (see [8]). Let be an entire function of infinite order with , and there exists a set which have a finite logarithmic measure. Then there exists a sequence such that , , , and , and such that(1)if , then, for any given , (2)if , then, for any given and for any large ,

Lemma 6 (see [7]). Suppose that , where are real numbers, , and that is a meromorphic function with . Set , , . Then, for any given , there exists a set of finite linear measure such that, for any , there exists such that, for and , we have(1)if , then (2)if , then where .

Lemma 7 (see [6]). Let be entire functions of finite order and if is a solution of equation then the hyperorder .

The proof of the lemma below follows the idea of Bergweiler et al.; see [9, Theorem 3.1].

Lemma 8. Let be an entire function, and for every . Set , and there exists a constant and a set with positive lower logarithmic density such that for all large enough and all such that .

Proof. Since is entire function, we know that is nondecrease, as , and is continuous on the circle for . Set as . denotes an annulus for and sufficiently large . Then, there exists a constant such that for , where for sufficiently large. Then the function is a positive harmonic in . So is a positive harmonic in the domain . Thus, if and satisfy and , where , , then and . So by Harnack's inequality; see [10, Theorem ]. Therefore, if and are in the domain , where is sufficiently large, then Set . Then, we have for . If let , and then the set of is of positive lower logarithmic density. Thus, the conclusion of this lemma holds.

Lemma 9. Let be an entire function with infinite order and let hyperorder , be an entire function with finite order . For , where is the infinite logarithmic measure set which is given in Lemma 8. Then, for any given , for all such that is sufficiently large and that .

Proof. Since, for the entire function , for any given , we have for all sufficiently large. Since the order of is infinite, for , there exists a sufficiently large real number such that Thus, for is sufficiently large, By (17) and (19), we conclude that for all satisfying such that is sufficiently large. Thus, the conclusion holds.

#### 3. Proofs of Theorems

Proof of Theorem 1. Suppose that is a solution of (4), and then is an entire function.
Step  1. We prove that . Since , set . Then for any given satisfying , when is sufficiently large, we have From the Wiman-Valiron theory, there is a set having finite logarithmic measure, such that whenever , where the is the central index of , and we know that as . When sufficiently large, we have . From (4) we have Substituting (21), (22), (23), and (24) into (25), we obtain where satisfies and sufficiently large. By (26) we get Since is arbitrary, by (27) and Lemma 4, we have .
Step  2. By Theorem B, we know that the order of is infinite, and, by the first step, we clear that the hyperorder of is less than one. Thus, by Lemma 9 and (23), we have for all satisfying such that is sufficiently large, where is of infinite logarithmic measure. Set , and we assert that . Now we assume that , and prove that results in contradictions. are the sets in Lemmas 5 and 6, respectively.
Since , we have that is infinite. Thus, by Lemma 5, we see that there exists a sequence of points such that , , , , , and if , then, for any given , if , then, for any given and for any large , Firstly, we prove the case when . It can separate into three cases to discuss.
Case  1. First assume that . From the continuity of , we have for sufficiently large . From (9), we deduce that for all sufficiently large.
From (4), we have
Subcase  1.1. We first assume that satisfies . From the continuity of , we also have for all sufficiently large. From (33), we get Substituting  (24), (28), (29), (32), and (34) into (35), we obtain Since , we see that (36) are contradictory as .
Subcase  1.2. Next assume that . Then, from (10), for large enough, we deduce that From (33), we get Substituting  (24), (28), (29), (32), and (37) into (38), we obtain as . Since , this implies that , , which is impossible.
Subcase  1.3. Assume finally that . Here, (12) may be used to construct another sequence of points with , such that . Indeed, we may suppose, without less of generality, that with . When is large enough, we have , where is a small constant. Choose now such that . Then . For sufficiently large , we have (12) for and . Therefore for sufficiently large . Taking now small enough, we have , by the continuity of . This yields Similarly as (36), a contradiction easily follows.
Case  2. Suppose now that . Then, from the continuity of and (10), we have for large enough.
Subcase  2.1. Assume first that . From the continuity of and (9), we deduce that for large enough. From (4), we have Substituting  (24), (28), (29), (44), and (45) into (46), we obtain Since , we see that (47) is contradictory as .
Subcase  2.2. Assume that . From the continuity of and (9), we deduce that for large enough. From (4), we have Substituting  (24), (28), (29), (44), and (48) into (49), we obtain as . Since , this implies that , , which is impossible.
Subcase  2.3. Assume that . Arguing similarly as in Subcase  1.3, we may again construct another sequence of points with , such that . Replace with in (44) and with in (45), respectively. We obtain (44) and (45) for the sequence of . Similarly as (47), we get a contradiction as .
Case  3. In this final case, we suppose that . We discuss three subcases according to as follows.
Subcase  3.1. Suppose that . By an argument similar to that in Subcase  1.3, we can choose another sequence of points with , and , such that . Similarly as in Subcase  2.3, a contradiction follows as .
Subcase  3.2. Suppose that . By an argument similar to the Subcase  3.2 of the proof of Theorem 1.1 in [7], we can choose another sequence of points , with , such that . From (4), for , we get Replace with in (32) and with in (48), respectively. We obtain (32) and (48) for the sequence of . Substituting them into (51), this implies that , , which is impossible.
Subcase  3.3. Finally, suppose that . We now have , , and so , .
If , we may choose another sequence such that . By an argument similar to that in Subcase 3.2, we can get , , a contradiction.
If , we similarly obtain and for another sequence. By an argument similar to that in Subcase  1.3, a contradiction follows.
Finally, if , we obtain for another sequence. Similarly as in Subcase  1.2, a contradiction again follows.
Thus, we complete the proof when . When , we have (30). Similarly as the case when , it results in contradiction. Hence, we get .

Proof of Theorem 3. Suppose that is a nontrivial solution of (5), and then is an entire function. Since , we have for any such that . Similarly as in the proof of Theorem 1, we may choose a sequence of points that satisfy , with , , .
Step  1. We will prove that . Since , set . Then for any given satisfying , when is sufficiently large, we have From the Wiman-Valiron theory, we have (24). By Theorem C, we know that . So we have (28). From (5) we have Substituting  (24), (28), and (53) into (55), we obtain (26) and (27); thus, we have .
Step  2. Set , and we assert that . Now we assume that , and prove that results in contradiction. By Lemma 5, we have (29) and (30). Next we only prove the case by using (29). The case also can be proved by the same method, a little different is that we use (30) instead of (29). Since is a real number, there are three cases to be discussed, according to the signs of and .
Case  1. First assume that , so we have (32) and (48). Combining (52), (32), and (48), we deduce
provided that is large enough. From (5), we have Substituting  (24), (28), (56), and (57) into (58), we obtain for large enough. Since , this implies that , , which is impossible.
Case  2. Next, assume that , so we have (44) and (45). Combining (52), (44), and (45), we deduce
for is large enough. From (5), we have Substituting  (24), (28), (60), and (61) into (62), we obtain for large enough. Since , , this leads to a contradiction.
Case  3. Finally, we have to assume that . Similarly as in Subcase  1.3 of the proof of Theorem 1, we may again construct a sequence of points , with , such that . Indeed, without loss of generality,
for all . Provided that is large enough, we have . Choosing now such that , then , thus, , and . Since now , a contradiction follows as in Case  2 above.

#### Acknowledgments

The author wishes to express his thanks to the referee for his/her valuable suggestions and comments. The present investigation was supported by the National Natural Science Foundation under Grant no. 11226088 and the Key Project of Natural Science Foundation of Educational Committee of Henan Province under Grant no. 12A110002 of the People's Republic of China.

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