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On meromorphic solutions of some linear differential equations with entire coefficients being Fabry gap series
- Shun-Zhou Wu^{1} and
- Xiu-Min Zheng^{1}Email author
https://doi.org/10.1186/s13662-015-0380-3
© Wu and Zheng; licensee Springer. 2015
- Received: 31 July 2014
- Accepted: 20 January 2015
- Published: 31 January 2015
Abstract
Keywords
- linear differential equation
- meromorphic solution
- growth
- exponent of convergence
- Fabry gap series
MSC
- 30D35
- 34M10
1 Introduction and main results
We make use of the standard notations of Nevanlinna’s value distribution theory (see, e.g., [1–3]). In the whole paper, let \(f(z)\) be a meromorphic function in the whole complex plane.
Firstly, let us recall the following definitions (see, e.g., [4–6]).
Definition 1.1
Definition 1.2
Definition 1.3
Further, we can get the definitions \(\lambda_{p}(f-\varphi)\) and \(\overline{\lambda}_{p}(f-\varphi)\), when a is replaced by a meromorphic function \(\varphi(z)\).
In 1993, Chen and Gao considered the growth of solutions of (1.1) and obtained the following theorem in [9].
Theorem A
(see [9])
- (i)
\(\sigma(A_{j})<\sigma(A_{0})<\infty\), \(j=1,\ldots,k-1\)
- (ii)
\(A_{0}(z)\) is a transcendental entire function with \(\sigma(A_{0})<\infty\), and \(A_{j}(z)\), \(j=1,\ldots,k-1\), are polynomials.
Generally, when \(A_{d}(z)\) (\(0\leq d\leq k-1\)) is dominant, Chen and Gao obtained the following theorem in [10] in 1997.
Theorem B
(see [10])
- (i)
\(\sigma(A_{j})<\sigma(A_{d})<\frac{1}{2} \) (\(j=0,\ldots,d-1,d+1,\ldots,k-1\))
- (ii)
\(A_{d}(z)\) is transcendental with \(\sigma(A_{d})=0\) and \(A_{j}(z)\) (\(j\neq d\)) are polynomials.
Theorems A and B give the properties of solutions of (1.1) when there is exactly one coefficient that has the maximal order. Thus, a natural question arises: how about the properties of solutions of (1.1) when there is more than one coefficient having the maximal order? In this paper, we proceed in this way.
Theorem 1.1
Remark 1.1
Suppose that \(A_{k}(z)=\sum_{n=0}^{\infty}c_{\lambda_{n}}z^{\lambda_{n}}\) is an entire function, and the sequence of exponents \(\{\lambda_{n}\}\) satisfies Fabry gap condition (1.3), then the series \(\sum_{n=0}^{\infty}c_{\lambda_{n}}z^{\lambda_{n}}\) is called a Fabry gap series. It follows by [16] that if \(A_{k}(z)\) is a Fabry gap series, then it has no deficient values. In particular, zero is not a deficient value of \(A_{k}(z)\), then the solutions of (1.2) are meromorphic in general.
Theorem 1.2
- (i)
\(\sigma(A_{j})<\sigma(A_{k})<\frac{1}{2}\), \(j=0,1,\ldots,k-1\)
- (ii)
\(A_{k}(z)\) is transcendental with \(\sigma(A_{k})=0\), and \(A_{j}(z)\), \(j=0,1,\ldots,k-1\), are polynomials.
Theorem 1.3
- (i)If \(\sigma(F)<\sigma(A_{k})\) (now \(A_{k}(z)\) does not satisfy (ii) of Theorem 1.2), then every rational solution \(f(z)\) of (1.4) is a polynomial with \(\deg f\leq k-1\), and every transcendental meromorphic solution \(f(z)\) of (1.4), whose poles are of uniformly bounded multiplicities such that \(\lambda(\frac{1}{f})<\mu(f)\), satisfieswhere \(\varphi(z)\) is a finite order meromorphic function and does not solve (1.4).$$\overline{\lambda}(f-\varphi)=\lambda(f-\varphi)=\sigma(f)=\infty,\qquad \overline{\lambda}_{2}(f-\varphi)=\lambda_{2}(f-\varphi)= \sigma_{2}(f)=\sigma(A_{k}), $$
- (ii)If \(\sigma(F)>\sigma(A_{k})\), then every infinite order meromorphic solution \(f(z)\) of (1.4) satisfieswhere \(\varphi(z)\) is a finite order meromorphic function and does not solve (1.4). And every finite order meromorphic solution \(f_{0}(z)\) satisfies$$\overline{\lambda}(f-\varphi)=\lambda(f-\varphi)=\sigma(f)=\infty, $$$$\sigma(F)\leq\sigma(f_{0})\leq\max\bigl\{ \sigma(F), \overline{ \lambda}(f_{0})\bigr\} . $$
For the case of entire solutions, we can deduce the following Corollary 1.1 easily.
2 Preliminary lemmas
Lemma 2.1
(see [17])
Lemma 2.2
(see [6])
Lemma 2.3
Proof
Lemma 2.4
We may deduce the following Remark 2.1 from Lemma 2.4 immediately.
Remark 2.1
Lemma 2.5
(see [6])
- (i)
\(\max\{i(F)=q, i(A_{j})\ (j=0,1,\ldots,k-1)\}< i(f)=p+1\) (\(0< p<\infty\)),
- (ii)
\(b=\max\{\sigma_{p+1}(F),\sigma_{p+1}(A_{j})\ (j=0,1,\ldots,k-1)\}<\sigma _{p+1}(f)=\sigma\).
Lemma 2.6
(see [12])
Suppose that \(k\geq2\), \(A_{j}(z)\), \(j=0,1,\ldots,k-1\), are meromorphic functions, \(\sigma=\max\{\sigma(A_{j}), j=0,1,\ldots,k-1\}\). If \(f(z)\) is a transcendental meromorphic solution of (1.1) and all poles of \(f(z)\) are of uniformly bounded multiplicity, then we have \(\sigma_{2}(f)\leq\sigma\).
Lemma 2.7
(see [17])
Lemma 2.8
(see [18])
Lemma 2.9
(see [19])
Lemma 2.10
(see [11])
Lemma 2.11
(see [20])
Let \(g: [0, +\infty)\rightarrow\mathbb{R}\) and \(h: [0, +\infty)\rightarrow\mathbb{R}\) be monotone nondecreasing functions such that \(g(r) \leq h(r)\) for all \(r\notin E\cup[0,1]\), where \(E\subset(1, +\infty)\) is a set of finite logarithmic measure. Let \(\alpha>1\) be a given constant. Then there exists \(r_{0}=r_{0}(\alpha)>0\) such that \(g(r) \leq h(\alpha r)\) for all \(r > r_{0}\).
Lemma 2.12
(see [6])
3 Proofs of Theorems 1.1-1.3
Proof of Theorem 1.1
Suppose that \(f(z)\) is a rational solution of (1.2). Since \(\sigma(A_{k})>\max\{\sigma(A_{j}), j=0,1,\ldots,k-1\}\), it is clear that \(f(z)\) is a polynomial with \(\deg f\leq k-1\).
Proof of Theorem 1.2
(i) Suppose that \(f(z)\) is a rational solution of (1.2). Since \(\sigma(A_{k})>\max\{\sigma(A_{j}), j=0,1,\ldots,k-1\}\), it is clear that \(f(z)\) is a polynomial with \(\deg f\leq k-1\).
(ii) Suppose that \(f(z)\) is a rational solution of (1.2). Since \(A_{k}(z)\) is transcendental and \(A_{j}(z)\), \(j=0,1,\ldots,k-1\), are polynomials, we can easily obtain that \(f(z)\) is a polynomial with \(\deg f\leq k-1\).
Proof of Theorem 1.3
We prove only the case under the hypotheses of Theorem 1.1, and the case under the hypotheses of Theorem 1.2 can be proved similarly. So, we omit the proof of the second case.
(i) Suppose that \(f(z)\) is a rational solution of (1.4). Since \(\sigma(A_{k})>\max\{\sigma(A_{j}), j=0, 1,\ldots, k-1, \sigma(F)\}\), it is clear that \(f(z)\) is a polynomial with \(\deg f\leq k-1\).
Since \(0<\varepsilon<\frac{\sigma-\delta}{2}\), (3.16) is a contradiction. Therefore, every transcendental meromorphic solution \(f(z)\) of (1.4), whose poles are of uniformly bounded multiplicities such that \(\lambda(\frac{1}{f})<\mu(f)\), satisfies \(\sigma(f)=\infty\).
Declarations
Acknowledgements
This work was supported by the National Natural Science Foundation of China (11301233, 11171119), the Youth Science Foundation of Education Bureau of Jiangxi Province (GJJ14271) and Sponsored Program for Cultivating Youths of Outstanding Ability in Jiangxi Normal University of China.
Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
Authors’ Affiliations
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