- Research
- Open Access
On the hyper-order of solutions of two class of complex linear differential equations
- Wen Ping Huang^{1},
- Jing Lun Zhou^{1},
- Jin Tu^{2}Email author and
- Ju Hong Ning^{2}
https://doi.org/10.1186/s13662-015-0571-y
© Huang et al. 2015
- Received: 28 February 2015
- Accepted: 10 July 2015
- Published: 29 July 2015
Abstract
We investigate the hyper-order of solutions of two class of complex linear differential equations. We investigate the growth of solutions of higher order and certain second order linear differential equations, and we obtain some results which improve and extend some previous results in complex oscillations.
Keywords
- linear differential equation
- entire function
- hyper-order
- maximum modulus
MSC
- 30D35
- 34M10
1 Introduction and results
Theorem A
(see [10])
Suppose that \(\rho(B)<\rho(A)<\infty\) and that \(T(r,A)\sim\log M(r,A)\) as \(r\rightarrow\infty\) outside a set of finite logarithmic measure. Then every nonconstant solution f of (1.1) is of infinite order.
Theorem B
(see [12])
Let \(A_{j}(z)\) (\(j=0,\ldots,k-1\)), \(F(z)\not\equiv0\) be entire functions. Suppose that there exists some \(d\in\{1,\ldots,k-1\}\) such that \(\max\{\rho(F), \rho(A_{j}): j\neq d\}=\rho<\rho(A_{d})<\infty\) and \(T(r,A_{d})\sim\log M(r,A_{d})\) as \(r\rightarrow\infty\) outside a set of upper logarithmic density less than \((\rho(A_{d})-\rho)/\rho(A_{d})\). Then every transcendental solution \(f(z)\) of (1.2) satisfies \(\overline{\lambda}(f)=\lambda(f)=\rho(f)=\infty\).
Theorem C
(see [12])
Let \(A_{j}(z)\) (\(j=0,\ldots,k-1\)), \(F(z)\equiv0\) be entire functions. Suppose that there exists some \(d\in\{1,\ldots,k-1\}\) such that \(\max\{\rho(A_{j}): j\neq0,d\}<\rho(A_{0})\leq\frac{1}{2}\) and that \(A_{d}(z)\) has a finite deficient value. Then every solution \(f(z)\not\equiv0\) of (1.2) satisfies \(\rho(A_{0})\leq\rho_{2}(f)\leq\rho(A_{d})\).
Then a natural question is: Can we estimate the hyper-order of the solutions of (1.1) and (1.2) under the same condition in Theorems A and B? And: Can we estimate the hyper-order of the solutions of (1.2) in Theorem C if \(\rho(A_{0})>\frac{1}{2}\)? Theorems 1.1 and 1.2 below give answers to the above questions.
Theorem D
(see [8])
- (i)If either \(\arg a_{n}\neq \arg b_{n}\) or \(a_{n}=cb_{n}\) with \(0< c<1\), then every nonconstant solution f ofhas infinite order with \(\rho_{2}(f)\geq n\).$$ f''+A_{1}(z)e^{P(z)}f'+A_{0}(z)e^{Q(z)}f=0 $$(1.4)
- (ii)
Let \(a_{n}=b_{n}\) and \(\operatorname{deg}(P-Q)=m\geq 1\), and let the orders of \(A_{1}(z)\) and \(A_{0}(z)\) be less than m. Then every nonconstant solution f of (1.4) has infinite order with \(\rho_{2}(f)\geq m\).
- (iii)
Let \(a_{n}=cb_{n}\) with \(c>1\) and \(\operatorname{deg}(P-cQ)=m\geq1\). Suppose \(\rho(A_{1})< m\) and \(0<\rho(A_{0})<\frac{1}{2}\). Then every nonconstant solution of (1.4) has infinite order with \(\rho_{2}(f)\geq\rho(A_{0})\).
- (iv)
Let \(a_{n}=cb_{n}\) with \(c>1\) and let \(P-cQ\) be a constant. Suppose that \(\rho(A_{1})<\rho(A_{0})<\frac{1}{2}\). Then every nonconstant solution of (1.4) has infinite order with \(\rho_{2}(f)\geq\rho(A_{0})\).
Theorem E
(see [13])
Theorem F
(see [14])
Theorem G
(see [15])
Theorem D left us a question: Can we have \(\rho_{2}(f)=n\) (n is a positive integer) for every nontrivial solution of (1.4) if \(a_{n}\neq b_{n}\)? Theorem E tells us that the question holds if \(n=1\). Many authors investigated the above question but none of them solve the question completely, and Theorem 1.3 completely solves this question. In the following, we give our results.
Theorem 1.1
- (i)
Every transcendental solution f of (1.2) satisfies \(\rho_{2}(f)=\rho(A_{d})\), and (1.2) may have polynomial solutions f of degree <d.
- (ii)
If \(F(z)\not\equiv0\), then every transcendental solution f of (1.2) satisfies \(\overline{\lambda}_{2}(f)=\lambda_{2}(f)=\rho_{2}(f)=\rho(A_{d})\).
- (iii)
If \(d=1\), then every nonconstant solution f of (1.2) satisfies \(\rho_{2}(f)=\rho(A_{1})\). Furthermore, if \(F(z)\not\equiv 0\), then every nonconstant solution f of (1.2) satisfies \(\overline{\lambda}_{2}(f)=\lambda_{2}(f)=\rho_{2}(f)=\rho(A_{1})\).
Theorem 1.2
Let \(A_{j}\) (\(j=0,\ldots,k-1\)), \(F(z)\equiv0\) be entire functions satisfying \(\max\{\rho(A_{j}): j\neq0,d\}<\rho(A_{0})<\rho(A_{d})<\infty\). Suppose that \(T(r,A_{0})\sim\log M(r,A_{0})\) as \(r\rightarrow\infty\) outside a set of r of finite logarithmic measure and that \(A_{d}\) has a finite deficient value. Then every solution \(f\not\equiv0\) of (1.2) satisfies \(\rho(A_{0})\leq\rho_{2}(f)\leq\rho (A_{d})\).
Theorem 1.3
2 Lemmas
Lemma 2.1
(see [16], p.399)
Let \(A_{j}(z)\) (\(j=0,\ldots,k-1\)), \(F(z)\) be entire functions satisfying \(\max\{\rho(A_{j}),\rho(F):j=0,\ldots,k-1\}\leq\rho<\infty\). Then every solution f of (1.2) satisfies \(\rho_{2}(f)\leq\rho\).
Lemma 2.2
(see [17])
- (i)There exist a constant \(B>0\) and a set \(E_{1}\subset(0,\infty)\) having finite logarithmic measure such that, for all z satisfying \(|z|=r\notin E_{1}\), we have$$ \biggl\vert \frac{f^{(j)}(z)}{f^{(i)}(z)}\biggr\vert \leq B\biggl[\frac{T(\alpha r,f)}{r}(\log r)^{\alpha}\log T(\alpha r,f)\biggr]^{j-i} \quad (0\leq i< j). $$(2.1)
- (ii)There exist a set \(H_{1}\subset[0,2\pi)\) that has linear measure zero and a constant \(B>0\) that depends only on α, for any \(\theta\in[0,2\pi)\backslash H_{1}\), there exists a constant \(R_{0}=R_{0}(\theta)>1\) such that, for all z satisfying \(\arg z=\theta\) and \(|z|=r>R_{0}\), we have$$ \biggl\vert \frac{f^{(j)}(z)}{f^{(i)}(z)}\biggr\vert \leq B\bigl[T(\alpha r,f)\log T( \alpha r,f)\bigr]^{j-i}\quad (0\leq i< j). $$(2.2)
Remark 2.1
We use \(E_{2}\subset(0,\infty)\) to denote a set of r of finite logarithmic measure throughout this paper, not necessarily the same at each occurrence.
Lemma 2.3
(see [18])
Lemma 2.4
(see [19])
Lemma 2.5
Proof
Remark 2.2
The following lemma is a special case of Lemma 29 in [12].
Lemma 2.6
(see [18])
Lemma 2.7
(see [20])
Lemma 2.8
(see [21])
Lemma 2.9
(see [22])
- (i)If \(\delta(P,\theta)>0\), then$$ \exp\bigl\{ (1-\varepsilon)\delta(P,\theta)r^{n}\bigr\} < \bigl\vert g \bigl(re^{i\theta}\bigr)\bigr\vert < \exp\bigl\{ (1+\varepsilon)\delta(P, \theta)r^{n}\bigr\} . $$(2.12)
- (ii)If \(\delta(P,\theta)<0\), then$$ \exp\bigl\{ (1+\varepsilon)\delta(P,\theta)r^{n}\bigr\} < \bigl\vert g \bigl(re^{i\theta}\bigr)\bigr\vert < \exp\bigl\{ (1-\varepsilon)\delta(P, \theta)r^{n}\bigr\} . $$(2.13)
3 Proofs of Theorems 1.1-1.3
Proof of Theorem 1.1
(ii) If \(F\not\equiv0\), by Lemma 2.8, we have that every transcendental solution \(f(z)\) of (1.2) satisfies \(\overline{\lambda}_{2}(f)=\lambda_{2}(f)=\rho_{2}(f)=\rho(A_{d})\).
(iii) If \(d=1\), it is easy to see that (1.2) cannot have polynomial solutions, and by (i) and (ii), we see that every nonconstant solution \(f(z)\) of (1.2) satisfies \(\rho_{2}(f)=\rho(A_{1})\) and \(\overline{\lambda}_{2}(f)=\lambda _{2}(f)=\rho_{2}(f)=\rho(A_{1})\) if \(F\not\equiv0\). □
Proof of Theorem 1.2
Proof of Theorem 1.3
(1) By Lemma 2.1, it is easy to see that every solution \(f\not\equiv0\) of (1.7) satisfies \(\rho _{2}(f)\leq n\) and that (1.7) has no polynomial solutions by the assumption. In the following, we need to show that every transcendental solution f of (1.7) satisfies \(\rho_{2}(f)\geq n\). We divide the proof into two parts: (i) \(\arg a_{n}\neq\arg b_{n}\) or \(a_{n}=cb_{n}\) (\(0< c<1\)), (ii) \(a_{n}=cb_{n}\) (\(c>1\)).
(i) \(\arg a_{n}\neq\arg b_{n}\) or \(a_{n}=cb_{n}\) (\(0< c<1\)). By Theorem D(i), every solution \(f\not\equiv0\) of (1.7) satisfies \(\rho_{2}(f)\geq n\).
Combining (i) and (ii), we have every solution \(f\not\equiv0\) of (1.7) satisfies \(\rho_{2}(f)= n\).
(2) By Lemma 2.1, it is easy to see that every solution f of (1.7) satisfies \(\rho_{2}(f)\leq n\). It is easy to know that (1.7) has no polynomial solutions by the assumption. In the following, we need to show that every transcendental solution f of (1.7) satisfies \(\rho_{2}(f)\geq n\). Since \(a_{n}=cb_{n}\) (\(c<0\)), then we have \(\{\theta:\delta(a_{n}z^{n},\theta)>0\}\cap \{\theta:\delta(b_{n}z^{n},\theta)>0\}=\emptyset\) and \(\{\theta:\delta(a_{n}z^{n},\theta)>0\}\cup \{\theta:\delta(b_{n}z^{n},\theta)>0\}\cup H_{3}=[0,2\pi)\), where \(H_{3}\subset[0,2\pi)\) is a set of linear measure zero. For each sufficiently large circle \(|z|=r\), we have if \(z_{r}=re^{i\theta_{r}}\) is a point satisfying \(|f(z_{r})|=M(r,f)\), for any given \(\delta_{4}>0\), set \(I=[\theta_{r}-\delta_{4},\theta_{r}+\delta_{4}]\), then we have either \(m(I\cap\{\theta:\delta(a_{n}z^{n},\theta)>0\})>0\) or \(m(I\cap\{\theta:\delta(a_{n}z^{n},\theta)<0\})>0\). We divide the proof into two cases: (i) \(m(I\cap\{\theta:\delta(a_{n}z^{n},\theta)>0\})>0\), (ii) \(m(I\cap\{\theta:\delta(a_{n}z^{n},\theta)<0\})>0\).
(ii) \(m(I\cap\{\theta:\delta(a_{n}z^{n},\theta)<0\})>0\). Replacing \(|A_{1}e^{P(z)}|\) with \(|A_{0}e^{Q(z)}|\) on the left side of (3.25) and by the same reasoning in case (i), we can obtain \(\rho_{2}(f)\geq n\) for every transcendental solution of (1.7).
Combining (i) and (ii), every solution f of (1.7) satisfies \(\rho_{2}(f)= n\). Since \(F(z)\not\equiv0\), by Lemma 2.8, every solution f of (1.7) satisfies \(\overline{\lambda}_{2}(f)=\lambda_{2}(f)=\rho_{2}(f)=n\). □
Declarations
Acknowledgements
This project is supported by the National Natural Science Foundation of China (11301232, 11301233, 11261024), the Natural Science Foundation of Jiangxi Province in China (20132BAB211002, 20151BAB201004) and the Foundation of Education Bureau of Jiangxi Province in China (GJJ14271, GJJ14272).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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