1. Introduction
In this paper, we use standard notations from the value
distribution theory of meromorphic functions (see [1,2,3]). We suppose that \(f\) is a meromorphic function in whole complex plane \(\mathbb{C}\).
In addition, we denote the order of growth of \(f\) by \(\rho (f)\), and use the notation \(\rho _{2}(f)\) to denote the hyper-order of \(f\),
defined by
\begin{equation*}
\rho _{2}(f)=\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log \log T(r,f)}{\log r},
\end{equation*}
where \(T(r,f)\) is the Nevanlinna characteristic function of \(f\).
To give the precise estimate of fixed points, we denote the exponent of convergence of fixed points by \(\tau (f)\), which is defined by
\begin{equation*}
\tau (f)=\lambda(f-z)=\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log N\left( r,\frac{1}{f-z}\right) }{\log r}
\end{equation*}
and the hyper-exponent of convergence of fixed points and distinct fixed
points are denoted by \(\tau _{2}(f)\) and \(\overline{\tau }_{2}(f)\) and are defined by
\begin{equation*}
\tau _{2}(f)=\lambda_{2}(f-z)=\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log \log N\left( r,\frac{1}{f-z}\right) }{\log r},
\end{equation*}
and
\begin{equation*}
\overline{\tau }_{2}(f)=\overline{\lambda}_{2}(f-z)=\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log \log \overline{N}\left( r,\frac{1}{f-z}\right) }{\log r},
\end{equation*}
respectively, where \(N\left( r,\frac{1}{f-z}\right) \) and \(\overline{N}\left( r,\frac{1}{f-z}\right) \) are respectively the integrated counting function of fixed
points and distinct fixed points of \(f\). We denote the exponent of convergence of zeros (distinct zeros) of \(f\) by \(\lambda (f)\) \((\overline{\lambda }(f))\) and the hyper-exponent of convergence of zeros (distinct zeros) of \(f\) by \(\lambda _{2}(f)\) \((\overline{\lambda }_{2}(f))\).
Consider the second-order homogeneous linear differential equation
\begin{equation}\label{1.1a}
f^{\prime \prime }+P(e^{z})f^{\prime }+Q(e^{z})f=0,
\end{equation}
(1)
where \(P(w)\) and \(Q(w)\) are not constants polynomials in \(w=e^{z}\) \((z\in\mathbb{C})\). It’s well-known that every solution of Equation (1) is entire.
Suppose \(f\not\equiv 0\) is a solution of (1). If \(f\) satisfies the condition
\begin{equation*}
\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log T(r,f)}{r}=0,
\end{equation*}
then we say that \(f\) is a nontrivial subnormal solution of (1), and if \(f\) satisfies the condition [
4],
\begin{equation*}
\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log T(r,f)}{r^{n}}=0,
\end{equation*}
then we say that \(f\) is a nontrivial \(n\)-subnormal solution of (1). In [
5], Wittich investigated the subnormal
solution of (1), and obtained the form of all subnormal solutions in the following theorem:
Theorem 1. [5]
If \(f\not\equiv 0\) is a subnormal solution of (1), then \(f\) must have the form
\begin{equation*}
f(z)=e^{cz}(a_{0}+a_{1}e^{z}+\cdots +a_{m}e^{mz}),
\end{equation*}
where \(m\geq 0\) is an integer and \(c\), \(a_{0}\), \(a_{1}\),…,\(a_{m}\) are constants with \(a_{0}a_{m}\neq 0\).
Gundersen and Steinbert [
6] refined Theorem 1 and got the following theorem:
Theorem 2. [6]
Under the assumption of Theorem 1, the following statements hold:
- (i) If \(\deg P>\deg Q\) and
\(Q\not\equiv 0\), then any subnormal solution \(f\not\equiv 0\) of (7) must have the form
\begin{equation*}
f(z)=\sum_{k=0}^{m}h_{k}e^{-kz},
\end{equation*}
where \(m\geq 1\) is an integer and \(h_{0}\), \(h_{1} \), …, \(h_{m}\) are constants with \(h_{0}\neq 0\) and \(h_{m}\neq 0\).
- (ii) If \(\deg P\geq 1\) and \(Q\equiv 0\), then any subnormal solution of Equation (7) must be constant.
- (iii) If \(\deg P< \deg Q\), then the only subnormal solution of (7) is \(f\equiv 0.\)
Chen and Shon [
7] investigated more general equation than (7), and got the following theorem: Set
\begin{equation}\label{1.2a}
a_{j}(z)=a_{jd_{j}}z^{d_{j}}+a_{j\left( d_{j}-1\right) }z^{d_{j}-1}+\cdots +a_{j1}z+a_{j0},\text{ }(j=0,\cdots ,n),
\end{equation}
(2)
\begin{equation}\label{1.3a}
b_{k}(z)=b_{km_{k}}z^{m_{k}}+b_{k\left( m_{k}-1\right) }z^{m_{k}-1}+\cdots +b_{k1}z+b_{k0},\text{ }(k=0,\cdots ,s),
\end{equation}
(3)
where \(d_{j}\geq 0\) \((j=0,\cdots ,n),\) \(m_{k}\geq 0\) \((k=0,\cdots ,s)\) are
integers, \(a_{jd_{j}},…,a_{j0};\) \(b_{km_{k}},…,b_{k0}\) are complex constants such that \(a_{jd_{j}}\neq 0,\) \(b_{km_{k}}\neq 0\).
Theorem 3. [7]
Let \(a_{n}(z),…,a_{1}(z),\) \(a_{0}\left( z\right) ,\) \(b_{s}(z),…,b_{1}(z),\) \(b_{0}\left( z\right) \) be polynomials and satisfy (2) and (3), and \(a_{n}(z)b_{s}(z)\neq 0\). Suppose that \(P^{\ast}(e^{z})=a_{n}(z)e^{nz}+\cdots +a_{1}(z)e^{z}+a_{0}(z)\), \(Q^{\ast}(e^{z})=b_{s}(z)e^{sz}+\cdots +b_{1}(z)e^{z}+b_{0}(z)\). If \(n< s\), then every solution \(f\) \((\not\equiv 0)\) of
equation
\begin{equation*}
f^{\prime \prime }+P^{\ast }(e^{z})f^{\prime }+Q^{\ast }(e^{z})f=0
\end{equation*}
satisfies \(\rho _{2}(f)=1\).
Many authors investigated the growth of solutions and
the existence of subnormal solutions for some class of higher order linear
differential equations (see [4,7,8,9,10,11,12,13]). For the higher-order linear homogeneous
differential equation
\begin{equation}\label{1.4a}
f^{\left( k\right) }+P_{k-1}(e^{z})f^{(k-1)}+\cdots +P_{0}(e^{z})f=0,
\end{equation}
(4)
where \(P_{j}(e^{z})\) \((j=0,\cdots ,k-1)\) are polynomials in \(z\), Yang and Li
[
11] generalized the result of Theorem 2 to the higher
order and obtained the following results:
Set
\begin{equation}\label{1.5a}
a_{jm_{j}}(z)=a_{jm_{j}d_{jmj}}z^{d_{jm_{j}}}+a_{jm_{j}(d_{jm_{j}}-1)}z^{d_{jm_{j}}-1}+\cdots +a_{jm_{j}1}z+a_{jm_{j}0},
\end{equation}
(5)
where \(d_{jm_{j}}\geq 0\) \((j=0,\cdots,k-1)\) are integers, \(a_{jm_{j}d_{jmj}},…,a_{jm_{j}0}\) are complex constants, \(a_{jm_{j}d_{jmj}}\neq 0\).
Theorem 4. [11]
Let \(a_{jm_{j}}\left(z\right) \) be polynomials and satisfy (5). Suppose that
\begin{equation*}
P_{j}(e^{z})=a_{jm_{j}}\left( z\right) e^{m_{j}z}+\cdots +a_{j1}\left(z\right) e^{z},
\end{equation*}
where \(a_{jm_{j}}\left( z\right) \not\equiv 0\). If there exists an integer \(s\) \((s\in \{0,\cdots ,k-1\})\)
satisfying
\begin{equation*}
m_{s}>\max \left\{ m_{j}:j=0,\cdots ,s-1,s+1,\cdots ,k-1\right\} =m,
\end{equation*}
then every solution \(f\not\equiv 0\) of Equation (4) satisfies \(\rho _{2}\left( f\right) =1\) if
one of the following condition holds:
- (i) \(s=0\) or \(1\).
- (ii) \(s\geq 2\) and \(\deg
a_{0j}\left( z\right) >\deg a_{ij}\left( z\right) \) \(\left( i\neq
0\right) \).
Theorem 5. [11]
Under the assumption of Theorem 4, if \(zP_{0}(e^{z})+P_{1}(e^{z})\)
\(\not\equiv0\), then we have every solution \(f\not\equiv 0\) of Equation (4) satisfies
\begin{equation*}
\tau _{2}(f)=\overline{\tau }_{2}(f)=\rho _{2}\left( f\right) =1.
\end{equation*}
In particular, they also investigated the exponents of
convergence of the fixed points of solutions and their first derivatives for
a second order Equation (1) and obtained the following
theorem:
Theorem 6. [11]
Let \(a_{n}(z)\),…, \(a_{1}(z)\), \(b_{s}(z)\),…, \(b_{1}(z)\) be
polynomials and satisfy (2) and (3), and \(a_{n}(z)b_{s}(z)\neq 0\). Suppose that \(P(e^{z})=a_{n}(z)e^{nz}+\cdots +a_{1}(z)e^{z}\),
\(Q(e^{z})=b_{s}(z)e^{sz}+\cdots +b_{1}(z)e^{z}\). If \(n\neq s\), then every solution \(f\) \((\not\equiv 0)\) of Equation (1) satisfy \(\lambda (f-z)=\lambda (f^{\prime}-z)=\rho \left( f\right) =\infty \) and \(\lambda _{2}(f-z)=\lambda_{2}(f^{\prime }-z)=\rho _{2}\left( f\right) =1\).
Thus, it is natural to ask what will happen if we change \(\exp \{z\}\) in the coefficients of (4) into \(\exp \{A(z)\}\)?
In this paper, we consider the above problem to Theorems 3, 4, 5 and 6, we obtain the following results:
We set
\begin{equation*}
A(z)=c_{n}z^{n}+c_{n-1}z^{n-1}+\cdots +c_{1}z+c_{0},
\end{equation*}
where \(n\geq 1\) is an integer and \(c_{0},…,c_{n}\) are complex constants
such that \(\mathit{Re}c_{n}>0\), throughout the rest of this paper.
Theorem 7.
Let \(a_{jm_{j}}\left( z\right) \) be polynomials and satisfy (5). Suppose that
\begin{equation}
P_{j}(e^{A(z)})=a_{jm_{j}}\left( z\right) e^{m_{j}A(z)}+\cdots +a_{j1}\left(
z\right) e^{A(z)}, \label{1.6}
\end{equation}
(6)
where \(a_{jm_{j}}\left( z\right) \not\equiv 0\). If there exists an integer \(s\) \((s\in \{0,\cdots ,k-1\})\) satisfying
\begin{equation*}
m_{s}>\max \left\{ m_{j}:j=0,\cdots ,s-1,s+1,\cdots ,k-1\right\} =m,
\end{equation*}
then every solution \(f\not\equiv 0\) of equation
\begin{equation}\label{1.7a}
f^{\left( k\right) }+P_{k-1}(e^{A\left( z\right) })f^{(k-1)}+\cdots
+P_{0}(e^{A\left( z\right) })f=0
\end{equation}
(7)
satisfies \(\rho \left( f\right) =\infty \) and \(\rho_{2}\left( f\right) =n\) if one of the following condition holds:
- (i) \(s=0\) or \(1\).
- (ii) \(s\geq 2\) and \(\deg a_{0j}\left( z\right) >\deg a_{ij}\left( z\right) \) \(\left( i\neq 0\right) \).
Example 1.
Let \(f=e^{e^{z^2}}\) be a solution of the equation
\begin{equation*}
f^{(4)}-2ze^{z^2}f^{(3)}-12z^2e^{z^2}f^{\prime \prime}-24z^3e^{z^2}f^{\prime}
-[24z^2e^{3z^2}+(96z^2+12)e^{2z^2}+(16z^{4}+48z^{2}+12)e^{z^2}]f=0.
\end{equation*}
Set
\begin{align*}P_3(e^{A(z)})&=a_{3,1}(z)e^{A(z)}=-2ze^{z^2},\\
P_2(e^{A(z)})&=a_{2,1}(z)e^{A(z)}=-12z^2e^{z^2},\\
P_1(e^{A(z)})&=a_{1,1}(z)e^{A(z)}=-24z^3e^{z^2}, \\
P_0(e^{A(z)})&=a_{0,3}(z)e^{3A(z)}+a_{0,2}(z)e^{2A(z)}+a_{0,1}(z)e^{A(z)}
=-24z^2e^{3z^2}-(96z^2+12)e^{2z^2}-(16z^{4}+48z^{2}+12)e^{z^2}.
\end{align*}
We remark that \(s=0\) and \(m_{0}=3>\max \left\{ m_{j}:j=1,2,3\right\} =m=1\).
Obviously, the conditions of Theorem 7 are satisfied, we see that \(\rho \left( f\right) =\infty \) and \(\rho_{2}\left( f\right)=n =2.\)
Remark 1.
Very recently, Li et al., [4] have investigated \(n\) subnormal solutions of the Equation (7) with
\begin{equation*}
P_{j}(e^{A(z)})=a_{jm_{j}}e^{m_{j}A(z)}+\cdots +a_{j1}e^{A(z)}\text{ }\left(
j=0,…,k-1\right) ,
\end{equation*}
where \(a_{jm_{j}},\cdots ,a_{j1}\) \(\left( j=0,…,k-1\right) \) are complex
constants instead of polynomials and obtained some results concerning their
growth.
Corollary 1.
Under the assumption of Theorem 7, if \(zP_{0}(e^{A(z)})+P_{1}(e^{A(z)})\not\equiv 0\), then we have every solution \(f\not\equiv 0\) of Equation (4) satisfies
\begin{equation*}
\tau (f)=\overline{\tau }(f)=\rho \left( f\right) =\infty \text{ and }\tau
_{2}(f)=\overline{\tau }_{2}(f)=\rho _{2}\left( f\right) =n.
\end{equation*}
In particular, we also investigate the exponents of
convergence of the fixed points of solutions and their first derivatives for
a second order equation
\begin{equation}
f^{\prime \prime }+P(e^{A(z)})f^{\prime }+Q(e^{A(z)})f=0, \label{1.8a}
\end{equation}
(8)
and we obtain the following theorems:
Theorem 8.
Let \(a_{p}(z)\),…, \(a_{1}(z)\), \(b_{s}(z)\),…, \(b_{1}(z)\) be polynomials and satisfy (2) and (3), and \(a_{p}(z)b_{s}(z)\neq 0\).
Suppose that \(P(e^{A(z)})=a_{p}(z)e^{pA(z)}+\cdots +a_{1}(z)e^{A(z)}\), \(Q(e^{A(z)})=b_{s}(z)e^{sA(z)}+\cdots +b_{1}(z)e^{A(z)}\). If \(p\neq s \), then every solution \(f\)
\((\not\equiv 0)\) of Equation (8) satisfies \(\lambda (f-z)=\lambda (f^{\prime }-z)=\rho \left( f\right) =\infty \) and \(\lambda_{2}(f-z)=\lambda _{2}(f^{\prime }-z)=\rho _{2}\left( f\right) =n\).
Example 2.
Let \(f=e^{e^{z^2}}\) be a solution of the equation
\begin{equation*}
f^{\prime \prime }-3ze^{z^2}f^{\prime }+[2z^2e^{2z^2}-(4{z^2}+2)e^{z^2}]f=0.
\end{equation*}
Set
\begin{align*}P(e^{A(z)})&=a_{1}(z)e^{A(z)}=-3ze^{z^2},\\
Q(e^{A(z)})&=b_{2}(z)e^{2A(z)}+b_{1}(z)e^{A(z)}=2z^2e^{2z^2}-(4z^2+2)e^{z^2}.\end{align*}
It is clear that the conditions of Theorem 8 are satisfied with \(p=1\neq s=2\), we see that \(\lambda (e^{e^{z^2}}-z)=\lambda
(2ze^{z^2}e^{e^{z^2}}-z)=\rho \left( f\right) =\infty \) and \(\lambda
_{2}(e^{e^{z^2}}-z)=\lambda _{2}(2ze^{z^2}e^{e^{z^2}}-z)=\rho _{2}\left(
f\right)=n =2\).
Remark 2.
If \(p=s,\) then the conclusions of Theorem 8 does not hold. For instance, consider the following equation
\begin{align}
f^{\prime \prime }+&\left( \left( z^{4}+2iz\right) e^{2\left( 1+5i\right)
z^{3}+2z}+\left( -z^{2}+\left( 2-i\right) z\right) e^{\left( 1+5i\right)
z^{3}+z}\right) f^{\prime }
\notag\\&-\left( \left( z^{3}+2i\right) e^{2\left( 1+5i\right) z^{3}+2z}+\left(
-z+2-i\right) e^{\left( 1+5i\right) z^{3}+z}\right) f=0. \label{1.9a}
\end{align}
(9)
We can easily see that (9) has solution \(f\left( z\right)
=z \) which satisfies \(\rho \left( f\right) =0\neq \infty \) and \(\rho
_{2}\left( f\right) =0\neq n=3.\ \) In this example, we have \(p=s=2, \) \(
A(z)=\left( 1+5i\right) z^{3}+z,\) \(a_{2}(z)=z^{4}+2iz,\) \(a_{1}(z)=-z^{2}+
\left( 2-i\right) z,\) \(b_{2}(z)=-\left( z^{3}+2i\right) \) and \(%
b_{1}(z)=-\left( -z+2-i\right).\)
Theorem 9.
Let \(a_{p}(z),…,a_{1}(z),a_{0}\left( z\right)
,b_{s}(z),…,b_{1}(z),b_{0}\left( z\right) \) be polynomials and
satisfy (2) and (3),
and \(a_{p}(z)b_{s}(z)\neq 0\). Suppose that
\[P^{\ast }(e^{A(z)})=a_{p}(z)e^{pA(z)}+\cdots +a_{1}(z)e^{A(z)}+a_{0}(z),\]
\[Q^{\ast}(e^{A(z)})=b_{s}(z)e^{sA(z)}+\cdots +b_{1}(z)e^{A(z)}+b_{0}(z).\] If \(p< s\), then every solution \(f\) \((\not\equiv 0)\) of equation
\begin{equation}
f^{\prime \prime }+P^{\ast }(e^{A(z)})f^{\prime }+Q^{\ast }(e^{A(z)})f=0
\label{1.10a}
\end{equation}
(10)
satisfies \(\rho \left( f\right) =\infty \) and \(\rho_{2}\left( f\right) =n\).
Example 3.
Let \(f=e^{z} e^{e^z}\) be a solution of the equation
\begin{equation*}
f^{\prime \prime }+(e^{z+1}-3)f^{\prime
}+[(-e^{-2}-e^{-1})e^{2(z+1)}-e^{z+1}+2]f=0.
\end{equation*}
Set
\begin{align*}
P^{\ast }(e^{A(z)})&=a_1(z)e^{A(z)}+a_0(z)=e^{z+1}-3,\\
Q^{\ast}(e^{A(z)})&=b_2(z)e^{2A(z)}+b_1(z)e^{A(z)}+b_0(z)=(-e^{-2}-e^{-1})e^{2(z+1)}-e^{z+1}+2.\end{align*}
It is clear that the conditions of Theorem 9 are satisfied with \(p=1< s=2\), here we have \(\rho \left( f\right) =\infty \) and \(\rho _{2}\left(
f\right) =n=1\).
Remark 3.
If \(p\geq s,\) then the conclusions of Theorem 9 does not hold. For instance, consider the following equation
\begin{align}
f^{\prime \prime }-&\left( \left( 2z^{2}+3z\right) e^{\left( 1-i\right)
z^{2}+2z+i}+iz^{3}-z^{2}+\left( 1+i\right) z\right) f^{\prime }\notag\\
&+\left( \left( 2z+3\right) e^{\left( 1-i\right)
z^{2}+2z+i}+iz^{2}-z+1+i\right) f=0. \label{1.11a}
\end{align}
(11)
It is easy to see that (11) has solution \(f\left( z\right)=z\) which satisfies \(\rho \left( f\right) =0\neq \infty \) and \(\rho_{2}\left( f\right) =0\neq n=2. \) In this example, we have \(p=s=1, \) \(A(z)=\left( 1-i\right) z^{2}+2z+i,\) \(a_{1}(z)=-\left( 2z^{2}+3z\right) ,\) \(a_{0}(z)=-\left( iz^{3}-z^{2}+\left( 1+i\right) z\right) ,\) \(b_{1}(z)=2z+3\)
and \(b_{0}(z)=iz^{2}-z+1+i.\)
Remark 4.
Setting \(c_{n}=1\), \(c_{n-1}=\cdots =c_{0}=0\) and \(n=1,\)
in Theorem 7, Corollary 1, Theorem 8 and
Theorem 9, we obtain Theorem 4,
Theorem 5, Theorem 6 and Theorem 3 respectively.
2. Auxiliary Lemmas
Recall that
\begin{equation*}
A(z)=c_{n}z^{n}+c_{n-1}z^{n-1}+\cdots +c_{0},,c_{l}=\alpha _{l}e^{i\theta
_{l}},\text{ }z=re^{i\theta },\text{ }\mathit{Re}c_{n}>0,
\end{equation*}
we set \(\delta _{l}(A,\theta )=\mathit{Re}(c_{l}(e^{i\theta })^{l})=\alpha_{l}\cos (\theta _{l}+l\theta )\), and \(H_{l,0}=\{\theta \in \lbrack 0,2\pi):\delta _{l}(A,\theta )=0\}\), \(H_{l,+}=\{\theta \in \lbrack 0,2\pi ):\delta_{l}(A,\theta )>0\}\), \(H_{l,-}=\{\theta \in \lbrack 0,2\pi ):\delta_{l}(A,\theta )< 0\}\), for \(l=1,\cdots ,n\), throughout the rest of this
paper. Obviously, if \(\delta _{n}(A,\theta )\neq 0\), as \(r\rightarrow+\infty \), we get
\begin{equation}
|e^{A(z)}|=e^{\delta _{n}(A,\theta )r^{n}+\cdots +\delta _{1}(A,\theta )r+%
\mathit{Re}{c_{0}}}=e^{\delta _{n}(A,\theta )r^{n}(1+o(1))}. \label{2.1}
\end{equation}
(12)
Lemma 1. [3]
Let \(f_{j}(z)\) \((j=1,\cdots ,n)\) \((n\geq 2)\) be meromorphic functions, \(g_{j}(z)\) \((j=1,\cdots ,n)\) be entire
functions, and satisfy
- (i) \( \sum_{j=1}^{n}e^{g_{j}(z)}\equiv 0;\)
- (ii) when \(1\leq j\leq k\leq n\), then \(g_{i}(z)-g_{k}(z)\) is not a constant;
- (iii) when \(1\leq j\leq n\), \(1\leq h\leq k\leq n\),
then
\begin{equation*}
T(r,f_{j})=o\{T(r,e^{g_{h}-g_{k}})\}\hspace{0.5cm}(r\rightarrow +\infty
,r\not\in E),
\end{equation*}
where \(E\subset (1,\infty )\) is of finite linear measure or logarithmic measure. Also, \(f_{j}(z)\equiv 0\) \((j=1,\cdots ,n)\).
Lemma 2.
Let \(A(z)\), \(P_{j}(e^{A(z)}) \), \(m_{j}\), \(m_{s}\), \(m\) and \(a_{ij}(z)\) satisfy the hypotheses of Theorem 7, then Equation (7) has no constant polynomial solution.
Proof.
Suppose that \(f_{0}\left( z\right) =b_{l}z^{l}+\cdots +b_{1}z+b_{0}\) \(\left(l\geq 1\right) \) is a nonconstant polynomial solution of (7), where \(b_{l}\neq 0,\cdots ,b_{0} \)
are complex constants.
If \(l\geq s\), then \(f^{(s)}\not\equiv 0\). Taking \(z=r\), we have
\begin{equation}
|e^{A(z)}|=\left\vert e^{A(r)}\right\vert =\left\vert e^{c_{n}r^{n}+c_{n-1}r^{n-1}+\cdots +c_{0}}\right\vert
=e^{\mathit{Re}c_{n}r^{n}+\mathit{Re}c_{n-1}r^{n-1}+\cdots +\mathit{Re} c_{0}}=e^{\mathit{Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right)}.
\label{2.2a}
\end{equation}
(13)
Substituting \(f_{0}\) into (7) and using (13), we conclude that
\begin{align}
&|a_{sm_{s}dsm_{s}}r^{d_{sm_{s}}}e^{m_{s}\mathit{Re}c_{n}r^{n}\left(1+o\left( 1\right) \right) }|b_{l}l(l-1)\cdots (l-s+1)|r^{l-s}(1+o(1))\leq |-P_{s}(e^{A\left( r\right) })f_{0}^{(s)}(r)|\notag\\
&\leq|f^{(k)}(r)|+|P_{k-1}(e^{A(r)})f_{0}^{(k-1)}(r)|
+\cdots+|P_{s+1}(e^{A(r)})f_{0}^{(s+1)}(r)|+|P_{s-1}(e^{A(r)})f_{0}^{(s-1)}(r)|+\cdots +|P_{0}(e^{A(r)})f_{0}(r)|\notag\\
&\leq M_{0}r^{d}e^{m\mathit{Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right)}(1+o(1)), \label{2.3a}
\end{align}
(14)
where \(d=\max \{d_{jm_{j}}:j=0,\cdots ,s-1,s+1,\cdots ,k-1\}\) and \(M_{0}>0\) is some constant. Since \(m_{s}>m\), we see that (3) is a
contradiction. Obviously, when \(s=0\) or \(1\), we can get that the Equation (7) has nonconstant polynomial solution from the above
process.
If \(l< s\), then
\begin{equation}
P_{l}(e^{A(z)})f_{0}^{(l)}(z)+\cdots +P_{0}(e^{A(z)})f_{0}(z)=0. \label{2.4a}
\end{equation}
(15)
Set \(\max \{m_{j}:j=0,\cdots ,l\}=h\). If \(m_{j}< h\), then we can rewrite
\begin{equation*}
P_{j}(e^{A(z)})=a_{jh}\left( z\right) e^{hA(z)}+\cdots +a_{j(m_{j}+1)}\left(
z\right) e^{(m_{j}+1)A(z)}
+a_{jm_{j}}\left( z\right) e^{m_{j}A(z)}+\cdots +a_{j1}\left( z\right)
e^{A(z)}
\end{equation*}
for \(j=0,\cdots ,l\), where \(a_{jh}\left( z\right) =\cdots=a_{j(m_{j}+1)}\left( z\right) =0\). Thus, we conclude by (15)
that
\begin{align}
&\lbrack a_{lh}\left( z\right) f_{0}^{(l)}+a_{(l-1)h}\left( z\right) f_{0}^{(l-1)}+\cdots +a_{0h}\left( z\right) f_{0}]e^{hA(z)}
+\cdots +[a_{lj}\left( z\right) f_{0}^{(l)}+a_{(l-1)j}\left( z\right)f_{0}^{(l-1)}\notag\\&+\cdots +a_{0j}\left( z\right) f_{0}]e^{jA(z)}
+\cdots +[a_{l1}\left( z\right) f_{0}^{(l)}+a_{(l-1)1}\left( z\right)f_{0}^{(l-1)}+\cdots +a_{01}\left( z\right) f_{0}]e^{A(z)}=0. \label{2.5a}
\end{align}
(16)
Set
\begin{equation}
Q_{j}(z)=a_{lj}(z)f_{0}^{(l)}+a_{(l-1)j}(z)f_{0}^{(l-1)}+\cdots +a_{0j}f_{0} \hspace{0.5cm}(j=1,\cdots ,h). \label{2.6a}
\end{equation}
(17)
Since \(f_{0}\) and \(a_{ij}(z)\) are polynomials, we see that
\begin{equation}
m(r,Q_{j})=o\{m(r,e^{(\alpha -\beta )A(z))}\},\hspace{0.5cm}(1\leq \beta
< \alpha \leq h). \label{2.7a}
\end{equation}
(18)
By Lemma 1 and (16)-(18), we conclude that
\begin{equation}
Q_{1}(z)\equiv Q_{2}(z)\equiv \cdots \equiv Q_{h}(z)\equiv 0. \label{2.8a}
\end{equation}
(19)
Since \(\deg f_{0}>\deg f_{0}^{\prime }>\cdots >\deg f_{0}^{(l)}\) and \(\deg a_{0j}(z)>\deg a_{ij}(z)\) \((i\neq 0)\), so by (16) and (19), we get a contradiction.
Lemma 3. [14,15]
Let \(f\left( z\right) \) be an entire function and suppose that \(|f^{(k)}(z)|\)
is unbounded on some ray \(\arg z=\theta \). Then, there
exists an infinite sequence of points \(z_{m}=r_{m}e^{i\theta }\)
\(\left( m=1,2,\cdots \right) \), where \(r_{m}\rightarrow +\infty\) such that \(f^{(k)}(z_{m})\rightarrow \infty \) and
\begin{equation*}
\left\vert \frac{f^{\left( j\right) }\left( z_{m}\right) }{f^{\left(
k\right) }\left( z_{m}\right) }\right\vert \leq \left\vert z_{m}\right\vert
^{k-j}(1+o(1))\hspace{0.5cm}\left( j=0,\cdots ,k-1\right) .
\end{equation*}
Lemma 4. [16]
Let \(f\left(z\right) \) be a transcendental meromorphic function of finite
order \(\rho .\) Let \(\Gamma =\left\{ \left( k_{1},j_{1}\right)
,\left( k_{2},j_{2}\right) ,\cdots ,\left( k_{m},j_{m}\right) \right\}\)
denote a set of distinct pairs of integers satisfying
\(k_{i}>j_{i}\geq 0\) \(\left( i=1,2,\cdots ,m\right) \) and
let \(\varepsilon >0\) be a given constant. Then, there exists a set \(E_{1}\subset \left[ 0,2\pi \right) \) that has linear measure zero
such that if \(\theta \in \left[ 0,2\pi \right) \diagdown E_{1}\), then there is a constant \(R_{1}=R_{1}\left( \theta \right) >1\)
such that for all \(z\) satisfying \(\arg z=\theta \) and
\(\left\vert z\right\vert \geq R_{1}\) and for all \(\left( k,j\right) \in \Gamma \), we have
\begin{equation*}
\left\vert \frac{f^{\left( k\right) }\left( z\right) }{f^{\left( j\right)}\left( z\right) }\right\vert \leq \left\vert z\right\vert ^{\left(k-j\right) \left( \rho -1+\varepsilon \right) }.
\end{equation*}
Lemma 5. [17]
Let \(f(z)\) be an entire function with \(\rho \left( f\right) =\rho 0\) is a constant
and \(k>0\) is a constant independent of \(\theta _{0}) \).
Then \(f(z)\) is a polynomial with \(\deg f\leq k\).
Lemma 6. [16]
Let \(f\) be a transcendental meromorphic function, and \(\alpha >1\) be a given
constant. Then, there exists a set \(E_{3}\subset (1,\infty )\) with
finite logarithmic measure and a constant \(C>0\) that depends only on \(\alpha \) and \(i\), \(j\)
\((i,j\in \mathbb{N})\), such that for all \(z\) satisfying \(|z|=r\not\in E_{3}\cup \lbrack 0,1]\),
\begin{equation}
\left\vert \frac{f^{\left( j\right) }\left( z\right) }{f^{\left( i\right)
}\left( z\right) }\right\vert \leq C\left( \frac{T(\alpha r,f)}{r}(\log
^{\alpha }r)\log T(\alpha r,f)\right) ^{j-i}. \label{2.9a}
\end{equation}
(20)
Remark 5.
From the proof of Lemma 6 ([16, Theorem 3]), we
can see that the exceptional set \(E_{4}\) equals \(\{|z|:z\in (\cup
_{n=1}^{+\infty }O(a_{n}))\}\), where \(a_{n}(n=1,2,\cdots )\) denote all zeros and poles of \(f^{(i)}\), and \(O(a_{n})\)
denote sufficiently small neighborhoods of \(a_{n}\). Hence, if \(f(z) \) is a transcendental entire function and \(z\) is a
point that satisfies \(|f(z)|\) to be sufficiently large, then the
point \(z\not\in E_{4}\) satisfies (20).
For details, see , [9, Remark 2.10].
Lemma 7. [10,18]
Let \(A_{0}\), \(\cdots \), \(A_{k-1}\) be entire functions of finite order.
If \(f(z)\) is a solution of equation
\begin{equation*}
f^{\left( k\right) }+A_{k-1}f^{\left( k-1\right) }+\cdots +A_{0}f=0,
\end{equation*}
then \(\rho _{2}\left( f\right) \leq \max \{\rho (A_{j}):j=0,\cdots ,k-1\}\).
Lemma 8. [19]
Let \(g(z)\) be an entire function of infinite order with the hyper-order \(\rho_{2}(g)=\rho \), and let \(\nu (r)\) be the central index of
\(g\). Then,
\begin{equation*}
\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log \log \nu (r)}{\log r}=\rho _{2}(g)=\rho .
\end{equation*}
Lemma 9. [7]
Let \(f(z)\) be an entire function that satisfies \(\rho \left( f\right) =\rho (n< \rho
< \infty )\); or \(\rho \left( f\right) =\infty \) and \(\rho
_{2}=0\); or \(\rho _{2}=\alpha (0< \alpha < \infty )\), and a
set \(E_{5}\subset \lbrack 1,\infty )\) has a finite logarithmic
measure. Then, there exists a sequence \(\{z_{k}=r_{k}e^{i\theta _{k}}\}\)
such that \(|f(z_{k})|=M(r_{k},f)\), \(\theta _{k}\in
\lbrack 0,2\pi )\), \(\lim_{k\rightarrow \infty }\theta _{k}=\theta
_{0}\in \lbrack 0,2\pi )\), \(r_{k}\not\in E_{5}\), and \(%
r_{k}\rightarrow \infty \), such that
- (i) if \(\rho \left( f\right) =\rho \) \((n< \rho < \infty )\), then for any given \(\varepsilon_{1}(0< \varepsilon _{1}< \frac{\rho -n}{2})\),
\begin{equation*}
{r_{k}}^{\rho -\varepsilon _{1}}< \nu (r_{k})< {r_{k}}^{\rho +\varepsilon_{1}};
\end{equation*}
- (ii) if \(\rho \left( f\right) =\infty \) and \(\rho _{2}\left( f\right) =0\), then for any given \(\varepsilon _{2}(0< \varepsilon _{2}0)\), we have, as \(r_{k}\) is sufficiently large,
\begin{equation*}
{r_{k}}^{M}< \nu (r_{k})< \exp \{{r_{k}}^{\varepsilon _{2}}\};
\end{equation*}
- (iii) if \(\rho _{2}\left( f\right) =\alpha
(0< \alpha < \infty )\), then for any given \(\varepsilon
_{3}(0< \varepsilon _{3}< \alpha )\),
\begin{equation*}
\exp \{{r_{k}}^{\alpha -\varepsilon _{3}}\}< \nu (r_{k})< \exp \{{r_{k}}^{\alpha +\varepsilon _{3}}\}.
\end{equation*}
Lemma 10. [20]
Let \(g\) be a non-constant entire function, and let \(0< \delta < 1\). There exists
a set \(E_{6}\subset \lbrack 1,\infty )\) of finite logarithmic
measure with the following property. For \(r\in \lbrack 1,\infty )\diagdown
E_{6}\), the central index \(\nu (r)\) of \(g\)
satisfies
\begin{equation*}
\nu (r)\leq (\log M(r,g))^{1+\delta }.
\end{equation*}
Lemma 11. [21,22]
Let \(A_{0}\), …, \(A_{k-1}\), \(F\not\equiv 0\) be finite order
meromorphic functions. If \(f\) is a meromorphic solution of the
equation
\begin{equation*}
f^{(k)}+A_{k-1}f^{(k-1)}+\cdots +A_{0}f=F,
\end{equation*}
with \(\rho \left( f\right) =+\infty \) and \(\rho_{2}\left( f\right) =\rho \), then \(f\) satisfies
\(\overline{\lambda }(f)=\lambda (f)=\rho \left( f\right) =\infty \) and \(\overline{\lambda }_{2}(f)=\lambda _{2}(f)=\rho _{2}\left( f\right)
=\rho \).
Lemma 12. [14]
Let \(\varphi :\left[ 0,+\infty \right) \rightarrow\mathbb{R}\) and \(\psi :\left[ 0,+\infty \right) \rightarrow\mathbb{R}\) be monotone non-decreasing functions such that
\(\varphi \left(r\right) \leq \psi \left( r\right) \) for all \(r\notin E_{7}\cup
\left[ 0,1\right] \), where \(E_{7}\subset \left( 1,+\infty \right) \)
is a set of finite logarithmic measure. Let \(\gamma >1\)
be a given constant. Then there exists a \(r_{1}=r_{1}\left( \gamma \right)
>0\) such that \(\varphi \left( r\right) \leq \psi \left( \gamma
r\right) \) for all \(r>r_{1}.\)
3. Proofs of the results
Proof of Theorem 7.
Suppose that \(f\not\equiv 0\) is a solution of (7), then \(f\)
is an entire function. By Lemma 2, we see that \(f\) is transcendental.
First step. We prove that \(\rho (f)=\infty \).
Suppose, to the contrary, that \(\rho (f)=\rho 0\), there exists a set \(E_{1}\subset \lbrack 0,2\pi
) \) with linear measure zero, such that if \(\theta \in \lbrack 0,2\pi
)\diagdown E_{1}\), then 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}
\left\vert \frac{f^{(j)}(z)}{f^{(s)}(z)}\right\vert \leq r^{\left( \rho
-1+\varepsilon \right) \left( j-s\right) }\hspace{0.5cm}j=s+1,\cdots ,k.
\label{3.1a}
\end{equation}
(21)
Case 1. Take a ray \(\arg z=\theta \in H_{n,+}\diagdown E_{1}\), then
\(\delta _{n}(A,\theta )>0\). We assume that \(|f^{(s)}(re^{i\theta })|\) is
bounded on the ray \(\arg z=\theta \). If \(|f^{(s)}(re^{i\theta })|\) is
unbounded on the ray \(\arg z=\theta \), then by Lemma 3, there exists a
sequence \(\{z_{t}=r_{t}e^{i\theta }\}\) such that as \(r_{t}\rightarrow
+\infty \), \(f^{(s)}(z_{t})\rightarrow \infty \) and
\begin{equation}
\left\vert \frac{f^{(i)}(z_{t})}{f^{(s)}(z_{t})}\right\vert \leq {r_{t}}%
^{s-i}(1+o(1))\leq 2{r_{t}}^{s},\hspace{0.5cm}i=0,\cdots ,s-1. \label{3.2a}
\end{equation}
(22)
By (7), we get
\begin{equation}
|P_{s}(e^{A(z_{t})})|\leq \left\vert \frac{f^{(k)}(z_{t})}{f^{(s)}(z_{t})}%
\right\vert +\sum_{{j=0}{j\neq s}}^{k-1}|P_{j}(e^{A(z_{t})})|\left\vert
\frac{f^{(j)}(z)}{f^{(s)}(z)}\right\vert . \label{3.3a}
\end{equation}
(23)
For \(r_{t}\rightarrow +\infty \), we have
\begin{align}
\left\vert P_{s}(e^{A(z_{t})})\right\vert
&=|a_{sm_{s}}(z_{t})e^{m_{s}A(z_{t})}+\cdots +a_{s1}(z_{t})e^{A(z_{t})}|\notag\\
&\geq |a_{sm_{s}}(z_{t})e^{m_{s}A(z_{t})}|-\left\vert a_{s\left(
m_{s}-1\right) }(z_{t})e^{m_{s-1}A(z_{t})}+\cdots
+a_{s1}(z_{t})e^{A(z_{t})}\right\vert\notag\\
&\geq |a_{sm_{s}}(z_{t})e^{m_{s}A(z_{t})}|-\left[ \left\vert a_{s\left(
m_{s}-1\right) }(z_{t})e^{m_{s-1}A(z_{t})}\right\vert +\cdots +\left\vert
a_{s1}(z_{t})e^{A(z_{t})}\right\vert \right]\notag\\
&=|a_{{sm_{s}}{d_{sm_{s}}}}|{r}_{t}^{d_{sm_{s}}}e^{m_{s}\delta _{n}(A,\theta )%
{r}_{t}^{n}(1+o(1))}(1+o(1))
-[|a_{{s(m_{s}-1)}{d_{s(m_{s}-1)}}}|{r}_{t}^{d_{s(m_{s}-1)}}e^{(m_{s}-1)%
\delta _{n}(A,\theta ){r}_{t}^{n}(1+o(1))}(1+o(1))\notag\\
&\;\;\;+\cdots +|a_{{s1}{d_{s1}}}|{r}_{t}^{d_{s1}}e^{\delta _{n}(A,\theta ){r}%
_{t}^{n}(1+o(1))}(1+o(1))]\notag\end{align}
\begin{align}
&\geq \frac{1}{2}|a_{{sm_{s}}{d_{sm_{s}}}}|{r}_{t}^{d_{sm_{s}}}e^{m_{s}\delta
_{n}(A,\theta ){r}_{t}^{n}(1+o(1))}(1+o(1)), \label{3.4a}
\end{align}
(24)
and
\begin{align}
\left\vert P_{j}(e^{A(z_{t})})\right\vert
&=|a_{jm_{j}}(z_{t})e^{m_{j}A(z_{t})}+\cdots +a_{j1}(z_{t})e^{A(z_{t})}|
\notag\\&\leq |a_{jm_{j}d_{jm_{j}}}|{r}_{t}^{d_{jm_{j}}}e^{m_{j}\delta _{n}(A,\theta )%
{r}_{t}^{n}(1+o(1))}(1+o(1))
+\cdots +|a_{jm_{j}1}|{r}_{t}^{d_{j1}}e^{\delta _{n}(A,\theta ){r}%
_{t}^{n}(1+o(1))}(1+o(1))\notag\\
&\leq 2|a_{{jm_{j}}{d_{jm_{j}}}}|{r}_{t}^{d}e^{m\delta _{n}(A,\theta ){r}%
_{t}^{n}(1+o(1))}(1+o(1)),\text{ }\left( j\neq s\right) , \label{3.5a}
\end{align}
(25)
where \(d=\max \{d_{jm_{j}}:j=0,\cdots ,s-1,s+1,\cdots ,k-1\}\). Substituting
(21), (22), (24), (25) into (23), we obtain that for sufficiently
large \(r_{t}\)
\begin{equation}
\frac{1}{2}|a_{sm_{s}d_{sm_{s}}}|{r_{t}}^{d_{sm_{s}}}e^{m_{s}\delta
_{n}(A,\theta )r_{t}^{n}(1+o(1))}(1+o(1))
\leq C_{0}r_{t}^{d+k\rho }e^{m\delta _{n}(A,\theta )(1+o(1)){r_{t}}%
^{n}}(1+o(1)), \label{3.6a}
\end{equation}
(26)
where \(C_{0}>0\) is a constant. From (26), we can get a
contradiction by \(m_{s}>m\) and \(\delta _{n}(A,\theta )>0\), so
\begin{equation}
|f(re^{i\theta })|\leq Mr^{s}\leq M_{1}r^{k},\hspace{0.5cm}M_{1}>0,
\label{3.7a}
\end{equation}
(27)
on the ray \(\arg z=\theta \in H_{n,+}\diagdown E_{1}\).
Case 2. Now, we take a ray \(\arg z=\theta \in H_{n,-}\),
then \(\delta _{n}(A,\theta )< 0\). If \(|f^{(k)}(re^{i\theta })|\) is unbounded
on the ray \(\arg z=\theta \), then by Lemma 3, there exists a sequence \(
\{z_{t}=r_{t}e^{i\theta }\}\) such that as \(r_{t}\rightarrow +\infty \), \(
f^{(s)}(z_{t})\rightarrow \infty \) and
\begin{equation}
\left\vert \frac{f^{(i)}(z_{t})}{f^{(k)}(z_{t})}\right\vert \leq {r_{t}}%
^{k-i}(1+o(1))\leq 2{r_{t}}^{k},\hspace{0.5cm}i=0,\cdots ,k-1. \label{3.8a}
\end{equation}
(28)
By (7), we get
\begin{equation}
-1=P_{k-1}(e^{A(z_{t})})\frac{f^{(k-1)}(z_{t})}{f^{(k)}(z_{t})}+\cdots
+P_{0}(e^{A(z_{t})})\frac{f(z_{t})}{f^{(k)}(z_{t})}. \label{3.9a}
\end{equation}
(29)
For \(r_{t}\rightarrow +\infty \), we have
\begin{align}
\left\vert P_{j}(e^{A(z_{t})})\right\vert
&=|a_{jm_{j}}(z_{t})e^{m_{j}A(z_{t})}+\cdots +a_{j1}(z_{t})e^{A(z_{t})}|\notag\\
&\leq |a_{jm_{j}d_{jm_{j}}}|{r}^{d_{jm_{j}}}e^{m_{j}\delta _{n}(A,\theta ){r}%
_{t}^{n}(1+o(1))}(1+o(1))
+\cdots +|a_{jm_{j}1}|{r}^{d_{j1}}e^{\delta _{n}(A,\theta ){r}%
_{t}^{n}(1+o(1))}(1+o(1))\notag\\
&\leq 2|a_{jm_{j}1}|{r}^{d}e^{\delta _{n}(A,\theta ){r}%
_{t}^{n}(1+o(1))}(1+o(1))\text{ }\left( j=0,…,k-1\right) . \label{3.10a}
\end{align}
(30)
Substituting (28) and (30) into (29), we obtain that for sufficiently large \(r_{t}\)
\begin{equation}
1\leq C_{1}{r_{t}}^{k+d}e^{\delta _{n}(A,\theta ){r_{t}}%
^{n}(1+o(1))}(1+o(1)),\hspace{0.5cm}C_{1}>0. \label{3.11a}
\end{equation}
(31)
Since \(\delta _{n}(A,\theta )< 0\), when \(r_{t}\rightarrow +\infty \), by (31), we get \(1\leq 0\), this is a contradiction. Hence
\begin{equation}
|f(re^{i\theta })|\leq M_{2}r^{k},\hspace{0.5cm}M_{2}>0, \label{3.12a}
\end{equation}
(32)
on the ray \(\arg z=\theta \in H_{n,-}\diagdown E_{1}\). From Lemma 5,
(27) and (32), we know that \(f(z)\) is a
polynomial, which contradicts the assertion that \(f(z)\) is transcendental.
Therefore, \(\rho (f)=\infty \).
Step 2. We prove that \(\rho _{2}(f)=n\). By Lemma 7 and \(\rho
(P_{j}(e^{A(z)}))=n\) \((j=0,\cdots ,k-1)\), we see that \(\rho _{2}(f)\leq \max
\{\rho (P_{j}(e^{A(z)}))\}=n\).
Now, we suppose that there exists a solution \(f_{0}\) satisfies \(\rho
_{2}(f_{0})=\alpha < n\). Then we have
\begin{equation}
\underset{r\rightarrow +\infty }{\lim \sup }\frac{\log T(r,f_{0})}{r^{n}}=0.
\label{3.13a}
\end{equation}
(33)
By Lemma 6, we see that there exists a subset \(E_{3}\subset (1,\infty )\)
having finite logarithmic measure such that for all \(z\) satisfying \(%
|z|=r\not\in E_{3}\cup \lbrack 0,1]\),
\begin{equation}
\left\vert \frac{{f_{0}}^{(j)}(z)}{f_{0}(z)}\right\vert \leq
C[T(2r,f_{0})]^{k+1},\hspace{0.5cm}j=1,\cdots ,k, \label{3.14a}
\end{equation}
(34)
where \(C(>0)\) is some constant. From the Wiman-Valiron theory, there is a
set \(E_{8}\subset (1,\infty )\) having finite logarithmic measure, such that
we can choose a \(z\) satisfying \(|z|=r\not\in \lbrack 0,1]\cup E_{8}\) and \(%
|f_{0}(z)|=M(r,f_{0})\), then we get
\begin{equation}
\frac{{f_{0}}^{(j)}(z)}{f_{0}(z)}=\left( \frac{\nu (r)}{z}\right)
^{j}(1+o(1)),\hspace{0.5cm}j=1,\cdots ,k-1, \label{3.15a}
\end{equation}
(35)
where \(\nu (r)\) is the central index of \(f_{0}(z)\). By Lemma 9, we see
that there exists a sequence \(\{z_{t}=r_{t}e^{i\theta _{t}}\}\) such that \(%
|f_{0}(z_{t})|=M(r_{t},f_{0})\), \(\theta _{t}\in \lbrack 0,2\pi )\), with \(%
r_{t}\not\in \lbrack 0,1]\cup E_{5}\cup E_{8}\), \(r_{t}\rightarrow +\infty \)
and for any sufficiently large \(M_{3}(>2k+3)\)
\begin{equation}
\nu (r_{t})>{r_{t}}^{M_{3}}>r_{t}. \label{3.16a}
\end{equation}
(36)
Case 1. Suppose \(\theta _{0}\in H_{n,+}\). Since \(\delta
_{n}(A,\theta )=\alpha _{n}\cos (\theta _{n}+n\theta )\) is a continuous
function of \(\theta \), by \(\theta _{t}\rightarrow \theta _{0}\) we get \(%
\lim_{t\rightarrow \infty }\delta _{n}(A,\theta _{t})=\delta _{n}(A,\theta
_{0})>0\). Therefore, there exists a constant \(N(>0)\) such that as \(t>N\),
\begin{equation*}
\delta _{n}(A,\theta _{t})\geq \frac{1}{2}\delta _{n}(A,\theta _{0})>0.
\end{equation*}
By (33), for any given \(\varepsilon _{1}(0< \varepsilon _{1}N\),
\begin{equation}
\lbrack T(2r_{t},f_{0})]^{k+1}\leq e^{\varepsilon _{1}(k+1)(2r_{t})^{n}}\leq
e^{\frac{1}{2}\delta _{n}(A,\theta _{t}){r_{t}}^{n}}. \label{3.17a}
\end{equation}
(37)
By (34), (35) and (37),
we have
\begin{equation}
\left( \frac{\nu (r_{t})}{r_{t}}\right) ^{k-s}(1+o(1))=\left\vert \frac{%
f_{0}^{(k-s)}(z_{t})}{f_{0}(z_{t})}\right\vert \leq
C[T(2r_{t},f_{0})]^{k+1}\leq Ce^{\frac{1}{2}\delta _{n}(A,\theta _{0}){r_{t}}%
^{n}}. \label{3.18a}
\end{equation}
(38)
By (7), we get
\begin{equation}
-\frac{f_{0}^{(s)}(z_{t})}{f_{0}(z_{t})}P_{s}(e^{A(z_{t})})=\frac{%
f_{0}^{(k)}(z_{t})}{f_{0}(z_{t})}+\sum_{j=0,j\neq s}^{k-1}P_{j}(e^{A(z_{t})})%
\frac{f_{0}^{(j)}(z_{t})}{f_{0}(z_{t})}. \label{3.19a}
\end{equation}
(39)
Substituting (24), (25) and (35) into (39), we get for sufficiently large \(r_{t}\),
\begin{align}
\left( \frac{\nu (r_{t})}{r_{t}}\right) ^{s}\frac{1}{2}&|a_{sm_{s}d_{sm_{s}}}|%
{r_{t}}^{d_{sm_{s}}}e^{m_{s}\delta _{n}(A,\theta _{t}){r_{t}}%
^{n}(1+o(1))}(1+o(1))\notag\\&\leq \left( \frac{\nu (r_{t})}{r_{t}}\right)
^{k}(1+o(1))
+\sum_{j=0,j\neq s}^{k-1}2|a_{{jm_{j}}{d_{jm_{j}}}}|{r}_{t}^{d}e^{m\delta
_{n}(A,\theta ){r}_{t}^{n}(1+o(1))}\left( \frac{\nu (r_{t})}{r_{t}}\right)
^{j}(1+o(1)). \label{3.20a}
\end{align}
(40)
By (36), (38) and (40),
we get
\begin{align}
|a_{sm_{s}d_{sm_{s}}}|&{r_{t}}^{d_{sm_{s}}}e^{m_{s}\delta _{n}(A,\theta _{t}){%
r_{t}}^{n}(1+o(1))}(1+o(1))\notag\\
&\leq 2\left( \frac{\nu (r_{t})}{r_{t}}\right)
^{k-s}(1+o(1))
+\sum_{j=0,j\neq s}^{k-1}4|a_{{jm_{j}}{d_{jm_{j}}}}|{r}_{t}^{d}e^{m\delta
_{n}(A,\theta ){r}_{t}^{n}(1+o(1))}\left( \frac{\nu (r_{t})}{r_{t}}\right)
^{j-s}(1+o(1))\notag\\
&\leq 2\left( \frac{\nu (r_{t})}{r_{t}}\right) ^{k-s}(1+o(1))
+\sum_{j=0}^{s-1}4|a_{{jm_{j}}{d_{jm_{j}}}}|{r}_{t}^{d}e^{m\delta
_{n}(A,\theta ){r}_{t}^{n}(1+o(1))}\left( \frac{\nu (r_{t})}{r_{t}}\right)
^{j-s}(1+o(1))\notag\\
&\;\;\;\;+\sum_{j=s+1}^{k-1}4|a_{{jm_{j}}{d_{jm_{j}}}}|{r}_{t}^{d}e^{m\delta
_{n}(A,\theta ){r}_{t}^{n}(1+o(1))}\left( \frac{\nu (r_{t})}{r_{t}}\right)
^{j-s}(1+o(1))\notag\end{align}
\begin{align}
&\leq C_{2}{r_{t}}^{d}e^{m\delta _{n}(A,\theta ){r}_{t}^{n}(1+o(1))}\left(
\frac{\nu (r_{t})}{r_{t}}\right) ^{k-s}(1+o(1)),\notag\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;
\end{align}
where \(C_{2}>0\) is a constant. From this inequality and (38), it follows that
\begin{align}
|a_{sm_{s}d_{sm_{s}}}|{r_{t}}^{d_{sm_{s}}}e^{\left( m_{s}-m\right) \delta
_{n}(A,\theta _{t}){r_{t}}^{n}(1+o(1))}(1+o(1))
&\leq C_{2}{r_{t}}^{d}\left( \frac{\nu (r_{t})}{r_{t}}\right) ^{k-s}(1+o(1))\leq CC_{2}|a_{{jm_{j}}{d_{jm_{j}}}}|{r_{t}}^{d}e^{\frac{1}{2}\delta
_{n}(A,\theta _{0}){r_{t}}^{n}}. \label{3.21a}
\end{align}
(41)
Since \(m_{s}-m\geq 1>\frac{1}{2}\) and \(\delta (A,\theta _{t})\geq \frac{1}{2}%
\delta _{n}(A,\theta _{0})>0\), we see that (41) is a
contradiction.
Case 2. Suppose \(\theta _{0}\in H_{n,-}\). Since \(\delta
_{n}(A,\theta )\) is a continuous function of \(\theta \), by \(\theta
_{t}\rightarrow \theta _{0}\) we get \(\lim_{t\rightarrow \infty }\delta
_{n}(A,\theta _{t})=\delta _{n}(A,\theta _{0})0)\) such that as \(t>N\),
\begin{equation*}
\delta _{n}(A,\theta _{t})\leq \frac{1}{2}\delta _{n}(A,\theta _{0})< 0.
\end{equation*}
By (7), we can write
\begin{equation}
e^{-m_{s}A(z_{t})}\frac{{f_{0}}^{(k)}(z_{t})}{f_{0}(z_{t})}%
=e^{-m_{s}A(z_{t})}P_{k-1}(e^{A(z_{t})})\frac{f^{(k-1)}(z_{t})}{f_{0}(z_{t})}
+\cdots +e^{-m_{s}A(z_{t})}P_{0}(e^{A(z_{t})}). \label{3.22a}
\end{equation}
(42)
From (6) and \(\delta _{n}(A,\theta _{t})< 0\), we get
\begin{align}
|e^{-m_{s}A(z_{t})}P_{j}(e^{A(z_{t})})|&=\left\vert e^{-m_{s}A(z_{t})}\left(
a_{jm_{j}}(z_{t})e^{m_{j}A(z_{t})}+\cdots +a_{j1}(z_{t})e^{A(z_{t})}\right)
\right\vert\notag\\
&=\left\vert a_{jm_{j}}(z_{t})e^{-\left( m_{s}-m_{j}\right) A(z_{t})}+\cdots
+a_{j1}(z_{t})e^{-\left( m_{s}-1\right) A(z_{t})}\right\vert\notag\\
&\leq C_{3}{r_{t}}^{d_{j1}}e^{-(m_{s}-1)\delta (A,\theta _{t}){r_{t}}%
^{n}(1+o(1))}(1+o(1)), \label{3.23a}
\end{align}
(43)
where \(C_{3}>0\) is a constant. Substituting (35) and (43) into (42), we get
\begin{equation}
e^{-m_{s}\delta (A,\theta _{t}){r_{t}}^{n}(1+o(1))}\nu (r_{t})\leq C_{4}{%
r_{t}}^{d+k}e^{-(m_{s}-1)\delta (A,\theta _{t}){r_{t}}^{n}(1+o(1))}(1+o(1)),
\label{3.24a}
\end{equation}
(44)
where \(C_{4}>0\) is a constant. By substituting (36) into (44), we have
\begin{equation}
{r_{t}}^{M_{3}}e^{-m_{s}\delta (A,\theta _{t}){r_{t}}^{n}(1+o(1))}\leq C_{4}{%
r_{t}}^{d+k}e^{-(m_{s}-1)\delta (A,\theta _{t}){r_{t}}^{n}(1+o(1))}(1+o(1)).
\label{3.25a}
\end{equation}
(45)
Since \(\delta (A,\theta _{t})\) \(\leq \frac{1}{2}\delta _{n}(A,\theta _{0})< 0\), we see (45) is also a contradiction.
Case 3. Suppose \(\theta _{0}\in H_{n,0}\). Since \(\theta
_{t}\rightarrow \theta _{0}\), for any given \(\varepsilon _{2}\) \(%
(0< \varepsilon _{2}0)\),
such that as \(t>N\), \(\theta _{t}\in \lbrack \theta _{0}-\varepsilon
_{2},\theta _{0}+\varepsilon _{2}]\), and
\begin{equation*}
z_{t}=r_{t}e^{i\theta _{t}}\in \overline{\Omega }=\{z:\theta
_{0}-\varepsilon _{2}\leq \arg z\leq \theta _{0}+\varepsilon _{2}\}.
\end{equation*}
By Lemma 6, we se that there exist a subset \(E_{3}\subset (1,\infty )\)
having logarithmic measure \(lmE_{3}0\) such that
for all \(z\) satisfying \(|z|=r\not\in E_{3}\cup \lbrack 0,1]\),
\begin{equation}
\left\vert \frac{{f_{0}}^{(i)}(z)}{{f_{0}}^{(s)}(z)}\right\vert \leq C[T(2r,{%
f_{0}}^{(s)})]^{k-s+1},\hspace{0.5cm}i=s+1,\cdots ,k, \label{3.26a}
\end{equation}
(46)
Now, we consider the growth of \(f_{0}(re^{i\theta })\) on a ray \(\arg
z=\theta \in \overline{\Omega }\diagdown \{\theta _{0}\}\). By the properties
of cosine function, we suppose without loss of generality that \(\delta
_{n}(A,\theta )>0\) for \(\theta \in \lbrack \theta _{0}-\varepsilon
_{2},\theta _{0})\) and \(\delta _{n}(A,\theta )< 0\) for \(\theta \in (\theta
_{0},\theta _{0}+\varepsilon _{2}]\).
Subcase 3.1 For a fixed \(\theta \in \lbrack \theta _{0}-\varepsilon
_{2},\theta _{0})\), we have \(\delta _{n}(A,\theta )>0\). Since \(\rho
_{2}(f_{0})< n\), we get that \(f_{0}\) satisfies (33). From
\(T(r,{f_{0}}^{(s)})< (s+1)T(r,f_{0})+S(r,f_{0}),\) where \(S(r,f)=o(T(r,f))\), as
\(r\rightarrow +\infty \) outside of a possible exceptional set of finite
logarithmic measure, we get that \({f_{0}}^{(s)}\) also satisfies (33). So for any given \(\varepsilon _{2}\) satisfying \(
0< \varepsilon _{2}< \frac{1}{2^{n+1}(k-s+1)}\delta _{n}(A,\theta ) \), we have
\begin{equation}
\lbrack T(2r_{t},f_{0}^{(s)})]^{k-s+1}\leq e^{\varepsilon
_{2}(k-s+1)(2r_{t})^{n}}\leq e^{\frac{1}{2}\delta _{n}(A,\theta _{0}){r_{t}}%
^{n}}. \label{3.27a}
\end{equation}
(47)
We assert that \(|f_{0}^{(s)}(re^{i\theta })|\) is bounded on the ray \(\arg
z=\theta \in \lbrack \theta _{0}-\varepsilon _{2},\theta _{0})\). If \(%
|f^{(s)}(re^{i\theta })|\) is unbounded on the ray \(\arg z=\theta \), then, by
Lemma 3, there exists a sequence \(\{y_{j}=R_{j}e^{i\theta }\}\) such that
as \(R_{j}\rightarrow \infty \), \(f_{0}^{(s)}(y_{j})\rightarrow \infty \) and
\begin{equation}
\left\vert \frac{f_{0}^{(i)}(y_{j})}{f_{0}^{(s)}(y_{j})}\right\vert \leq {%
R_{j}}^{s-i}(1+o(1))\leq 2{R_{j}}^{s},\hspace{0.5cm}i=0,\cdots ,s-1.
\label{3.28a}
\end{equation}
(48)
By Remark 5 and \(f_{0}^{(s)}(y_{j})\rightarrow \infty \), we know that \(%
|y_{j}|=R_{j}\not\in E_{4}\). By (46) and (47), we have for sufficiently large \(j\),
\begin{equation}
\left\vert \frac{f_{0}^{(j)}(y_{j})}{f_{0}^{(s)}(y_{j})}\right\vert \leq
C[T(2R_{j},f_{0}^{(s)})]^{k-s+1}\leq Ce^{\frac{1}{2}\delta _{n}(A,\theta
_{0})R_{j}^{n}},\hspace{0.5cm}j=s+1,\cdots ,k. \label{3.29a}
\end{equation}
(49)
Substituting (24), (25), (48) and (49) into (23)
\begin{align}
\frac{1}{2}|a_{sm_{s}d_{sm_{s}}}|&{R_{j}}^{d_{sm_{s}}}e^{m_{s}\delta
_{n}(A,\theta )R_{j}^{n}(1+o(1))}(1+o(1))\notag\\=&|P_{s}(e^{A(y_{j})})|\notag\\
\leq& \left\vert \frac{f^{(k)}(y_{j})}{f^{(s)}(y_{j})}\right\vert +\sum_{{j=0}%
{j\neq s}}^{k-1}|P_{j}(e^{A(y_{j})})|\left\vert \frac{f^{(j)}(y_{j})}{%
f^{(s)}(y_{j})}\right\vert\notag\\
=&\left\vert \frac{f^{(k)}(y_{j})}{f^{(s)}(y_{j})}\right\vert +\sum_{{j=0}%
}^{s-1}|P_{j}(e^{A(y_{j})})|\left\vert \frac{f^{(j)}(y_{j})}{f^{(s)}(y_{j})}%
\right\vert +\sum_{{j=s+1}}^{k-1}|P_{j}(e^{A(y_{j})})|\left\vert \frac{%
f^{(j)}(y_{j})}{f^{(s)}(y_{j})}\right\vert\notag\\
\leq& Ce^{\frac{1}{2}\delta _{n}(A,\theta
_{0})R_{j}^{n}}+\sum_{j=0}^{s-1}4|a_{{jm_{j}}{d_{jm_{j}}}}|{R}%
_{j}^{d}e^{m\delta _{n}(A,\theta ){R}_{j}^{n}(1+o(1))}{R_{j}}^{s}(1+o(1))\notag\\
&+\sum_{j=s+1}^{k-1}2|a_{{jm_{j}}{d_{jm_{j}}}}|{R}_{j}^{d}e^{m\delta
_{n}(A,\theta ){R}_{j}^{n}(1+o(1))}Ce^{\frac{1}{2}\delta _{n}(A,\theta _{0}){%
R}_{j}^{n}}\notag\\&\leq C_{5}{R}_{j}^{d}e^{(\frac{1}{2}+m)\delta _{n}(A,\theta ){%
R_{j}}^{n}},
\end{align}
(50)
where \(C_{5}>0\) is a constant, which yields a contradiction by \(m_{s}-m\geq
1>\frac{1}{2}\) and \(\delta _{n}(A,\theta )>0\). Hence \(|f_{0}^{(s)}(re^{i
\theta })|\) is bounded on the ray \(\arg z=\theta \), so
\begin{equation}
|f_{0}(re^{i\theta })|\leq M_{4}r^{s},\hspace{0.5cm}M_{4}>0, \label{3.30a}
\end{equation}
(51)
on the ray \(\arg z=\theta \in \lbrack \theta _{0}-\varepsilon _{4},\theta
_{0})\).
Subcase 3.2 For a fixed \(\theta \in (\theta _{0},\theta
_{0}+\varepsilon _{2}]\), we have \(\delta _{n}(A,\theta )< 0\). Using a
reasoning similar to that in Subcase 3.1, we obtain
\begin{equation}
|f_{0}(re^{i\theta })|\leq M_{5}r^{k},\hspace{0.5cm}M_{5}>0, \label{3.31a}
\end{equation}
(52)
on the ray \(\arg z=\theta \in (\theta _{0},\theta _{0}+\varepsilon _{4}]\).
By (51) and (52), we see that on the
ray \(\arg z=\theta \in \overline{\Omega }\diagdown \{\theta _{0}\}\),
\begin{equation}
|f_{0}(re^{i\theta })|\leq M_{5}r^{k},\hspace{0.5cm}M_{5}>0. \label{3.32a}
\end{equation}
(53)
But since \(\rho (f_{0}(re^{i\theta }))=\infty \) and \(\{z_{t}=r_{t}e^{i\theta
_{t}}\}\) satisfies \(|f_{0}(z_{t})|=M(r_{t},f_{0})\), we see that, for any
large \(M_{6}(>k)\), as \(t\) is sufficiently large,
\begin{equation}
|f_{0}(z_{t})|=|f_{0}(z_{t})|=|f_{0}(z_{t})|=|f_{0}(r_{t}e^{i\theta
_{t}})|\geq \exp \{r_{t}^{M_{6}}\}. \label{3.33a}
\end{equation}
(54)
Since \(z_{t}\in \overline{\Omega }\), by (53) and (54), we see that \(\theta _{t}=\theta _{0}\) as
\(t\rightarrow \infty \). Therefore, \(\delta _{n}(A,\theta _{t})=0\) as \(t\rightarrow \infty \). Thus, for sufficiently large \(t\),
\begin{align}
|P_{j}(e^{z_{t}})|&=|a_{jm_{j}}(z_{t})e^{m_{j}A(z_{t})}+a_{jm_{j-1}}(z_{t})e^{m_{j-1}A(z_{t})}+\cdots +a_{j1}(z_{t})e^{A(z_{t})}|\notag\\
&\leq |a_{jm_{j}}(z_{t})|+|a_{jm_{j-1}}(z_{t})|+\cdots +|a_{j1}(z_{t})|\leq
C_{6}r^{d}, \label{3.34a}
\end{align}
(55)
where \(j=0,\cdots ,k-1\) and \(C_{6}>0\) is a constant. By (7), (35) and (55), we get that
\begin{equation*}
|-(\frac{\nu (r_{t})}{z_{t}})^{k}(1+o(1))|=|-\frac{f_{0}^{(k)}(z_{t})}{%
f_{0}(z_{t})}|\leq C_{7}r^{d}(\frac{\nu (r_{t})}{z_{t}})^{k-1}(1+o(1)),
\end{equation*}
i.e.,
\begin{equation}
\nu (r_{t})(1+o(1))\leq C_{7}r^{d+1}(1+o(1)), \label{3.35a}
\end{equation}
(56)
where \(C_{7}>0\) is a constant. Substituting (36) into (56), we obtain also a contradiction. So we have
\(\rho_{2}(f)=n\).
Proof of Corollary 1.
From Theorem 7, we get \(\rho (f)=\infty \) and \(\rho _{2}(f)=n\). Let \(%
g=f-z\), then \(f=g+z\). Substituting it into (7), we have
\begin{equation*}
g^{(k)}+P_{k-1}(e^{A(z)})g^{(k-1)}+\cdots
+P_{0}(e^{A(z)})g=-zP_{0}(e^{A(z)})-P_{1}(e^{A(z)}).
\end{equation*}
Since \(-zP_{0}(e^{A(z)})-P_{1}(e^{A(z)})\not\equiv 0\), from Lemma 11, \(
\rho (g)=\infty \) and \(\rho _{2}(g)=n\) we conclude \(\overline{\lambda }%
(g)=\lambda (g)=\rho (g)=\infty \) and \(\overline{\lambda }_{2}(g)=\lambda
_{2}(g)=\rho _{2}(g)=n\). So \(\overline{\tau }(f)=\tau (f)=\rho (f)=\infty \)
and \(\overline{\tau }_{2}(f)=\tau _{2}(f)=\rho _{2}(f)=n\).
Proof of Theorem 8.
From Theorem 7, we get \(\rho (f)=\infty \) and \(\rho _{2}(f)=n\).
- (i) Let \(g=f-z\), then \(f=g+z\). Substituting
it into (8), we have
\begin{equation*}
g^{\prime \prime }+P(e^{A(z)})g^{\prime
}+Q(e^{A(z)})g=-P(e^{A(z)})-zQ(e^{A(z)}).
\end{equation*}
Since \(p\neq s\), we get \(-P(e^{A(z)})-Q(e^{A(z)})z\not\equiv 0\). From Lemma 11, we obtain \({\lambda }(g)=\rho (g)=\rho (f)=\infty \) and \(\lambda
_{2}(g)=\rho _{2}(g)=\rho _{2}(f)=n\). So \(\lambda (f-z)=\infty \) and \(\lambda _{2}(f-z)=n\).
- (ii) Differentiating both sides of (8), we get that
\begin{equation}
f^{\prime \prime \prime }+P(e^{A(z)})f^{\prime \prime
}+[(P(e^{A(z)}))^{\prime }+Q(e^{A(z)})]f^{\prime }+(Q(e^{A(z)}))^{\prime
}f=0. \label{3.36a}
\end{equation}
(57)
By (8), we have
\begin{equation}
f=-\frac{f^{\prime \prime }+P(e^{A(z)})f^{\prime }}{Q(e^{A(z)})}. \label{3.37a}
\end{equation}
(58)
Substituting (58) into (57), we get
\begin{equation}
f^{\prime \prime \prime }+[(P(e^{A(z)}))^{\prime }-\frac{(Q(e^{A(z)}))^{%
\prime }}{Q(e^{A(z)})}]f^{\prime \prime }
+[(P(e^{A(z)}))^{\prime }+Q(e^{A(z)})-\frac{(Q(e^{A(z)}))^{\prime }}{%
Q(e^{A(z)})}P(e^{A(z)})]f^{\prime }=0. \label{3.38a}
\end{equation}
(59)
Let \(g=f^{\prime }-z\), then \(f^{\prime }=g+z\), \(f^{\prime \prime }=g^{\prime
}+1\), \(f^{\prime \prime \prime }=g^{\prime \prime }\). Substituting these
into (59), we get that
\begin{align}
g^{\prime \prime }&+[(P(e^{A(z)}))^{\prime }-\frac{(Q(e^{A(z)}))^{\prime }}{%
Q(e^{A(z)})}]g^{\prime }+[(P(e^{A(z)}))^{\prime }+Q(e^{A(z)})-\frac{%
(Q(e^{A(z)}))^{\prime }}{Q(e^{A(z)})}P(e^{A(z)})]g\notag\\
&=-P(e^{A(z)})+\frac{(Q(e^{A(z)}))^{\prime }}{Q(e^{A(z)})}
-[(P(e^{A(z)}))^{\prime }+Q(e^{A(z)})-\frac{(Q(e^{A(z)}))^{\prime }}{%
Q(e^{A(z)})}P(e^{A(z)})]z\notag\\&=h(z). \label{3.39}
\end{align}
(60)
Next, we prove that \(h(z)\not\equiv 0\). If \(h(z)\equiv 0\), then
\begin{equation*}
-P(e^{A(z)})+\frac{(Q(e^{A(z)}))^{\prime }}{Q(e^{A(z)})}\equiv \lbrack
(P(e^{A(z)}))^{\prime }+Q(e^{A(z)})-\frac{(Q(e^{A(z)}))^{\prime }}{%
Q(e^{A(z)})}P(e^{A(z)})]z.
\end{equation*}
Since \(Q(z)\not\equiv 0\), we have
\begin{equation}
(Q(e^{A(z)}))^{\prime }-(Q(e^{A(z)}))^{2}z\equiv P(e^{A(z)})Q(e^{A(z)})
+[(P(e^{A(z)}))^{\prime }Q(e^{A(z)})-(Q(e^{A(z)}))^{\prime }P(e^{A(z)})]z.
\label{3.40a}
\end{equation}
(61)
Suppose \(p>s.\) By taking \(z=r\), we have
\begin{equation*}
P(e^{A(r)})=a_{p}(r)e^{pA(r)}+\cdots +a_{1}(r)e^{A(r)},\;\;\;\;\;\text{and}\;\;\;\;\;\;
Q(e^{A(r)})=b_{s}(r)e^{sA(r)}+\cdots +b_{1}(r)e^{A(r)}.
\end{equation*}
We get
\begin{align*}
(P(e^{A(r)}))^{\prime }&=\overset{p}{\underset{j=1}{\sum }}(a_{j}^{\prime
}(r)+jA^{\prime }(r)a_{j}(r))e^{jA(r)}\\
&=(a_{p}^{\prime }(r)+pA^{\prime }(r)a_{p}(r))e^{pA(r)}+\cdots
+(a_{1}^{\prime }(r)+A^{\prime }(r)a_{1}(r))e^{A(r)}
\end{align*}
and
\begin{align*}
(Q(e^{A(r)}))^{\prime }&=\overset{s}{\underset{j=1}{\sum }}(b_{j}^{\prime
}(r)+jA^{\prime }(r)b_{j}(r))e^{jA(r)}\\
&=(b_{s}^{\prime }(r)+sA^{\prime }(r)b_{s}(r))e^{sA(r)}+\cdots
+(b_{1}^{\prime }(r)+A^{\prime }(r)b_{1}(r))e^{A(r)}.
\end{align*}
So, we obtain
\begin{align*}
&|P(e^{A(r)})Q(e^{A(r)})+(P(e^{A(r)}))^{\prime
}Q(e^{A(r)})r-(Q(e^{A(r)}))^{^{\prime }}P(e^{A(r)})r|\\
&=|a_{p}(r)b_{s}(r)+(p-s)rA^{\prime }(r)a_{p}(r)b_{s}(r)
+(a_{p}^{\prime }(r)b_{s}(r)-a_{p}(r)b_{s}^{\prime }(r))r|e^{(p+s)\mathit{Re}%
c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1)).
\end{align*}
Since \(a_{p}(r)\), \(b_{s}(r)\) and \(A(r)\) are polynomials and \(p>s\), we get
\begin{equation*}
\deg ((p-s)rA^{\prime }(r)a_{p}(r)b_{s}(r))>\deg
[a_{p}(r)b_{s}(r)+(a_{p}^{\prime }(r)b_{s}(r)-a_{p}(r)b_{s}^{\prime }(r))r].
\end{equation*}
So, we have
\begin{equation*}
|(p-s)rA^{\prime }(r)a_{p}(r)b_{s}(r)+a_{p}(r)b_{s}(r)+(a_{p}^{\prime
}(r)b_{s}(r)-a_{p}(r)b_{s}^{\prime }(r))r|
=Mr^{d_{1}}(1+o(1))\not\equiv 0,
\end{equation*}
where \(M>0\) and \(d_{1}>0\) are some constants. It follows that
\begin{equation*}
|P(e^{A(r)})Q(e^{A(r)})+(P(e^{A(r)}))^{\prime
}Q(e^{A(r)})r-(Q(e^{A(r)}))^{^{\prime }}P(e^{A(r)})r|
=Mr^{d_{1}}e^{(p+s)\mathit{Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right)
}(1+o(1)).
\end{equation*}
From (61), we have
\begin{align*}
Mr^{d_{1}}e^{(p+s)\mathit{Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right)
}(1+o(1))
&=|P(e^{A(r)})Q(e^{A(r)})+(P(e^{A(r)}))^{^{\prime
}}Q(e^{A(r)})r-(Q(e^{A(r)}))^{^{\prime }}P(e^{A(r)})r|\\
&=|(Q(e^{A(r)}))^{^{\prime }}-(Q(e^{A(r)}))^{2}r|\leq M_{1}r^{d_{2}}e^{2s%
\mathit{Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1)),
\end{align*}
where \(M_{1}>0\) and \(d_{2}>0\) are some constants, which is a
contradiction. So we have \(h(z)\not\equiv 0.\) If \(p0,\) {\(d_{3}>0,\) }\(M_{3}>0\) and
\(d_{4}>0\) are some constants.
This is a contradiction. So, we obtain \(h(z)\not\equiv 0.\) Hence, if \(%
p\neq s\) we have \(h(z)\not\equiv 0.\) From Lemma 11, we get \(\lambda
(g)=\rho (g)=\rho (f^{\prime }-z)=\rho (f)=\infty \) and \(\lambda
_{2}(g)=\rho _{2}(g)=\rho _{2}(f^{\prime }-z)=\rho _{2}(f)=n\).
Proof of Theorem 9.
Suppose that \(f\not\equiv 0\) is a solution of (10). Since
\(\rho (P^{\ast })=\rho (Q^{\ast })=n,\) then by Lemma 7, we see that
\begin{equation}
\rho _{2}(f)\leq \max \left\{ \rho (P^{\ast }),\rho (Q^{\ast })\right\} =n.
\label{3.41a}
\end{equation}
(62)
By Lemma 6, we se that there exist a subset \(E_{3}\subset (1,\infty )\)
having logarithmic measure \(lmE_{3}0\) such that
for all \(z\) satisfying \(|z|=r\not\in E_{3}\cup \lbrack 0,1]\),
\begin{equation}
\left\vert \frac{{f}^{(j)}(z)}{{f}(z)}\right\vert \leq C[T(2r,{f})]^{j+1},%
\hspace{0.5cm}j=1,2. \label{3.42a}
\end{equation}
(63)
Taking \(z=r\), in (2) and (3), we obtain
that for sufficiently large \(r\)
\begin{align}
\left\vert P^{\ast }\left( e^{A\left( r\right) }\right) \right\vert
&=\left\vert a_{p}(r)e^{pA(r)}+\cdots +a_{1}(r)e^{A(r)}+a_{0}(r)\right\vert
\notag\\&
\leq 2\left\vert a_{pd_{p}}\right\vert r^{d_{p}}e^{p\mathit{Re}
c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1)), \label{3.43a}
\end{align}
(64)
and
\begin{align}
\left\vert Q^{\ast }\left( e^{A\left( r\right) }\right) \right\vert
&=\left\vert b_{s}(r)e^{sA(r)}+\cdots +b_{1}(r)e^{A(r)}+b_{0}(r)\right\vert\notag\\
&\geq \frac{1}{2}\left\vert b_{sm_{s}}\right\vert r^{m_{s}}e^{s\mathit{Re}
c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1)). \label{3.44a}
\end{align}
(65)
Substituting (63)-(65) into (10), we deduce that for all \(z\) satisfying \(|z|=r\not\in
E_{3}\cup \lbrack 0,1]\)
\begin{align}
\frac{1}{2}\left\vert b_{sm_{s}}\right\vert r^{m_{s}}e^{s\mathit{Re}
c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1))&\leq\left\vert \frac{{f}
^{\prime \prime }(z)}{{f}(z)}+P^{\ast }\left(e^{A\left( z\right) }\right)
\frac{{f}^{\prime }(z)}{{f}(z)}\right\vert\notag\\
&\leq \left\vert \frac{{f}^{\prime \prime }(z)}{{f}(z)}\right\vert
+\left\vert P^{\ast }\left( e^{A\left( z\right) }\right) \right\vert
\left\vert \frac{{f}^{\prime }(z)}{{f}(z)}\right\vert\notag\\
&\leq C[T(2r,{f})]^{3}+2\left\vert a_{pd_{p}}\right\vert r^{d_{p}}e^{p\mathit{
Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right) }C[T(2r,{f})]^{2}(1+o(1))\notag\\
&\leq 3C\left\vert a_{pd_{p}}\right\vert r^{d_{p}}e^{p\mathit{Re}
c_{n}r^{n}\left( 1+o\left( 1\right) \right) }[T(2r,{f})]^{3}(1+o(1)).
\label{3.45a}
\end{align}
(66)
By (66), we deduce that for all \(z\) satisfying \(|z|=r\not\in E_{3}\cup \lbrack 0,1]\)
\begin{equation}
\left\vert b_{sm_{s}}\right\vert r^{m_{s}-d_{p}}e^{\left( s-p\right) \mathit{%
Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1))\leq 6C\left\vert
a_{pd_{p}}\right\vert[T(2r,{f})]^{3}(1+o(1)). \label{3.46a}
\end{equation}
(67)
Since \(s-p>0\), by (67) and Lemma 12, we get
\begin{equation}
\rho (f)\geq \underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log T(r,f)%
}{\log r}=+\infty ,\text{ }\rho _{2}(f)\geq \underset{r\rightarrow +\infty }{%
{\lim \sup }}\frac{\log \log T(r,f)}{\log r}=n. \label{3.47a}
\end{equation}
(68)
From (62) and (68) we obtain \(\rho (f)=+\infty \) and \(\rho _{2}(f)=n.\)
Acknowledgments
This paper was supported by Directorate-General for Scientific Research and Technological Development(DGRSDT).
Author Contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Conflicts of Interest
The authors declare no conflict of interest.
