\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 164, pp. 1--7.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/164\hfil Growth of solutions] {Growth of solutions to higher order linear homogeneous differential equations in angular domains} \author[N. Wu\hfil EJDE-2010/164\hfilneg] {Nan Wu} \address{Nan Wu \newline Department of Mathematical Sciences, Tsinghua University, Beijing, 100084, China} \email{wunan07@gmail.com} \thanks{Submitted February 23, 2010. Published November 17, 2010.} \thanks{Supported by grant 10871108 from the NSF of China} \subjclass[2000]{30D10, 30D20, 30B10, 34M05} \keywords{Meromorphic solutions; Nevanlinna theory; order} \begin{abstract} In this article, we discuss the growth of meromorphic solutions to higher order homogeneous differential equations in some angular domains, instead of the whole complex plane. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \section{Introduction and statement of main results} By a transcendental meromorphic function, we mean a function that is meromorphic on the whole complex plane, and is not a rational function; in other words, $\infty$ is an essential singular point. We assume the reader is familiar with the Nevanlinna theory of meromorphic functions and basic notation such as: Nevanlinna characteristic $T(r,f)$, integrated counting function $N(r,f)$, and proximity function $m(r,f)$, and the deficiency $\delta(a,f)$ of $f(z)$. For the details, see \cite{Hayman,Yang}. The order $\lambda$ and the lower order $\mu$ are defined as follows: $$\lambda(f)=\limsup_{r\to\infty}\frac{\log T(r,f)}{\log r},\quad \mu(f)=\liminf_{r\to\infty}\frac{\log T(r,f)}{\log r}.$$ It is known the growth of meromorphic solutions of differential equations with meromorphic coefficients in the complex plane $\mathbb{C}$ attracted a lot research. In this article, we discuss the growth of meromorphic solutions of differential equations with transcendental meromorphic coefficients in a proper subset of $\mathbb{C}$. Let $f(z)$ be a meromorphic function in an angular region $\overline{\Omega}(\alpha,\beta)=\{z:\alpha\leq\operatorname{arg} z\leq\beta\}$. Recall the definition of Ahlfors-Shimizu characteristic in an angle (see \cite{Tsuji}). Set $$\Omega(r)=\Omega(\alpha,\beta)\cap\{z:1<|z|0. For q pair of real numbers \{\alpha_j,\beta_j\} satisfying \eqref{1.1} and $$\label{thmeq1.1} \sum_{j=1}^q(\alpha_{j+1}-\beta_j)<\frac{4}{\sigma} \arcsin\sqrt{\delta/2}$$ where \sigma>0 with \mu\leq\sigma\leq\lambda. If A_j(z)(j=1,2,\dots,n) are meromorphic functions in \mathbb{C} with T(r,A_j)=o(T(r,A_0)), then every solution f\not\equiv0 to the equation$$ A_nf^{(n)}+A_{n-1}f^{(n-1)}+\dots+A_0f=0 $$has the order \sigma_X(f)=+\infty in X=\cup_{j=1}^q\{z: \alpha_j\leq\operatorname{arg} z\leq\beta_j\}. \end{theorem} If we remove the condition \mu(A_0)<\infty in Theorem \ref{thm1.1}, we can establish the following result. \begin{theorem} \label{thm1.2} Let A_0(z) be a meromorphic function in \mathbb{C} with nonzero order 0<\lambda\leq\infty and \delta(\infty,A_0)>0. Suppose that for q directions \operatorname{arg} z=\alpha_j(1\leq j\leq q), satisfying$$ -\pi\leq\alpha_1<\alpha_2<\dots<\alpha_q<\pi, \alpha_{q+1} =\alpha_1+2\pi, $$A_j(z), j=1,2,\dots,n, are meromorphic functions in \mathbb{C} with finite lower order and T(r, A_j)=o(T(r, A_0)). Then every solution f\not\equiv0 to the equation$$ A_nf^{(n)}+A_{n-1}f^{(n-1)}+\dots+A_0f=0 $$has order \sigma_X(f)=+\infty in X=\mathbb{C}\backslash\cup_{j=1}^q\{z:\operatorname{arg} z=\alpha_j\}. \end{theorem} The method in this paper was firstly used by Zheng \cite{Zheng} to investigate the growth of transcendental meromorphic functions with radially distributed values. \section{Some auxiliary results} To prove the theorems, we give some lemmas. The following result is from \cite{Yang01,JH01,Zheng}. \begin{lemma}\label{lem1.1} Let f(z) be a transcendental meromorphic function with lower order \mu<\infty and order 0<\lambda\leq\infty, then for any positive number \mu\leq\sigma\leq\lambda and any set E with finite measure, there exist a sequence \{r_n\}, such that \begin{itemize} \item[(1)] r_n\notin E, \lim_{n\to\infty}\frac{r_n}{n}=\infty; \item[(2)] \liminf_{n\to\infty}\frac{\log T(r_n,f)}{\log r_n}\geq\sigma; \item[(3)] T(t,f)<(1+o(1))(\frac{2t}{r_n})^\sigma T(r_n/2,f), t\in[r_n/n,nr_n]; \item[(4)] T(t,f)/t^{\sigma-\varepsilon_n} \leq2^{\sigma+1}T(r_n,f)/r_n^{\sigma-\varepsilon_n}, 1\leq t\leq nr_n, \varepsilon_n=[\log n]^{-2}. \end{itemize} \end{lemma} We recall that \{r_n\} is called the P\'{o}lya peaks of order \sigma outside E. Given a positive function \Lambda(r) satisfying \lim_{r\to\infty}\Lambda(r)=0. For r>0 and a\in\mathbb{C}, define$$ D_\Lambda(r,a)=\{\theta\in[-\pi,\pi):\log^+ \frac{1}{|f(re^{i\theta})-a|}>\Lambda(r)T(r,f)\}, $$and$$ D_\Lambda(r,\infty)=\{\theta\in[-\pi,\pi):\log^+ |f(re^{i\theta})|>\Lambda(r)T(r,f)\}. $$The following result is called the spread relation, which was conjectured by Edrei \cite{Edrei} and proved by Baernstein \cite{Baernstein}. \begin{lemma}\label{lem1.2} Let f(z) be transcendental and meromorphic in \mathbb{C} with the finite lower order \mu<\infty and the positive order 0<\lambda\leq\infty and has one deficient values a\in\widehat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}. Then for any sequence of P\'{o}lya peaks \{r_n\} of order \sigma>0,\mu\leq\sigma\leq\lambda and any positive function \Lambda(r)\to0 as r\to+\infty, we have$$ \liminf_{n\to\infty} \operatorname{meas} D_\Lambda(r_n,a) \geq\min\{2\pi, \frac{4}{\sigma}\arcsin\sqrt{\delta(a,f)/2}\}. $$\end{lemma} To make it clearly, we give the definition of \mathbb{R}-set on the complex plane \mathbb{C}. \begin{definition} \label{def2.2} \rm Let B(z_n, r_n)=\{z: |z-z_n|m_i\geq0 for i=1,2,\dots,j, and let \varepsilon>0 and \delta>0 be given constants. Then there exists K>0 depending only on f,\varepsilon,\delta such that \label{2.1a} \Big|\frac{f^{(n)}(z)}{f^{(m)}(z)}\Big| 0. Then given \varepsilon>0, we have$$ \operatorname{meas} E(r,f)>\frac{1}{T^\varepsilon(r,f)[\log r]^{1+\varepsilon}}, r\notin F, $$where$$ E(r,f)=\{\theta\in[-\pi,\pi): \log^+|f(re^{i\theta})|>\frac{\delta}{4}T(r,f)\} $$and F is a set of positive real numbers with finite logarithmic measure depending on \varepsilon. \end{lemma} \section{Proof of the Theorems} \begin{proof}[Proof of Theorem \ref{thm1.1}] We suppose that there exists a nontrival meromorphic solution f such that \sigma_{\alpha_j,\beta_j}(f)<+\infty, j=1,2,\dots, q. In view of Lemma \ref{lem1.4}, there exists a constant M>0 not depending on z such that $\left|\frac{f^{(j)}(z)}{f(z)}\right|<|z|^M,\quad j=1,2,\dots,n\,.$ for all z\in\Omega(\alpha_j+\varepsilon,\beta_j-\varepsilon), j=1,2,\dots,q, except for a \mathbb{R}-set E. For E, we can define a set F=\{r>0| \exists z\in E, s.t. |z|=r\} thus$$\operatorname{meas} F<\infty.$$(I) \lambda(A_0)>\mu(A_0). Then \lambda(A_0)>\sigma\geq\mu(A_0). By the inequality \eqref{thmeq1.1}, we can take a real number \varepsilon>0 such that $$\label{2.1} \sum_{j=1}^q(\alpha_{j+1}-\beta_j+2\varepsilon)+2\varepsilon <\frac{4}{\sigma+2\varepsilon}\arcsin\sqrt{\delta/2},$$ where \alpha_{q+1}=2\pi+\alpha_1, and$$ \lambda(A_0)>\sigma+2\varepsilon>\mu(A_0). $$Applying Lemma \ref{lem1.1} to A(z) gives the existence of the P\'{o}lya peak \{r_n\} of order \sigma+2\varepsilon of A(z) such that r_n\notin F, and then from Lemma \ref{lem1.2} for sufficiently large n we have $$\label{2.2} \operatorname{meas} D(r_n,\infty)>\frac{4}{\sigma+2\varepsilon}\arcsin \sqrt{\delta/2}-\varepsilon.$$ We can assume for all the n, above holds. Set$$ K:=\operatorname{meas} (D(r_n,\infty)\cap\cup_{j=1}^q(\alpha_j+\varepsilon, \beta_j-\varepsilon)). Then from \eqref{2.1} and \eqref{2.2} it follows that \begin{align*} K&\geq \operatorname{meas}(D(r_n,\infty))-\operatorname{meas}([0,2\pi)\backslash \cup_{j=1}^q(\alpha_j+\varepsilon,\beta_j-\varepsilon))\\ &=\operatorname{meas}(D(r_n,\infty))-\operatorname{meas}(\cup_{j=1}^q(\beta_j -\varepsilon,\alpha_{j+1}+\varepsilon))\\ &=\operatorname{meas}(D(r_n,\infty))-\sum_{j=1}^q(\alpha_{j+1}-\beta_j +2\varepsilon)>\varepsilon>0. \end{align*} It is easy to see that there exists a j_0 such that for infinitely many n, we have $$\label{2.3} \operatorname{meas}(D(r_n,\infty)\cap(\alpha_{j_0}+\varepsilon,\beta_{j_0} -\varepsilon))>\frac{K}{q}.$$ We can assume for all the n, \eqref{2.3} holds. We define a real function by $\Lambda(r)^2=\max\left\{\frac{T(r_n,A_j)}{T(r_n,A_0)}, \frac{\log r_n}{T(r_n,A_0)}; j=1,2,\dots,n\right\},$ for  r_n\leq r\frac{K}{q}\Lambda(r_n)T(r_n, A_0). \end{aligned} Thus, we have $$\label{2.5} \begin{split} &\int_{\alpha_{j_0}+\varepsilon}^{\beta_{j_0}-\varepsilon} \log^+|A_0(r_ne^{i\theta})|d\theta\\ &\leq\int_{\alpha_{j_0}+\varepsilon}^{\beta_{j_0}-\varepsilon} \sum_{j=1}^n\left(\log^+\left|\frac{f^{(j)}(r_ne^{i\theta})} {f(r_ne^{i\theta})}\right|+\log^+|A_j(r_ne^{i\theta})|\right) d\theta\\ &=\Big(\int_{D_\Lambda'(r_n)}+\int_{D_n}\Big) \sum_{j=1}^n\left(\log^+\left|\frac{f^{(j)}(r_ne^{i\theta})}{f(r_n e^{i\theta})}\right|+\log^+|A_j(r_ne^{i\theta})|\right)d\theta\\ &\leq\int_{\alpha_{j_0}+\varepsilon}^{\beta_{j_0}-\varepsilon} \sum_{j=1}^n\log^+|A_j(r_ne^{i\theta})|d\theta+O(\log r_n)\\ &\leq \sum_{j=1}^nT(r_n, A_j)+O(\log r_n)\\ &\leq \Lambda^2(r_n)T(r_n, A_0). \end{split}$$ Therefore, \frac{K}{q}\Lambda(r_n)<\Lambda^2(r_n). $$This contradicts that \Lambda(r)\to 0. (II) \lambda(A_0)=\mu(A_0). Then \lambda(A_0)=\sigma=\mu(A_0). By the same argument as in (I) with all the \sigma+2\varepsilon replaced by \sigma, we can derive a contradiction. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.2}] Applying Lemma \ref{lem1.1} to A_0(z) confirms the existence of a sequence \{r_n\} of positive numbers such that r_n\notin E and $$\label{3.6} \operatorname{meas} E(r_n, A_0)>\frac{1}{T^\varepsilon(r_n,A_0)[\log r_n]^{1+\varepsilon}},$$ where E(r_n, A_0) is defined as in Lemma \ref{lem1.5}. Set$$ \varepsilon_n=\frac{1}{2q+1}\frac{1}{T^\varepsilon(r_n,A_0)[\log r_n]^{1+\varepsilon}}. Then for \eqref{3.6} it follows that \begin{align*} &\operatorname{meas} (E(r_n, A_0)\cap\cup_{j=1}^q(\alpha_j+\varepsilon_n, \alpha_{j+1}-\varepsilon_n))\\ &\geq \operatorname{meas} E(r_n, A_0) -\operatorname{meas}(\cup_{j=1}^q(\alpha_j+\varepsilon_n, \alpha_{j+1}-\varepsilon_n))\\ &\geq(2q+1)\varepsilon_n-2q\varepsilon_n=\varepsilon_n>0. \end{align*} so that there exists a j such that for infinitely many n, we have $$\label{3.7} \operatorname{meas} E_n>\frac{\varepsilon_n}{q},$$ where E_n=E(r_n, A_0)\cap(\alpha_j+\varepsilon_n, \alpha_{j+1}-\varepsilon_n). We can assume that \eqref{3.7} holds for all the n. Thus $$\label{3.8} \begin{split} \int_{\alpha_j+\varepsilon_n}^{\alpha_{j+1}-\varepsilon_n}\log^+ |A_0(r_ne^{i\theta})|d\theta&\geq\int_{E_n}\log^+|A_0(r_ne^{i\theta})|d\theta\\ &\geq \operatorname{meas}(E_n)\frac{\delta}{4}T(r_n, A_0)\\ &\geq \frac{\delta\varepsilon_n}{4q}T(r_n, A_0). \end{split}$$ On the other hand, $$\label{3.9} \int_{\alpha_j+\varepsilon_n}^{\alpha_{j+1}-\varepsilon_n}\log^+ |A_0(r_ne^{i\theta})|d\theta<\sum_{j=1}^n T(r_n, A_j)+O(\log r_n)$$ Combining \eqref{3.8} and \eqref{3.9} gives \varepsilon_n T(r_n, A_0)\leq \frac{4q}{\delta}\sum_{j=1}^n T(r_n, A_j)+O(\log r_n), $$so that$$ T^{1-\varepsilon}(r_n, A_0)\leq\frac{4q(2q+1)}{\delta} [\log r_n]^{1+\varepsilon}\sum_{j=1}^n T(r_n, A_j)+O(\log^{2+\varepsilon} r_n),  we have $\mu(A_0)\leq\max_{1\leq j\leq q}(\mu(A_j))/(1-\varepsilon)$. By the same method as in Theorem \ref{thm1.1}, we obtain a contradiction, which completes the proof. \end{proof} \begin{thebibliography}{00} \bibitem{Baernstein} A. 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