# On meromorphic solutions of a second order differential equation proposed by
Hayman

Yueyang Zhang  School of Mathematics and Physics, University of Science and
Technology Beijing, No. 30 Xueyuan Road, Haidian, Beijing, 100083, P.R. China
<EMAIL_ADDRESS>
###### Abstract.

This paper mainly consists of two parts. First, we use the idea of Halburd and
Wang to give the form of all admissible meromorphic solutions of the second
order differential equation: $w^{\prime\prime}w-w^{\prime
2}+aww^{\prime}+bw^{2}=\alpha w+\beta w^{\prime}+\gamma$, where
$a,b,\alpha,\beta,\gamma$ are small meromorphic functions of $w$ in the sense
of Nevanlinna. This allows us to find all transcendental meromorphic solutions
of an equation proposed by Hayman: $a_{1}(f^{\prime\prime}f-f^{\prime
2})+a_{2}f^{\prime}f+a_{3}f^{2}+b_{1}f^{\prime\prime}+b_{2}f^{\prime}+b_{3}f+b_{4}=0$,
$({\dagger})$ where $a_{i}$ and $b_{j}$ are polynomials, all of which have
finite hyper order at most. Second, we consider an autonomous equation related
to $({\dagger})$, namely: $f^{\prime\prime}f=p_{1}f^{\prime
2}+p_{2}f^{\prime}f+p_{3}f^{\prime}+p_{4}f^{4}+p_{5}f^{3}+p_{6}f^{2}+p_{7}f+p_{8}$,
$({\ddagger})$ where
$p_{1}\in\\{\frac{1}{2},1,2,\frac{3}{2},\cdots,\frac{m+1}{m},\cdots\\}$,
$m\in\mathbb{Z}^{+}$ and $p_{2},\cdots,p_{8}$ are constants. We use the
Ahlfors–Shimizu characteristic to prove that all meromorphic solutions of
$({\ddagger})$ have hyper order not exceeding $2$. The method also yields that
all meromorphic solutions of the third Painlevé equation $P_{III}$ have order
not exceeding $4$.

###### Key words and phrases:

Hayman’s equation, Meromorphic solutions, Ahlfors-Shimizu characteristic,
Painlevé equation

###### 2010 Mathematics Subject Classification:

Primary 30D35; Secondary 34M10

##  1\. Introduction

Meromorphic functions are functions analytic in the complex plane $\mathbb{C}$
apart from some isolated poles. In this paper, we assume that the readers are
familiar with the standard notation and fundamental results of Nevanlinna’s
value distribution theory of meromorphic functions; see e.g. [7]. In the last
several decades, Nevanlinna theory has been widely used in the study on the
growth and value distribution of meromorphic solutions of ordinary
differential equations (ODEs); see e.g. [12] and references therein. For first
order ODEs, a major result due to Gol’dberg [4] is the following

######  Theorem 1.1 ([4]).

All meromorphic solutions of the first order ODE: $\Omega(z,f,f^{\prime})=0$,
where $\Omega$ is polynomial in all of its arguments, are of finite order.

We refer to [12] for more results on first order ODEs. A natural question is
if there is an uniform upper growth estimate for all mermorphic solutions of a
second order ODE:

(1.1) $\Omega(z,f,f^{\prime},f^{\prime\prime})=0,$

where $\Omega$ is polynomial in all of its arguments. Bank [1] conjectured
that the Nevanlinna characteristic $T(r,f)$ for meromorphic solutions of
equation (1.1) would satisfy $T(r,f)\leq O(\exp(r^{c}))$ as $r\to+\infty$ for
some $c\geq 0$. Bank [1] himself proved that his conjecture is true with an
additional assumption that $N(r,\mu,f)=O(\exp(r^{c}))$ as $r\to+\infty$ for
some $c\geq 0$ for two distinct values of $\mu\in\mathbb{C}\cup\\{\infty\\}$.
Steinmetz [15] proved that Bank’s conjecture is true in the case that (1.1) is
homogeneous with respect to $f$, $f^{\prime}$ and $f^{\prime\prime}$. In fact,
Steinmetz [15] showed that in this case the meromorphic solutions of equation
(1.1) are expressed in terms of finite order entire functions.

######  Theorem 1.2 ([15]; see also [12]).

Suppose that in (1.1), $\Omega$ is homogeneous in $f$, $f^{\prime}$,
$f^{\prime\prime}$. Then all meromorphic solutions of (1.1) take the form

$f(z)=\frac{g_{1}(z)}{g_{2}(z)}e^{g_{3}(z)},$

where $g_{j}(z)$, $j=1,2,3$ are entire functions of finite order.

For higher order ODEs, little is known about the growth of their meromorphic
solutions. We refer to Hayman’s paper [8] for a generalized description of
Bank’s conjecture for the $n$-th order ODEs.

In [8], Hayman also considered some non-homogeneous second order differential
equations. Write the second order ODE in (1.1) in the form

(1.2)
$\Omega=\sum_{\lambda\in{\Lambda}}d_{\lambda}(z)f^{i_{0}}(f^{\prime})^{i_{1}}(f^{\prime\prime})^{i_{2}}=0,$

where $\Lambda=\\{(i_{0},\cdots,i_{2})\\}\in{\mathbb{N}^{2}}$ is a set and
$d_{\lambda}$ are polynomials in $z$. Define the _degree_ of the terms of
(1.2) by $|\lambda|=i_{0}+i_{1}+i_{2}$ and the _weight_ of the terms of (1.2)
by $||\lambda||=i_{0}+2i_{1}+3i_{3}$, respectively. Consider meromorphic
solutions of the second order differential equation:

(1.3) $a_{0}f^{\prime\prime}f+a_{1}f^{\prime
2}+a_{2}f^{\prime}f+a_{3}f^{2}+b_{1}f^{\prime\prime}+b_{2}f^{\prime}+b_{3}f+b_{4}=0,$

where $a_{i}$, $i=0,1,2,3$, $b_{j}$, $j=1,2,3,4$ are polynomials. Hayman [8]
pointed out that the entire solutions of a second order differential equation
(1.1) with the degree of the terms being at most $2$ have finite order unless
it is equation (1.3) with $a_{1}\equiv-a_{0}\not\equiv 0$, i.e.,

(1.4) $a_{1}(f^{\prime\prime}f-f^{\prime
2})+a_{2}f^{\prime}f+a_{3}f^{2}+b_{1}f^{\prime\prime}+b_{2}f^{\prime}+b_{3}f+b_{4}=0.$

When $a_{2}=a_{3}=0$, equation (1.4) is in some sense the simplest equation
not covered by Steinmetz’s theorem and Hayman’s theorem [8, Theorem C].
Rewrite equation (1.4) as

(1.5) $ff^{\prime\prime}-f^{\prime
2}+\iota_{1}ff^{\prime}+\iota_{2}f^{2}=\kappa_{0}+\kappa_{1}f+\kappa_{2}f^{\prime}+\kappa_{3}f^{\prime\prime},$

where $\iota_{i}$, $i=1,2$, $\kappa_{j}$, $j=0,1,2,3$ are rational functions.
Chiang and Halburd [2] found all meromorphic solutions of equation (1.5) in
the autonomous case. Letting $w=f-\kappa_{3}$, equation (1.5) becomes

(1.6) $w^{\prime\prime}w-w^{\prime
2}+aw^{\prime}w+bw^{2}=\alpha{w}+\beta{w^{\prime}}+\gamma,$

where $a=\iota_{1}$, $b=\iota_{2}$,
$\alpha=\kappa_{1}-2\iota_{2}\kappa_{3}-\iota_{1}\kappa_{3}^{\prime}-\kappa_{3}^{\prime\prime}$,
$\beta=\kappa_{2}-\iota_{1}\kappa_{3}+2\kappa_{3}^{\prime}$,
$\gamma=\kappa_{0}+\kappa_{1}\kappa_{3}+\kappa_{2}\kappa_{3}^{\prime}+\kappa_{3}\kappa_{3}^{\prime\prime}-\iota_{2}\kappa_{3}^{2}-\iota_{1}\kappa_{3}\kappa_{3}^{\prime}+\kappa_{3}^{\prime
2}-\kappa_{3}\kappa_{3}^{\prime\prime}$. From now on, we suppose that $w$ is
an _admissible_ meromorphic solution of of equation (1.6). Precisely, this
means that the coefficients $a$, $b$, $\alpha$, $\beta$ and $\gamma$ satisfy

(1.7) $T(r,a)+T(r,b)+T(r,\alpha)+T(r,\beta)+T(r,\gamma)=S(r,w),$

where the notation $S(r,w)$ denotes any quantity satisfying
$S(r,w)=o(T(r,w))$, $r\rightarrow{\infty}$, possibly outside of an exceptional
set of finite linear measure. In the following, we also say that $a$, $b$,
$\alpha$, $\beta$ and $\gamma$ are _small_ functions of $w$. In particular,
when $a$, $b$, $\alpha$, $\beta$ and $\gamma$ are all rational functions, all
transcendental meromorphic solutions of (1.6) are admissible. Halburd and Wang
[6] found all admissible meromorphic solutions of equation (1.6) with $a=b=0$
and in so doing they confirmed Hayman’s conjecture that all mermorphic
(entire) solutions of the simplest equation have finite order. Following the
idea of Halburd and Wang [6], the author and the coauthors [23] considered
equation (1.6) with $a$ and $b$ being both constants but not necessarily both
zero and found all its admissible meromorphic solutions.

The main purpose of this paper is to complete the two theorems in the author’s
previous paper [23]. First, in section 2 we will give the form of all
admissible meromorphic solutions of equation (1.6). The idea is still from
Halburd and Wang [6], i.e., to construct small functions of $w$ and
$w^{\prime}$ by using the first one or two terms in the local series expansion
for $w$ at zeros. The results of theorem 2.1 in section 2 allow us to find all
transcendental meromorphic solutions of equation (1.4) proposed by Hayman. We
will not present explicitly these transcendental solutions, but it is easy to
see that all of them have finite hyper order at most. Second, we will show
that all meromorphic solutions of equation (1.3) in the autonomous case have
hyper order 1 at most. When $a_{0}\not=0$, a simple local series analysis
shows that meromorphic solutions with some poles of equation (1.3) occurs only
when $-\frac{a_{1}}{a_{0}}=\frac{m+1}{m}$ for some integer $m\geq 1$. On the
other hand, when $a_{0}=0$, by Gold’berg’s theorem, equation (1.3) can have
infinite order meromorphic solutions only when $a_{1}\equiv 0$ and
$b_{1}\not\equiv 0$. In the latter case, a transformation $f\to 1/f$ to
equation (1.3) results in the equation:

(1.8) $b_{1}(ff^{\prime\prime}-2f^{\prime
2})+b_{2}ff^{\prime}-b_{3}f^{2}-b_{4}f^{3}=-a_{2}f^{\prime}+a_{3}f.$

With these observations, in section 3 we will consider a more general second
order differential equation:

(1.9) $\begin{split}f^{\prime\prime}f=p_{1}f^{\prime
2}+p_{2}f^{\prime}f+p_{3}f^{\prime}+p_{4}f^{4}+p_{5}f^{3}+p_{6}f^{2}+p_{7}f+p_{8},\end{split}$

where $p_{1},p_{2},\cdots,p_{8}$ are constants and
$p_{1}\in\\{\frac{1}{2},1,2,\frac{3}{2},\cdots,\frac{m+1}{m},\cdots\\}$,
$m\in\mathbb{Z}^{+}$. Instead of the Nevanlinna characteristic, we will use
the Ahlfors–Shimizu characteristic (see [7]) to prove that all meromorphic
solutions of (1.9) have hyper order not exceeding $2$. This then yields that
all meromorphic solutions of equation (1.1) with the degree of the terms being
at most $2$, and hence of equation (1.3) in the autonomous case, have hyper
order $1$ at most, which improves the results of [23, Theorem 1.5]. Finally,
in section 4, we give some comments on the results in sections 2 and 3. If we
suppose that the coefficients $p_{j}$ in (1.9) are rational functions, then
equation (1.9) includes the first four Painlevé equations $P_{I}$, $P_{II}$,
$P_{III}$ and $P_{IV}$ (see, e.g., [5]). We point out the method in section 3
also applies to $P_{III}$ and show that all meromorphic solutions of $P_{III}$
have order not exceeding $4$.

##  2\. Meromorphic solutions of Haymna’s equation

In this section, we show how to find transcendental meromorphic solutions of
equation (1.4). From the discussions in the introduction, we only need to
consider equation (1.6). Recall that the coefficients $a$, $b$, $\alpha$,
$\beta$ and $\gamma$ in equation (1.6) are assumed to be small functions of
$w$. We will give the form of all admissible solutions of equation (1.6) and
prove the following

######  Theorem 2.1.

Suppose that $w$ is an admissible meromorphic solution of (1.6). Then $w$
assumes one of the form described in the following list, where $c_{1}$,
$c_{2}$, $k_{1}$ and $k_{2}$ are constants:

  * (1)

If $\beta\equiv{\gamma}\equiv 0$, $\alpha\not\equiv 0$, then $a=b=0$ and the
results are in [6, Theorem 1.1].

  * (2)

If $\gamma\equiv 0$, $\beta\not\equiv 0$ and
$(-\alpha/\beta)^{\prime}+a(-\alpha/\beta)+b=0$, then $w=e^{\int h_{1}dz}$,
where $h_{1}=-\alpha/\beta=e^{-\int adz}\left(k_{1}-\int{be^{\int
adz}}dz\right)$.

  * (3)

If $\gamma\equiv 0$ and $\alpha+\beta^{\prime}+a\beta\equiv 0$, then
$w=e^{\int h_{1}dz}\left[c_{1}-\int\beta e^{-\int h_{1}dz}dz\right]$, where
$h_{1}=e^{-\int adz}\left(k_{1}-\int{be^{\int adz}}dz\right)$.

  * (4)

If $\gamma\not\equiv 0$ and there are meromorphic functions $h_{1}$ and
$h_{2}$ such that $h_{1}^{\prime}+ah_{1}+b=0$ and
${h_{2}^{2}+\beta{h_{2}}+\gamma}\equiv 0$, then
$w=e^{\int{h_{1}}dz}(c_{1}+\int{h_{2}}e^{-\int{h_{1}}dz}dz)$.

  * (5)

If $\gamma\not\equiv 0$ and
$A=\frac{\beta(\alpha+\beta^{\prime})-\gamma^{\prime}-a(2\gamma-\beta^{2})}{\gamma}$
satisfies $A^{\prime}+aA-2b=0$, denoting $B=2\alpha+\beta^{\prime}+a\beta$,
then

    * (a)

if there exists a nonzero constant $k_{1}$ such that
$B^{\prime}+2aB+A\alpha+\beta(k_{1}^{2}e^{-2\int adz}-\frac{A^{2}}{4}-b)=0$,
then

$k_{2}^{2}=\frac{1}{k_{1}^{2}}\left[\frac{1}{4k_{1}^{2}}\left(\frac{\beta}{2}A-B\right)^{2}e^{2\int
adz}+\left(\gamma-\frac{\beta^{2}}{4}\right)\right]e^{\int(A+2a)dz}$

is a constant. If $k_{2}\not=0$, then $w=[\pm{k_{2}}e^{\int adz}\cosh
g-\frac{h}{2k_{1}^{2}}e^{2\int adz}]e^{-\int(\frac{A}{2}+a)dz}$, where
$g=c_{1}-\int{k_{1}}e^{-\int adz}dz$ and
$h=(\frac{\beta}{2}A-B)e^{\int(\frac{A}{2}+a)dz}$.

    * (b)

if $k_{2}=0$, then $w=e^{\int
g_{2}dz}-\frac{1}{2k_{1}^{2}}(\frac{\beta}{2}A-B)e^{2\int adz}$, where
$g_{2}=-\frac{A}{2}+{k_{1}e^{-\int adz}}$.

    * (c)

if $h\not\equiv 0$ satisfies $h^{\prime}+ah=0$, then $w=\frac{(\int
hdz+c_{1})^{2}}{4h^{2}}(2a\beta+B)+(\frac{\beta^{2}}{4}-\gamma)\frac{1}{2a\beta+B}$.

    * (d)

if $k_{1}^{2}=(\frac{\beta^{2}}{4}-\gamma)e^{\int(A+2a)dz}\not=0$, then
$w=e^{-\int\frac{A}{2}dz}(\int[k_{1}e^{-\int
adz}-\frac{\beta}{2}e^{\int\frac{A}{2}dz}]dz+c_{1})$. In particular, if
$A+2a=a^{\prime}+b+a^{2}\equiv 0$, then $A$ is a constant and $w=e^{\int
adz}[c_{1}-\int(k_{1}-\frac{\beta}{2})e^{-\int adz}dz]$.

    * (e)

if $\frac{\beta^{2}}{4}-\gamma\equiv 0$, then
$w=e^{-\int\frac{A}{2}dz}(c_{1}-\frac{\beta}{2}e^{\int\frac{A}{2}dz})$.

In theorem 2.1, we have used $c_{1}$, $c_{2}$ to denote the integration
constants of the solutions of (1.6) and $k_{1}$, $k_{2}$ to denote the
integration constants of the solutions of certain differential equations in
terms of $a$, $b$, $\alpha$, $\beta$ and $\gamma$ and their derivatives or
some other functions we construct in the proof, respectively.

We note that in theorem 2.1 there are some first order differential equations.
We remark that a first order differential equation $f^{\prime}-af=\tau$, where
$a\not=0$ is a constant and $\tau$ is a non-vanishing entire function of order
less than $1$, always has a solution of order less than $1$. This is a result
from an application of Phragmén–Lindelöf theorem to [24, Lemma 1.1] (seel also
[22, Theorem 1.1]).

Unfortunately, unlike in the proof of [23, Theorem 1.5], for some solutions in
theorem 2.1 we are unable to determine precisely the admissible ones due to
the possible complicated structure of the two functions $a$ and $b$. We are
satisfied with only giving the form of these solutions because, in the case
that $a$, $b$, $\alpha$, $\beta$ and $\gamma$ are rational functions, it is
not difficult to determine the transcendental meromorphic solutions of
equation (1.5) precisely by using Nevanlinna theory to give some discussions
on the integrals $\int adz$, $\int(\frac{A}{2}+a)dz$ and $\int\frac{A}{2}dz$
as in the proof of [23, Theorem 1.5]. It is easy to see that all
transcendental meromorphic solutions of equation (1.4) have finite hyper order
at most.

###### Proof of Theorem 2.1.

Let $w\not\equiv 0$ be a nonconstant meromorphic solution of equation (1.6).
If $\alpha\equiv\beta\equiv\gamma\equiv 0$, then equation (1.6) becomes
$(w^{\prime}/w)^{\prime}+a(w^{\prime}/w)+b=0$, which has the general solution
$w=e^{\int h_{1}dz}$, where $h_{1}=e^{-\int adz}\left(k_{1}-\int{be^{\int
adz}}dz\right)$ and $k_{1}$ is a constant. This is the special case of part
(3) of theorem 2.1. From now on, we suppose that at least one of $\alpha$,
$\beta$, $\gamma$ is nonzero.

As in the proof of [6, Theorem 1.1], we define the set $\Phi_{f}$ for any
meromorphic function $f$ as follows: if $f\equiv 0$, then
$\Phi_{f}=\emptyset$; if $f\not\equiv 0$, then $\Phi_{f}$ denotes the set of
all zeros and poles of $f$. Let
$\Phi=\Phi_{a}\cup\Phi_{b}\cup\Phi_{\alpha}\cup\Phi_{\beta}\cup\Phi_{\gamma}$
and denote by $N_{\Phi}(r,f)$ and $\bar{N}_{\Phi}(r,f)$ the counting functions
(with and without multiplicities, respectively) where we only count the poles
of $f$ in the set $\Phi$. Let $z_{0}\in\Psi:=\mathbb{C}\backslash\Phi$ be
either a zero or a pole of $w$. Then, in a neighborhood of $z_{0}$, $w$ has a
Laurent series expansion of the form
$w(z)=a_{0}\xi^{p}+a_{1}\xi^{p+1}+O(\xi^{p+2})$, where $\xi=z-z_{0}$,
$a_{0}\not=0$, $a_{1}$ are constants and $p$ is a nonzero integer.
Substitution into (1.6) gives

$\begin{split}&(-pa_{0}^{2}\xi^{2p-2}+\cdots)+(a(z_{0})+\cdots)(pa_{0}^{2}\xi^{2p-1}+\cdots)+(b(z_{0})+\cdots)(a_{0}^{2}\xi^{2p}+\cdots)\\\
=&\,(\alpha(z_{0})+\cdots)(a_{0}\xi^{p}+\cdots)+(\beta(z_{0})+\cdots)(a_{0}p\xi^{p-1}+\cdots)+(\gamma(z_{0})+\cdots).\end{split}$

It follows that $p=2$ if $\beta\equiv\gamma\equiv 0$ and $p=1$ in other cases.
In particular, $w$ is analytic on $\Psi$. We see that $N(r,w)=N_{\Phi}(r,w)$
and from the proof of [6, Theorem 1.1], we know that
$\bar{N}(r,w)=\bar{N}_{\Phi}(r,w)=S(r,w)$. We distinguish the following cases:

Case 1: $\alpha\not\equiv 0$, $\beta\equiv\gamma\equiv 0$.

Substituting $w(z)=a_{0}\xi^{2}+a_{1}\xi^{3}+O(\xi^{4})$ into (1.6) shows that
about any $z_{0}\in\Psi$ such that $w(z_{0})=0$, we have

$w(z)=-\frac{\alpha(z_{0})}{2}\xi^{2}-\left(\frac{\alpha^{\prime}(z_{0})+a(z_{0})\alpha(z_{0})}{2}\right)\xi^{3}+O(\xi^{4}).$

Since $w$ is analytic on $\Psi$, it follows that

(2.1)
$f(z):=\left(\frac{w^{\prime}}{w}-\frac{\alpha^{\prime}+a\alpha}{\alpha}\right)^{2}+2\frac{\alpha}{w}$

is also analytic on $\Psi$. Since $a$ and $b$ are both small functions of $w$,
then in the same way as that in the proof of [23, Theorem 1.4], we may use
equation (1.6) with $\beta\equiv\gamma\equiv 0$ and the lemma on the
logarithmic derivative to show that $T(r,f)=S(r,w)$. Differentiating (2.1) and
using (1.6) to eliminate $(w^{\prime}/w)^{\prime}$ from the resulting equation
and then use (2.1) to eliminate $(w^{\prime}/w)^{2}$ from the resulting
equation gives

(2.2) $Pw^{\prime}=Qw+2a\alpha,$

where

$\begin{split}P=&2\left[b+a\frac{\alpha^{\prime}+a\alpha}{\alpha}+\left(\frac{\alpha^{\prime}+a\alpha}{\alpha}\right)^{\prime}\right],\\\
Q=&2b\frac{\alpha^{\prime}+a\alpha}{\alpha}+2\left(\frac{\alpha^{\prime}+a\alpha}{\alpha}\right)^{\prime}\frac{\alpha^{\prime}+a\alpha}{\alpha}+2a\left(\frac{\alpha^{\prime}+a\alpha}{\alpha}\right)^{2}-f(z)^{\prime}-2af(z).\end{split}$

Since $a$, $b$ are both small functions of $w$, then by similar discussions as
in the proof of [23, Theorem 1.4], we may show that equation (1.6) has no
admissible meromorphic solutions when $P\not\equiv 0$, and that $a=b=0$ when
$P\equiv 0$. Moreover, in the latter case, either
$w=-\frac{1}{2}(z+c_{2})^{2}\alpha$ or
$w=c_{1}^{-2}(\cosh(c_{1}z+c_{2})+1)\alpha$, where $k_{1}=\alpha\not=0$ is a
constant. We omit those details. This gives part (1) of theorem 2.1.

Case 2: $\beta\not\equiv 0$, $\gamma\equiv 0$.

Recall that in this case $w$ has only simple zeros in $\Psi$. Substituting
$w(z)=a_{0}\xi+a_{1}\xi^{2}+O(\xi^{3})$ into (1.6) yields at the leading order
$a_{0}=-\beta(z_{0})$ and at the next-to-leading order that
$\alpha(z_{0})+\beta^{\prime}(z_{0})+a(z_{0})\beta(z_{0})=0$.

Case 2a: $\alpha+\beta^{\prime}+a\beta\not\equiv{0}$, $\gamma\equiv 0$.

Let $f=w^{\prime}/w$. Since $a$ and $b$ are both small functions of $w$, then
in the same way as that in the proof of [23, Theorem 1.4], we may use equation
(1.6) with $\gamma\equiv 0$ and the lemma on the logarithmic derivative to
show that $T(r,f)=S(r,w)$. Then substituting $w^{\prime}=fw$ and
$w^{\prime\prime}=(f^{\prime}+f^{2})w$ into (1.6) with $\gamma\equiv 0$ yields

$(f^{\prime}+af+b)w=\alpha+\beta{f}.$

Since $T(r,f^{\prime}+af+b)=S(r,w)$ and $T(r,\alpha+f\beta)=S(r,w)$, we must
have $f^{\prime}+af+b=\alpha+\beta{f}=0$. So $\alpha,\beta$ satisfy
$(-\alpha/\beta)^{\prime}+a(-\alpha/\beta)+b=0$ and thus $w$ has the form
$w=e^{\int h_{1}dz}$, where $h_{1}=-\alpha/\beta=e^{-\int
adz}(k_{1}-\int{be^{\int adz}}dz)$ and $k_{1}$ is a constant. These results
give part (2) of the theorem.

Case 2b: $\alpha+\beta^{\prime}+a\beta\equiv{0}$, $\gamma\equiv 0$.

Then (1.6) takes the form
$((w^{\prime}+\beta)/w)^{\prime}+a((w^{\prime}+\beta)/w)+b=0$. Therefore,
$w=e^{\int h_{1}dz}(c_{1}-\int\beta e^{-\int h_{1}dz}dz)$, where $c_{1}$ is a
constant, $h_{1}=e^{-\int adz}(k_{1}-\int{be^{\int adz}}dz)$ and $k_{1}$ is a
constant. These results give part (3) of the theorem.

Case 3: $\gamma\not\equiv 0$.

Recall that in this case $w$ is analytic in $\Psi$ and any zero $z_{0}$ of $w$
in $\Psi$ is simple. Denote

$\begin{split}\delta_{1}=\frac{\gamma^{\prime}+a(\gamma-\beta^{2})-\beta(\alpha+\beta^{\prime})}{2\gamma},\quad\delta_{2}=\frac{\alpha+\beta^{\prime}+a\beta}{2}.\end{split}$

On substituting $w(z)=a_{0}\xi+a_{1}\xi^{2}+O(\xi^{3})$ into (1.6), we find
that $a_{0}^{2}+\beta(z_{0})a_{0}+\gamma(z_{0})=0$ and
$a_{1}=\delta_{1}(z_{0})a_{0}-\delta_{2}(z_{0})$. Let

$f(z)=\frac{w^{\prime 2}+\beta{w^{\prime}}+\gamma}{w^{2}},$

and denote

(2.3)
$A=\frac{\beta(\alpha+\beta^{\prime})-\gamma^{\prime}-a(2\gamma-\beta^{2})}{\gamma},\
\ B=2\alpha+\beta^{\prime}+a\beta.$

Denote

(2.4) $g(z)=\frac{w^{\prime
2}+\beta{w^{\prime}}+\gamma}{w^{2}}+A\frac{w^{\prime}}{w}+B\frac{1}{w}.$

Since $a$ and $b$ are both small functions of $w$, then in the same way as
that in the proof of [23, Theorem 1.4], we may show that $g$ is analytic on
$\Psi$ and then, together with equation (1.6) with $\gamma\not\equiv 0$ and
the lemma on the logarithmic derivative show that $T(r,g)=S(r,w)$. Multiplying
by $w^{2}$ on both sides of (2.4) gives

(2.5) $g(z)w^{2}=w^{\prime 2}+Aw^{\prime}w+Bw+\beta{w^{\prime}}+\gamma.$

As in the proof of [23, Theorem 1.4], by eliminating $w^{\prime 2}$ from (2.5)
and (1.6) and then using (2.5) to eliminate the terms $ww^{\prime\prime}$ and
$w^{\prime 2}$ together with the expressions of $A$ and $B$, we finally obtain

(2.6) $(A^{\prime}+aA-2b)w^{\prime}=(g^{\prime}+2ag+Ab)w-E,$

where $E=B^{\prime}+2aB+A\alpha+\beta(g-b)$.

Case 3a: $A^{\prime}+aA-2b\not\equiv 0$.

For convenience, we denote

$h_{1}=\frac{g^{\prime}+2ag+Ab}{A^{\prime}+aA-2b},\ \
h_{2}=\frac{-E}{A^{\prime}+aA-2b}.$

Then we have $w^{\prime}=h_{1}w+h_{2}$. Since $a$ and $b$ are both small
functions of $w$, then by substituting this equation into (2.5) and giving
similar discussions as in the proof of [23, Theorem 1.4], we get
$g-h_{1}^{2}-Ah_{1}\equiv 0$, which leads to that $G=g+A^{2}/4$ satisfies
$(G^{\prime}+2aG)^{2}=(A^{\prime}+aA-2b)^{2}G$. Obviously, $G\equiv 0$ is a
solution of this equation. Further, if $G\equiv 0$, then we may show that
equation (1.6) has no admissible meromorphic solutions; when $G\not\equiv 0$,
we may show that $h_{1}^{\prime}+ah_{1}+b=0$ and
$h_{2}^{2}+\beta{h_{2}}+\gamma=0$. Now
$w=e^{\int{h_{1}}dz}(c_{1}+\int{h_{2}}e^{-\int{h_{1}}dz}dz)$, where $c_{1}$ is
a constant. We omit those details. These results give part (4) of the theorem.

Case 3b: $A^{\prime}+aA-2b\equiv 0$.

Since $g^{\prime}+2ag+Ab$ and $E$ are both small functions of $w$, it follows
from (2.6) that $g^{\prime}+2ag+Ab\equiv 0$ and $E\equiv 0$. Equation (2.5)
can be rewritten as

(2.7)
$\left(w^{\prime}+\frac{1}{2}[Aw+\beta]\right)^{2}=\left(g+\frac{A^{2}}{4}\right)w^{2}+\left(\frac{\beta}{2}A-B\right)w+\left(\frac{\beta^{2}}{4}-\gamma\right).$

Let $h(z)=(\frac{\beta}{2}A-B)e^{\int(\frac{A}{2}+a)dz}$. Note that $h(z)$ may
not be a meromorphic function. Then the first equation of (2.3) yields

(2.8)
$\left(\left[\frac{\beta^{2}}{4}-\gamma\right]e^{\int(A+2a)dz}\right)^{\prime}=\frac{\beta}{2}e^{\int(\frac{A}{2}+a)dz}h$

and the condition $E\equiv 0$ is equivalent to

(2.9)
$h^{\prime}+ah=\left(g+\frac{A^{2}}{4}\right)\beta{e^{\int(\frac{A}{2}+a)dz}}.$

Clearly, if $g=-A^{2}/4$, then $h^{\prime}+ah=0$ and it follows that
$\frac{\beta}{2}A-B=e^{-\int(\frac{A}{2}+2a)dz}$ if $h\not\equiv 0$.

Case 3b(i): $g+A^{2}/4\not\equiv 0$.

Equations $A^{\prime}+aA-2b\equiv 0$ and $g^{\prime}+2ag+Ab\equiv 0$ lead to
that $G=g+\frac{A^{2}}{4}$ satisfies $G^{\prime}+2aG=0$. So
$G=g+\frac{A^{2}}{4}=k_{1}^{2}e^{-2\int adz}$ for some nonzero constant
$k_{1}$. Note that $T(r,G)=S(r,w)$. It follows from (2.8) and (2.9) that

$\left(\left[\frac{\beta^{2}}{4}-\gamma\right]e^{\int(A+2a)dz}\right)^{\prime}=\frac{h}{2k_{1}^{2}}(h^{\prime}+ah)e^{2\int
adz}.$

Integration shows that

(2.10)
$k_{2}^{2}=\frac{1}{k_{1}^{2}}\left[\frac{1}{4k_{1}^{2}}\left(\frac{\beta}{2}A-B\right)^{2}e^{2\int
adz}+\left(\gamma-\frac{\beta^{2}}{4}\right)\right]e^{\int(A+2a)dz}$

is a constant. Let $u=we^{\int(\frac{A}{2}+a)dz}+\frac{h}{2k_{1}^{2}}e^{2\int
adz}$. Then (2.7) becomes

(2.11) $(u^{\prime}-au)^{2}=k_{1}^{2}e^{-2\int adz}(u^{2}-k_{2}^{2}e^{2\int
adz})=k_{1}^{2}e^{-2\int adz}u^{2}-k_{1}^{2}k_{2}^{2}.$

When $k_{2}\not=0$, since $e^{-2\int adz}$ is a meromorphic function, we see
from (2.10) that $e^{\int(A+2a)dz}$ is a meromorphic function. Then we obtain
from (2.10) that $T(r,e^{\int(A+2a)dz})=S(r,w)$ and also that $T(r,h)=S(r,w)$.
Therefore, we have $T(r,u)=T(r,w)+S(r,w)$. The function $v=e^{-\int adz}u$
leads the above equation to

(2.12) $v^{\prime 2}e^{2\int adz}=k_{1}^{2}(v^{2}-k_{2}^{2}).$

Rewrite (2.12) as

$(v^{\prime}e^{\int adz}+k_{1}v)(v^{\prime}e^{\int
adz}-k_{1}v)=-k_{1}^{2}k_{2}^{2}.$

In general, $v$ may have some branch points of degree 2. Denote
$\kappa:=v^{\prime}e^{\int adz}+k_{1}v$. It follows that $v^{\prime}e^{\int
adz}-k_{1}v=-k_{1}^{2}k_{2}^{2}\kappa^{-1}$ and further that

(2.13)
$\begin{split}v=\frac{1}{2k_{1}}\left(\kappa+k_{1}^{2}k_{2}^{2}\kappa^{-1}\right),\quad
v^{\prime}=\frac{1}{2}e^{-\int adz}\left(\kappa-
k_{1}^{2}k_{2}^{2}\kappa^{-1}\right).\end{split}$

By taking derivatives on both sides of the first equation in (2.13) and then
comparing the resulting equation with the second equation in (2.13), we obtain
$\kappa=k_{1}\exp({k_{1}\int e^{-\int adz}dz})$. Therefore,

$\begin{split}u&=e^{\int adz}v=\frac{1}{2}e^{\int
adz}\left[\exp\left({k_{1}\int e^{-\int
adz}dz}\right)+k_{2}^{2}\exp\left({-k_{1}\int e^{-\int
adz}dz}\right)\right],\\\ w&=\left[\pm{k_{2}}e^{\int
adz}\cosh\left(c_{1}-\int{k_{1}}e^{-\int
adz}dz\right)-\frac{h}{2k_{1}^{2}}e^{2\int
adz}\right]e^{-\int(\frac{A}{2}+a)dz},\end{split}$

where $c_{1}$ is a constant. Note that $e^{\int adz}$ may not be a meromorphic
function in general. This gives part (5)(a) of the theorem. On the other hand,
when $k_{2}=0$, from (2.11) we have the first order differential equation
$u^{\prime}-au=k_{1}e^{-\int adz}u$. Thus, Part (5)(b) corresponds to the case
in which $k_{2}=0$, where

$\begin{split}w&=\left[\exp\left(\int{(k_{1}e^{-\int
adz}}+a)dz\right)-\frac{h}{2k_{1}^{2}}e^{2\int
adz}\right]e^{-\int(\frac{A}{2}+a)dz}\\\
&=\exp\left(\int{\left[-\frac{A}{2}+k_{1}e^{-\int
adz}\right]}dz\right)-\frac{1}{2k_{1}^{2}}\left(\frac{\beta}{2}A-B\right)e^{2\int
adz}.\end{split}$

Case 3b(ii): $g+A^{2}/4\equiv 0$, $h\not\equiv 0$.

Let $\lambda$ be a function such that
$\lambda^{\prime}-a\lambda=\frac{\beta}{2}e^{\int(\frac{A}{2}+a)dz}$. It
follows from (2.8) and (2.9) that $h^{\prime}+ah=0$ and

$\frac{C}{h}:=\frac{1}{h}\left(\frac{\beta^{2}}{4}-\gamma\right)e^{\int(A+2a)dz}-\lambda,$

where $C$ is a constant. Let $u=we^{\int(\frac{A}{2}+a)dz}+\lambda$, then
(2.7) becomes $(u^{\prime}-au)^{2}=h(u+\frac{C}{h})$. By combining the
equation $h^{\prime}+ah=0$, we get the general solution
$u=\frac{(\int{h}dz+c_{1})^{2}}{4h}-\frac{C}{h}$, where $c_{1}$ is some
constant. Therefore,

(2.14)
$\begin{split}w=(u-\lambda)e^{-\int(\frac{A}{2}+a)dz}=\frac{(\int{h}dz+c_{1})^{2}}{4h^{2}}\left(\frac{\beta}{2}A-B\right)-\left(\frac{\beta^{2}}{4}-\gamma\right)\frac{1}{\frac{\beta}{2}A-B}.\end{split}$

These results give part (5)(c) of the theorem.

Case 3b(iii): $g=-A^{2}/4$, $h\equiv 0$.

From (2.7) and (2.8), it follows that
$w^{\prime}+\frac{1}{2}(Aw+\beta)=k_{1}e^{-\int(\frac{A}{2}+a)dz}$, where
$k_{1}^{2}=(\frac{\beta^{2}}{4}-\gamma)e^{\int(A+2a)dz}$ is a constant, giving

$w=e^{-\int\frac{A}{2}dz}\left(\int\left[k_{1}e^{-\int
adz}-\frac{\beta}{2}e^{\int\frac{A}{2}dz}\right]dz+c_{1}\right),$

where $c_{1}$ is a constant. We see that $T(r,e^{\int(A+2a)dz})=S(r,w)$. When
$k_{1}$ is a nonzero constant, if $a^{\prime}+a^{2}+2b\equiv 0$, then
$(A+2a)^{\prime}+a(A+2a)=0$ and thus $T(r,e^{\int adz})=S(r,w)$. Since
$T(r,e^{\int adz})=T(r,e^{\int\frac{A}{2}})+S(r,w)$, it follows that
$T(r,e^{\int Adz})=T(r,e^{\int(A+2a)dz}e^{\int-2adz})=S(r,w)$. Therefore, if
$\int(k_{1}e^{-\int adz}-\frac{\beta}{2}e^{\int\frac{A}{2}}dz)dz$ is a
rational function, then we have $T(r,w)=T(r,e^{-\int\frac{A}{2}dz})+O(\log
r)+O(1)=S(r,w)$, a contradiction; if $\int(k_{1}e^{-\int
adz}-\frac{\beta}{2}e^{\int\frac{A}{2}dz})dz$ is transcendental, then we use
the theorem from Hayman and Miles [9] as in the proof of [23, Theorem 1.4] and
also derive a contradiction. So we must have $A+2a\equiv 0$ and hence
$w=e^{\int adz}[c_{1}-\int(k_{1}-\frac{\beta}{2})e^{-\int adz}dz]$. These
results correspond to part (5)(d) of the theorem. Otherwise, we have $k_{1}=0$
and it follows that $w^{\prime}+\frac{1}{2}(Aw+\beta)=0$. So
$w=e^{-\int\frac{A}{2}dz}(c_{1}-\frac{\beta}{2}e^{\int\frac{A}{2}dz})$. These
results give part (5)(e) of the theorem and also complete the proof.

∎

##  3\. Growth of meromorphic solutions of equation (1.9)

In this section, we consider the growth of meromorphic solutions of equation
(1.9). The main device we use is the Ahlfors–Shimizu characteristic defined as
(see [7]):

(3.1)
$\begin{split}T_{0}(r,f)=\int_{0}^{r}\frac{\mathcal{A}(t,f)}{t}dt,\end{split}$

where

(3.2)
$\begin{split}\mathcal{A}(t,f)=\frac{1}{\pi}\int_{0}^{t}\varrho\int_{0}^{2\pi}\left(\frac{|f^{\prime}(\varrho
e^{i\theta})|}{1+|f(\varrho e^{i\theta})|^{2}}\right)^{2}d\theta
d\varrho.\end{split}$

It is well known that the difference of the Ahlfors–Shimizu characteristic
$T_{0}(r,f)$ and the Nevanlinna characteristic $T(r,f)$ is a bounded term. By
estimating $T_{0}(r,f)$ of the meromorphic solutions of equation (1.9) in a
similar way as in the proof of [12, Theorem 9.2.1], we prove the following

######  Theorem 3.1.

Let $p_{1},p_{2},\cdots,p_{8}$ be constants and
$p_{1}\in\\{\frac{1}{2},1,2,\frac{3}{2},\cdots,\frac{m+1}{m},\cdots\\}$,
$m\in\mathbb{Z}^{+}$. Then all meromorphic solutions of equation (1.9) have
hyper order not exceeding $2$.

To prove theorem 3.1, we first prove the following preliminary

######  Lemma 3.2.

Suppose that equation (1.9) has a transcendental meromorphic solution. If
$p_{1}=2$, then $p_{4}=0$; if $p_{1}=\frac{3}{2}$, then $p_{5}=0$; if
$p_{1}=\frac{1}{2}$, then either $p_{3}\not=0$ or $p_{3}=0$ and $p_{7}=0$.

###### Proof.

Let $f$ be a transcendental meromorphic solution of (1.9). We rewrite equation
(1.9) as

(3.3) $\begin{split}f^{\prime\prime}f-p_{1}f^{\prime
2}-p_{2}f^{\prime}f-p_{3}f^{\prime}=p_{4}f^{4}+p_{5}f^{3}+p_{6}f^{2}+p_{7}f+p_{8}.\end{split}$

Consider first the case when $p_{1}=2$. Suppose that $p_{4}\not=0$. Then an
application of Clunie’s lemma (see [3] or [12]) to equation (3.3) yields
$m(r,f)=S(r,f)$. This implies that $f$ has infinitely many poles. Then, around
a pole of $f$, say $z_{0}$, $f$ has a Laurent series expansion
$f(z)=a_{0}\xi^{-n}+O(\xi^{-n+1})$, where $\xi=z-z_{0}$, $a_{0}\not=0$ is a
constant and $n$ is a positive integer. By substituting this series into (3.3)
and comparing the highest-order terms on both sides of (3.3) yields $n=1$.
However, since $p_{1}=2$, the terms $\xi^{-4}$ on the left of (3.3) cancels
out, which gives a contradiction when $p_{4}\not=0$. Therefore, $p_{4}=0$.

Consider next the case when $p_{1}=\frac{3}{2}$. If $p_{4}\not=0$, then
Clunie’s lemma again implies that $f$ has infinitely many poles. Then, around
a pole of $f$, say $z_{0}$, $f$ has a Laurent series expansion
$f(z)=a_{0}\xi^{-n}+a_{1}\xi^{-n+1}+O(\xi^{-n+2})$, where $\xi=z-z_{0}$,
$a_{0}\not=0$ and $a_{1}$ are constants and $n$ is a positive integer. By
substituting this series into (3.3) and comparing the highest-order terms on
both sides of equation (3.3) yields that $n=1$ and $2p_{4}a_{0}^{2}=1$. Then
by looking at the coefficients of terms $\xi^{-3}$ on both sides of equation
(3.3), we get $p_{5}=0$. On the other hand, if $p_{4}=0$ and $p_{5}\not=0$,
then Clunie’s lemma also yields that $f$ has infinitely many poles. Similarly
as in the case $p_{1}=2$, by substituting a series around a pole of $f$ into
(1.9) and comparing the highest-order terms on both sides of equation (3.3)
yields that $f$ has a double pole and then we have a contradiction since
$p_{5}\not=0$. Therefore, when $p_{1}=\frac{3}{2}$, we always have $p_{5}=0$.

Consider finally the case when $p_{1}=\frac{1}{2}$. When $p_{3}=0$, if
$p_{7}\not=0$, then by doing a transformation $f\to 1/g$, we obtain from (3.3)
that $g^{\prime\prime}g-\frac{3}{2}g^{\prime
2}-p_{2}g^{\prime}g=-p_{4}-p_{5}g-p_{6}g^{2}-p_{7}g^{3}-p_{8}g^{4}$. Since
$p_{7}\not=0$, we have a contradiction from the results in the case when
$p_{1}=\frac{3}{2}$. Thus the assertion follows.

∎

Now we begin to prove theorem 3.1.

###### Proof of Theorem 3.1.

Suppose that $f$ is a transcendental meromorphic solution of equation (1.9).
We do a linear transformation $f\to\alpha f$ with an analytic function
$\alpha$ so that equation (1.9) becomes

(3.4) $\begin{split}f^{\prime\prime}f=p_{1}f^{\prime
2}+\hat{p}_{2}f^{\prime}f+\hat{p}_{3}f^{\prime}+\hat{p}_{4}f^{4}+\hat{p}_{5}f^{3}+\hat{p}_{6}f^{2}+\hat{p}_{7}f+\hat{p}_{8},\end{split}$

where $\hat{p}_{2}=2(p_{1}-1)\alpha^{\prime}/\alpha+p_{2}$,
$\hat{p}_{3}=p_{3}/\alpha$, $\hat{p}_{4}=p_{4}\alpha^{2}$,
$\hat{p}_{5}=p_{5}\alpha$,
$\hat{p}_{6}=-\alpha^{\prime\prime}/\alpha+p_{1}(\alpha^{\prime}/\alpha)^{2}+p_{2}\alpha^{\prime}/\alpha+p_{6}$,
$\hat{p}_{7}=p_{3}\alpha^{\prime}/\alpha^{2}+p_{7}/\alpha$,
$\hat{p}_{8}=p_{8}/\alpha^{2}$. By Lemma 3.2, the function $\alpha$ can be
chosen in the following way: when $p_{1}=1$, $\alpha$ satisfies
$\hat{p}_{6}\equiv 0$; when $p_{1}=\frac{1}{2}$, $\alpha$ satisfies
$\hat{p}_{7}\equiv 0$ where we choose $\alpha\equiv 1$ when $p_{3}=p_{7}=0$;
when $p_{1}=2$, $\alpha=1$ and satisfies $\hat{p}_{4}\equiv 0$; when
$p_{1}=\frac{3}{2}$, $\alpha=1$ and satisfies $\hat{p}_{5}\equiv 0$; when
$p_{1}=\frac{m+1}{m}$ and $m\geq 3$, $\alpha=1$. With this choice of $\alpha$,
we multiply by $f^{-2p_{1}-1}f^{\prime}$ on both sides of (3.4) and integrate
the resulting equation to define an auxiliary function $F$ such that

(3.5) $\begin{split}f^{2p_{1}}F=\alpha_{1}f^{\prime
2}+\alpha_{2}ff^{\prime}+\alpha_{3}f^{\prime}+\alpha_{4}f^{4}+\alpha_{5}f^{3}+\alpha_{6}f^{2}+\alpha_{7}f+\alpha_{8}+\alpha_{9}/f^{\prime},\end{split}$

where $\alpha_{j}$ are analytic functions to be determined later. In general,
we may choose the function $\alpha_{9}$ to be $0$. Here we have refined the
function $F$ by adding the term $1/f^{\prime}$ with a coefficient
$\alpha_{9}$. The choice of $\alpha_{9}$ depends on $p_{3}$ and $p_{8}$ and
allows us to estimate $\mathcal{A}(t,1/f^{\prime})$ instead of
$\mathcal{A}(r,1/f)$, as will be seen later.

We always choose $\alpha_{1}\not\equiv 0$. By the definition of $F$, we see
that: when $p_{1}=1$, we have $\alpha_{6}=0$; when $p_{1}=\frac{1}{2}$, we
have $\alpha_{7}=0$; when $p_{1}=2$, we have $\alpha_{4}=0$; when
$p_{1}=\frac{3}{2}$, we have $\alpha_{5}=0$. By taking the derivatives on both
sides of (3.5) and then combining (3.5) with the resulting equation, we obtain

(3.6) $\begin{split}f^{2p_{1}+1}F^{\prime}=&\,(\alpha_{1}^{\prime}f^{\prime
2}+2\alpha_{1}f^{\prime}f^{\prime\prime}+\alpha_{2}^{\prime}ff^{\prime}+\alpha_{2}f^{\prime
2}+\alpha_{2}ff^{\prime\prime}+\alpha_{3}^{\prime}f^{\prime}+\alpha_{3}f^{\prime\prime}+\alpha_{4}^{\prime}f^{4})f\\\
&+(4\alpha_{4}f^{3}f^{\prime}+\alpha_{5}^{\prime}f^{3}+3\alpha_{5}f^{2}f^{\prime}+\alpha_{6}^{\prime}f^{2}+2\alpha_{6}ff^{\prime}+\alpha_{7}^{\prime}f+\alpha_{7}f^{\prime}+\alpha_{8}^{\prime})f\\\
&-2p_{1}(\alpha_{1}f^{\prime
2}+\alpha_{2}ff^{\prime}+\alpha_{6}f^{2}+\alpha_{3}f^{\prime}+\alpha_{4}f^{4}+\alpha_{5}f^{3}+\alpha_{7}f+\alpha_{8})f^{\prime}\\\
&+\alpha_{9}^{\prime}f/f^{\prime}-\alpha_{9}ff^{\prime\prime}/f^{\prime
2}-2\alpha_{9}p_{1}.\end{split}$

Using the expression for $ff^{\prime\prime}$ in (3.4), we finally obtain

(3.7)
$\begin{split}f^{2p_{1}+1}F^{\prime}=&\,[\alpha_{1}^{\prime}+2\alpha_{1}\hat{p}_{2}-(p_{1}-1)\alpha_{2}]ff^{\prime
2}+[2\alpha_{1}\hat{p}_{3}-\alpha_{3}p_{1}]f^{\prime 2}\\\
&+[2\alpha_{1}\hat{p}_{4}-(2p_{1}-4)\alpha_{4}]f^{4}f^{\prime}+[2\alpha_{1}\hat{p}_{5}-(2p_{1}-3)\alpha_{5}]f^{3}f^{\prime}\\\
&+[\alpha_{2}^{\prime}+\alpha_{2}\hat{p}_{2}+2\alpha_{1}\hat{p}_{6}-2(p_{1}-1)\alpha_{6}]f^{2}f^{\prime}\\\
&+[\alpha_{2}\hat{p}_{3}+2\alpha_{1}\hat{p}_{7}+\alpha_{3}^{\prime}+\alpha_{3}\hat{p}_{2}-(2p_{1}-1)\alpha_{7}]ff^{\prime}\\\
&+[2\alpha_{1}\hat{p}_{8}+\alpha_{3}\hat{p}_{3}-2\alpha_{8}p_{1}]f^{\prime}+(\alpha_{4}^{\prime}+\alpha_{2}\hat{p}_{4})f^{5}+(\alpha_{5}^{\prime}+\alpha_{2}\hat{p}_{5}+\alpha_{3}\hat{p}_{4})f^{4}\\\
&+(\alpha_{6}^{\prime}+\alpha_{2}\hat{p}_{6}+\alpha_{3}\hat{p}_{5})f^{3}+(\alpha_{2}\hat{p}_{7}+\alpha_{7}^{\prime}+\alpha_{3}\hat{p}_{6})f^{2}+(\alpha_{2}\hat{p}_{8}+\alpha_{3}\hat{p}_{7}+\alpha_{8}^{\prime})f\\\
&+\alpha_{3}\hat{p}_{8}-2\alpha_{9}p_{1}+\alpha_{9}^{\prime}f/f^{\prime}-\alpha_{9}ff^{\prime\prime}/f^{\prime
2}.\end{split}$

Below we first consider the case when $p_{1}=1$. Recall that in this case
$\alpha_{6}=0$ and also that $\alpha$ is chosen so that $\hat{p}_{6}=0$, i.e.,
$(\alpha^{\prime}/\alpha)^{\prime}=p_{2}\alpha^{\prime}/\alpha+p_{6}$. If
$p_{2}$ is zero, then $\alpha=c_{2}e^{\frac{p_{6}}{2}z^{2}+c_{1}z}$, where
$c_{1}$ and $c_{2}\not=0$ are constants; if $p_{2}$ is nonzero, then
$\alpha=c_{2}e^{(c_{1}/p_{2})e^{p_{2}z}-(p_{6}/p_{2})z}$, where $c_{1}$ and
$c_{2}\not=0$ are constants. We choose $c_{1}=0$ in both these two expressions
of $\alpha$.

Now we begin to determine $\alpha_{j}$. We choose $\alpha_{2}$ and
$\alpha_{1}$ to be solutions of the first order differential equations
$\alpha_{1}^{\prime}+2\alpha_{1}\hat{p}_{2}=0$ and
$\alpha_{2}^{\prime}+\alpha_{2}\hat{p}_{2}=0$, respectively. Noting that
$\hat{p}_{2}=p_{2}$, we have $\alpha_{1}=c_{4}e^{-2p_{2}z}$ and
$\alpha_{2}=c_{3}e^{-p_{2}z}$ for two constants $c_{3}$ and $c_{4}\not=0$.
Recall that $\alpha_{6}\equiv 0$. We choose $\alpha_{3}$, $\alpha_{4}$,
$\alpha_{5}$, $\alpha_{7}$, $\alpha_{8}$ and $\alpha_{9}$ such that
$\alpha_{3}=2\hat{p}_{3}\alpha_{1}$, $\alpha_{4}=-\hat{p}_{4}\alpha_{1}$,
$\alpha_{5}=-2\hat{p}_{5}\alpha_{1}$,
$\alpha_{7}=\alpha_{2}\hat{p}_{3}+2\alpha_{1}\hat{p}_{7}+\alpha_{3}^{\prime}+\alpha_{3}\hat{p}_{2}$,
$\alpha_{8}=(\hat{p}_{8}+\hat{p}_{3}^{2})\alpha_{1}$ and
$\alpha_{9}=\hat{p}_{3}\hat{p}_{8}\alpha_{1}$. With these choices of
$\alpha_{j}$, equation (3.7) becomes

(3.8)
$\begin{split}F^{\prime}f^{2}=&\,(\alpha_{4}^{\prime}+\alpha_{2}\hat{p}_{4})f^{4}+(\alpha_{4}^{\prime}+\alpha_{2}\hat{p}_{5}+\alpha_{3}\hat{p}_{4})f^{3}+(\alpha_{3}\hat{p}_{5})f^{2}\\\
&+(\alpha_{2}\hat{p}_{7}+\alpha_{7}^{\prime})f+(\alpha_{2}\hat{p}_{8}+\alpha_{3}\hat{p}_{7}+\alpha_{8}^{\prime})+\alpha_{9}^{\prime}/f^{\prime}-\alpha_{9}f^{\prime\prime}/f^{\prime
2}.\end{split}$

We see that the function $|\alpha_{1}|$, as well as $|\alpha_{2}|$,
$|\alpha_{3}|$, $|\alpha_{4}|$, $|\alpha_{5}|$, $|\alpha_{7}|$, $|\alpha_{8}|$
and $|\alpha_{9}|$, is bounded above by a function with growth of type
$O(e^{cr^{2}})$ for some positive real number $c$. We may choose this function
as $G(r)=Ce^{cr^{2}}$ with two large positive numbers $C$ and $c$.

With the above preparations, we are ready to estimate $T_{0}(r,f)$.

(3.9) $\begin{split}|f^{\prime}|^{2}\leq
A_{1}|F||f|^{2}+A_{2}|f||f^{\prime}|+A_{3}|f^{\prime}|+A_{4}|f|^{4}+A_{5}|f|^{3}+A_{7}|f|+A_{8}+A_{9}/|f^{\prime}|,\end{split}$

where $A_{1}=1/|\alpha_{1}|$, $A_{i}=|\alpha_{i}|/|\alpha_{1}|$,
$i=2,3,4,5,7,8,9$. By the expressions of $\alpha_{j}$, we see that $A_{i}\leq
G(r)$ for $i=1,2,3,4,5,7,8,9$. We divide the circle $\\{z:|z|=\varrho\\}$ in
two parts by

(3.10) $\begin{split}\Delta_{1}&=\left\\{|z|=\varrho\ \Big{|}\
|f^{\prime}|\leq 1+|f|^{2}\right\\},\\\ \Delta_{2}&=\left\\{|z|=\varrho\
\Big{|}\ |f^{\prime}|\geq 1+|f|^{2}\right\\}.\end{split}$

By the definition of $\mathcal{A}(r,f)$ in (3.2), we have

(3.11)
$\begin{split}\mathcal{A}(t,f)&=\frac{1}{\pi}\int_{0}^{t}\varrho\int_{\Delta_{1}}\frac{|f^{\prime}|^{2}}{(1+|f|^{2})^{2}}d\theta
d\varrho+\frac{1}{\pi}\int_{0}^{t}\varrho\int_{\Delta_{2}}\frac{|f^{\prime}|^{2}}{(1+|f|^{2})^{2}}d\theta
d\varrho\\\ &\leq
t^{2}+\frac{1}{\pi}\int_{0}^{t}\varrho\int_{\Delta_{2}}\frac{|f^{\prime}|^{2}}{(1+|f|^{2})^{2}}d\theta
d\varrho.\end{split}$

Denote
$B=A_{2}|f||f^{\prime}|+A_{3}|f^{\prime}|+A_{4}|f|^{4}+A_{5}|f|^{3}+A_{7}|f|+A_{8}+A_{9}/|f^{\prime}|$
for simplicity. Observing that

(3.12)
$\begin{split}\frac{|f|^{n}}{(1+|f|^{2})^{2}}=\frac{|f|^{n}}{1+2|f|^{2}+|f|^{4}}\leq
1\end{split}$

holds for all $n=1,2,3,4$, we obtain from (3.2), (3.9), (3.10), (3.11) and
(3.12) that

(3.13) $\begin{split}\mathcal{A}(t,f)&\leq
t^{2}+\frac{1}{\pi}\int_{0}^{t}\varrho\int_{\Delta_{2}}\frac{B}{(1+|f|^{2})^{2}}d\theta
d\varrho+\frac{1}{\pi}\int_{0}^{t}\varrho\int_{\Delta_{2}}\frac{|A_{1}||f^{2}||F|}{(1+|f|^{2})^{2}}d\theta
d\varrho\\\ &\leq
t^{2}+5t^{2}G(t)+\frac{2}{\pi}\int_{0}^{t}\varrho\int_{\Delta_{2}}\frac{G(\varrho)|f^{\prime}|}{1+|f|^{2}}d\theta
d\varrho+\frac{1}{\pi}\int_{0}^{t}\varrho\int_{\Delta_{2}}\frac{|A_{1}||f^{2}||F|}{(1+|f|^{2})^{2}}d\theta
d\varrho.\end{split}$

Firstly, we have

(3.14) $\begin{split}\int_{0}^{r}\frac{t^{2}+5t^{2}G(t)}{t}dt\leq
r^{2}+5r^{2}G(r).\end{split}$

Secondly, we estimate the last integral in (3.13). To this end, we divide the
circle $\\{z:|z|=\varrho\\}$ in $\Delta_{2}$ in two parts by

(3.15) $\begin{split}\Delta_{21}&=\left\\{|z|=\varrho\ \Big{|}\
|F^{\prime}|\leq\frac{1}{\varrho}|F|,\ |f^{\prime}|\geq 1+|f|^{2}\right\\},\\\
\Delta_{22}&=\left\\{|z|=\varrho\ \Big{|}\
|F^{\prime}|\geq\frac{1}{\varrho}|F|,\ |f^{\prime}|\geq
1+|f|^{2}\right\\},\end{split}$

and so

(3.16)
$\begin{split}\int_{\Delta_{2}}\frac{|A_{1}||f|^{2}|F|}{(1+|f|^{2})^{2}}d\theta=\int_{\Delta_{21}}\frac{|A_{1}||f|^{2}|F|}{(1+|f|^{2})^{2}}d\theta+\int_{\Delta_{22}}\frac{|A_{1}||f|^{2}|F|}{(1+|f|^{2})^{2}}d\theta.\end{split}$

By (3.8), (3.12) and (3.15) and noting that $A_{1}=1/|\alpha_{1}|$ and
$(1/f^{\prime})^{\prime}=-f^{\prime\prime}/f^{\prime 2}$, we see that

(3.17)
$\begin{split}\int_{\Delta_{22}}\frac{|A_{1}||f|^{2}|F|}{(1+|f|^{2})^{2}}d\theta\leq\int_{\Delta_{22}}\frac{|f|^{2}|\varrho
A_{1}F^{\prime}|}{(1+|f|^{2})^{2}}d\theta\leq k\varrho G(\varrho)+\varrho
G(\varrho)\int_{\Delta_{22}}|(1/f^{\prime})^{\prime}|d\theta,\end{split}$

where $k>0$ is some large integer. Obviously, for $z\in\Delta_{22}$,

(3.18)
$\begin{split}|(1/f^{\prime})^{\prime}|=\frac{|(1/f^{\prime})^{\prime}|}{1+|1/f^{\prime}|^{2}}(1+|1/f^{\prime}|^{2})\leq\frac{2|(1/f^{\prime})^{\prime}|}{1+|1/f^{\prime}|^{2}}.\end{split}$

Recall the Schwarz inequality in the form of integration:

(3.19)
$\left(\int_{X}g_{1}g_{2}dx\right)^{2}\leq\left(\int_{X}g_{1}^{2}dx\right)\left(\int_{X}g_{2}^{2}dx\right),$

where $X$ is a measurable space and $g_{1}$ and $g_{2}$ are two functions on
$X$ with the range in $[0,\infty]$. By applying this inequality three times,
we deduce that

(3.20)
$\begin{split}&\,\int_{0}^{r}\frac{1}{t}\left[\frac{1}{\pi}\int_{0}^{t}\varrho\left(k\varrho
G(\varrho)+2\varrho
G(\varrho)\int_{\Delta_{22}}\frac{|(1/f^{\prime})^{\prime}|}{1+|1/f^{\prime}|^{2}}d\theta\right)d\varrho\right]dt\\\
\leq&kr^{3}G(r)+2r^{2}G(r)\int_{0}^{r}\frac{1}{t}\left[\frac{1}{\pi}\int_{0}^{t}\left(\int_{\Delta_{22}}\frac{|(1/f^{\prime})^{\prime}|}{1+|1/f^{\prime}|^{2}}d\theta\right)d\varrho\right]dt,\\\
\leq&kr^{3}G(r)+4r^{2}G(r)\int_{0}^{r}\frac{1}{t}\left[\frac{1}{\pi}\int_{0}^{t}\left(\int_{0}^{2\pi}\frac{|(1/f^{\prime})^{\prime}|^{2}}{(1+|1/f^{\prime}|^{2})^{2}}d\theta\right)^{1/2}d\varrho\right]dt,\\\
\leq&kr^{3}G(r)+4r^{2}G(r)\int_{0}^{r}\sqrt{t}\frac{1}{\sqrt{t}}\left[\frac{1}{\pi}\int_{0}^{t}\left(\int_{0}^{2\pi}\frac{|(1/f^{\prime})^{\prime}|^{2}}{(1+|1/f^{\prime}|^{2})^{2}}d\theta\right)d\varrho\right]^{1/2}dt,\\\
\leq&kr^{3}G(r)+4r^{4}G(r)T_{0}(r,1/f^{\prime})^{1/2}.\end{split}$

Denote now

(3.21) $\begin{split}L(\varrho)=\int_{\Delta_{21}}|F(\varrho
e^{i\theta})|d\theta.\end{split}$

Since $|f|^{t}\leq 1+2|f|^{2}+|f|^{4}$ for all positive rational number $t$
such that $t\leq 4$, we then have

(3.22)
$\begin{split}\int_{\Delta_{21}}\frac{|A_{1}||f|^{2}|F|}{(1+|f|^{2})^{2}}d\theta\leq
G(\varrho)L(\varrho).\end{split}$

Hence it suffices to obtain a suitable estimate of growth for $L(\varrho)$. To
this end, we differentiate $L(\varrho)$ under the integral sign. Dividing
$F(z)=u(z)+iv(z)$ into its real and imaginary parts, we first observe that

(3.23)
$\begin{split}\frac{\partial}{\partial\varrho}|F|=\frac{\partial}{\partial\varrho}(|u|+|v|)^{1/2}=\frac{u\frac{\partial
u}{\partial\varrho}+v\frac{\partial
v}{\partial\varrho}}{(|u|+|v|)^{1/2}}.\end{split}$

By the Schwarz inequality,

(3.24) $\begin{split}\left(u\frac{\partial u}{\partial\varrho}+v\frac{\partial
v}{\partial\varrho}\right)^{2}\leq(u^{2}+v^{2})\left(\left(\frac{\partial
u}{\partial\varrho}\right)^{2}+\left(\frac{\partial
v}{\partial\varrho}\right)^{2}\right),\end{split}$

we obtain

(3.25) $\begin{split}\frac{\partial
L(\varrho)}{\partial\varrho}&=\int_{\Delta_{21}}\frac{\partial}{\partial\varrho}|F(\varrho
e^{i\theta})|d\theta=\int_{\Delta_{21}}\frac{\partial}{\partial\varrho}|F(\varrho
e^{i\theta})|d\theta\\\ &=\int_{\Delta_{21}}\frac{u\frac{\partial
u}{\partial\varrho}+v\frac{\partial
v}{\partial\varrho}}{(|u|+|v|)^{1/2}}d\theta\leq\int_{\Delta_{21}}\left(\left(\frac{\partial
u}{\partial\varrho}\right)^{2}+\left(\frac{\partial
v}{\partial\varrho}\right)^{2}\right)^{1/2}d\theta\\\
&=\int_{\Delta_{21}}\left|\frac{\partial
w}{\partial\varrho}\right|d\theta=\int_{\Delta_{21}}\left|F^{\prime}\right|d\theta\leq\frac{2}{\varrho}\int_{\Delta_{21}}\left|F\right|d\theta=\frac{1}{\varrho}L(\varrho).\end{split}$

Therefore, $L^{\prime}(\varrho)\leq\frac{1}{\varrho}L(\varrho)$. Integrating
this differential inequality we get

(3.26) $\begin{split}L(\varrho)\leq BL(1)\end{split}$

for some constant $B>1$. Combining equations (3.22) and (3.26), we have

(3.27)
$\begin{split}\int_{0}^{r}\frac{1}{t}\left[\frac{1}{\pi}\int_{0}^{t}\varrho
G(\varrho)BL(1)d\varrho\right]dt\leq BL(1)r^{2}G(r).\end{split}$

Thirdly, for the remaining integral in (3.13), by using the Schwarz inequality
(3.19) as in (3.20), we obtain

(3.28)
$\begin{split}\int_{0}^{r}\frac{1}{t}\left[\frac{2}{\pi}\int_{0}^{t}\varrho\left(\int_{\Delta_{2}}\frac{G(\varrho)|f^{\prime}|}{1+|f|^{2}}d\theta\right)d\varrho\right]dt\leq
4r^{4}G(r)T_{0}(r,f)^{1/2}.\end{split}$

Finally, we combine (3.1), (3.2) and the estimates in (3.14), (3.20), (3.27)
and (3.28) together and get

(3.29) $\begin{split}T_{0}(r,f)\leq&\,r^{2}+[BL(1)+5]r^{2}G(r)+kr^{3}G(r)\\\
&+4r^{4}G(r)T_{0}(r,1/f^{\prime})^{1/2}+4r^{4}G(r)T_{0}(r,f)^{1/2}\end{split}$

for all sufficiently large $r$. Now, by the first fundamental theorem, we have
$T_{0}(r,1/f^{\prime})=T_{0}(r,f^{\prime})+O(1)=T(r,f^{\prime})+O(1)$, where
$T(r,f^{\prime})$ is the Nevanlinna characteristic of $f^{\prime}$. Since
$T(r,f^{\prime})\leq 2T(r,f)+S(r,f)$ and thus $T(r,f^{\prime})\leq
2T_{0}(r,f)+S(r,f)$, we finally obtain

(3.30) $\begin{split}T_{0}(r,f)\leq
r^{2}+[BL(1)+4]r^{2}G(r)+kr^{3}G(r)+12r^{4}H(r)T_{0}(r,f)^{1/2}.\end{split}$

This implies that $T_{0}(r,f)\leq O(e^{2cr^{k}})$ as $r\to\infty$, possibly
outside a set of $r$ with finite linear measure. Thus $f$ has hyper order not
exceeding 2.

When $p_{1}\in\\{\frac{1}{2},2,\frac{3}{2},\cdots,\frac{m+1}{m},\cdots\\}$,
$m\in\mathbb{Z}^{+}$, the assertion can be proved similarly. With the choice
of $\alpha$, we also define the auxiliary function $F$ in (3.6) and then
choose $\alpha_{j}$ such that
$\alpha_{1}^{\prime}+2\alpha_{1}\hat{p}_{2}=(p_{1}-1)\alpha_{2}$,
$\alpha_{2}^{\prime}+\alpha_{2}\hat{p}_{2}=2(p_{1}-1)\alpha_{6}$,
$2\alpha_{1}\hat{p}_{3}=p_{1}\alpha_{3}$,
$2\alpha_{1}\hat{p}_{4}=2(p_{1}-2)\alpha_{4}$,
$2\alpha_{1}\hat{p}_{5}=(2p_{1}-3)\alpha_{5}$,
$\alpha_{2}\hat{p}_{3}+2\alpha_{1}\hat{p}_{7}+\alpha_{3}^{\prime}+\alpha_{3}\hat{p}_{2}=(2p_{1}-1)\alpha_{7}$,
$2\alpha_{1}\hat{p}_{8}+\alpha_{3}\hat{p}_{3}=2p_{1}\alpha_{8}$,
$\alpha_{3}\hat{p}_{8}=2p_{1}\alpha_{9}$. Note that when $m\geq 3$ the
function $F$ is in general an algebroid function (see [10] for the theory of
algebroid functions), but the calculations in the case $p_{1}=1$ obviously
applies to the algebroid functions. Recall that
$\hat{p}_{2}=2(p_{1}-1)\alpha^{\prime}/\alpha+p_{2}$,
$\hat{p}_{3}=p_{3}/\alpha$, $\hat{p}_{4}=p_{4}\alpha^{2}$,
$\hat{p}_{5}=p_{5}\alpha$,
$\hat{p}_{6}=-\alpha^{\prime\prime}/\alpha+p_{1}\alpha^{\prime}/\alpha+p_{2}\alpha^{\prime}/\alpha+p_{6}$,
$\hat{p}_{7}=p_{3}\alpha^{\prime}/\alpha^{2}+p_{7}/\alpha$,
$\hat{p}_{8}=p_{8}/\alpha^{2}$ and also recall that the function $\alpha$ is
chosen in the following way: when $p_{1}=\frac{1}{2}$, $\alpha$ satisfies
$\hat{p}_{4}=0$ where we choose $\alpha\equiv 1$ when $p_{3}=p_{7}=0$; when
$p_{1}=2$, $\alpha=1$ satisfies $\hat{p}_{4}=0$; when $p_{1}=\frac{3}{2}$, we
have shown that $p_{5}=0$ and thus we choose $\alpha=1$; when
$p_{1}=\frac{m+1}{m}$ and $m\geq 3$, $\alpha=1$. Therefore, when
$p_{1}=\frac{1}{2}$, $\alpha_{1}$ can be chosen such that
$\alpha_{2}\hat{p}_{3}+2\alpha_{1}\hat{p}_{7}+\alpha_{3}^{\prime}+\alpha_{3}\hat{p}_{2}=0$,
which implies that $\alpha_{1}$ satisfies a second order linear differential
equation with constant coefficients. When $p_{1}=\frac{m+1}{m}$ and $m\geq 1$,
we may solve $\alpha_{2}$, $\alpha_{3}$, $\alpha_{4}$, $\alpha_{5}$,
$\alpha_{6}$, $\alpha_{7}$, $\alpha_{8}$ and $\alpha_{9}$ in terms of
$\alpha_{1}$ and its derivatives and then choose $\alpha_{1}$ such that one of
the coefficients
$\alpha_{6}^{\prime}+\alpha_{2}\hat{p}_{5}+\alpha_{3}\hat{p}_{4}$ or
$\alpha_{6}^{\prime}+\alpha_{3}\hat{p}_{5}$ vanishes identically. By looking
at the expressions of $\hat{p}_{j}$, we easily find that for each
$p_{1}=\frac{m+1}{m}$ we have that $\alpha_{1}$ satisfies a linear
differential equation of the form
$C_{k}\alpha_{1}^{(k)}+\cdots+C_{1}\alpha_{1}^{\prime}+C_{0}=0$, where $k\leq
3$, $C_{k},\cdots,C_{0}$ are constants. In each of these cases, we may choose
the function $G(r)=Ce^{cr}$ with two large positive numbers $C$ and $c$. Then
by the same arguments as in the case $p_{1}=1$, we may prove that all
meromorphic solutions of equation (1.9) have hyper order not exceeding $1$
when $p_{1}\in\\{\frac{1}{2},2,\frac{3}{2},\cdots,\frac{m+1}{m},\cdots\\}$,
$m\in\mathbb{Z}^{+}$. We omit those details.

∎

From the discussions in [23] and theorems 2.1 and 3.1, we have the following
corollary 3.3 which improves [23, Theorem 1.5].

######  Corollary 3.3.

All meromorphic solutions of the autonomous equation (1.1) with the degree of
the terms being at most $2$ have hyper order $1$ at most.

We note that theorem 3.1 or corollary 3.3 applies to the autonomous case of
the Malmquist type differential equation:

(3.31) $\begin{split}f^{\prime\prime}=N(z,f)+M(z,f)f^{\prime},\end{split}$

where $N(z,f)$ and $M(z,f)$ are polynomials in $f$ with rational functions.
According to [12, Theorem 12.2], if equation (3.31) admits a transcendental
meromorphic solution, then either $f$ satisfies a Riccati equation with
rational coefficients, or (3.31) reduces into

(3.32)
$\begin{split}f^{\prime\prime}=q_{1}+q_{2}f+q_{3}f^{2}+q_{4}f^{3}+q_{5}f^{\prime}+q_{6}ff^{\prime},\end{split}$

where $q_{j}$ are rational functions. When $q_{4}=0$ and $q_{6}\not=0$, the
meromorphic solutions of (3.32) are explicitly determined and are of finite
hyper at most; see [12, pp. 254-256]. When $q_{4}\not\equiv 0$, by doing the
transformation $f\to 1/f$, we obtain

(3.33) $\begin{split}ff^{\prime\prime}=2f^{\prime
2}+q_{5}ff^{\prime}+q_{6}f^{\prime}-q_{1}f^{3}-q_{2}f^{2}-q_{3}f-q_{4}.\end{split}$

When $q_{j}$ are all constants, equation (3.33) is exactly equation (1.9) with
$p_{1}=2$. Note that in this case we may suppose that $q_{1}=0$ after doing a
linear transformation $f\to f+c$ with a suitable constant $c$ in equation
(3.32). Since meromorphic solutions of a Riccati equation have finite order at
most by Gold’berg’s theorem, all meromorphic solutions of (3.33) in the
autonomous case have hyper order 1 at most. We remark that Ng and Wu [14] have
found some meromorphic solutions of (3.32), some of which are found in the
case when $q_{4}q_{6}\not=0$ and are of hyper order equal to 1.

##  4\. Discussions

In section 2, we use the idea of Halburd and Wang [6] to give the form of all
admissible meromorphic solutions of equation (1.6). This allows us to find all
transcendental meromorphic solutions of equation (1.4) proposed by Hayman,
which all have finite hyper order at most. It seems that the construction
method in section 2 is not enough to determine transcendental meromorphic
solutions of equation (1.3) that has too many poles. Instead of searching for
all the meromorphic solutions of equation (1.3), in section 3 we use the
Ahlfors–Shimizu characteristic to show that all meromorphic solutions of
equation (1.3) in the autonomous case have hyper order 1 at most. We actually
consider a more general differential equation (1.9). The order estimate
obtained by using the Ahlfors–Shimizu characteristic is not sharp in general,
as is pointed out by Steinmetz [19]. However, for infinite order solutions,
the hyper order 1 in corollary 3.3 is sharp.

If we assume that the coefficients of equation (1.9) are rational functions,
then the method in section 3 also applies to the third Painlevé equation
$P_{III}$. Recall that the first four Painlevé equations $P_{I}$, $P_{II}$,
$P_{III}$ and $P_{IV}$ (see, e.g., [5]) are:

(4.1) $\displaystyle f^{\prime\prime}$ $\displaystyle=$ $\displaystyle
6f^{2}+z,$ (4.2) $\displaystyle f^{\prime\prime}$ $\displaystyle=$
$\displaystyle 2f^{3}+zf+\alpha,$ (4.3) $\displaystyle f^{\prime\prime}$
$\displaystyle=$ $\displaystyle\frac{f^{\prime
2}}{f}-\frac{1}{z}f^{\prime}+\frac{1}{z}(\alpha f^{2}+\beta)+\gamma
f^{3}+\frac{\delta}{f},$ (4.4) $\displaystyle f^{\prime\prime}$
$\displaystyle=$ $\displaystyle\frac{f^{\prime
2}}{2f}+\frac{3}{2}f^{3}+4zf^{2}+2(z^{2}-\alpha)f+\frac{\beta}{f},$

respectively, where $\alpha,\beta,\gamma,\delta$ are free parameters. All
solutions of $P_{I}$, $P_{II}$ and $P_{IV}$ are meromorphic and their growth
and value distribution has attracted much attention; see [16, 17, 18, 19, 20,
21] and also the two monographs [5, 12]. All solutions of $P_{I}$ have order
exactly $\frac{5}{2}$ and they are of regular growth, and therefore the lower
order is $\frac{5}{2}$; all solutions of $P_{II}$ have order either $3$ or
$\frac{3}{2}$; all solutions of $P_{IV}$ have order not exceeding $4$. The
general solution of $P_{III}$ is not meromorphic, but $P_{III}$ can indeed
have some meromorphic solutions; see e.g. [11]. By doing the transformation
$g(\zeta)=zf(z)=e^{\zeta}w(e^{\zeta/2})$, $z=e^{\zeta/2}$, $P_{III}$ becomes
$\tilde{P}_{III}$:

(4.5) $\begin{split}g^{\prime\prime}=\frac{g^{\prime 2}}{g}+\alpha
g^{2}+\gamma g^{3}+\beta e^{z}+\frac{\delta e^{2z}}{g},\end{split}$

all solutions of which are meromorphic and are of finite hyper order at most
(see [5]). Below we use the method in the proof of theorem 3.1 to study the
growth of meromorphic solutions of the third Painlevé equation $P_{III}$ and
also $\tilde{P}_{III}$. We prove the following

######  Theorem 4.1.

All meromorphic solutions of $P_{III}$ have order not exceeding $4$.

###### Proof.

Here we suppose that the coefficients of (1.9) are rational functions. As in
the proof of theorem 3.1, we multiply $f^{-3}f^{\prime}$ on both sides of
(1.9) with $p_{6}\equiv 0$ and integrate the resulting equation to define an
auxiliary function $F$ in (3.5) such that:

(4.6) $\begin{split}f^{2p_{0}}F=\alpha_{1}f^{\prime
2}+\alpha_{2}ff^{\prime}+\alpha_{4}f^{4}+\alpha_{5}f^{3}+\alpha_{6}f^{2}+\alpha_{7}f+\alpha_{8},\end{split}$

where $\alpha_{j}$ are differentiable functions to be determined later and
$\alpha_{1}\not\equiv 0$. In particular, for $P_{III}$, we need to solve the
following system of equations:

(4.7)
$\left\\{\begin{array}[]{llllll}\alpha_{1}^{\prime}+2\alpha_{1}p_{2}&=0,\\\
\alpha_{2}^{\prime}+\alpha_{2}p_{2}+2\alpha_{1}p_{6}&=0,\\\
2\alpha_{1}p_{4}&=-2\alpha_{4},\\\ 2\alpha_{1}p_{5}&=-\alpha_{5},\\\
2\alpha_{1}p_{7}&=\alpha_{7},\\\
2\alpha_{1}p_{8}&=2\alpha_{8}.\end{array}\right.$

Now, for $P_{III}$ in (4.3), we use the relations in (4.7) to find that
$z\alpha_{2}^{\prime}=\alpha_{2}$ and $z\alpha_{1}^{\prime}=2\alpha_{1}$ and
the function $F$ in (3.5) satisfies:

(4.8) $\left\\{\begin{array}[]{ll}f^{2}F&=-d_{1}\gamma z^{2}f^{4}-2d_{1}\alpha
zf^{3}+d_{2}zff^{\prime}+d_{1}z^{2}f^{\prime 2}+2d_{1}\beta zf+d_{1}\delta
z^{2},\\\ f^{2}F^{\prime}&=(d_{2}-2d_{1})\gamma zf^{4}+(d_{2}-2d_{1})\alpha
f^{3}+(d_{2}+2d_{1})\beta f+(d_{2}+2d_{1})\delta z,\end{array}\right.$

where $d_{1}\not=0$ and $d_{2}$ are integration constants. We may choose
$d_{2}$ so that $d_{2}-2d_{1}=0$ or $d_{2}+2d_{1}=0$. Then by looking at the
proof in [12, Theorem 9.2.1], we easily obtain that all meromorphic solutions
of $P_{III}$ satisfy $T(r,f)=O(r^{4})$. Thus the assertion of theorem 4.1
follow.

∎

When $\alpha=\gamma=0$, the general solution of equation of (4.5) can be
integrated; see [5, p. 148]. Using the auxiliary function $F$ in (4.8), below
we solve the general solution of $P_{III}$ with $\alpha=\gamma=0$ directly. By
choosing $d_{2}=-2d_{1}$ in (4.8), we see that $F$ is a constant. In this
case, we rewrite the first equation in (4.8) as

(4.9)
$\begin{split}\left(f^{\prime}-\frac{f}{z}\right)^{2}-\left(\frac{F}{d_{1}}+1\right)\left(\frac{f}{z}\right)^{2}+2\beta\frac{f}{z}=-\delta.\end{split}$

The general solution of equation (4.9) can be obtained as follows. When
$F/d_{1}+1=0$, the function $g=f/z$ satisfies

(4.10) $\begin{split}(zg^{\prime})^{2}+2\beta g=-\delta,\end{split}$

which is solved by $g=c$ or $g=-\frac{1}{2}\beta\ln^{2}z\pm\sqrt{-2\beta
c-\delta}\ln z+c$, where $c$ is a constant. Thus $f=cz$ or
$f=-\frac{1}{2}\beta z\ln^{2}z\pm\sqrt{-2\beta c-\delta}z\ln z+cz$. On the
other hand, when $F/d_{1}+1\not=0$, letting $A$ and $B$ be such that
$A^{2}=-(F/d_{1}+1)$ and $B^{2}=(\beta/A)^{2}-\delta$, we rewrite equation
(4.9) as

(4.11)
$\begin{split}\left(f^{\prime}-\frac{f}{z}\right)^{2}+\left(A\frac{f}{z}+\frac{\beta}{A}\right)^{2}=B^{2}.\end{split}$

If $B=0$, then we have $f^{\prime}-f/z=\pm i(Af/z+\beta/A)$, which gives
$f=cz^{1\pm iA}-\frac{\beta z}{\pm iA^{2}}$. If $B\not=0$, we put

(4.12)
$\begin{split}f^{\prime}-\frac{f}{z}=\frac{B}{2}\left(h+\frac{1}{h}\right),\quad
A\frac{f}{z}+\frac{\beta}{A}=\frac{B}{2i}\left(h-\frac{1}{h}\right),\end{split}$

where $h$ is a function. A direct computation shows that $h=cz^{iA}$, where
$c$ is a nonzero constant, and thus
$f=\frac{Bz}{2iA}(cz^{iA}-\frac{1}{cz^{iA}})-\frac{\beta z}{A^{2}}$.

## Acknowledgements

The author is supported by China Postdoctoral Science Foundation
(2020M680334), the Fundamental Research Funds for the Central Universities
(FRF-TP-19-055A1) and the Ministry of Science and Technology of the People’s
Republic of China (G2021105019L). The author would like to thank professor Rod
Halburd of the University College London for giving a lot of valuable
suggestions during the preparation of this paper.

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