\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 157, 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} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/157\hfil Entire coefficients having the same order and type] {Linear differential equations with entire coefficients having the same order and type} \author[N. Berrighi, S. Hamouda \hfil EJDE-2011/157\hfilneg] {Nacera Berrighi, Saada Hamouda} % in alphabetical order \address{Nacera Berrighi \newline Laboratory of Pure and Applied Mathematics\\ University of Mostaganem, B.P. 227 Mostaganem, Algeria} \email{n\_berrighi@yahoo.fr} \address{Saada Hamouda \newline Laboratory of Pure and Applied Mathematics\\ University of Mostaganem, B.P. 227 Mostaganem, Algeria} \email{hamouda\_saada@yahoo.fr} \thanks{Submitted October 20, 2011. Published November 21, 2011.} \subjclass[2000]{34M10, 30D35} \keywords{Linear differential equations; growth of solutions; hyper-order} \begin{abstract} In this article, we study the growth of solutions to the differential equation \begin{align*} &f^{k}+(A_{k-1}(z)e^{P_{k-1}(z)}e^{\lambda z^m} +B_{k-1}(z))f^{k-1}+\dots \\ &+(A_0(z)e^{P_0(z)} e^{\lambda z^m}+B_0(z))f=0, \end{align*} where $\lambda \in \mathbb{C}^{\ast}$, $m\geq 2$ is an integer and $\max_{j=0,\dots ,k-1}\{ \deg P_j(z)\} 1)$ and $\deg (P-cQ)=m\geq 1$, $\sigma (A_j)0$, $\delta (P_0,\theta )>0$ and $\delta (P_j,\theta )<0$ $(j=1,\dots ,k-1)$ for all $\theta \in (\theta _1,\theta_2)$. Then every non trivial solution $f$ of \eqref{eq2} is of infinite order with $n\leq \sigma _2(f)\leq m$, where $n=\deg P_0$. \end{theorem} \begin{corollary}\label{cor1} Let $P_j(z)=\sum_{i=0}^{n}a_{i,j}z^i$ $(j=0,\dots ,k-1)$ be non constant polynomials where $a_{i,j}$ are complex numbers such that $a_{n,j}\neq 0$ $(j=0,\dots ,k-1)$, $\arg a_{n,j}=\arg a_{n,1}$ $(j=2,\dots ,k-1)$ and $\arg a_{n,1}\neq \arg a_{n,0}$; $A_j(z)$ $(\not\equiv 0)$, $B_j(z)$, $(j=0,\dots ,k-1)$ be entire functions such that $\sigma (A_j)1)$ and obtain the following results. \begin{theorem}\label{thm3} Let $P_j(z)=\sum_{i=0}^{n}a_{i,j}z^i$ $(j=0,\dots ,k-1)$ be non constant polynomials where $a_{i,j}$ are complex numbers such that $a_{n,0}\neq 0$. Suppose that there exists $s\in \{ 1,\dots ,k-1\} $ such that \begin{equation} \arg (a_{n,j}-a_{n,s})=\varphi \neq \arg ( a_{n,0}-a_{n,s})\quad \text{\ for all }j\neq 0,s. \label{eg1} \end{equation} Then every non trivial solution $f$ of \eqref{eq2} is of infinite order with $n\leq \sigma _2(f)\leq m$. \end{theorem} \begin{theorem}\label{thm4} Let $P_j(z)=\sum_{i=0}^{n}a_{i,j}z^i$ $(j=0,\dots ,k-1)$ be non constant polynomials where $a_{i,j}$ are complex numbers such that $a_{n,0}\neq 0$. Suppose that there exists $s\in \{ 1,\dots ,k-1\} $ such that \begin{equation} a_{n,j}-a_{n,s}=c_j(a_{n,0}-a_{n,s})\ (01$. There exist a set $E\subset [0,2\pi )$ that has linear measure zero and a constant $M>0$ that depends only on $\alpha $ such that for any $\theta \in [0,2\pi )\backslash E $ 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 \[ |\frac{f^{(k)}(z)}{f(z)} |\leq M\big[T(\alpha r,f)\frac{(\log ^{\alpha }r)}{r}\log T(\alpha r,f)\big] ^{k},\quad k\in \mathbb{N}. \] \end{lemma} \begin{lemma}[\cite{gund1}]\label{lem1b} Let $f(z)$ be a transcendental meromorphic function of finite order $\sigma $, and let $\varepsilon >0$ be a given constant. Then there exists a set $E\subset [0,2\pi )$ of linear measure zero such that for all $z=re^{i\theta }$ with $|z|$ sufficiently large and $\theta \in [0,2\pi )\backslash E$, and for all $k,j$, $0\leq j\leq k$, we have \[ |\frac{f^{(k)}(z)}{f^{(j)}(z)}|\leq |z|^{(k-j) (\sigma -1+\varepsilon )}. \] \end{lemma} Using the Wiman-Valiron theory, we can easily prove the following lemma (see \cite{chen1}). \begin{lemma}\label{lem2} Let $A,B$ be entire functions of finite order. If $f$ is a solution of the differential equation \[ f^{(k)}+A_{k-1}f^{(k-1)}+\dots +A_1f'+A_0f=0, \] then $\sigma _2(f)\leq \max_{j=0,\dots ,k-1}\{ \sigma(Aj)\} $. \end{lemma} \begin{lemma}[\cite{chen1}]\label{lem3} Let $P(z)=a_{n}z^{n}+\dots $, $(a_{n}=\alpha +i\beta \neq 0)$ be a polynomial with degree $n\geq 1$ and $A(z)$ $(\not\equiv 0)$ be entire function with $\sigma (A)0$, there exists a set $H\subset [0,2\pi)$ that has linear measure zero, such that for any $\theta \in [0,2\pi )\backslash H$, where $H=\{ \theta \in [0,2\pi ):\delta (P,\theta )=0\} $ is a finite set, there is $R>0$ such that for $|z|=r>R$, we have \begin{itemize} \item[(i)] if $\delta (P,\theta )>0$, then \[ \exp \{ (1-\varepsilon )\delta (P,\theta ) r^{n}\} \leq |f(z)|\leq \exp \{ (1+\varepsilon )\delta (P,\theta )r^{n}\} , \] \item[(ii)] if $\delta (P,\theta )<0$, then \[ \exp \{ (1+\varepsilon )\delta (P,\theta ) r^{n}\} \leq |f(z)|\leq \exp \{ (1-\varepsilon )\delta (P,\theta )r^{n}\} . \] \end{itemize} \end{lemma} \begin{lemma}[\cite{chen2}]\label{lem5} Let $f(z)$ be a entire function with $\sigma (f)=+\infty $ and $\sigma _2(f)=\alpha <+\infty $, let a set $E_2\subset [1,+\infty )$ has finite logarithmic measure. Then there exists a sequence $\{z_{p}=r_{p}e^{i\theta _{p}}\} $ such that $f(z_{p}) =M(r_{p},f)$, $\theta _{p}\in [0,2\pi )$, $\lim_{p\to \infty }\theta _{p}=\theta _0\in [0,2\pi )$, $r_{p}\notin E_2$, and for any given $\epsilon >0$, as $r_{p}\to \infty $, we have \[ \exp \{ r_{p}^{\alpha -\epsilon }\} \leq \nu (r_{p}) \leq \exp \{ r_{p}^{\alpha +\epsilon }\} , \] where $\nu (r)$ is the central index of $f$. \end{lemma} \section{Proofs of theorems} \begin{proof}[Proof of theorem \protect\ref{thm1}] From \eqref{eq2}, we obtain \begin{equation} \label{f1} \begin{split} &|A_0(z)e^{P_0(z)}+B_0(z)e^{-\lambda z^m}|\\ &\leq |e^{-\lambda z^m}||\frac{f^{(k)}}{f}|+ \sum_{j=1}^{k-1}|A_j(z)e^{P_j(z)}+B_j(z)e^{-\lambda z^m}|\,| \frac{f^{(j)}}{f}|. \end{split} \end{equation} Since $\delta (\lambda z^m$, $\theta )>0$, $\delta ( P_0,\theta )>0$ and $\delta (P_j,\theta )<0$ $(j=1,\dots ,k-1)$ for all $\theta \in (\theta _1,\theta _2)$, by Lemma \ref{lem3}, for any $\theta \in ( \theta _1,\theta _2)$ there is $R_0(\theta )>0$ such that for $|z|=r>R_0$, we have \begin{gather} \exp \{ (1-\varepsilon )\delta (P_0,\theta )r^{n}\} \leq |A_0(z)e^{P_0(z) }+B_0(z)e^{-\lambda z^m}|, \label{f2} \\ |A_j(z)e^{P_j(z)}+B_j(z)e^{-\lambda z^m}| \leq C_1\quad (j=1,\dots ,k-1). \label{f3} \end{gather} From Lemma \ref{lem1}, there exist a set $E\subset [0,2\pi )$ that has linear measure zero and a constant $M>0$ such that for any $ \theta \in [0,2\pi )\backslash E$ there exists a constant $R_1=R_1(\theta )>1$ such that for all $z$ satisfying $\arg z=\theta $ and $|z|=r>R_1$, we have \begin{equation} |\frac{f^{(j)}}{f}|\leq C_2[T(2r,f)] ^{2k}\quad (j=1,\dots ,k). \label{f4} \end{equation} By using \eqref{f2}-\eqref{f4} in \eqref{f1}, for $r>\max \{R_0,R_1\} $, we obtain \[ \exp \{ (1-\varepsilon )\delta (P_0,\theta ) r^{n}\} \leq C_{3}[T(2r,f)] ^{2k}, \] which implies that $\sigma _2(f)\geq n$. By Lemma \ref{lem2}, we obtain $n\leq \sigma _2(f)\leq m$. %\label{f5} \end{proof} \begin{proof}[Proof of Corollary \ref{cor1}] In these conditions, there exist $\theta _1,\theta _2$ such that $\theta _1<\theta _2$, $\delta (\lambda z^m,\theta )>0$, $\delta (P_0,\theta )>0$ and $\delta (P_j,\theta )<0$ $(j=1,\dots ,k-1)$ for all $\theta \in (\theta _1,\theta _2)$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2}] Since $m>n$ and $a_{n,j}=c_ja_{n,0}$ $(00$ and $\delta (P_j,\theta )>0$ $(j=0,\dots ,k-1)$ for all $\theta \in (\theta _1,\theta _2)$. In this case from Lemma \ref{lem3}, for sufficiently large $r$, we have \begin{gather} \exp \{ (1-\varepsilon )\delta (P_0,\theta )r^{n}\} \leq |A_0(z)e^{P_0(z)}+B_0(z)e^{-\lambda z^m}|, \label{f6} \\ |A_j(z)e^{P_j(z)}+B_j(z)e^{-\lambda z^m}| \leq \exp \{ ( 1+\varepsilon )c\delta (P_0,\theta )r^{n}\} ,\label{f7} \end{gather} where $c=\max \{ c_j\}$. Using \eqref{f6}, \eqref{f7} and \eqref{f4} in \eqref{f1}, for $r$ large enough, \begin{equation} \exp \{ (1-\varepsilon )\delta (P_0,\theta ) r^{n}\} \leq C_{4}\exp \{ (1+\varepsilon )c\delta (P_0,\theta )r^{n}\} [T(2r,f)]^{2k}, \label{f7b} \end{equation} and thus \begin{equation} \exp \{ (1-\varepsilon -(1+\varepsilon )c) \delta (P_0,\theta )r^{n}\} \leq C_{4}[T( 2r,f)] ^{2k}. \label{f8} \end{equation} Taking $0<\varepsilon <\frac{1-c}{1+c}$, we obtain, from \eqref{f8} and Lemma \ref{lem2}, the desired estimate $n\leq \sigma _2(f)\leq m$. \end{proof} \begin{proof}[Proof of Corollary \ref{cor2}] In this case also there exist $\theta _1<\theta _2$ such that $\delta (\lambda z^m,\theta )>0$, $\delta (P_0,\theta ) >0$ and $\delta (P_{s},\theta )<0$ for all $\theta \in (\theta _1,\theta _2)$. We have \begin{equation} |A_{s}(z)e^{P_{s}(z)}+B_{s}( z)e^{-\lambda z^m}|\leq C_5 \label{f8b} \end{equation} and also for $j$ such that $\arg a_{n,j}=\arg a_{n,s}$ \begin{equation} |A_j(z)e^{P_j(z)}+B_j(z)e^{-\lambda z^m}|\leq C_{6}. \label{f8c} \end{equation} Now for $j$ such that $a_{n,j}=c_ja_{n,0}$ $(00$, $\delta (P_0-P_{s},\theta )>0$ and $\delta (P_j-P_{s},\theta )>0$ for every $\theta \in (\theta _1,\theta _2)$. Thus for $r$ sufficiently large, we have \begin{gather} \exp \{ (1-\varepsilon )\delta (P_0-P_{s},\theta)r^{n}\} \leq |A_0(z)e^{P_0(z)-P_{s}(z)}+B_0(z)e^{-\lambda z^m-P_{s}(z)}|, \label{f9} \\ |A_{s}(z)+B_{s}(z)e^{-\lambda z^m-P_{s}(z)}| \leq \exp \{ r^{\sigma (A_{s})+\varepsilon }\} , \label{f10} \\ |A_0(z)e^{P_j(z)-P_{s}(z)}+B_0(z)e^{-\lambda z^m-P_{s}(z)}| \leq C_{7}. \label{f10b} \end{gather} From \eqref{eq2}, we obtain \begin{equation} \begin{split} &|A_0(z)e^{P_0(z)-P_{s}( z)}+B_0(z)e^{-\lambda z^m-P_{s}(z)}|\\ &\leq |e^{-\lambda z^m-P_{s}(z)}||\frac{f^{(k)}}{f}| +\sum_{j=1}^{k-1}|A_j(z)e^{P_j( z)-P_{s}(z)}+B_j(z)e^{-\lambda z^m-P_{s}(z)}||\frac{f^{(j)}}{f}|. \end{split}\label{f11} \end{equation} Substituting \eqref{f9}-\eqref{f10b} and \eqref{f4} in \eqref{f11}, we obtain \[ \exp \{ (1-\varepsilon )\delta (Q-P,\theta ) r^{n}\} \leq C_{8}\exp \{ r^{\sigma (A_{s}) +\varepsilon }\} [T(2r,f)] ^{2k}. \] Which implies $n\leq \sigma _2(f)$, and by Lemma \ref{lem2}, we obtain $n\leq \sigma _2(f)\leq m$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm4}] There exist $\theta _1<\theta _2$ such that $\delta (\lambda z^m,\theta )>0$ and $\delta (P_j,\theta )>0$ $(j=0,\dots ,k-1)$ for all $\theta \in (\theta _1,\theta _2)$. In this case from Lemma \ref{lem3}, for sufficiently large $r$, we have \begin{gather} \exp \{ (1-\varepsilon )\delta (P_0-P_{s},\theta)r^{n}\} \leq |A_0(z)e^{P_0(z)-P_{s}(z)}+B_0(z)e^{-\lambda z^m-P_{s}(z)}|, \label{f41} \\ |A_{s}(z)+B_{s}(z)e^{-\lambda z^m-P_{s}(z)}|\leq \exp \{ r^{\sigma ( A_{s})+\varepsilon }\} \label{f42} \end{gather} and for $j\neq 0,s$ \begin{equation} |A_j(z)e^{P_j(z)-P_{s}(z)}+B_j(z)e^{-\lambda z^m-P_{s}(z)}| \leq \exp \{ (1+\varepsilon )c\delta ( P_0-P_{s},\theta )r^{n}\} , \label{f43} \end{equation} where $c=\max \{ c_j\} $. Using \eqref{f41}-\eqref{f43} and \eqref{f4} in \eqref{f11}, for $r$ large enough, we obtain %\begin{align*} \[ \exp \{ (1-\varepsilon )\delta (P_0,\theta )r^{n}\} \leq C_{9}\exp \{ r^{\sigma (A_{s})+\varepsilon }\} \exp \{ (1+\varepsilon )c\delta (P_0,\theta ) r^{n}\} [T(2r,f)] ^{2k}, \] %\end{align*} and thus \begin{equation} \exp \{ (1-\varepsilon -(1+\varepsilon )c) \delta (P_0,\theta )r^{n}\} \leq C_{9}\exp \{r^{\sigma (A_{s})+\varepsilon }\} [T( 2r,f)] ^{2k} \label{f44} \end{equation} By taking $0<\varepsilon <\frac{1-c}{1+c}$, from \eqref{f44} and Lemma \ref{lem2}, we obtain $n\leq \sigma _2(f)\leq m$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm5}] By taking $B_j(z)\equiv 0$ $(j=0,\dots ,k-1)$ in previous theorems, we obtain that every solution $f(z) \not\equiv 0$ of \eqref{eq3} is of infinite order with $n\leq \sigma_2(f)\leq m$. It remains to show that $\sigma _2(f)=m$ or $\sigma _2(f)=n$. For that we suppose the contrary, i.e. $n<\sigma _2(f)0$, as $r_{p}\to \infty $, we have \begin{equation} \exp \{ r_{p}^{\gamma -\epsilon }\} \leq \nu (r_{p}) \leq \exp \{ r_{p}^{\gamma +\epsilon }\} . \label{f112} \end{equation} From \eqref{eq3}, we can write \begin{equation} -\frac{f^{(k)}}{f}=\Big(A_{k-1}(z) e^{P_{k-1}(z)}\frac{f^{(k-1)}}{f}+\dots +A_0( z)e^{P_0(z)}\Big)e^{\lambda z^m}. \label{f113} \end{equation} Using \eqref{f111} in \eqref{f113}, we obtain \begin{equation} \label{f114} \begin{split} -\nu ^{k}(r_{p})(1+o(1)) &=(z_{p}A_{k-1}e^{P_{k-1}}\nu ^{k-1}(r_{p})(1+o( 1))+\dots \\ &\quad +z_{p}^{k-1}A_1e^{P_1}\nu (r_{p})(1+o( 1))+z_{p}^{k}A_0e^{Q})e^{\lambda z_{p}^m}. \end{split} \end{equation} Now we prove three cases separately. \textbf{Case 1.} $\delta (\lambda z^m,\theta _0)=:\delta >0$. From \eqref{f112}, for $p$ sufficiently large, we obtain \begin{equation} |-\nu ^{k}(r_{p})(1+o(1))|\leq 2\exp \{ kr_{p}^{\gamma +\epsilon }\} . \label{f115} \end{equation} From Lemma \ref{lem3} and by taking account $\gamma +\epsilon 0$ such that \begin{align*} &(z_{p}A_{k-1}e^{P_{k-1}}\nu ^{k-1}(r_{p})(1+o(1))+\dots +z_{p}^{k-1}A_1e^{P_1}\nu (r_{p})(1+o(1))+z_{p}^{k}A_0e^{Q}) e^{\lambda z_{p}^m}\\ &\leq \exp \{\alpha r_{p}^{n}\} \nu ^{k-1}(r_{p}). \end{align*} Combining this with \eqref{f114}, we obtain \[ \frac{1}{2}|\nu (r_{p})|\leq \exp \{\alpha r_{p}^{n}\} ; \] and with \[ \exp \{ r_{p}^{\gamma -\epsilon }\} \leq |\nu ( r_{p})|, \] and by taking account $n<\gamma -\epsilon $, provided $\epsilon $ small enough, a contradiction follows. \end{proof} \begin{thebibliography}{99} \bibitem{ozaw} I. Amemiya, M. Ozawa; \emph{Non-existence of finite order solutions of $w''+e^{-z}w'+Q(z)w=0$}, Hokkaido Math. J., \textbf{10} (1981), 1-17. \bibitem{chen1} Z. X. 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