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# On the Exponent of Convergence for the Zeros of the Solutions of

*Journal of Inequalities and Applications*
**volumeÂ 2010**, ArticleÂ number:Â 428936 (2010)

## Abstract

Let and be entire functions of order less than 1 with and transcendental. We prove that every solution of the equation , , being has zeros with infinite exponent of convergence.

## 1. Introduction

It is assumed that the reader of this paper is familiar with the basic concepts of Nevanlinna theory [1, 2] such as , and . Suppose that is a meromorphic function, then the order of growth of the function and the exponent of convergence of the zeros of are defined, respectively, as

Let be a measurable subset of . The lower logarithmic density and the upper logarithmic density of are defined, respectively, by

where is the characteristic function of defined as

Now let us recall some of the previous results on the linear differential equation

where is an entire function of finite order, When is polynomial, many authors [3â€“6] have studied the properties of the solutions of (1.4). If is a transcendental entire function with , Gundersen [7] proved that every nontrivial solution of (1.4) has infinite order of growth. In [8], Wang and Laine considered the nonhomogeneous equation of type

where are entire functions of order less than one and are complex numbers. In fact, they have proved the following theorem.

Theorem 1.1.

Suppose that are entire functions of order less than one, and suppose that with and . Then every nontrivial solution of (1.5) is of infinite order.

Corollary 1.2.

Suppose that , where is a nonvanishing entire function with and with . Then every nontrivial solution of (1.4) is of infinite order.

## 2. Results

We observe that all the above results concern the order of growth only. In this paper, we are going to prove the following theorem which concerns the exponent of convergence.

Theorem 2.1.

Let and be entire functions of order less than 1 with and being transcendental. Then every solution of the equation

has zeros with infinite exponent of convergence.

The hypothesis that is transcendental is not redundant since Frei [4] has shown that

has solutions of finite order, for certain constants .

We notice that Theorem 2.1 fails for . For any entire function the function solves (2.1) with

## 3. Some Lemmas

Throughout this paper we need the following lemmas. In 1965, Hayman [9] proved the following lemma.

Lemma 3.1.

Let the function be meromorphic of finite order in the plane and let . Then

for all outside a set of upper logarithmic density , where the positive constant depends only on and .

In 1962, Edrei and Fuchs [10] proved the following lemma.

Lemma 3.2.

Let be a meromorphic function in the complex plane and let have measure . Then

In 2007, Bergweiler and Langley [11] proved the following lemma.

Lemma 3.3.

Let be a transcendental entire function of order . For large define to be the length of the longest arc of the circle on which , with if the minimum modulus satisfies . Then at least one of the following is true:

(i)there exists a set of positive upper logarithmic density such that for ;

(ii)for each the set has lower logarithmic density at least

We deduce the following.

Lemma 3.4.

Let , let be a positive integer, and let have logarithmic density . Let be a transcendental entire function such that on a path tending to infinity and for all with and . Then has order at least .

Proof.

Assume that and choose a polynomial of degree at most such that

is transcendental entire. Then we have for all and for all with and . With the notation of Lemma 3.3, we see that for all large , and so we must have case (ii), as well as for . Define by

Since has logarithmic density this gives

## 4. Proof of Theorem 2.1

Let , and be as in the hypotheses. We can assume that . Suppose that is a solution of (2.1) having zeros with finite exponent of convergence. Then we can write

where and are entire functions with . We can assume that , since if is constant we can replace by and by . Using (2.1) and (4.1), we get

Lemma 4.1.

One has .

Proof.

Suppose that . Dividing (4.2) by , we get

Hence, provided lies outside a set of finite measure,

and so, using the fact that and have order less than , we obtain

This holds outside a set of finite measure and so for all large , since we may take with so that

Lemma 4.1 is proved.

Let denote large positive constants. Choose with

There exists an -set [2, page 84] such that for all large outside , we have

and using the Poisson-Jensen formula,

Moreover, there exists a set of logarithmic density such that for the circle does not meet the -set .

Lemma 4.2.

The functions and are both transcendental.

Proof.

Let be small and positive and suppose that or is a polynomial. Let be large with and . Since is small it follows from (4.2) and (4.8) that . Choose with such that the intersection of with the ray given by is bounded. Applying Lemma 3.4 to the function , with a subpath of , gives , but may be chosen arbitrarily small, and this contradicts (4.7).

The next step is to estimate in the right half-plane.

Lemma 4.3.

Let be a large positive integer and let . Then for large with

one has, either

or

Proof.

Let be large and satisfy (4.10), and assume that (4.11) does not hold. Then (4.8) implies that

Also, (4.7), and (4.9) give

Here denote positive constants which may depend on but not on . Using (4.8), (4.12) and (4.14) we get, from (4.2),

Now divide (4.2) by . We obtain, using (4.15),

which gives and (4.12). This proves Lemma 4.3.

Lemma 4.4.

Let and be as in Lemma 4.3. Choose such that the ray has bounded intersection with the -set . Let be the union of the ray and the arcs , where is the set chosen following (4.9). Then one of the following holds:

(i)one has (4.11) for all large in ;

(ii)one has (4.12) for all large in .

Proof.

This follows simply from continuity. For each large in we have (4.11) or (4.12), but we cannot have both because of (4.14). This proves Lemma 4.4.

Lemma 4.5.

Let . Then for large with , one has

Proof.

Let be as in the hypotheses. Since we only need to prove (4.17) for . Assume that . Then dividing (4.2) by gives

by (4.8), and so (4.17) follows using (4.7). This proves Lemma 4.5.

Lemma 4.6.

If conclusion (i) of Lemma 4.4 holds then , while if conclusion (ii) of Lemma 4.4 holds then .

Proof.

Suppose that conclusion (i) of Lemma 4.4 holds. Choose such that

and let be small compared to . Assume that in Lemma 4.4 is small compared to , in particular so small that

where is the positive constant from Lemma 3.1. Let

and let be the exceptional set of Lemma 3.1, with . Then for large we have, using (4.20) and Lemmas 3.1, 3.2, and 4.5,

We then have

for large . Now take any large . Since has logarithmic density , while has upper logarithmic density at most , and since is small, there exists with

which proves Lemma 4.6 in this case. The alternative case, in which we have conclusion (ii) in Lemma 4.4, is proved the same way, using in place of .

To finish the proof suppose first that conclusion (ii) of Lemma 4.4 holds. Then Lemma 3.4 implies that has order at least . Since may be chosen arbitrarily small, this contradicts Lemma 4.6. The same contradiction, with replaced by , arises if conclusion (i) of Lemma 4.4 holds, and the proof of the theorem is complete.

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## Acknowledgment

The author thanks Professor J. K. Langley for the invaluable discussions on the results of this paper during his visit in summer 2008 and summer 2010 to the University of Nottingham in the U.K.

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Alotaibi, A. On the Exponent of Convergence for the Zeros of the Solutions of .
*J Inequal Appl* **2010**, 428936 (2010). https://doi.org/10.1155/2010/428936

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DOI: https://doi.org/10.1155/2010/428936