\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 $$\arg (a_{n,j}-a_{n,s})=\varphi \neq \arg ( a_{n,0}-a_{n,s})\quad \text{\ for all }j\neq 0,s. \label{eg1}$$ 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 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 $$\label{f1} \begin{split} &|A_0(z)e^{P_0(z)}+B_0(z)e^{-\lambda z^m}|\\ &\leq |e^{-\lambda z^m}||\frac{f^{(k)}}{f}|+ \sum_{j=1}^{k-1}|A_j(z)e^{P_j(z)}+B_j(z)e^{-\lambda z^m}|\,| \frac{f^{(j)}}{f}|. \end{split}$$ 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 $$|\frac{f^{(j)}}{f}|\leq C_2[T(2r,f)] ^{2k}\quad (j=1,\dots ,k). \label{f4}$$ 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, $$\exp \{ (1-\varepsilon )\delta (P_0,\theta ) r^{n}\} \leq C_{4}\exp \{ (1+\varepsilon )c\delta (P_0,\theta )r^{n}\} [T(2r,f)]^{2k}, \label{f7b}$$ and thus $$\exp \{ (1-\varepsilon -(1+\varepsilon )c) \delta (P_0,\theta )r^{n}\} \leq C_{4}[T( 2r,f)] ^{2k}. \label{f8}$$ 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 $$|A_{s}(z)e^{P_{s}(z)}+B_{s}( z)e^{-\lambda z^m}|\leq C_5 \label{f8b}$$ and also for$j$such that$\arg a_{n,j}=\arg a_{n,s}$$$|A_j(z)e^{P_j(z)}+B_j(z)e^{-\lambda z^m}|\leq C_{6}. \label{f8c}$$ 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 $$\begin{split} &|A_0(z)e^{P_0(z)-P_{s}( z)}+B_0(z)e^{-\lambda z^m-P_{s}(z)}|\\ &\leq |e^{-\lambda z^m-P_{s}(z)}||\frac{f^{(k)}}{f}| +\sum_{j=1}^{k-1}|A_j(z)e^{P_j( z)-P_{s}(z)}+B_j(z)e^{-\lambda z^m-P_{s}(z)}||\frac{f^{(j)}}{f}|. \end{split}\label{f11}$$ 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$$$|A_j(z)e^{P_j(z)-P_{s}(z)}+B_j(z)e^{-\lambda z^m-P_{s}(z)}| \leq \exp \{ (1+\varepsilon )c\delta ( P_0-P_{s},\theta )r^{n}\} , \label{f43}$$ where$c=\max \{ c_j\} $. Using \eqref{f41}-\eqref{f43} and \eqref{f4} in \eqref{f11}, for$rlarge 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 $$\exp \{ (1-\varepsilon -(1+\varepsilon )c) \delta (P_0,\theta )r^{n}\} \leq C_{9}\exp \{r^{\sigma (A_{s})+\varepsilon }\} [T( 2r,f)] ^{2k} \label{f44}$$ By taking0<\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 $$\exp \{ r_{p}^{\gamma -\epsilon }\} \leq \nu (r_{p}) \leq \exp \{ r_{p}^{\gamma +\epsilon }\} . \label{f112}$$ From \eqref{eq3}, we can write $$-\frac{f^{(k)}}{f}=\Big(A_{k-1}(z) e^{P_{k-1}(z)}\frac{f^{(k-1)}}{f}+\dots +A_0( z)e^{P_0(z)}\Big)e^{\lambda z^m}. \label{f113}$$ Using \eqref{f111} in \eqref{f113}, we obtain $$\label{f114} \begin{split} -\nu ^{k}(r_{p})(1+o(1)) &=(z_{p}A_{k-1}e^{P_{k-1}}\nu ^{k-1}(r_{p})(1+o( 1))+\dots \\ &\quad +z_{p}^{k-1}A_1e^{P_1}\nu (r_{p})(1+o( 1))+z_{p}^{k}A_0e^{Q})e^{\lambda z_{p}^m}. \end{split}$$ 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 $$|-\nu ^{k}(r_{p})(1+o(1))|\leq 2\exp \{ kr_{p}^{\gamma +\epsilon }\} . \label{f115}$$ From Lemma \ref{lem3} and by taking account$\gamma +\epsilon 0such 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 accountn<\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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