\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 65, pp. 1--8.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/65\hfil Finite order solutions] {Finite order solutions of \\ complex linear differential equations} \author[I. Laine \& R. Yang\hfil EJDE-2004/65\hfilneg] {Ilpo Laine \& Ronghua Yang } % in alphabetical order \address{Ilpo Laine \hfill\break University of Joensuu, Department of Mathematics \\ P.O. Box 111, FIN-80101 Joensuu, Finland} \email{ilpo.laine@joensuu.fi} \address{Ronghua Yang \hfill\break University of Joensuu, Department of Mathematics \\ P.O. Box 111, FIN-80101 Joensuu, Finland} \email{ronghua.yang@joensuu.fi} \date{} \thanks{Submitted December 10, 2003. Published April 28, 2004.} \thanks{Partially supported by grant 50981 from the Academy of Finland.} \subjclass[2000]{30D35, 34M10} \keywords{Linear differential equations, growth of solutions, iterated order} \begin{abstract} We shall consider the growth of solutions of complex linear homogeneous differential equations $$f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots +A_1(z)f'+A_0(z)f=0$$ with entire coefficients. If one of the intermediate coefficients in exponentially dominating in a sector and $f$ is of finite order, then a derivative $f^{(j)}$ is asymptotically constant in a slightly smaller sector. We also find conditions on the coefficients to ensure that all transcendental solutions are of infinite order. This paper extends previous results due to Gundersen and to Bela\"{\i}di and Hamani. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem*{remark}{Remark} \section{Introduction}\label{intro} It is well known that all solutions of the linear differential equation $$\label{0lin} f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots +A_{1}(z)f'+A_{0}(z)f=0$$ are entire functions, provided the coefficients $A_{0}(z)\not\equiv 0$, $A_{1}(z),\ldots ,A_{k-1}(z)$ are entire. A classical result, due to Wittich, tells that all solutions of (\ref{0lin}) are of finite order of growth if and only if all coefficients $A_{j}(z)$, $j=0,\ldots ,k-1,$ are polynomials. For a complete analysis of possible orders in the polynomial case, see \cite{GSW}. If some (or all) of the coefficients are transcendental, a natural question is to ask when and how many solutions of finite order may appear. Partial results have been available since a paper of Frei \cite{Fr}. In its all generality, however, the problem remains open. Our starting point for this paper is a result due to Gundersen in \cite{Gary1}: \begin{theorem}\label{thma} Let $A_{0}(z)\not\equiv 0, A_{1}(z)$ be entire functions such that for some real constants $\alpha >0$, $\beta >0$, $\theta_{1}<\theta_{2}$ we have \begin{gather*} |A_{1}(z)|\geq \exp ((1+o(1))\alpha |z|^{\beta}),\\ |A_{0}(z)|\leq \exp (o(1)|z|^{\beta}) \end{gather*} as $z\to\infty$ in the sector $S(0):\theta_{1}\leq \arg z\leq \theta_{2}$. Given $\varepsilon >0$ small enough, let $S(\varepsilon )$ denote the sector $\theta_{1}+\varepsilon \leq \arg z \leq \theta_{2}-\varepsilon$. If $f$ is a nontrivial solution of $$\label{2lin} f''+A_{1}(z)f'+A_{0}(z)f=0$$ of finite order, then the following conditions hold: \begin{itemize} \item[(i)] There exists a constant $b\neq 0$ such that $f(z)\to b$ as $z\to\infty$ in $S(\varepsilon)$. Indeed, $$|f(z)-b|\leq \exp (-(1+o(1))\alpha |z|^{\beta}).$$ \item[(ii)] For each integer $k\geq 1$, $$|f^{(k)}(z)|\leq \exp (-(1+o(1))\alpha|z|^{\beta})$$ as $z\to\infty$ in $S(\varepsilon)$. \end{itemize} \end{theorem} This result has been recently generalized to the higher order case (\ref{0lin}) by Bela\"{\i}di and Hamani, see \cite{BEL}, as follows: \begin{theorem}\label{thmb} Let $A_{0}(z)\not\equiv 0,A_{1}(z),\ldots ,A_{k-1}(z)$ be entire functions such that for some real constants $\alpha >0,\beta >0, \theta_{1}<\theta_{2}$, we have \begin{gather*} |A_{1}(z)|\geq \exp((1+o(1))\alpha |z|^{\beta}),\\ |A_{j}(z)|\leq \exp (o(1)|z|^{\beta}),\quad j=0,2,3,\ldots ,k-1, \end{gather*} as $z\to\infty$ in $S(0)$. If $f$ is a nontrivial solution of (\ref{0lin}) of finite order, then the following conditions hold, provided $\varepsilon >0$ is small enough: \begin{itemize} \item[(i)] There exists a constant $b\neq 0$ such that $f(z)\to b$ as $z\to\infty$ in $S(\varepsilon )$. Indeed, $$|f(z)-b|\leq \exp (-(1+o(1))\alpha |z|^{\beta}).$$ \item[(ii)] For each integer $k\geq 1$, $$|f^{(k)}(z)|\leq \exp (-(1+o(1))\alpha |z|^{\beta})$$ as $z\to\infty$ in $S(\varepsilon )$. \end{itemize} \end{theorem} A natural question is now to ask about a counterpart of Theorem \ref{thmb} in the case that the coefficient in the same sense as in Theorem \ref{thmb} is $A_{s}(z)$ instead of $A_{1}(z)$. We are going to present such a counterpart in this paper, see Theorem~\ref{Gary} below. As the proof now is more complicated than the corresponding proof of Theorem \ref{thmb}, see \cite{BEL}, we express the growth conditions for the coefficients more explicitly. In fact, making use of $o(1)$ only as in Theorem \ref{thmb} might leave some doubts on the necessary uniformity in the course of the proof. \section{Notation and results}\label{notats} Given $\varepsilon >0$, and $\theta_{1},\theta_{2}\in [0,2\pi ), \theta_{1}<\theta_{2}$, we denote by $S(\theta_{1},\theta_{2},\varepsilon )$, resp. $S(R,\theta_{1},\theta_{2},\varepsilon )$, the sector $\{ z|\theta_{1}+\varepsilon \leq \arg z\leq \theta_{2}-\varepsilon\}$, resp. the truncated sector $S(\theta_{1},\theta_{2},\varepsilon )\cap \{ |z|\geq R\}$. If the sector boundaries are clear, and there is no possibility of confusion, we apply the shorter notations $S(\varepsilon)$, resp. $S(R,\varepsilon )$. In the proof of Theorem~1 below, we agree that whenever stating that $r\geq r_{j}$ to indicate that $|z|=r$ has to be large enough, we always assume that $r_{j}\geq r_{j-1}$. Hence, in such a situation, all corresponding previous conditions for $r\geq r_{j-1}$ remain valid, without saying this explicitly in what follows. \begin{theorem}\label{Gary} Let $\theta_{1}<\theta_{2}$ be given to fix a sector $S(0)$, let $k\geq 2$ be a natural number, and let $\delta >0$ be any real number such that $k\delta < 1$. Suppose that $A_{0}(z),\ldots A_{k-1}(z)$ with $A_{0}(z)\not\equiv 0$ are entire functions such that for real constants $\alpha >0$, $\beta >0$, we have, for some $s=1,\ldots ,k-1$, \begin{gather}\label{large} |A_{s}(z)|\geq \exp ((1+\delta )\alpha |z|^{\beta}),\\ \label{small} |A_{j}(z)|\leq \exp (\delta \alpha |z|^{\beta}) \end{gather} for all $j=0,\ldots ,s-1,s+1,\ldots ,k-1$ whenever $|z|=r\geq r_{\delta}$ in the sector $S(0)$. Given $\varepsilon >0$ small enough, if $f$ is a transcendental solution of finite order $\rho <\infty$ of the linear differential equation $$\label{lin} f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots +A_{1}(z)f'+A_{0}(z)f=0,$$ then the following conditions hold: \begin{itemize} \item[(i)] There exists $j\in \{ 0,\ldots ,s-1\}$ and a complex constant $b_{j}\neq 0$ such that $f^{(j)}(z)\to b_{j}$ as $z\to\infty$ in the sector $S(\varepsilon)$. More precisely, $$\label{almostb} |f^{(j)}(z)-b_{j}|\leq \exp (-(1-k\delta )\alpha |z|^{\beta})$$ in $S(\varepsilon )$, provided $|z|$ is large enough. \item[(ii)] For each integer $m\geq j+1$, $$\label{dersmall} |f^{(m)}(z)|\leq \exp (-(1-k\delta )\alpha |z|^{\beta})$$ in $S(3\varepsilon )$ for all $|z|$ large enough. \end{itemize} \end{theorem} \begin{remark} \rm In Theorem \ref{Gary}, it may happen that $j0$, $\beta>0$ be given constants, let $\delta >0$ be a real number such that $k\delta <1$ and let $s$ be an integer such that $1\leq s\leq k-1$. Suppose that (i) $\rho (A_{j})<\beta$ for $j\neq s$ and (ii) for any given $\varepsilon>0$, there exists two finite collections of real numbers $(\phi_{m})$ and $(\theta_{m})$ that satisfy $\phi_{1}<\theta_{1}<\phi_{2}<\theta_{2}<\dots <\phi_{n}<\theta_{n}<\phi_{n+1}=\phi_{1}+2\pi$, such that \begin{gather}\label{epsilon} \sum_{m=1}^{n}(\phi_{m+1}-\theta_{m})<\varepsilon,\\ \label{suuri} |A_{s}(z)|\geq \exp ((1+\delta )\alpha |z|^{\beta}) \end{gather} as $z\to\infty$ in $\phi_{m}\leq \arg z \leq \theta_{m}$, $m=1,\ldots ,n$. Then every transcendental solution $f$ of (\ref{lin}) is of infinite order. \end{theorem} Invoking the iterated order $\rho_{p}(f):=\limsup_{r\to\infty}(\log_{p}T(r,f))/\log r$ for entire functions, we add our final theorem which is a simple extension of \cite[Theorem 1.7]{BEL}. \begin{theorem}\label{triv} Let $A_{0}(z),\ldots ,A_{k-1}(z)$ be entire functions such that for some integer $s$, $1\leq s \leq k-1$, we have $\rho_{p} (A_{j})\leq \alpha <\beta =\rho_{p} (A_{s})\leq +\infty$ for all $j\neq s$. Then every transcendental solution $f$ of (\ref{lin}) satisfies $\rho_{p} (f)\geq \rho_{p} (A_{s})$. \end{theorem} \section{Preparations for proofs}\label{preps} To prove Theorem \ref{Gary}, we need two preparatory lemmas. The first one is a simple extension of \cite[Lemma 4]{Gary1}. See also \cite[Lemma 3]{GS}. \begin{lemma}\label{logderline} Let $f(z)$ be an entire function, and suppose that $|f^{(k)}(z)|$ is unbounded on a ray $\arg z=\theta$. Then there exists a sequence $z_{n}=r_{n}e^{i\theta}$ tending to infinity such that $f^{(k)}(z_{n})\to\infty$ and that $$\label{quotient} \left|\frac{f^{(j)}(z_{n})}{f^{(k)}(z_{n})}\right|\leq \frac{1}{(k-j)!}(1+o(1))|z_{n}|^{k-j},$$ provided $j0$, $\beta >0$, $K\geq \frac{1}{2}$ and $0<\eta <\frac{1}{2}$, the integral $I(r):=\int_{r}^{+\infty}\exp (-K\alpha t^{\beta})dt$ converges. More precisely, if $\beta >1$, then $I(r)\leq \exp (-K\alpha r^{\beta})$, whenever $r^{\beta -1}\geq \frac{1}{K\alpha\beta}$ and if $\beta \leq 1$, then $I(r)\leq \exp (-(K-\eta )\alpha r^{\beta})$ for any given $\eta \in (0,1/2)$, provided $\eta\alpha r^{\beta}\geq (1-\beta )\log r + \log \frac{2}{\alpha\beta}$. \end{lemma} \begin{remark} \rm Observe that the lower bound obtained for $r$ above is independent of $K$, in both cases. Moreover, if we take $r$ large enough, say $r\geq r_{0}\geq r_{\delta }$, then $I(r)\leq \exp (-(K-\eta )\alpha r^{\beta})$, in both cases again. \end{remark} \begin{proof} Clearly, $$\label{arvio} \begin{split} I(r)&=\int_{r}^{+\infty}\exp (-K\alpha t^{\beta})dt = \frac{K\alpha\beta r^{\beta -1}}{K\alpha\beta r^{\beta -1}}\int_{r}^{+\infty}\exp (-K\alpha t^{\beta})dt \\ &\leq -\frac{1}{K\alpha\beta r^{\beta -1}}\int_{r}^{+\infty}-K\alpha\beta t^{\beta -1}\exp (-K\alpha t^{\beta})dt \\ &= \frac{1}{K\alpha\beta r^{\beta -1}}\exp (-K\alpha r^{\beta}). \end{split}$$ If now $\beta >1$, it suffices to have $K\alpha\beta r^{\beta -1}\geq 1$, as required. If $\beta \leq 1$, we may write the above estimate for $I(r)$ as $$\label{next} I(r)\leq \frac{2}{\alpha\beta}r^{1-\beta}\exp (-\eta\alpha r^{\beta})\exp(-(K-\eta )\alpha r^{\beta}).$$ Provided $\frac{2}{\alpha\beta}r^{1-\beta}\exp (-\eta\alpha r^{\beta}) \leq 1$, we again have the required estimate. \end{proof} \section{Proofs of theorems}\label{proofs} \begin{proof}[Proof of Theorem $\ref{Gary}$] \textbf{Boundedness of $f^{(s)}$ in $S(\varepsilon )$.} Recall first that by \cite{Gary2}, Corollary 1, there exists a set $E\subset [0,2\pi )$ of linear measure zero such that for all $k\geq s\geq 0$, all $j=s+1,\ldots ,k$, and all $r\geq r_{1}$, $$\label{dersmall2} \Big|\frac{f^{(j)}(z)}{f^{(s)}(z)}\Big|\leq |z|^{(j-s)(\rho -1+\varepsilon)} \leq |z|^{k\rho}$$ along any ray $\arg z = \psi$ such that $\psi \in [0,2\pi)\setminus E$, provided $0<\varepsilon <1$. Suppose now, for a while, that $|f^{(s)}(z)|$ is unbounded on some ray $\arg z = \phi \in S(0)\setminus E$. By Lemma \ref{logderline}, there exists a sequence of points $z_{n}=r_{n}e^{i\phi}$, $r_{n}\to\infty$ such that $f^{(s)}(z_{n})\to\infty$ and so $$\label{dersmall3} \Big|\frac{f^{(j)}(z_{n})}{f^{(s)}(z_{n})}\Big|\leq \frac{1}{(s-j)!}(1+o(1))|z_{n}|^{s-j}\leq 2|z_{n}|^{k}$$ for all $j=0,\ldots ,s-1$ and all $n$ large enough, say $|z_{n}|\geq r_{2}$. From (\ref{lin}), we next conclude that $$\label{estAs} \begin{split} |A_{s}|\leq& \Big|\frac{f^{(k)}}{f^{(s)}}\Big| +|A_{k-1}|\Big|\frac{f^{(k-1)}}{f^{(s)}}\Big|+\dots +|A_{s+1}|\Big|\frac{f^{(s+1)}}{f^{(s)}}\Big| \\ &+|A_{s-1}|\Big|\frac{f^{(s-1)}}{f^{(s)}}\Big|+\dots +|A_{1}|\Big|\frac{f'}{f^{(s)}}\Big| +|A_{0}|\Big|\frac{f}{f^{(s)}}\Big|. \end{split}$$ Combining now (\ref{small}), (\ref{dersmall2}) and (\ref{dersmall3}) with the above estimate (\ref{estAs}) for $A_{s}$, it is straightforward to see that $|A_{s}(z_{n})|\leq \exp (3\delta\alpha r_{n}^{\beta})$ for all $n$ large enough in the sequence $z_{n}$ on the ray $\arg z=\phi$, contradicting (\ref{large}). Therefore, $|f^{(s)}(z)|$ remains bounded on all rays $\arg z =\phi \in S(0)\setminus E$. By a standard application of the Phragm\'{e}n--Lindel\"{o}f principle, we conclude that $f^{(s)}(z)$ is bounded, say $|f^{(s)}(z)|\leq M$, in the whole sector $S(\varepsilon)$. \smallskip \noindent \textbf{Preliminary estimate for $|f^{(m)}(z)|$, $m\leq s$.} We now proceed to show that $|f^{(m)}(z)|=O(|z|^{s-m})$ on any ray $\arg z = \phi \in S(0)\setminus E$, for all $m\leq s$. Of course, it suffices to consider $ms$.} We may now restrict ourselves to the sector $S(3\varepsilon )$. We assume that $r\geq r_{4}$ is large enough to satisfy that for an arbitrary $z=re^{i\theta}\in S(3\varepsilon )$, the disk $\Gamma (z)$ of radius at most $\rho = \max_{ss$, we may use (\ref{estders}) in the Cauchy formula to see that $$\label{Cauchy} |f^{(m)}(z)|\leq \frac{(m-s)!}{2\pi}\int_{\Gamma (z)}\frac{|f^{(s)}(\zeta )|}{|z-\zeta |^{m-s+1}}d\zeta .$$ By the selection of $\rho$ above, we may combine (\ref{estders}) and (\ref{Cauchy}) to conclude that $$\label{fm} |f^{(m)}(z)|\leq \exp (-(1-\delta )\alpha |z|^{\beta}).$$ \noindent\textbf{Proof of the first assertion for $j=s-1$.} Fix now $\theta \in S(2\varepsilon)$, and define $$\label{as} a_{s}:=\int_{0}^{+\infty}f^{(s)}(te^{i\theta})e^{i\theta}dt = \lim_{R\to\infty}\int_{0}^{R}f^{(s)}(te^{i\theta })e^{i\theta }dt.$$ By (\ref{estders}), it is not difficult to see that $a_{s}\in\mathbb{C}$. Moreover, the definition of $a_{s}$ is independent of $\theta$. Indeed, integrating $f^{(s)}(\zeta )$ along the sector boundary $0\to Re^{i\phi }\to Re^{i\theta }\to 0$, and using (\ref{estders}) to conclude that the integral of $f^{(s)}(\zeta )$ over the arc $[Re^{i\phi },Re^{i\theta }]$ tends to zero as $R\to\infty$, the independence from $\theta$ immediately follows. Define now $b_{s-1}:=f^{(s)}(0)+a_{s}$, and suppose that $b_{s-1}\neq 0$. Let $z=re^{i\phi}$ be an arbitrary point in $S(2\varepsilon )$ such that $r\geq r_{4}$. Then, since $$\label{erotus} f^{(s-1)}(z)-b_{s-1}=\int_{0}^{z}f^{(s)}(\zeta )d\zeta - \int_{0}^{+\infty}f^{(s)}(te^{i\phi })e^{i\phi }dt,$$ we may apply (\ref{estders}) and Lemma \ref{integral} to conclude that $$\label{estb} \begin{split} |f^{(s-1)}(z)-b_{s-1}| & = \Bigl|\int_{0}^{z}f^{(s)}(\zeta )d\zeta -\int_{0}^{\infty}f^{(s)}(te^{i\phi })e^{i\phi }dt\Bigr| \\ & = \Bigl|\int_{\infty }^{z}f^{(s)}(te^{i\phi })e^{i\phi }dt\Bigr|\leq \int_{|z|}^{\infty}|f^{(s)}(te^{i\phi })|dt \\ & \leq \int_{r}^{\infty}\exp (-(1-\delta )\alpha t^{\beta})dt \\ &\leq \exp (-(1-2\delta )\alpha r^{\beta}), \end{split}$$ provided that $r\geq r_{4}$. Since we assumed that $b_{s-1}\neq 0$, we have completed the proof of the first assertion in this case. \smallskip \noindent\textbf{Proof of the first assertion for $j0$, let $(\phi_{m})$ and $(\theta_{m})$ be as in the assumptions. From (\ref{suuri}) and the supposition that $\rho (A_{j})<\beta$ whenever $j\neq s$, we conclude by using Theorem \ref{Gary}(ii) that $|f^{(s)}(z)|$ is bounded in each of the sectors $\phi_{m}+3\varepsilon\leq\arg z\leq\theta_{m}-3\varepsilon, m=1,\ldots ,n$. As $\varepsilon$ is arbitrarily small, we infer from (\ref{epsilon}) and the Phragm\'{e}n--Lindel\"{o}f principle that $|f^{(s)}(z)|$ must be bounded in the whole complex plane. By the Liouville theorem, $f$ has to be a polynomial, a contradiction. \end{proof} \begin{proof}[Proof of Theorem \ref{triv}] Suppose $f$ is a transcendental solution of (\ref{lin}) such that $\rho_{p}(f)<\rho_{p}(A_{s})$. Writing (\ref{lin}) in the form $$\label{mod} A_{s}(z)f^{(s)}=-\sum_{j=0,j\neq s}^{k}A_{j}(z)f^{(j)},$$ and making use of the elementary iterated order (in)equalities and the invariance of the iterated order under differentiation, we immediately observe that the left hand side of (\ref{mod}) is of iterated order $\rho_{p}(A_{s})$, while the right hand side must be of iterated order $\leq \max_{j\neq s}(\rho_{p}(f),\rho_{p}(A_{j}))<\rho_{p}(A_{s})$, a contradiction. \end{proof} \noindent\textbf{Acknowledgement.} The authors would like to thank the anonymous referee for his/her careful reading of the manuscript. \begin{thebibliography}{99} \bibitem{BEL} Bela\"{\i}di, B. and Hamani, K., \emph{Order and hyper-order of entire solutions of linear differential equations with entire coefficients}, Electronic J. Differ. 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