\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 114, pp. 1--17.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/114\hfil Growth of solutions] {Growth of solutions to higher-order linear differential equations with entire coefficients} \author[H. Habib, B. Bela\"idi \hfil EJDE-2014/114\hfilneg] {Habib Habib, Benharrat Bela\"idi} % in alphabetical order \address{Habib Habib \newline Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem (UMAB), B. P. 227 Mostaganem, Algeria} \email{habibhabib2927@yahoo.fr} \address{Benharrat Bela\"idi \newline Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem (UMAB), B. P. 227 Mostaganem, Algeria} \email{belaidi@univ-mosta.dz} \thanks{Submitted November 22, 2013. Published April 21, 2014.} \subjclass[2000]{34M10, 30D35} \keywords{Linear differential equation; entire solution; order of growth; \hfill\break\indent hyper-order of growth; fixed point} \begin{abstract} In this article, we discuss the order and hyper-order of the linear differential equation $f^{(k) }+\sum_{j=1}^{k-1} (B_je^{b_jz}+D_je^{d_jz}) f^{(j) }+( A_1e^{a_1z}+A_2e^{a_2z}) f=0,$ where $A_j(z), B_j(z), D_j(z)$ are entire functions $(\not\equiv 0)$ and $a_1,a_2, d_j$ are complex numbers $(\neq 0)$, and $b_j$ are real numbers. Under certain conditions, we prove that every solution $f\not\equiv 0$ of the above equation is of infinite order. Then, we obtain an estimate of the hyper-order. Finally, we give an estimate of the exponent of convergence for distinct zeros of the functions $f^{(j) }-\varphi$ $(j=0,1,2)$, where $\varphi$ is an entire function $(\not\equiv 0)$ and of order $\sigma (\varphi ) <1$, while the solution $f$ of the differential equation is of infinite order. Our results extend the previous results due to Chen, Peng and Chen and others. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction and statement of results} Throughout this paper, we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna's value distribution theory (see \cite{h2,y1}). Let $\sigma (f)$ denote the order of growth of an entire function $f$ and the hyper-order $\sigma _2(f)$ of $f$ is defined by (see \cite{y1}) $\sigma _2(f) =\limsup_{r\to +\infty} \frac{\log \log T(r,f) }{\log r} =\limsup_{r\to +\infty} \frac{\log \log \log M(r,f) }{\log r},$ where $T(r,f)$ is the Nevanlinna characteristic function of $f$ and $M(r,f) =\max_{| z| =r}| f(z) |.$ To give some estimates of fixed points, we recall the following definition. \begin{definition}[\cite{c2,l2}] \label{def1.1} \rm Let $f$ be a meromorphic function. Then the exponent of convergence of the sequence of distinct fixed points of $f(z)$ is defined by $\overline{\tau }(f) =\overline{\lambda }(f-z) = \limsup_{r\to +\infty} \frac{\log \overline{N}(r, \frac{1}{f-z}) }{\log r},$ where $\overline{N}(r,1/f)$ is the counting function of distinct zeros of $f(z)$ in $\{z:| z|\leq r\}$. We also define $\overline{\lambda }(f-\varphi ) =\limsup_{r\to +\infty } \frac{\log \overline{N}(r,\frac{1}{f-\varphi })}{\log r}$ for any meromorphic function $\varphi (z)$. \end{definition} For the second-order linear differential equation $$f''+e^{-z}f'+B(z) f=0, \label{e1.1}$$ where $B(z)$ is an entire function, it is well-known that each solution $f$ of equation \eqref{e1.1} is an entire function, and that if $f_1$ and $f_2$ are two linearly independent solutions of \eqref{e1.1}, then by \cite{f1}, there is at least one of $f_1$, $f_2$ of infinite order. Hence, most'' solutions of \eqref{e1.1} will have infinite order. But equation \eqref{e1.1} with $B(z)=-(1+e^{-z})$ possesses a solution $f(z) =e^{z}$ of finite order. A natural question arises: What conditions on $B(z)$ will guarantee that every solution $f\not\equiv 0$ of \eqref{e1.1} has infinite order? Many authors, Frei \cite{f2}, Ozawa \cite{o1}, Amemiya-Ozawa \cite{a1} and Gundersen \cite{g2}, Langley \cite{l1} have studied this problem. They proved that when $B(z)$ is a nonconstant polynomial or $B(z)$ is a transcendental entire function with order $\sigma (B)\neq 1$, then every solution $f\not\equiv 0$ of \eqref{e1.1} has infinite order. In 2002, Chen \cite{c3} considered the question: What conditions on $B(z)$ when $\sigma (B)$ $=1$ will guarantee that every nontrivial solution of \eqref{e1.1} has infinite order? He proved the following result, which improved results of Frei, Amemiya-Ozawa, Ozawa, Langley and Gundersen. \begin{theorem}[\cite{c3}] \label{thmA} Let $A_j(z)$ $(\not\equiv 0)$ $(j=0,1)$ be entire functions with $\max \{\sigma (A_j)$ $(j=0,1) \}<1$. and let $a,b$ be complex constants that satisfy $ab\neq 0$ and $a\neq b$. Then every solution $f\not\equiv 0$ of the differential equation $f''+A_1(z) e^{az}f'+A_0(z) e^{bz}f=0$ is of infinite order. \end{theorem} In \cite{p1}, Peng and Chen investigated the order and hyper-order of solutions of some second order linear differential equations and have proved the following result. \begin{theorem}[\cite{p1}] \label{thmB} Let $A_j(z)$ $(\not\equiv 0)$ $(j=1,2)$ be entire functions with $\sigma (A_j) <1$, $a_1$, $a_2$ be complex numbers such that $a_1a_2\neq 0$, $a_1\neq a_2$ (suppose that $| a_1| \leq |a_2|$). If $\arg a_1\neq \pi$ or $a_1<-1$, then every solution $f(\not\equiv 0)$ of the differential equation $f''+e^{-z}f'+(A_1e^{a_1z}+A_2e^{a_2z}) f=0$ has infinite order and $\sigma _2(f) =1$. \end{theorem} Recently in \cite{h1}, the authors extend and improve the results of Theorem \ref{thmB} to some higher order linear differential equations as follows. \begin{theorem}[\cite{h1}] \label{thmC} Let $A_j(z)$ $(\not\equiv 0)$ $(j=1,2)$, $B_{l}(z)$ $(\not\equiv 0)$ $(l=1,\dots ,k-1)$, $D_{m}$ $(m=0,\dots ,k-1)$ be entire functions with $\max \{\sigma (A_j) ,\sigma (B_{l}) ,\sigma (D_{m}) \} <1$, $b_{l}$ $(l=1,\dots ,k-1)$ be complex constants such that \begin{itemize} \item[(i)] $\arg b_{l}=\arg a_1$ and $b_{l}=c_{l}a_1$ $(0\frac{| a_1| }{1-\beta }$ or (ii) $| a_2| <(1-\alpha ) |a_1|$; or \item[(3)] $a_1<0$ and $\arg a_1\neq \arg a_2$; or \item[(4)] (i) $(1-\beta ) a_2-b0$\ be a given constant, and let $k$, $j$ be integers satisfying $k>j\geq 0$. Then, there exists a set $E_1\subset [-\frac{\pi }{2},\frac{3\pi }{2})$ with linear measure zero, such that, if $\psi \in [-\frac{\pi }{2},\frac{3\pi }{2}) \setminus E_1$, then there is a constant $R_0=R_0(\psi ) >1$, such that for all $z$ satisfying $\arg z=\psi$ and $| z| \geq R_0$, we have $$| \frac{f^{(k) }(z) }{f^{(j)}(z) }| \leq | z| ^{(k-j) (\sigma -1+\varepsilon ) }. \label{e2.1}$$ \end{lemma} \begin{lemma}[\cite{c3}] \label{lem2.2} Suppose that $P(z) =(\alpha +i\beta ) z^n+\dots$ ($\alpha ,\beta$ are real numbers, $| \alpha |+| \beta | \neq 0$) is a polynomial with degree $n\geq 1$, that $A(z)$ $(\not\equiv 0)$ is an entire function with $\sigma (A) 0$, there is a set $E_2\subset [0,2\pi )$ that has linear measure zero, such that for any $\theta \in [0,2\pi ) \setminus (E_2\cup E_3)$, there is $R>0$, such that for $| z| =r>R$, we have \begin{itemize} \item[(i)] If $\delta (P,\theta) >0$, then $$\exp \{(1-\varepsilon ) \delta (P,\theta ) r^n\} \leq | g(re^{i\theta }) | \leq \exp \{(1+\varepsilon ) \delta (P,\theta )r^n\} . \label{e2.2}$$ \item[(ii)] If $\delta (P,\theta ) <0$, then $$\exp \{(1+\varepsilon ) \delta (P,\theta ) r^n\} \leq | g(re^{i\theta }) | \leq \exp \{(1-\varepsilon ) \delta (P,\theta )r^n\} , \label{e2.3}$$ where $E_3=\{\theta \in [0,2\pi ) :\delta (P,\theta ) =0\}$ is a finite set. \end{itemize} \end{lemma} \begin{lemma}[\cite{p1}] \label{lem2.3} Suppose that $n\geq 1$ is a natural number. Let $P_j( z) =a_{jn}z^n+\dots$ $(j=1,2)$ be nonconstant polynomials, where $a_{jq}$ ($q=1,\dots ,n$) are complex numbers and $a_{1n}a_{2n}\neq 0$. Set $z=re^{i\theta }$, $a_{jn}=| a_{jn}| e^{i\theta _j}$, $\theta _j\in [-\frac{\pi }{2},\frac{3\pi }{2})$, $\delta (P_j,\theta ) =| a_{jn}| \cos (\theta _j+n\theta )$, then there is a set $E_4\subset[-\frac{\pi }{2n},\frac{3\pi }{2n})$ that has linear measure zero such that if $\theta _1\neq \theta _2$, then there exists a ray $\arg z=\theta$, $\theta \in (-\frac{\pi }{2n}, \frac{\pi }{2n}) \setminus (E_4\cup E_5)$, satisfying $$\delta (P_1,\theta ) >0,\quad \delta (P_2,\theta) <0 \label{e2.4}$$ or $$\delta (P_1,\theta ) <0,\quad \delta (P_2,\theta) >0, \label{e2.5}$$ where $E_5=\{\theta \in [-\frac{\pi }{2n},\frac{3\pi }{2n}) :\delta (P_j,\theta ) =0\}$ is a finite set, which has linear measure zero. \end{lemma} \begin{remark}[\cite{p1}] \label{rmk2.1} \rm In Lemma \ref{lem2.3}, if $\theta \in (-\frac{\pi }{2n},\frac{\pi }{2n}) \setminus (E_4\cup E_5)$ is replaced by $\theta \in (\frac{\pi }{2n},\frac{3\pi }{2n}) \setminus (E_4\cup E_5)$, then we obtain the same result. \end{remark} \begin{lemma}[\cite{c4}] \label{lem2.4} Suppose that $k\geq 2$ and $B_0,B_1,\dots ,B_{k-1}$ are entire functions of finite order and let $\sigma =\max \{\sigma ( B_j) :j=0,\dots ,k-1\}$. Then every solution $f$ of the differential equation $$f^{(k) }+B_{k-1}f^{(k-1) }+\dots +B_1f'+B_0f=0 \label{e2.6}$$ satisfies $\sigma _2(f) \leq \sigma$. \end{lemma} \begin{lemma}[\cite{g3}] \label{lem2.5} Let $f(z)$ be a transcendental meromorphic function, and let $\alpha >1$ be a given constant. Then there exist a set $E_6\subset (1,\infty )$ with finite logarithmic measure and a constant $B>0$ that depends only on $\alpha$ and $i,j$ $(0\leq i1$ be a given constant. Then there exists an $r_1=r_1(\gamma ) >0$ such that $\varphi (r) \leq \psi (\gamma r)$ for all $r>r_1$. \end{lemma} \begin{lemma}[\cite{c1}] \label{lem2.7} Let $A_0,A_1,\dots ,A_{k-1}$, $F\not\equiv 0$ be finite order meromorphic functions. If $f(z)$ is an infinite order meromorphic solution of the equation $$f^{(k) }+A_{k-1}f^{(k-1) }+\dots +A_1f'+A_0f=F, \label{e2.8}$$ then $f$ satisfies $\overline{\lambda }(f)=\lambda (f) =\sigma (f) =\infty$. \end{lemma} The following lemma, due to Gross \cite{g1}, is important in the factorization and uniqueness theory of meromorphic functions, playing an important role in this paper as well. \begin{lemma}[\cite{g1,y1}] \label{lem2.8} Suppose that $f_1(z) ,f_2(z),\dots ,f_{n}(z) (n\geq 2)$ are meromorphic functions and $g_1(z) ,g_2(z),\dots ,g_{n}(z)$ are entire functions satisfying the following conditions: \begin{itemize} \item[(i)] $\sum_{j=1}^nf_j(z) e^{g_j(z) }\equiv 0$; \item[(ii)] $g_j(z) -g_{k}(z)$ are not constants for $1\leq j1$, such that for all $z$ satisfying $\arg z=\theta$ and $| z| =r\geq R_0$, we have $$| \frac{f^{(j) }(z) }{f(z) } | \leq r^{j(\sigma -1+\varepsilon ) }\quad (j=1,\dots ,k) . \label{e3.3}$$ Let $z=re^{i\theta }$, $a_1=| a_1| e^{i\theta _1}$, $a_2=| a_2| e^{i\theta _2}$, $\theta _1,\theta_2\in [-\frac{\pi }{2},\frac{3\pi }{2})$. We know that $\delta (\alpha _ja_1z,\theta ) =\alpha _j\delta (a_1z,\theta )$, $\delta (\beta _ja_2z,\theta ) =\beta _j\delta (a_2z,\theta )$ $(j=1,\dots ,k-1)$ and $\alpha <1$, $\beta <1$. \smallskip \noindent \textbf{Case 1.} Assume that $\arg a_1\neq \pi$ and $\arg a_1\neq \arg a_2$, which is $\theta _1\neq \pi$ and $\theta _1\neq \theta _2$. By Lemma \ref{lem2.2} and Lemma \ref{lem2.3}, for any given $\varepsilon$, $0<\varepsilon <\min \{1-\gamma ,\frac{1-\alpha }{2(1+\alpha ) },\frac{1-\beta }{2(1+\beta ) }\} ,$ there is a ray $\arg z=\theta$ such that $\theta \in (-\frac{\pi }{2}, \frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5)$ (where $E_4$ and $E_5$ are defined as in Lemma \ref{lem2.3}, $E_1\cup E_4\cup E_5$ is of the linear measure zero), and satisfying $\delta (a_1z,\theta ) >0\text{, }\delta (a_2z,\theta) <0$ or $\delta (a_1z,\theta ) <0\text{, }\delta (a_2z,\theta) >0.$ (a) When $\delta (a_1z,\theta ) >0$, $\delta (a_2z,\theta ) <0$, for sufficiently large $r$, we obtain by Lemma \ref{lem2.2}, \begin{gather} | A_1e^{a_1z}| \geq \exp \{(1-\varepsilon) \delta (a_1z,\theta ) r\} , \label{e3.4} \\ | A_2e^{a_2z}| \leq \exp \{(1-\varepsilon) \delta (a_2z,\theta ) r\} <1, \label{e3.5} \\ \begin{aligned} | D_je^{\alpha _ja_1z}| &\leq \exp \{(1+\varepsilon ) \alpha _j\delta (a_1z,\theta )r\} \\ &\leq \exp \{(1+\varepsilon ) \alpha \delta ( a_1z,\theta ) r\} \quad (j=1,\dots ,k-1) , \end{aligned} \label{e3.6} \\ | e^{\beta _ja_2z}| \leq \exp \{( 1-\varepsilon ) \beta _j\delta (a_2z,\theta ) r\} <1\ (j=1,\dots ,k-1) . \label{e3.7} \end{gather} By \eqref{e3.6} and \eqref{e3.7}, we obtain $$| D_je^{(\alpha _ja_1+\beta _ja_2) z}| =| D_je^{\alpha _ja_1z}| | e^{\beta _ja_2z}| \leq \exp \{(1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\} , \label{e3.8}$$ where $j=1,\dots ,k-1$. For $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$, by \eqref{e3.2}, we have $$| B_je^{b_jz}| =| B_j|\,| e^{b_jz}| \leq \exp \{r^{\gamma +\varepsilon }\} e^{b_jr\cos \theta }\leq \exp \{r^{\gamma +\varepsilon }\} \label{e3.9}$$ because $b_j<0$ and $\cos \theta >0$ $(j=1,\dots ,k-1)$. By \eqref{e3.1}, we obtain $$| A_1e^{a_1z}| \leq | \frac{f^{(k) }}{f}| +\sum_{j=1}^{k-1} \Big(| B_je^{b_jz}| +| D_je^{(\alpha _ja_1+\beta _ja_2) z}| \Big) | \frac{f^{(j) }}{f}| +| A_2e^{a_2z}| . \label{e3.10}$$ Substituting \eqref{e3.3} -\eqref{e3.5}, \eqref{e3.8} and \eqref{e3.9} in \eqref{e3.10}, we have \begin{aligned} \exp \{(1-\varepsilon ) \delta (a_1z,\theta) r\} & \leq | A_1e^{a_1z}| \\ & \leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} \exp \{(1+\varepsilon ) \alpha \delta (a_1z,\theta) r\} , \end{aligned} \label{e3.11} where $M_1>0$ and $M_2>0$ are some constants. By $0<\varepsilon <\frac{1-\alpha }{2(1+\alpha ) }$ and \eqref{e3.11}, we obtain $$\exp \{\frac{1-\alpha }{2}\delta (a_1z,\theta )r\} \leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} . \label{e3.12}$$ By $\delta (a_1z,\theta ) >0$ and $\gamma +\varepsilon <1$ we know that \eqref{e3.12} is a contradiction. \smallskip \noindent(b) When $\delta (a_1z,\theta ) <0$, $\delta (a_2z,\theta ) >0$, for sufficiently large $r$, we obtain \begin{gather} | A_2e^{a_2z}| \geq \exp \{(1-\varepsilon) \delta (a_2z,\theta ) r\},\label{e3.13} \\ | A_1e^{a_1z}| \leq \exp \{(1-\varepsilon) \delta (a_1z,\theta ) r\} <1,\label{e3.14} \\ | D_je^{\alpha _ja_1z}| \leq \exp \{(1-\varepsilon ) \alpha _j\delta (a_1z,\theta ) r\} <1\quad (j=1,\dots ,k-1) , \label{e3.15} \\ \begin{aligned} | e^{\beta _ja_2z}| &\leq \exp \{(1+\varepsilon ) \beta _j\delta (a_2z,\theta ) r\}\\ &\leq \exp \{(1+\varepsilon ) \beta \delta ( a_2z,\theta ) r\} \quad (j=1,\dots ,k-1) . \end{aligned} \label{e3.16} \end{gather} By \eqref{e3.15} and \eqref{e3.16}, we have $$| D_je^{(\alpha _ja_1+\beta _ja_2)z}| =| D_je^{\alpha _ja_1z}|\, |e^{\beta _ja_2z}| \leq \exp \{(1+\varepsilon ) \beta \delta (a_2z,\theta ) r\} , \label{e3.17}$$ where $j=1,\dots ,k-1$. By \eqref{e3.1}, we obtain $$| A_2e^{a_2z}| \leq | \frac{f^{(k) }}{f}| +\sum_{j=1}^{k-1} \Big(| B_je^{b_jz}| +| D_je^{(\alpha _ja_1+\beta _ja_2) z}| \Big) | \frac{ f^{(j) }}{f}| +| A_1e^{a_1z}| . \label{e3.18}$$ Substituting \eqref{e3.3}, \eqref{e3.9}, \eqref{e3.13}, \eqref{e3.14} and \eqref{e3.17} in \eqref{e3.18}, we have \begin{aligned} \exp \{(1-\varepsilon ) \delta (a_2z,\theta) r\} &\leq | A_2e^{a_2z}| \\ & \leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} \exp \{(1+\varepsilon ) \beta \delta (a_2z,\theta) r\} . \end{aligned}\label{e3.19} By $0<\varepsilon <\frac{1-\beta }{2(1+\beta ) }$ and \eqref{e3.19}, we obtain $$\exp \big\{\frac{1-\beta }{2}\delta (a_2z,\theta ) r\big\} \leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} . \label{e3.20}$$ By $\delta (a_2z,\theta ) >0$ and $\gamma +\varepsilon <1$ we know that \eqref{e3.20} is a contradiction. \smallskip \noindent\textbf{Case 2.} Assume that $\arg a_1\neq \pi$, $\arg a_1=\arg a_2$, which is $\theta _1\neq \pi$, $\theta _1=\theta _2$. By Lemma \ref{lem2.3}, for any given $\varepsilon$ $0<\varepsilon <\min \big\{1-\gamma ,\frac{(1-\alpha ) | a_1| -| a_2| }{2[(1+\alpha ) | a_1| +| a_2|] }, \frac{(1-\beta ) | a_2| -| a_1| }{2[(1+\beta ) |a_2| +| a_1| ] }\big\} ,$ there is a ray $\arg z=\theta$ such that $\theta \in (-\frac{\pi }{2}, \frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5)$ and $\delta (a_1z,\theta ) >0$. Since $\theta _1=\theta _2$, then $\delta (a_2z,\theta ) >0$. \smallskip \noindent (i) $| a_2| >\frac{| a_1| }{1-\beta }$. For sufficiently large $r$, we have \eqref{e3.6}, \eqref{e3.13} , \eqref{e3.16} hold and $$| A_1e^{a_1z}| \leq \exp \{(1+\varepsilon ) \delta (a_1z,\theta ) r\} . \label{e3.21}$$ By \eqref{e3.6} and \eqref{e3.16}, we obtain $$| D_je^{(\alpha _ja_1+\beta _ja_2) z}| \leq \exp \{(1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\} \exp \{(1+\varepsilon ) \beta \delta (a_2z,\theta ) r\} , \label{e3.22}$$ where $j=1,\dots ,k-1$. Substituting \eqref{e3.3}, \eqref{e3.9}, \eqref{e3.13}, \eqref{e3.21} and \eqref{e3.22} in \eqref{e3.18}, we have \begin{aligned} &\exp \{(1-\varepsilon ) \delta (a_2z,\theta) r\} \\ &\leq | A_2e^{a_2z}| \\ & \leq k\exp \{r^{\gamma +\varepsilon }\} \exp \{( 1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\} \exp \{(1+\varepsilon ) \beta \delta (a_2z,\theta ) r\} r^{k(\sigma -1+\varepsilon ) } \\ &\quad +\exp \{(1+\varepsilon ) \delta (a_1z,\theta) r\} \\ &\leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} \exp \{(1+\varepsilon ) \delta (a_1z,\theta ) r\} \exp \{(1+\varepsilon ) \beta \delta ( a_2z,\theta ) r\} . \end{aligned} \label{e3.23} From \eqref{e3.23}, we obtain $$\exp \{\eta _1r\} \leq M_1r^{M_2}\exp \{r^{\gamma+\varepsilon }\} , \label{e3.24}$$ where $\eta _1=(1-\varepsilon ) \delta (a_2z,\theta ) -(1+\varepsilon ) \delta (a_1z,\theta ) -(1+\varepsilon ) \beta \delta (a_2z,\theta ) .$ Since $0<\varepsilon <\frac{(1-\beta ) |a_2| -| a_1| }{2[(1+\beta) | a_2| +| a_1| ] },$ $\theta _1=\theta _2$ and $\cos (\theta _1+\theta ) >0$, we have \begin{align*} \eta _1&=[1-\beta -\varepsilon (1+\beta ) ] \delta (a_2z,\theta ) -(1+\varepsilon ) \delta (a_1z,\theta ) \\ &=[1-\beta -\varepsilon (1+\beta ) ] |a_2| \cos (\theta _1+\theta ) -(1+\varepsilon ) | a_1| \cos (\theta_1+\theta )\\ &=\{[1-\beta -\varepsilon (1+\beta ) ] | a_2| -(1+\varepsilon ) |a_1| \} \cos (\theta _1+\theta )\\ &=\{(1-\beta ) | a_2| -|a_1| -\varepsilon [(1+\beta ) |a_2| +| a_1| ] \} \cos (\theta _1+\theta )\\ & >\frac{(1-\beta ) | a_2| -|a_1| }{2}\cos (\theta _1+\theta ) >0. \end{align*} Since $\eta _1>0$ and $\gamma +\varepsilon <1$, we know that \eqref{e3.24} is a contradiction. \smallskip \noindent (ii) $| a_2| <(1-\alpha ) | a_1|$. For sufficiently large $r$, we have \eqref{e3.4}, \eqref{e3.6}, \eqref{e3.16} and \eqref{e3.22} hold; then we obtain $$| A_2e^{a_2z}| \leq \exp \{(1+\varepsilon ) \delta (a_2z,\theta ) r\} . \label{e3.25}$$ Substituting \eqref{e3.3}, \eqref{e3.4}, \eqref{e3.9}, \eqref{e3.22} and \eqref{e3.25} in \eqref{e3.10}, we have \begin{aligned} &\exp \{(1-\varepsilon ) \delta (a_1z,\theta ) r\} \\ &\leq | A_1e^{a_1z}| \\ &\leq k\exp \{r^{\gamma +\varepsilon }\} \exp \{( 1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\} \exp \{(1+\varepsilon ) \beta \delta (a_2z,\theta ) r\} r^{k(\sigma -1+\varepsilon ) }\\ &\quad +\exp \{(1+\varepsilon ) \delta (a_2z,\theta) r\} \\ & \leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} \exp \{(1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\} \exp \{(1+\varepsilon ) \delta ( a_2z,\theta ) r\} . \end{aligned} \label{e3.26} From the above inequality we obtain $$\exp \{\eta _2r\} \leq M_1r^{M_2}\exp \{r^{\gamma+\varepsilon }\} , \label{e3.27}$$ where $\eta _2=(1-\varepsilon ) \delta (a_1z,\theta ) -(1+\varepsilon ) \alpha \delta (a_1z,\theta ) -(1+\varepsilon ) \delta (a_2z,\theta ) .$ Since $0<\varepsilon <\frac{(1-\alpha ) |a_1| -| a_2| }{2[(1+\alpha) | a_1| +| a_2| ] }$, $\theta _1=\theta _2$ and $\cos (\theta _1+\theta ) >0$, then we obtain \begin{align*} \eta _2&=\{(1-\alpha ) | a_1|-| a_2| -\varepsilon [(1+\alpha ) | a_1| +| a_2| ] \}\cos (\theta _1+\theta ) \\ &>\frac{(1-\alpha ) | a_1| -| a_2| }{2}\cos (\theta _1+\theta ) >0. \end{align*} By $\eta _2>0$ and $\gamma +\varepsilon <1$ we know that \eqref{e3.27} is a contradiction. \smallskip \noindent\textbf{Case 3.} Assume that $a_1<0$ and $\arg a_1\neq \arg a_2$, which is $\theta _1=\pi$ and $\theta _2\neq \pi$. By Lemma \ref{lem2.2}, for the above $\varepsilon$, there is a ray $\arg z=\theta$ such that $\theta \in (-\frac{\pi }{2},\frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5)$ and $\delta (a_2z,\theta ) >0$. Because $\cos \theta >0$, $\delta (a_1z,\theta )=| a_1| \cos (\theta _1+\theta )=-| a_1| \cos \theta <0$. Using the same reasoning as in Case 1 (b), we can get a contradiction. \smallskip \noindent \textbf{Case 4.} Assume that (i) $(1-\beta ) a_2-b0$ and $\delta (a_2z,\theta ) =| a_2| \cos (\theta _2+\theta ) =-| a_2| \cos \theta >0.$ (i) $(1-\beta ) a_2-b0$. We can see that $0<(1-\beta ) |a_2| -| a_1| +b<(1-\beta )| a_2| -| a_1| <2[(1+\beta ) | a_2| +| a_1|]$. Therefore, $0<\frac{(1-\beta ) | a_2| -|a_1| +b}{2[(1+\beta ) |a_2| +| a_1| ] }<1.$ From $0<\varepsilon <\frac{(1-\beta ) | a_2|-| a_1| +b}{2[(1+\beta ) | a_2| +| a_1| ] }$, $\theta _1=\theta _2=\pi$ and $\cos \theta <0$, we obtain \begin{align*} \eta _3 &=[1-\beta -\varepsilon (1+\beta ) ] \delta (a_2z,\theta ) -(1+\varepsilon ) \delta (a_1z,\theta ) -b\cos \theta\\ &=-[1-\beta -\varepsilon (1+\beta ) ] |a_2| \cos \theta +(1+\varepsilon ) | a_1| \cos \theta -b\cos \theta \\ &=(-\cos \theta ) \{[1-\beta -\varepsilon ( 1+\beta ) ] | a_2| -(1+\varepsilon) | a_1| +b\} \\ &=(-\cos \theta ) \{(1-\beta ) |a_2| -| a_1| +b-\varepsilon [(1+\beta ) | a_2| +| a_1| ] \} \\ &>\frac{-1}{2}[(1-\beta ) | a_2|-| a_1| +b] \cos \theta >0. \end{align*} From $\eta _3>0$ and $\gamma +\varepsilon <1$ we know that \eqref{e3.30} is a contradiction. \smallskip \noindent (ii) $a_1<\frac{a_2+b}{1-\alpha }$ and $a_2<0$. For sufficiently large $r$, we obtain \eqref{e3.4}, \eqref{e3.6}, \eqref{e3.16}, \eqref{e3.22}, and \eqref{e3.25} hold. Substituting \eqref{e3.3}, \eqref{e3.4}, \eqref{e3.22}, \eqref{e3.25} and \eqref{e3.28} in \eqref{e3.10}, we obtain \begin{aligned} \exp \{(1-\varepsilon ) \delta (a_1z,\theta) r\} &\leq | A_1e^{a_1z}|\\ &\leq M_1r^{M_2}e^{br\cos \theta } \exp \{r^{\gamma +\varepsilon}\} \exp \{(1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\} \\ &\quad\times \exp \{(1+\varepsilon ) \delta (a_2z,\theta ) r\} . \end{aligned} \label{e3.31} From this inequality we have $$\exp \{\eta _4r\} \leq M_1r^{M_2}\exp \{r^{\gamma +\varepsilon }\} , \label{e3.32}$$ where $\eta _4=(1-\varepsilon ) \delta (a_1z,\theta ) -(1+\varepsilon ) \alpha \delta (a_1z,\theta ) -(1+\varepsilon ) \delta (a_2z,\theta ) -b\cos \theta .$ Since $a_1<\frac{a_2+b}{1-\alpha },a_2=-| a_2|$ and $a_1=-| a_1|$, then we obtain $(1-\alpha) | a_1| -| a_2| +b>0$. We can see that $0<(1-\alpha ) | a_1| -| a_2| +b<(1-\alpha ) |a_1| -| a_2| <2[(1+\alpha) | a_1| +| a_2| ]$. Therefore, $0<\frac{(1-\alpha ) | a_1| -|a_2| +b}{2[(1+\alpha ) |a_1| +| a_2| ] }<1.$ From $0<\varepsilon <\frac{(1-\alpha ) | a_1| -| a_2| +b}{2[(1+\alpha ) |a_1| +| a_2| ] },$ $\theta _1=\theta_2=\pi$ and $\cos \theta <0$, we obtain \begin{align*} \eta _4 &=(-\cos \theta ) \{(1-\alpha )| a_1| -| a_2| +b-\varepsilon [(1+\alpha ) | a_1| +|a_2| ] \}\\ &>\frac{-1}{2}[(1-\alpha ) | a_1| -| a_2| +b] \cos \theta >0. \end{align*} By $\eta _4>0$ and $\gamma +\varepsilon <1$ we know that \eqref{e3.32} is a contradiction. Concluding the above proof, we obtain $\sigma (f) =+\infty$. \smallskip \noindent\textbf{Second step.} We prove that $\sigma _2(f) =1$. By $\max \{\sigma (B_je^{b_jz}+D_je^{d_jz})\; (j=1,\dots ,k-1) , \sigma (A_1e^{a_1z}+A_2e^{a_2z}) \} =1$ and Lemma \ref{lem2.4}, we obtain $\sigma _2(f) \leq 1$. By Lemma \ref{lem2.5}, we know that there exists a set $E_6\subset (1,+\infty )$ with finite logarithmic measure and a constant $C>0$, such that for all $z$ satisfying $| z| =r\notin [0,1] \cup E_6$, we obtain $$| \frac{f^{(j) }(z)}{f(z)}| \leq C[ T(2r,f)] ^{j+1}\quad (j=1,\dots ,k) . \label{e3.33}$$ \noindent\textbf{Case 1.} $\arg a_1\neq \pi$ and $\arg a_1\neq \arg a_2$. In first step, we have proved that there is a ray $\arg z=\theta$ where $\theta \in (-\frac{\pi }{2},\frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5)$, satisfying $\delta (a_1z,\theta ) >0,\quad \text{ }\delta (a_2z,\theta ) <0\quad\text{or}\quad \delta (a_1z,\theta ) <0,\quad \delta (a_2z,\theta ) >0.$ (a) When $\delta (a_1z,\theta ) >0$, $\delta ( a_2z,\theta ) <0$, for sufficiently large $r$, we obtain \eqref{e3.4}--\eqref{e3.8} hold. Substituting \eqref{e3.4}, \eqref{e3.5}, \eqref{e3.8}, \eqref{e3.9} and \eqref{e3.33} in \eqref{e3.10}, we obtain that for all $z=re^{i\theta }$ satisfying $| z| =r\notin [0,1] \cup E_6$, $\theta \in (-\frac{\pi }{2},\frac{\pi }{2}) \setminus ( E_1\cup E_4\cup E_5)$, \begin{aligned} \exp \{(1-\varepsilon ) \delta (a_1z,\theta) r\} &\leq | A_1e^{a_1z}| \\ &\leq M\exp \{r^{\gamma +\varepsilon }\} \exp \{( 1+\varepsilon ) \alpha \delta (a_1z,\theta ) r\}[T(2r,f) ] ^{k+1}, \end{aligned} \label{e3.34} where $M>0$ is a constant. From \eqref{e3.34} and $0<\varepsilon <\frac{1-\alpha }{2(1+\alpha ) }$, we obtain $$\exp \big\{\frac{1-\alpha }{2}\delta (a_1z,\theta )r\big\} \leq M\exp \{r^{\gamma +\varepsilon }\} [T(2r,f) ] ^{k+1}. \label{e3.35}$$ Since $\delta (a_1z,\theta ) >0$ and $\gamma +\varepsilon <1$, then by using Lemma \ref{lem2.6} and \eqref{e3.35}, we obtain $\sigma_2(f) \geq 1$. Hence $\sigma _2(f) =1$. \smallskip (b) When $\delta (a_1z,\theta ) <0,\delta (a_2z,\theta ) >0$, for sufficiently large $r$, we obtain \eqref{e3.13}--\eqref{e3.17} hold. By using the a same reasoning as above, we can get $\sigma _2(f) =1$. \smallskip \noindent\textbf{Case 2.} $\arg a_1\neq \pi$, $\arg a_1=\arg a_2$. In the first step, we have proved that there is a ray $\arg z=\theta$ where $\theta \in (-\frac{\pi }{2},\frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5)$, satisfying $\delta (a_1z,\theta) >0$ and $\delta (a_2z,\theta ) >0$. \smallskip \noindent (i) $| a_2| >\frac{| a_1| }{1-\beta }$. For sufficiently large $r$, we have \eqref{e3.6}, \eqref{e3.13}, \eqref{e3.16}, \eqref{e3.21} and \eqref{e3.22} hold. Substituting \eqref{e3.9}, \eqref{e3.13}, \eqref{e3.21}, \eqref{e3.22} and \eqref{e3.33} in \eqref{e3.18}, we obtain that for all $z=re^{i\theta }$ satisfying $| z| =r\notin [0,1] \cup E_6$, $\theta \in (-\frac{\pi }{2},\frac{\pi }{2} ) \setminus (E_1\cup E_4\cup E_5)$, \begin{aligned} \exp \{(1-\varepsilon ) \delta (a_2z,\theta) r\} &\leq | A_2e^{a_2z}| \\ &\leq M\exp \{r^{\gamma +\varepsilon }\} \exp \{(1+\varepsilon ) \delta (a_1z,\theta ) r\} \\ &\quad \times \exp \{(1+\varepsilon ) \beta \delta (a_2z,\theta ) r\} [T(2r,f) ] ^{k+1}. \end{aligned} \label{e3.36} From this inequality, we obtain $$\exp \{\eta _1r\} \leq M\exp \{r^{\gamma +\varepsilon}\} [T(2r,f) ] ^{k+1}, \label{e3.37}$$ where $\eta _1=(1-\varepsilon ) \delta (a_2z,\theta ) -(1+\varepsilon ) \delta (a_1z,\theta ) -(1+\varepsilon ) \beta \delta (a_2z,\theta ) .$ Since $\eta _1>0$ and $\gamma +\varepsilon <1$, then by using Lemma \ref{lem2.6} and \eqref{e3.37}, we obtain $\sigma _2(f) \geq 1$. Hence $\sigma _2(f) =1$. \smallskip \noindent (ii) $| a_2| <(1-\alpha ) | a_1|$. For sufficiently large $r$, we have \eqref{e3.4}, \eqref{e3.6}, \eqref{e3.16}, \eqref{e3.22} and \eqref{e3.25} hold. By using the same reasoning as above, we can get $\sigma _2(f) =1$. \smallskip \noindent\textbf{Case 3.} $a_1<0$ and $\arg a_1\neq \arg a_2$. In the first step, we have proved that there is a ray $\arg z=\theta$ where $\theta \in (-\frac{\pi }{2},\frac{\pi }{2}) \setminus (E_1\cup E_4\cup E_5)$, satisfying $\delta (a_2z,\theta) >0$ and $\delta (a_1z,\theta ) <0$. Using the same reasoning as in second step (Case 1 (b)), we can get $\sigma _2(f) =1$. \smallskip \noindent \textbf{Case 4.} (i) $(1-\beta )a_2-b0$ and $\delta (a_1z,\theta ) >0$. \smallskip \noindent(i) $(1-\beta ) a_2-b0$ and $\gamma +\varepsilon <1$, then by using Lemma \ref{lem2.6} and \eqref{e3.39}, we obtain $\sigma _2(f) \geq 1$. Hence $\sigma _2(f) =1$. \smallskip \noindent(ii) $a_1<\frac{a_2+b}{1-\alpha }$ and $a_2<0$. For sufficiently large $r$, we obtain \eqref{e3.4}, \eqref{e3.6}, \eqref{e3.16}, \eqref{e3.22} and \eqref{e3.25} hold. By using the same reasoning as above, we can get $\sigma _2(f) =1$. Concluding the above proof, we obtain that every solution $f(\not\equiv 0)$ of \eqref{e1.2} satisfies $\sigma _2(f) =1$. The proof of Theorem \ref{thm1.1} is complete. \section{Proof of Theorem \ref{thm1.2}} Set $R_0(z) =A_1e^{a_1z}+A_2e^{a_2z}$ and $R_i(z) =B_ie^{b_iz}+D_ie^{d_iz}$ $(i=1,\dots,k-1)$. Assume $f(\not\equiv 0)$ is a solution of \eqref{e1.2}. Then $\sigma (f) =+\infty$ by Theorem \ref{thm1.1}. Set $g_0(z) =f(z) -\varphi (z)$. Then we have $\sigma (g_0) =\sigma (f) =\infty$. Substituting $f=g_0+\varphi$ into \eqref{e1.2}, we obtain \begin{aligned} &g_0^{(k) }+R_{k-1}g_0^{(k-1) }+\dots +R_2g_0''+R_1g_0'+R_0g_0\\ &=-[\varphi ^{(k) }+R_{k-1}\varphi ^{(k-1) }+\dots +R_2\varphi ''+R_1\varphi '+R_0\varphi ] . \end{aligned} \label{e4.1} We can rewrite \eqref{e4.1} in the form $$g_0^{(k) }+h_{0,k-1}g_0^{(k-1) }+\dots +h_{0,2}g_0''+h_{0,1}g_0'+h_{0,0}g_0=h_0, \label{e4.2}$$ where $h_0=-[\varphi ^{(k) }+R_{k-1}\varphi ^{(k-1) }+\dots +R_2\varphi ''+R_1\varphi '+R_0\varphi ] .$ We prove that $h_0\not\equiv 0$. In fact, if $h_0\equiv 0$, then $\varphi ^{(k) }+R_{k-1}\varphi ^{(k-1) }+\dots +R_2\varphi ''+R_1\varphi '+R_0\varphi =0.$ Hence, $\varphi \not\equiv 0$ is a solution of \eqref{e1.2} with $\sigma (\varphi ) =+\infty$ by Theorem \ref{thm1.1}, which is a contradiction. Hence, $h_0\not\equiv 0$ is proved. By Lemma \ref{lem2.7} and \eqref{e4.2} we know that $\overline{\lambda }(g_0) = \overline{\lambda }(f-\varphi ) =\sigma (g_0) =\sigma (f) =\infty$. Now we prove that $\overline{\lambda }(f'-\varphi) =\infty$. Set $g_1(z) =f'(z) -\varphi (z)$. Then we have $\sigma (g_1) =\sigma (f') =\sigma (f) =\infty$. Differentiating both sides of equation \eqref{e1.2}, we obtain \begin{aligned} f^{(k+1) }+R_{k-1}f^{(k) }+(R_{k-1}'+R_{k-2}) f^{(k-1) }+(R_{k-2}'+R_{k-3}) f^{(k-2) } \\ +\dots +(R_3'+R_2) f'''+(R_2'+R_1) f''+(R_1'+R_0) f'+R_0'f=0. \end{aligned} \label{e4.3} By \eqref{e1.2}, we have $$f=-\frac{1}{R_0}[f^{(k) }+R_{k-1}f^{(k-1) }+\dots +R_2f''+R_1f'] . \label{e4.4}$$ Substituting \eqref{e4.4} into \eqref{e4.3}, we have \begin{aligned} &f^{(k+1) }+\Big(R_{k-1}-\frac{R_0'}{R_0}\Big) f^{(k) }+\Big(R_{k-1}'+R_{k-2}-R_{k-1}\frac{R_0'}{R_0}\Big) f^{(k-1) }\\ &+\Big(R_{k-2}'+R_{k-3}-R_{k-2}\frac{R_0'}{R_0} \Big) f^{(k-2) }+\dots +\Big(R_3'+R_2-R_3 \frac{R_0'}{R_0}\Big) f'''\\ &+\Big(R_2'+R_1-R_2\frac{R_0'}{R_0}\Big) f''+\Big(R_1'+R_0-R_1\frac{R_0'}{ R_0}\Big) f'=0. \end{aligned} \label{e4.5} We can write equation \eqref{e4.5} in the form $$f^{(k+1) }+h_{1,k-1}f^{(k) }+h_{1,k-2}f^{( k-1) }+\dots +h_{1,2}f'''+h_{1,1}f''+h_{1,0}f'=0, \label{e4.6}$$ where \begin{gather*} h_{1,i}=R_{i+1}'+R_i-R_{i+1}\frac{R_0'}{R_0}\quad (i=0,1,\dots ,k-2) , \\ h_{1,k-1}=R_{k-1}-\frac{R_0'}{R_0}. \end{gather*} Substituting $f^{(j+1) }=g_1^{(j) }+\varphi ^{(j) }$ $(j=0,\dots ,k)$ into \eqref{e4.6}, we obtain $$g_1^{(k) }+h_{1,k-1}g_1^{(k-1)}+h_{1,k-2}g_1^{(k-2) }+\dots +h_{1,2}g_1''+h_{1,1}g_1'+h_{1,0}g_1=h_1, \label{e4.7}$$ where $h_1=-[\varphi ^{(k) }+h_{1,k-1}\varphi ^{( k-1) }+h_{1,k-2}\varphi ^{(k-2) }+\dots +h_{1,2}\varphi ''+h_{1,1}\varphi '+h_{1,0}\varphi ] .$ We can get $$h_{1,i}(z) =\frac{N_i(z) }{R_0(z) }\quad (i=0,1,\dots ,k-1) , \label{e4.8}$$ where \begin{gather} N_0=R_1'R_0+R_0^{2}-R_1R_0', \label{e4.9} \\ N_i=R_{i+1}'R_0+R_iR_0-R_{i+1}R_0'\quad (i=1,2,\dots ,k-2) , \label{e4.10}\\ N_{k-1}=R_{k-1}R_0-R_0'. \label{e4.11} \end{gather} Now we prove that $h_1\not\equiv 0$. In fact, if $h_1\equiv 0$, then $\frac{h_1}{\varphi }\equiv 0$. Hence, by \eqref{e4.8} we obtain $$\frac{\varphi ^{(k) }}{\varphi }R_0+\frac{\varphi ^{(k-1) }}{\varphi }N_{k-1} +\frac{\varphi ^{(k-2) }}{\varphi } N_{k-2}+\dots +\frac{\varphi ''}{\varphi }N_2+\frac{ \varphi '}{\varphi }N_1+N_0=0. \label{e4.12}$$ Obviously, $\frac{\varphi ^{(j) }}{\varphi }$ $(j=1,\dots ,k)$ are meromorphic functions with $\sigma (\frac{\varphi ^{(j) }}{\varphi }) <1$. By \eqref{e4.9}--\eqref{e4.11} we can rewrite \eqref{e4.12} in the form $$A_1^{2}e^{2a_1z}+A_2^{2}e^{2a_2z}+\underset{\lambda \in I_1'}{\sum }f_{\lambda }e^{\lambda z}=0, \label{e4.13}$$ where $I_1'=I_1\setminus \{2a_1,2a_2\}$ and $f_{\lambda }$ $(\lambda \in I_1')$ are meromorphic functions with order less than $1$. \smallskip \noindent(1) If $(2a_1) \notin I_1\setminus \{2a_1\}$, then we write \eqref{e4.13} in the form $A_1^{2}e^{2a_1z}+\underset{\lambda \in \Gamma _1}{\sum }g_{1,\lambda }e^{\lambda z}=0,$ where $\Gamma _1\subseteq I_1\setminus \{2a_1\}$, $g_{1,\lambda }$ $(\lambda \in \Gamma _1)$ are meromorphic functions with order less than $1$ and $2a_1$, $\lambda$ $(\lambda \in \Gamma _1)$ are distinct numbers. By Lemmas \ref{lem2.8} and \ref{lem2.9}, we obtain $A_1\equiv 0$, which is a contradiction. \smallskip \noindent (2) If $(2a_2) \notin I_1\setminus \{2a_2\}$, then we write \eqref{e4.13} in the form $A_2^{2}e^{2a_2z}+\underset{\lambda \in \Gamma _2}{\sum }g_{2,\lambda}e^{\lambda z}=0,$ where $\Gamma _2\subseteq I_1\setminus \{2a_2\}$, $g_{2,\lambda }$ $(\lambda \in \Gamma _2)$ are meromorphic functions with order less than $1$ and $2a_2$, $\lambda$ $(\lambda \in \Gamma _2)$ are distinct numbers. By Lemmas \ref{lem2.8} and \ref{lem2.9}, we obtain $A_2\equiv 0$, which is a contradiction. Hence, $h_1\not\equiv 0$ is proved. By Lemma \ref{lem2.7} and \eqref{e4.7} we know that $\overline{ \lambda }(g_1) =\overline{\lambda }(f'-\varphi ) =\sigma (g_1) =\sigma (f) =\infty$. Now we prove that $\overline{\lambda }(f''-\varphi ) =\infty$. Set $g_2(z) =f''(z) -\varphi (z)$. Then we have $\sigma (g_2) =\sigma (f'') =\sigma (f) =\infty$. Differentiating both sides of equation \eqref{e4.3}, we have \begin{aligned} &f^{(k+2) }+R_{k-1}f^{(k+1) }+(2R_{k-1}'+R_{k-2}) f^{(k) } +(R_{k-1}''+2R_{k-2}'+R_{k-3}) f^{(k-1) } \\ & +(R_{k-2}''+2R_{k-3}'+R_{k-4}) f^{(k-2) }+\dots +(R_3''+2R_2'+R_1) f'''\\ &+(R_2''+2R_1'+R_0) f''+(R_1''+2R_0') f'+R_0''f=0. \end{aligned} \label{e4.14} By \eqref{e4.4} and \eqref{e4.14}, we have \begin{aligned} &f^{(k+2) }+R_{k-1}f^{(k+1) }+\Big(2R_{k-1}'+R_{k-2}-\frac{R_0''}{R_0}\Big) f^{(k) }\\ &+\Big(R_{k-1}''+2R_{k-2}'+R_{k-3}-R_{k-1}\frac{R_0''}{R_0}\Big) f^{(k-1) } +\dots \\ &+\Big(R_4''+2R_3'+R_2-R_4\frac{ R_0''}{R_0}\Big) f^{(4) }+\Big(R_3''+2R_2'+R_1-R_3\frac{R_0''}{R_0}\Big) f'''\\ &+\Big(R_2''+2R_1'+R_0-R_2\frac{R_0''}{R_0}\Big) f''+\Big( R_1''+2R_0'-R_1\frac{R_0''}{R_0}\Big) f'=0. \end{aligned} \label{e4.15} Now we prove that $R_1'+R_0-R_1\frac{R_0'}{R_0}\not\equiv 0$. Suppose that $R_1'+R_0-R_1\frac{R_0'}{R_0}\equiv 0$, then we have $$A_1^{2}e^{2a_1z}+A_2^{2}e^{2a_2z}+\underset{\lambda \in I_2'}{\sum }f_{\lambda }e^{\lambda z}=0, \label{e4.16}$$ where $I_2'=I_2\setminus \{2a_1,2a_2\}$ and $f_{\lambda }$ $(\lambda \in I_2')$ are meromorphic functions with order less than $1$. By using the same reasoning as above, we can get a contradiction. Hence, $R_1'+R_0-R_1\frac{R_0'}{R_0}\not\equiv 0$ is proved. Set $$\psi (z) =R_1'R_0+R_0^{2}-R_1R_0' \text{ and }\phi (z) =R_1''R_0+2R_0'R_0-R_1R_0''. \label{e4.17}$$ By \eqref{e4.5} and \eqref{e4.17}, we obtain \begin{aligned} f'&=\frac{-R_0}{\psi (z) }\Big\{f^{(k+1) }+\Big(R_{k-1}-\frac{R_0'}{R_0}\Big) f^{(k) }+\Big(R_{k-1}'+R_{k-2}-R_{k-1}\frac{R_0'}{R_0}\Big) f^{(k-1) }\\ &\quad +\Big(R_{k-2}'+R_{k-3}-R_{k-2}\frac{R_0'}{R_0} \Big) f^{(k-2) }+\dots +\Big(R_2'+R_1-R_2 \frac{R_0'}{R_0}\Big) f''\Big\} . \end{aligned} \label{e4.18} Substituting \eqref{e4.17} and \eqref{e4.18} into \eqref{e4.15}, we obtain \begin{aligned} &f^{(k+2) }+[R_{k-1}-\frac{\phi }{\psi }] f^{(k+1) } +\Big[2R_{k-1}'+R_{k-2}-\frac{R_0''}{ R_0}-\frac{\phi }{\psi }\big(R_{k-1}-\frac{R_0'}{R_0} \big) \Big] f^{(k) }\\ &+\Big[R_{k-1}''+2R_{k-2}'+R_{k-3}-R_{k-1}\frac{ R_0''}{R_0}-\frac{\phi }{\psi }\big(R_{k-1}'+R_{k-2}-R_{k-1}\frac{R_0'}{R_0}\big) \Big] f^{(k-1) } \\ &+\dots +\Big[R_3''+2R_2'+R_1-R_3\frac{ R_0''}{R_0}-\frac{\phi }{\psi }\big(R_3'+R_2-R_3\frac{R_0'}{R_0}\big) \Big] f''' \\ &+\Big[R_2''+2R_1'+R_0-R_2\frac{ R_0''}{R_0}-\frac{\phi }{\psi }\big(R_2'+R_1-R_2\frac{R_0'}{R_0}\big) \Big] f''=0. \end{aligned}\label{e4.19} We can write \eqref{e4.19} in the form $$f^{(k+2) }+h_{2,k-1}f^{(k+1) }+h_{2,k-2}f^{( k) }+\dots +h_{2,2}f^{(4) }+h_{2,1}f'''+h_{2,0}f''=0, \label{e4.20}$$ where \begin{gather*} \begin{aligned} h_{2,i}&=R_{i+2}''+2R_{i+1}'+R_i-R_{i+2}\frac{R_0''}{R_0} \\ &\quad -\frac{\phi (z) }{\psi (z) }\Big(R_{i+2}'+R_{i+1}-R_{i+2}\frac{R_0'}{R_0}\Big) \quad (i=0,1,\dots ,k-3), \end{aligned} \\ h_{2,k-2}=2R_{k-1}'+R_{k-2}-\frac{R_0''}{R_0}- \frac{\phi (z) }{\psi (z) }\Big(R_{k-1}-\frac{R_0'}{R_0}\Big) , \\ h_{2,k-1}=R_{k-1}-\frac{\phi (z) }{\psi (z) }. \end{gather*} Substituting $f^{(j+2) }=g_2^{(j) }+\varphi^{(j) }$ $(j=0,\dots ,k)$ in \eqref{e4.20} we have $$g_2^{(k) }+h_{2,k-1}g_2^{(k-1) }+h_{2,k-2}g_2^{(k-2) }+\dots +h_{2,1}g_2'+h_{2,0}g_2=h_2, \label{e4.21}$$ where $h_2=-[\varphi ^{(k) }+h_{2,k-1}\varphi ^{( k-1) }+h_{2,k-2}\varphi ^{(k-2) }+\dots +h_{2,2}\varphi ''+h_{2,1}\varphi '+h_{2,0}\varphi ] .$ We obtain $$h_{2,i}=\frac{L_i(z) }{\psi (z) }\quad (i=0,1,\dots ,k-1) , \label{e4.22}$$ where \begin{gather} \begin{aligned} L_0(z) &=R_2''R_1'R_0+R_2''R_0^{2}-R_2''R_1R_0'+2{R_1'}^2 R_0+3R_1'R_0^{2} -2R_1'R_1R_0'+R_0^{3} \\ &\quad -3R_1R_0'R_0-R_2R_1'R_0''-R_2R_0''R_0-R_2'R_1''R_0-2R_2'R_0'R_0+R_2'R_1R_0'' \\ &\quad -R_1''R_1R_0+R_1^{2}R_0''+R_2R_1''R_0'+2R_2{R_0'}^2, \end{aligned} \label{e4.23} \\ \begin{aligned} L_i&=R_{i+2}''R_1'R_0+R_{i+2}''R_0^{2}-R_{i+2}''R_1R_0'+2R_{i+1}'R_1'R_0 +2R_{i+1}'R_0^{2}-2R_{i+1}'R_1R_0' \\ &\quad +R_iR_1'R_0+R_iR_0^{2}-R_iR_1R_0'-R_{i+2}R_1'R_0''-R_{i+2}R_0''R_0 -R_{i+2}'R_1''R_0 \\ &\quad -2R_{i+2}'R_0'R_0+R_{i+2}'R_1R_0''-R_{i+1}R_1''R_0-2R_{i+1}R_0'R_0+R_{i+1}R_1R_0''\\ &\quad +R_{i+2}R_1''R_0'+2R_{i+2}{R_0'}^2\quad (i=1,2,\dots ,k-3) , \end{aligned} \label{e4.24} \\ \begin{aligned} L_{k-2}&=2R_{k-1}'R_1'R_0+2R_{k-1}'R_0^{2}-2R_{k-1}'R_1R_0'+R_{k-2}R_1'R_0 +R_{k-2}R_0^{2} \\ &\quad -R_{k-2}R_1R_0'-R_1'R_0''-R_0''R_0-R_{k-1}R_1''R_0-2R_{k-1}R_0'R_0 \\ &\quad +R_{k-1}R_1R_0''+R_1''R_0'+2{R_0'}^2, \end{aligned} \label{e4.25} \\ L_{k-1}=R_{k-1}R_1'R_0+R_{k-1}R_0^{2}-R_{k-1}R_1R_0'-R_1''R_0-2R_0'R_0+R_1R_0''. \label{e4.26} \end{gather} Therefore, $$\frac{-h_2}{\varphi }=\frac{1}{\psi }[\frac{\varphi ^{( k) }}{\varphi }\psi +\frac{\varphi ^{(k-1) }}{\varphi } L_{k-1}+\dots +\frac{\varphi ''}{\varphi }L_2+\frac{ \varphi '}{\varphi }L_1+L_0] . \label{e4.27}$$ Now we prove that $h_2\not\equiv 0$. In fact, if $h_2\equiv 0$, then $\frac{-h_2}{\varphi }\equiv 0$. Hence, by \eqref{e4.27} we obtain $$\frac{\varphi ^{(k) }}{\varphi }\psi +\frac{\varphi ^{( k-1) }}{\varphi }L_{k-1}+\dots +\frac{\varphi ''}{ \varphi }L_2+\frac{\varphi '}{\varphi }L_1+L_0=0. \label{e4.28}$$ Obviously, $\frac{\varphi ^{(j) }}{\varphi }$ $(j=1,\dots,k)$ are meromorphic functions with $\sigma (\frac{\varphi^{(j) }}{\varphi }) <1$. By \eqref{e4.17} and \eqref{e4.23}--\eqref{e4.26}, we can rewrite \eqref{e4.28} in the form $$A_1^{3}e^{3a_1z}+A_2^{3}e^{3a_2z}+\sum_{\lambda \in I_3'} f_{\lambda }e^{\lambda z}=0, \label{e4.29}$$ where $I_3'=I_3\setminus \{3a_1,3a_2\}$ and $f_{\lambda }$ $(\lambda \in I_3')$ are meromorphic functions with order less than $1$. \smallskip \noindent(1) If $(3a_1) \notin I_3\setminus \{3a_1\}$, then we write \eqref{e4.29} in the form $A_1^{3}e^{3a_1z}+\underset{\lambda \in \Gamma _1}{\sum }g_{1,\lambda }e^{\lambda z}=0,$ where $\Gamma _1\subseteq I_3\setminus \{3a_1\}$, $g_{1,\lambda }$ $(\lambda \in \Gamma _1)$ are meromorphic functions with order less than $1$ and $3a_1$, $\lambda$ $(\lambda \in \Gamma _1)$ are distinct numbers. By Lemmas \ref{lem2.8} and \ref{lem2.9}, we obtain $A_1\equiv 0$, which is a contradiction. \smallskip \noindent(2) If $(3a_2) \notin I_3\setminus \{3a_2\}$, then we write \eqref{e4.29} in the form $A_2^{3}e^{3a_2z}+\underset{\lambda \in \Gamma _2}{\sum }g_{2,\lambda }e^{\lambda z}=0,$ where $\Gamma _2\subseteq I_3\setminus \{3a_2\}$, $g_{2,\lambda }$ $(\lambda \in \Gamma _2)$ are meromorphic functions with order less than $1$ and $3a_2$, $\lambda$ $(\lambda \in \Gamma _2)$ are distinct numbers. By Lemmas \ref{lem2.8} and \ref{lem2.9}, we obtain $A_2\equiv 0$, which is a contradiction. Hence, $h_2\not\equiv 0$ is proved. By Lemma \ref{lem2.7} and \eqref{e4.21}, we have $\overline{ \lambda }(g_2) =\overline{\lambda }(f''-\varphi ) =\sigma (g_2) =\sigma (f) =\infty$. 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