#### Abstract

This paper is devoted to studying growth of solutions of linear differential equations of type where and are entire functions of finite order.

#### 1. Introduction and Main Results

We assume that the reader is familiar with the usual notations and basic results of the Nevanlinna theory [1–3]. Let now be a nonconstant meromorphic function in the complex plane. We remark that will be used to denote the order of , and

We now recall some previous results concerning nonhomogeneous linear differential equations of typewhere and , are entire functions of finite-order, . In the case that the coefficients are polynomials, growth properties of solutions of (1.2) have been extensively studied, see, for example, [4]. In (1.2), if is the largest integer such that is transcendental, it is well known that there exist at most linearly independent finite-order solutions of the corresponding homogeneous equationThus, when at least one of the coefficients is transcendental, most of the solutions of (1.2) and (1.3) are of infinite-order. In the case when Hellerstein et al. [5] proved that every transcendental solution of (1.2) is of infinite-order. As for sectorial growth conditions on the coefficients of (1.2) that imply that all solutions are of infinite-order, see, for example, [6]. As for the special case of , Wang and Laine studied equations of typewhere are entire functions of order less than one, and the complex numbers satisfy . They proved that every nontrivial solution of (1.5) is of infinite-order if , see [7]. We remark that (1.2) may indeed have solutions of finite-order as soon as , as shown by the next examples.

*Example 1.1. *The exponential function satisfies the equationwhere can be any entire function. Choosing shows that (1.2) may admit a solution of
finite-order even if .
On the other hand, taking ,
we have the case that in (1.2).

*Example 1.2. * The function satisfies the equation

In this paper, we continue to consider (1.2) in the
case when .
Recently, Tu and Yi investigated the growth of solutions of (1.3) when
most coefficients have the same order, see [8]. We next prove two results of (1.2), which
generalize Theorems 2 and 4 in [8] and Theorem 1.1 in [7].

Theorem 1.3. *Suppose that where are polynomials with degree , are entire functions of order less than ,
not all vanishing, and is an entire function of order less than .
If are distinct complex numbers, then every
solution of (1.2) is of infinite-order.*

Theorem 1.4. *Suppose that where are polynomials with degree , and are entire functions of order less than .
Moreover, suppose that there are two coefficients so that for and , where
, , , , and for all , satisfies either or .
Then every transcendental solution of (1.2) is of infinite-order.*

In the case when where are polynomials, Chen considered the growth of solutions of (1.3) with some additional conditions imposed upon on , see [9]. Our last results generalizes his result and [7, Theorem 1.3].

Theorem 1.5. *Suppose that where are polynomials with degree , and are entire functions of order less than .
Moreover, suppose that there exist and with and such that for , or ,
and .
If ,
then every transcendental solution of (1.2) is of infinite-order.*

*Remark 1.6. * Under the
assumptions of Theorem 1.4, respectively, of Theorem 1.5, polynomial solutions may
exist. However, such possible polynomial solutions must be of degree less than .
If not, a contradiction immediately follows by combining (5.1) with Lemma 2.1,
if ,
respectively, with Lemma 2.2, if .

*Remark 1.7. *In the
preceding three theorems, if ,
then we also have for the exponent of convergence of the
zero-sequence of .
Indeed, rewriting (1.2) in the formwe havefor some finite .
Therefore, must be of infinite-order.

#### 2. Preliminary Lemmas

Lemma 2.1 (see [10]). * Suppose that are meromorphic functions and are entire functions satisfying the following
conditions:*

(i)(ii)* are not constants for ,*(iii)*for ,where is a set with finite linear measure.** Then .*

Lemma 2.2 (see [10]). * Suppose that are linearly independent meromorphic functions
satisfying the following identity:**Then for ,
one has
**where is the Wronskian determinant ,** is a set with finite linear measure.*

Lemma 2.3 (see [11, 12]). * Suppose that ( are real numbers, ) is a polynomial with degree ,
and that is an entire function with .
Set .
Then for any given ,
there exists a set of finite linear measure such that for any ,
there is such that for ,
one has
*

(i)*if ,
then*(ii)* if ,
thenwhere .*

Lemma 2.4 (see [13]). * Let be a transcendental meromorphic function of
finite-order ,
and let be a given constant. Then there exists a set that has finite logarithmic measure, such that
for all satisfying and for all ,
one has
**Similarly, there exists a set of linear measure zero such that for all with sufficiently large and ,
and for all ,
the inequality (2.7) holds.*

Lemma 2.5. *Let be an entire function and suppose that**is unbounded on some ray with constant .
Then there exists an infinite sequence of points where ,
such that and**as .*

*Proof. *The first assertion is trivial.
Denotingwe may take the sequence in the first assertion so that .
Sinceas ,
we immediately see thatas .
Using now the same reasoning as in the proof of [14, Lemma 4], see also [15,
Lemma 3.1], the second assertion (2.9) follows.

Lemma 2.6. *Let be an entire function with .
Suppose that there exists a set which has linear measure zero, such that for any ray ,
where is a positive constant depending on ,
while is a positive constant independent of .
Then .*

*Proof. *Clearly, we may assume that .
Since has linear measure zero, we may choose such that ,
andWe first treat the sectordefiningwhere and is a positive constant, to be determined
in what follows. Then is a holomorphic inside the sector .
By (2.13), we have .
Therefore,on the ray ,
and, respectively,on the ray .
Hence, we may now fix so thatBy elementary computation, on the boundary of ,
where is a bounded constant, not the same at each
occurrence. By the definition of in (2.15), it is immediate to see that is of order at most .
By the Phragmén-Lindelöf theorem, we conclude that holds on the whole sector .
Henceon .
Repeating the same reasoning for all the sectors where are determined in (2.13), the assertion
immediately follows.

#### 3. Proof of Theorem 1.3

Suppose, contrary to the assertion, that is a solution of (1.2) with , then . Indeed, if , we may apply Lemma 2.1 to conclude that for some , such that . Then has to be a polynomial of degree less than , so , a contradiction. Therefore, we may assume that . By Lemma 2.2, it is easy to see that since the exponential functions are linearly independent.

By Lemma 2.3, there is a set of linear measure such that whenever , then for all and for all with . If, moreover, has large enough, then each satisfies either (2.5) or (2.6). By Lemma 2.4, we may assume, at the same time, thatSince are distinct complex numbers, then for any fixed , there exists exactly one such thatDenoting , then and . We now discuss two cases separately.

*Case 1. *Assume first that .
By Lemma 2.3, for any given with ,
we havefor ,
provided that is sufficiently large. We now proceed to show
thatis bounded on the ray .
Supposing that this is not the case, then by Lemma 2.5, there is a sequence of
points ,
such that ,
and thatFrom (3.5) and the definition of
order, it is easy to see thatfor is large enough. From (1.2), we
obtainUsing inequalities (3.1), (3.3), (3.6), and the limit (3.7), we conclude from the preceding inequality
thatwhere is a bounded constant, which is a
contradiction. Therefore, is bounded, and we have on the ray .
By the same reasoning as in the proof of [15, Lemma 3.1], we immediately
conclude thaton the ray .

*Case 2. *Suppose now that .
From (1.2), we getAgain by Lemma 2.3, for any
given with ,
we havefor sufficiently large. As in Case 1, we prove
thatis bounded on the ray .
If not, similarly as in Case 1, it follows from Lemma 2.5 that there is a
sequence of points ,
such thatfor all large enough. Substituting the inequalities
(3.12) and (3.14) into (3.11), a contradiction immediately follows. Hence,
we have on the ray .
This implies, as in Case 1, that Therefore, for
any given , of linear measure zero, we have got (3.15) on
the ray ,
provided that is large enough. Then by Lemma 2.6, ,
a contradiction. Hence, every transcendental solution of (1.2) must be of
infinite-order.

#### 4. Proof of Theorem 1.4

Suppose that is a transcendental solution of (1.2) with .

If and , it follows from (1.2) thatwhere are entire functions of order less than . Collecting terms of the same type together, if needed, we may assume that the coefficients , respectively, , are distinct. Since and , we conclude that are polynomials of degree . Indeed, if , we havewhich is impossible. Similarly, , and are also polynomials of degree . Therefore, applying Lemma 2.1 to (4.1), we infer that . Since , has to be a polynomial of degree less than , then , a contradiction.

Therefore, we may proceed under the assumption that . By Lemma 2.2, if , then since the exponential functions and are linearly independent.

Since , by Lemmas 2.3 and 2.4, there exists a set of linear measure zero such that whenever then satisfies either (2.5) or (2.6), (3.1) holds, andIn what follows, we apply the notations from the proof of Theorem 1.5 as well.

*Case 1. *Firstly assume that .
Without loss of generality, we may assume that .
From the hypothesis of ,
we know that .
Therefore, (3.3) holds by Lemma 2.3. Using the same reasoning as in
Case 1
of the proof of Theorem 1.3, we obtain the inequality (3.15) on the
ray .

*Case 2. *Finally, assume that .
Again by the condition on ,
we see that .
Then the same argument as in Case 2
of the proof of Theorem 1.3 applies, and
we again obtain (3.15).

Therefore, by Lemma 2.6, we obtain a contradiction, so .

#### 5. Proof of Theorem 1.5

Contrary to the assertion, suppose that is a transcendental solution of (1.2) of finite-order. If , then it follows from (1.2) thatwhere , and are entire functions of order less than , are distinct, and are also distinct. Similarly as in the proof of Theorem 1.4, we may assume that . Since , we havefor any with , and for sufficiently large. Since and in and are strictly positive, the set is of linear measure zero. Therefore, again by Lemmas 2.3 and 2.4, there exists a set of linear measure zero such that for any given , satisfies either (2.5) or (2.6), and (3.1) holds. Moreover, . Without loss of generality, we may assume that where . Then from (2.5) and (5.2), for any also satisfying , we obtain for sufficiently large thatFor all other coefficients , considering the hypothesis of , we havewhen is large enough. It follows from (1.2) thatSimilarly as in Case 1 of the proof of Theorem 1.4, and using Lemma 2.5, we may prove thatis bounded on the ray . Therefore, the inequality (3.15) always holds on the ray . Then, by Lemma 2.6, a contradiction follows, and so .

#### Acknowledgments

The authors wish to express their thanks to the referee for his valuable suggestions and comments. The authors have been partially supported by the National Natural Science Foundation of China (Grant No. 10871047) and the Academy of Finland (Grant no. 210245 and no. 124954).