# On meromorphic solutions of Malmquist type difference equations

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>and  Risto Korhonen  Department of Physics and
Mathematics, University of Eastern Finland, P.O. Box 111, FI-80101 Joensuu,
Finland,
and
School of Mathematics and Physics, University of Science and Technology
Beijing, No. 30 Xueyuan Road, Haidian, Beijing, 100083, P.R. China
<EMAIL_ADDRESS>
###### Abstract.

Recently, the present authors used Nevanlinna theory to provide a
classification for the Malmquist type difference equations of the form
$f(z+1)^{n}=R(z,f)$ $({\dagger})$ that have transcendental meromorphic
solutions, where $R(z,f)$ is rational in both arguments. In this paper, we
first complete the classification for the case $\deg_{f}(R(z,f))=n$ of
$({\dagger})$ by identifying a new equation that was left out in our previous
work. We will actually derive all the equations in this case based on some new
observations on $({\dagger})$. Then, we study the relations between
$({\dagger})$ and its differential counterpart $(f^{\prime})^{n}=R(z,f)$. We
show that most autonomous equations, singled out from $({\dagger})$ with
$n=2$, have a natural continuum limit to either the differential Riccati
equation $f^{\prime}=a+f^{2}$ or the differential equation
$(f^{\prime})^{2}=a(f^{2}-\tau_{1}^{2})(f^{2}-\tau_{2}^{2})$, where $a\not=0$
and $\tau_{i}$ are constants such that $\tau_{1}^{2}\not=\tau_{2}^{2}$. The
latter second degree differential equation and the symmetric QRT map are
derived from each other using the bilinear method and the continuum limit
method.

###### Key words and phrases:

Malmquist type difference equations; Nevanlinna theory; Meromorphic solutions;
Differential equations; Continuum limit

###### 2010 Mathematics Subject Classification:

Primary 39A10; Secondary 30D35 and 39A12

The first author is supported by a Project funded by China Postdoctoral
Science Foundation (2020M680334) and the Fundamental Research Funds for the
Central Universities (FRF-TP-19-055A1). The second author is supported by a
Visiting Professorship for Senior Foreign Experts by Ministry of Science and
Technology of the People’s Republic of China (G2021105019L)

##  1\. Introduction

The classical Malmquist theorem [18] states that: If the first order
differential equation $f^{\prime}=R(z,f)$, where $R(z,f)$ is rational in both
arguments, has a transcendental meromorphic solution, then this equation
reduces into the Riccati equation

(1.1) $f^{\prime}=a_{2}f^{2}+a_{1}f+a_{0},$

where $a_{0}$, $a_{1}$ and $a_{2}$ are rational functions. Generalizations of
Malmquist’s theorem for the equation

(1.2) $(f^{\prime})^{n}=R(z,f),\qquad n\in\mathbb{N},$

have been given by Yosida [29] and Laine [16]. Steinmetz [24], and Bank and
Kaufman [2] proved that if (1.2) has rational coefficients and a
transcendental meromorphic solution, then by a suitable Möbius transformation,
(1.2) can be either mapped to the Riccati equation (1.1), or to one of the
equations in the following list:

(1.3) $\displaystyle(f^{\prime})^{2}$ $\displaystyle=$ $\displaystyle
a(f-b)^{2}(f-\tau_{1})(f-\tau_{2}),$ (1.4) $\displaystyle(f^{\prime})^{2}$
$\displaystyle=$ $\displaystyle
a(f-\tau_{1})(f-\tau_{2})(f-\tau_{3})(f-\tau_{4}),$ (1.5)
$\displaystyle(f^{\prime})^{3}$ $\displaystyle=$ $\displaystyle
a(f-\tau_{1})^{2}(f-\tau_{2})^{2}(f-\tau_{3})^{2},$ (1.6)
$\displaystyle(f^{\prime})^{4}$ $\displaystyle=$ $\displaystyle
a(f-\tau_{1})^{2}(f-\tau_{2})^{3}(f-\tau_{3})^{3},$ (1.7)
$\displaystyle(f^{\prime})^{6}$ $\displaystyle=$ $\displaystyle
a(f-\tau_{1})^{3}(f-\tau_{2})^{4}(f-\tau_{3})^{5},$

where $a$ and $b$ are rational functions, and $\tau_{1},\ldots,\tau_{4}$ are
distinct constants. See [17, Chapter 10] for more information about
Malmquist–Yosida–Steinmetz type theorems.

Recently, the present authors [15, 30] used Nevanlinna theory to provide a
classification for a natural difference analogue of equation (1.2), i.e., the
first-order difference equation

(1.8) $f(z+1)^{n}=R(z,f),$

where $n\in\mathbb{N}$ and $R(z,f)$ is rational in both arguments. In
particular, it is shown in [15] that if the difference equation (1.8) has a
transcendental meromorphic solution $f$ of hyper-order $<1$, then either $f$
satisfies a difference linear or Riccati equation

(1.9) $\displaystyle f(z+1)$ $\displaystyle=$ $\displaystyle
a_{1}(z)f(z)+a_{2}(z),$ (1.10) $\displaystyle f(z+1)$ $\displaystyle=$
$\displaystyle\frac{b_{1}(z)f(z)+b_{2}(z)}{f(z)+b_{3}(z)},$

where $a_{i}(z)$ and $b_{j}(z)$ are rational functions, or, by implementing a
transformation $f\rightarrow\alpha f$ or $f\rightarrow 1/(\alpha f)$ with an
algebraic function $\alpha$ of degree at most 2, (1.8) reduces into one of the
following equations:

(1.11) $\displaystyle f(z+1)^{2}$ $\displaystyle=$ $\displaystyle 1-f(z)^{2},$
(1.12) $\displaystyle f(z+1)^{2}$ $\displaystyle=$ $\displaystyle
1-\left(\frac{\delta(z)f(z)-1}{f(z)-\delta(z)}\right)^{2},$ (1.13)
$\displaystyle f(z+1)^{2}$ $\displaystyle=$ $\displaystyle
1-\left(\frac{f(z)+3}{f(z)-1}\right)^{2},$ (1.14) $\displaystyle f(z+1)^{2}$
$\displaystyle=$ $\displaystyle\frac{f(z)^{2}-\kappa^{2}}{f(z)^{2}-1},$ (1.15)
$\displaystyle f(z+1)^{3}$ $\displaystyle=$ $\displaystyle 1-f(z)^{-3},$

where $\delta(z)\not\equiv\pm 1$ is an algebraic function of degree $2$ at
most and $\kappa^{2}\not=0,1$ is a constant. Finite-order meromorphic
solutions of the autonomous forms of the equations (1.9)–(1.15) are presented
explicitly in [15]. These results provide a natural difference analogue of
Steinmetz’ generalization of Malmquist’s theorem in the sense of Ablowitz,
Halburd and Herbst [1], who suggested that the existence of sufficiently many
finite-order meromorphic solutions of a difference equation is a good
candidate for a difference analogue of the Painlevé property [4]. It was shown
that the finite-order condition of the proposed difference Painlevé property
can be relaxed to hyper-order strictly less than one in [6], and recently to
hyper-order equal to one limitedly in [14, 31]. Further, by discarding the
assumption that the meromorphic solution is of hyper-order $<1$ and
considering transcendental meromorphic solutions of (1.8) with
$\deg_{f}(R(z,f))=n$, it was shown in [15] that either $f$ satisfies (1.9) or
(1.10), or (1.8) can be transformed into one of the equations (1.11)–(1.15),
or one of the following equations:

(1.16) $\displaystyle f(z+1)^{2}$ $\displaystyle=$
$\displaystyle\eta^{2}(f(z)^{2}-1),$ (1.17) $\displaystyle f(z+1)^{2}$
$\displaystyle=$ $\displaystyle 2(1-f(z)^{-2}),$ (1.18) $\displaystyle
f(z+1)^{2}$ $\displaystyle=$ $\displaystyle\frac{1+f(z)^{2}}{1-f(z)^{2}},$
(1.19) $\displaystyle f(z+1)^{2}$ $\displaystyle=$
$\displaystyle\theta\frac{f(z)^{2}-\kappa_{1}f(z)+1}{f(z)^{2}+\kappa_{1}f(z)+1},$
(1.20) $\displaystyle f(z+1)^{3}$ $\displaystyle=$ $\displaystyle 1-f(z)^{3},$

where $\theta=\pm 1$, $\eta\not=1$ is the cubic root of $1$ and $\kappa_{1}$
is a constant such that
$\kappa_{1}^{2}(\kappa_{1}^{2}-4)=2(1-\theta)\kappa_{1}^{2}-8(1+\theta)$.
Transcendental meromorphic solutions of the five equations above are elliptic
functions composed with entire functions and have hyper order $\geq 1$.

This paper has two purposes. The first one is to complete the classification
for the case $\deg_{f}(R(z,f))=n$ of (1.8). When classifying equation (1.8)
for the case $\deg_{f}(R(z,f))\not=n$ in [30], we made some new observations
that also apply to equation (1.8) in the case where $\deg_{f}(R(z,f))=n$. In
the next section 2, we will use these new observations to derive the ten
equations (1.11)–(1.20) in a straightforward manner and, at the same time,
identify the following equation which was omitted in[15]:

(1.21)
$f(z+1)^{2}=\frac{1}{2}\frac{(1+\delta)^{2}}{1+\delta^{2}}\frac{(f-1)(f-\delta^{2})}{(f-\delta)^{2}},$

where $\delta\not=0,\pm 1,\pm i$ is a constant such that

(1.22) $8\delta^{5}(\delta^{2}+1)-(\delta+1)^{4}=0.$

Transcendental meromorphic solutions of (1.21) are Jacobi elliptic functions
composed with entire functions and have hyper-order at least 1, as is shown in
section 2 below. With this new equation (1.21), the results in [15, 30] can be
summarised as: If equation (1.8), where $R(z,f)$ is rational in both
arguments, has a transcendental meromorphic solution, then (1.8) can be
reduced into one out of $30$ equations. Moreover, the autonomous versions of
all these $30$ equations can be solved in terms of elliptic functions,
exponential type functions or functions which are solutions to a certain
autonomous first-order difference equation having meromorphic solutions with
preassigned asymptotic behavior. We mention that Nakamura and Yanagihara [19]
and Yanagihara [28] have already classified equation (1.8) in the case where
$R(z,f)$ is a polynomial in $f$ with constant coefficients.

The second purpose of this paper is to describe relations between meromorphic
solutions of (1.8) and (1.2) when $n=2$. In section 3, we will consider
relations between meromorphic solutions of the seven equations (1.9)–(1.14)
and equations (1.1), (1.3) and (1.4) in the autonomous case. Equations (1.11)
and (1.14) are, in fact, special cases of the symmetric
Quispel–Roberts–Thompson (QRT) map [20, 21], which will be introduced in
section 3 below. The symmetric QRT map and (1.3) or (1.4) are derived from
each other by using the _bilinear method_ and the _continuum limit method_.
The bilinear method was first used by Hirota [8, 9, 10] to find nonlinear
partial difference equations that are difference analogues of some basic
partial differential equations. The application of this method here implies
that each of the four difference equations (1.11)–(1.14) has a natural
continuum limit to equations (1.1), (1.3) or (1.4). Moreover, in section 3, we
also show that each of the five equations (1.16)–(1.19) and (1.21) can be
mapped to the symmetric QRT map by doing suitable transformations and thus
have a continuum limit to the differential equation (1.4). Finally, in section
4, we will provide some comments on our results.

##  2\. Derivations of the equations (1.11)–(1.21)

In this section, we use some new observations on equation (1.8) to derive the
eleven equations (1.11)–(1.21). We shall assume that the reader is familiar
with the standard notation and fundamental results of Nevanlinna theory (see,
e.g., [7]). For a nonconstant meromorphic function $f(z)$, recall that a value
$a\in\mathbb{C}\cup\\{\infty\\}$ is said to be a _completely ramified value_
of $f(z)$ when $f(z)-a=0$ has no simple roots. A direct consequence of
Nevanlinna’s second main theorem is that a transcendental meromorphic function
can have at most four completely ramified values in
$\mathbb{C}\cup\\{\infty\\}$.

###  2.1. Derivations of the eleven equations (1.11)–(1.21)

To prove the theorems of [15, 30], we have actually used Yamanoi’s second main
theorem for small functions as targets [25, 26]. Denote the field of rational
functions by $\mathcal{R}$ and set
$\hat{\mathcal{R}}=\mathcal{R}\cup\\{\infty\\}$. Throughout this section, we
say that $c(z)\in\hat{\mathcal{R}}$ is a _completely ramified rational
function_ of a transcendental meromorphic function $f(z)$ when the equation
$f(z)=c(z)$ has at most finitely many simple roots and that $c(z)$ is a
_Picard exceptional rational function_ of $f(z)$ when $N(r,c,f)=O(\log r)$. We
also say that $c(z)$ has multiplicity $m$ if all the roots of $f(z)=c(z)$ have
multiplicity at least $m$ with at most finitely many exceptions. Yamanoi’s
second main theorem yields that a transcendental meromorphic function can have
at most two Picard exceptional rational functions and also that the inequality

(2.1) $\sum_{i=1}^{q}\Theta(c_{i},f)\leq 2$

holds for any collection of $c_{1},\cdots,c_{q}\in\hat{\mathcal{R}}$ when $f$
is transcendental. Moreover, we have

######  Theorem 2.1.

A non-constant transcendental meromorphic function $f(z)$ can have at most
four completely ramified rational functions.

All the above statements hold when the field $\mathcal{R}$ is extended
slightly to include algebraic functions. For simplicity, in the following we
will not distinguish algebraic functions and rational functions. For example,
we always use the terms ’completely ramified rational function’ and ’Picard
exceptional rational function’ of $f$ even though sometimes they may actually
refer to algebraic functions.

We will restrict ourselves to consider equation (1.8) with
$n=\deg_{f}(R(z,f))\geq 2$. Moreover, from now on we use the suppressed
notation $\overline{f}=f(z+1)$ and $\underline{f}=f(z-1)$. We write equation
(1.8) as

(2.2) $\overline{f}^{n}=R(z,f)=\frac{P(z,f)}{Q(z,f)},$

where $P(z,f)$ and $Q(z,f)$ are two polynomials in $f$ with polynomial
coefficients having no common factors and of degrees $p$ and $q$,
respectively. Then $\deg_{f}(R(f))=\max\\{p,q\\}=n$. For simplicity, we may
also write $P(z,f)$ and $Q(z,f)$ as

(2.3) $P(z,f)=a_{p}(f-\alpha_{1})^{k_{1}}\cdots(f-\alpha_{\mu})^{k_{\mu}}$

and

(2.4) $Q(z,f)=(f-\beta_{1})^{l_{1}}\cdots(f-\beta_{\nu})^{l_{\nu}},$

where $a_{p}$ now denotes a rational function, $\alpha_{i}$ and $\beta_{j}$
are in general algebraic functions, distinct from each other, and
$k_{i},l_{j}\in\mathbb{N}$ denote the orders of the roots $\alpha_{i}$ and
$\beta_{j}$, respectively. We may suppose that the _greatest common divisor_
of $k_{1},\ldots,k_{\mu},l_{1},\ldots,l_{\mu}$, denoted by
$k=(k_{1},\ldots,k_{\mu},l_{1},\ldots,l_{\mu})$, is $1$. Denote $N_{c}$ to be
the total number of $\alpha_{i}$ with $k_{i}<n$ and $\beta_{j}$ with
$l_{j}<n$. As in [30], the classification for equation (2.2) will be according
to the number $N_{c}$ of the roots $\alpha_{i}$ in (2.3) and $\beta_{j}$ in
(2.4) and whether some of them is zero.

First, we summarize the analysis on the roots $\alpha_{i}$ of $P(z,f)$ and
$\beta_{j}$ of $Q(z,f)$ in the proof of [15, Theorem 2] and formulate the
following Lemmas 2.2 and 2.3.

######  Lemma 2.2.

Let $f$ be a transcendental meromorphic solution of equation (2.2). Then
$\alpha_{i}$ is either a Picard exceptional rational function of $f$ or a
completely ramified rational function of $f$ with multiplicity $n/(n,k_{i})$
and $\beta_{j}$ is either a Picard exceptional rational function of $f$ or a
completely ramified rational function of $f$ with multiplicity $n/(n,l_{j})$.
Moreover, if $q=0$, then $N_{c}\in\\{2,3\\}$; if $q\geq 1$, then
$N_{c}\in\\{2,3,4\\}$.

###### Proof..

The first two assertions are direct results from the proof of [15]. Now the
inequality (2.1) implies that $N_{c}\leq 4$. In particular, when $q=0$, if
$N_{c}=4$, then by the inequality (2.1) we see that the roots $\alpha_{1}$,
$\alpha_{2}$, $\alpha_{3}$ and $\alpha_{4}$ are all completely ramified
functions of $f$ with multiplicity $2$, implying that the order $k_{i}=n/2$
for all $\alpha_{i}$, which is impossible. Therefore, when $q=0$ we must have
$N_{c}=2$ or $N_{c}=3$. ∎

######  Lemma 2.3 ([15]).

Let $f$ be a transcendental meromorphic solution of equation (2.2). Then none
of $\alpha_{i}$ in (2.3) such that $k_{i}<n$ is $0$. Moreover, if $q\geq 1$,
then after doing a bilinear transformation $f\to 1/f$, if necessary, we may
suppose that $p=q=n$.

By Lemma 2.3, we may only consider the two cases that $p=n$, $q=0$ or that
$p=q=n$ below. Moreover, if $P(z,f)$ has two or more distinct roots, then none
of them vanishes identically. We use the idea in the proof of [30, Lemma 2.3]
to prove the following

######  Lemma 2.4.

Let $f$ be a transcendental meromorphic solution of equation (2.2). If
$\gamma$ is a nonzero rational function, then $\gamma$ cannot be a Picard
exceptional rational function of $f$. Moreover, if $\gamma\not\equiv 0$ is a
completely ramified rational function of $f$ with multiplicity $m$, then
$\omega\gamma$ is a completely ramified rational function of $f$ with
multiplicity $m$, where $\omega$ is the $n$-th root of 1. In particular, if
$0$ is a root of $Q(z,f)$ of order less than $n$, then $0$ is not a Picard
exceptional rational function of $f$.

###### Proof..

First, we suppose that $\gamma\not\equiv 0$ is a Picard exceptional rational
function of $f$. Under our assumptions on equation (2.2), we see that at least
one of $\alpha_{i}$ and $\beta_{j}$ in (2.3) and (2.4) is non-zero and of
order less than $n$. Denote this $\alpha_{i}$ or $\beta_{j}$ by $\delta$ and
the order of this root by $t_{1}$. As in the proof of [30, Lemma 2.3], we put

(2.5) $u=\frac{\overline{f}}{f-\delta},\quad v=\frac{1}{f-\delta}.$

Then $u$ and $v$ are two functions meromorphic apart from at most finitely
branch points and we have

(2.6) $\overline{f}=\frac{u}{v},\quad f=\frac{1}{v}+\delta,$

and it follows that (2.2) becomes

(2.7) $u^{n}=\frac{P_{1}(z,v)}{Q_{1}(z,v)}v^{n_{1}},$

where $n_{1}\in\mathbb{Z}$, $P_{1}(z,v)$ and $Q_{1}(z,v)$ are two polynomials
in $v$ having no common factors and none of the roots of $P_{1}(z,v)$ or
$Q_{1}(z,v)$ is zero. Denote by $p_{1}=\deg_{v}(P_{1}(z,v))$ the degree of
$P_{1}(z,v)$ in $v$ and by $q_{1}=\deg_{v}(Q_{1}(z,v))$ the degree of
$Q_{1}(z,v)$ in $v$, respectively. By simple calculations, when $p=n$, $q=0$
we get $n_{1}=0$, $p_{1}=p-t_{1}$ and $q_{1}=0$; when $p=q=n$ and
$\delta=\alpha_{i}$ we get $n_{1}=n$, $p_{1}=p-t_{1}$ and $q_{1}=n$; when
$p=q=n$ and $\delta=\beta_{j}$ we get $n_{1}=n$, $p_{1}=n$ and
$q_{1}=n-t_{1}$. Therefore, we always have $p_{1}-q_{1}+n_{1}\not=n$. Then by
the same arguments in the proof of [30, Lemma 2.3], we may consider the roots
of $\overline{f}-\overline{\gamma}=0$ and also
$\overline{f}-\omega\overline{\gamma}=0$, where $\omega$ is the $n$-th root of
1, and finally obtain that the equation
$\overline{f}-\omega\overline{\gamma}=0$ can have at most finitely many roots,
i.e., $\omega\gamma$ is a Picard exceptional rational function of $f$. This
implies that $n=2$. Moreover, $N_{c}=2$ and the two roots of $P(z,f)$ or
$Q(z,f)$ are $\pm\gamma$ since for otherwise by Lemma 2.2 $f$ would have $3$
Picard exceptional rational functions or $2$ Picard exceptional rational
functions with one more completely ramified rational function, a contradiction
to the inequality (2.1). But then it follows from (2.2) that either $0$ or
$\infty$ is a Picard exceptional rational function of $f$, a contradiction.
Therefore, $\gamma$ cannot be a Picard exceptional rational function of $f$.

Next, we suppose that $\gamma\not\equiv 0$ is a completely ramified rational
function of $f$ with multiplicity $m$. We also do the transformations in (2.5)
and get the equation in (2.7). Then by the same arguments as in the proof of
[30, Lemma 2.3], we obtain that the equation
$\overline{f}-\omega\overline{\gamma}=0$ can have at most finitely many roots
with multiplicities less than $m$, i.e., $\omega\gamma$ is also a completely
ramified rational function of $f$ with multiplicity $m$.

Finally, we suppose that $0$ is a root of $Q(z,f)$ of order less than $n$.
Then there is a $\beta_{j}$ such that $\omega\beta_{j}$ is a completely
ramified rational function of $f$ with multiplicity $m\geq 2$, where $\omega$
is the $n$-th root of 1. If $0$ is a Picard exceptional rational function of
$f$, then it follows from (2.2) that the roots of $P(z,f)$ are all Picard
exceptional rational functions of $f$, which together with previous
discussions shows that $f$ has at least $3$ Picard exceptional rational
functions, which is impossible. Thus $0$ cannot be a Picard exceptional
rational function of $f$. The proof is complete.

∎

Corresponding to [30, Lemma 2.4] in the case $\deg(R(z,f))\not=n$ of equation
(2.2), we have the following

######  Lemma 2.5.

Let $f$ be a transcendental meromorphic solution of equation (2.2). Then $n=2$
or $n=3$. Moreover, $\alpha_{i}$ in (2.3) with $k_{i}<n$ and $\beta_{j}$ in
(2.4) with $l_{j}<n$ are all simple.

###### Proof..

We consider the cases $q=0$ and $p=q=n$, respectively. If $n\geq 4$, then by
Lemma 2.4 a nonzero rational function $\gamma$ cannot be Picard exceptional
rational function of $f$. When $q=0$, if $n\geq 4$, then at least one of
$\alpha_{i}$ in (2.3) has order $k_{i}$ not dividing $n$, which with Lemma 2.2
shows that $\alpha_{i}$ is a completely ramified rational function of $f$ with
multiplicity $n$. However, since $\alpha_{i}\not\equiv 0$, by Lemma 2.4 $f$
would have $4$ completely ramified rational functions with multiplicity $n$, a
contradiction to the inequality (2.1). Therefore, when $q=0$ we have $n=2$ or
$n=3$. When $p=q=n$, we suppose that $n\geq 4$. Recall that $N_{c}\leq 4$. If
some $\alpha_{i}$ in (2.3) has order $k_{i}$ such that $(n,k_{i})<n/2$, then
we get a similar contradiction as in the case $q=0$. Since $n\geq 4$, this
implies that either $P(z,f)$ has only one root or that $P(z,f)$ has two
distinct $\alpha_{i}$ with two orders $k_{i}$ satisfying $k_{i}=n/2$. In the
first case, $Q(z,f)$ has at least two distinct roots and none of $\beta_{j}$
in (2.4) is zero for otherwise by Lemma 2.4 it follows that $f$ has $5$
completely ramified rational functions, a contradiction to Theorem 2.1; but
then we also have a contradiction as in the case $q=0$ since at least one
$\beta_{j}$ is a completely ramified rational function with multiplicity $n$.
In the latter case, $Q(z,f)$ must have two distinct roots and none of
$\beta_{j}$ in (2.4) is zero for otherwise $f$ would have $5$ completely
ramified rational function of $f$, a contradiction to Theorem 2.1; but then we
also have a contradiction as in the case $q=0$ since at least one $\beta_{j}$
has order $l_{j}$ not dividing $n$ and thus is a completely ramified rational
function with multiplicity $n$. Therefore, when $p=q=n$, we also have $n=2$ or
$n=3$.

Clearly, when $n=2$, $\alpha_{i}$ in (2.3) with $k_{i}<n$ and $\beta_{j}$ in
(2.4) with $l_{j}<n$ are all simple. We claim that $\alpha_{i}$ in (2.3) with
$k_{i}<n$ and $\beta_{j}$ in (2.4) with $l_{j}<n$ are also simple when $n=3$.
In fact, when $n=3$, since $f$ has $3$ non-zero completely ramified rational
functions with multiplicities $3$ it follows by Yamanoi’s second main theorem
that $0$ and $\infty$ are both not completely ramified rational functions of
$f$. If one $\alpha_{i}$ or $\beta_{j}$ in (2.3) and (2.4) is not simple, then
by a simple analysis as in the proof of [30, Lemma 2.4] we conclude that there
are at least $T(r,f)+o(T(r,f))$ many points $z_{0}$ such that $f(z_{0}+1)=0$
or $f(z_{0}+1)=\infty$ with multiplicity $m_{0}\geq 2$ and then by computing
$\overline{N}(r,1/\overline{f})$ or $\overline{N}(r,\overline{f})$ as in the
proof of [30, Lemma 2.4] we will get a contradiction. We omit those details.

∎

Let $\gamma\not\equiv 0$ be a completely ramified rational function of $f$
with multiplicity $m\geq 2$. We further consider the roots of the equation
$\overline{f}^{n}-\overline{\gamma}^{n}=0$. In particular, we may choose
$\gamma=\alpha_{i}$ or $\gamma=\beta_{j}$. By Lemma 2.4, $\omega\gamma$ is a
completely ramified rational function of $f$ with multiplicity $m$, where
$\omega$ is an $n$-th root of 1. By (2.2), when $q=0$, we have

(2.8)
$\overline{f}^{n}-\overline{\gamma}^{n}=P(z,f)-\overline{\gamma}^{n}=a_{p}(f-\gamma_{1})^{t_{1}}\cdots(f-\gamma_{\tau})^{t_{\tau}},$

or, when $q\geq 1$, we have

(2.9)
$\overline{f}^{n}-\overline{\gamma}^{n}=\frac{P(z,f)-\overline{\gamma}^{n}Q(z,f)}{Q(z,f)}=\frac{a_{p_{\tau}}(f-\gamma_{1})^{t_{1}}\cdots(f-\gamma_{\tau})^{t_{\tau}}}{Q(z,f)},$

where $\gamma_{1}$, $\cdots$, $\gamma_{\tau}$ are in general algebraic
functions distinct from each other and $t_{1},\cdots,t_{\tau}\in\mathbb{N}$
denote the orders of the roots $\gamma_{1},\ldots,\gamma_{\tau}$,
respectively, and $t_{1}+\cdots+t_{\tau}=p_{\tau}\in\mathbb{N}$. In (2.8) we
have $p_{\tau}=n$ and in (2.9) we have either $p_{\tau}<n$ when $p=q=n$ and
$a_{p}=\overline{\gamma}^{n}$ or $p_{\tau}=n$ otherwise. We apply the analysis
in the proof of [30, Lemma 2.3] to equations (2.8) and (2.9), respectively,
and get the following

######  Lemma 2.6.

Let $f$ be a transcendental meromorphic solution of equation (2.2). Suppose
that $\gamma\not\equiv 0$ is a completely ramified rational function of $f$
with multiplicity $m\geq 2$. If some $\gamma_{i}$ in (2.8) or (2.9) has order
$t_{i}<m$, then $\gamma_{i}$ is completely ramified rational function of $f$.
In particular, in (2.9) if $0<q-p_{\tau}<m$, then $\infty$ is a completely
ramified rational function of $f$. Further, suppose that
$\zeta_{i},\ldots,\zeta_{t}$ are completely ramified rational functions of $f$
such that $\sum_{i=1}^{t}\Theta(\zeta_{i},f)=2$. Then, for each $\gamma_{i}$
in (2.8) or (2.9), if $\gamma_{i}$ is not a completely ramified rational
function of $f$, then $t_{i}=m$; if $\gamma_{i}$ is a completely ramified
rational function of $f$ with multiplicity $m_{i}\geq 2$, then $t_{i}m_{i}=m$.
In particular, for (2.9), when $1\leq p_{\tau}<q$, if $\infty$ is not a
completely ramified rational function of $f$, then $q-p_{\tau}=m$; if $\infty$
is a completely ramified rational function of $f$ with multiplicity
$m_{\infty}\geq 2$, then $(q-p_{\tau})m_{\infty}=m$.

With the above five lemmas, we are ready to derive the eleven equations
(1.11)–(1.21) from (2.2). We make two remarks. First, when $p=q=n=3$, if
$P(z,f)$ has only one root $\alpha$ and $Q(z,f)$ has three distinct roots
$\beta_{1}$, $\beta_{2}$ and $\beta_{3}$, then by Lemma 2.4 and the inequality
(2.1) we see that none of $\beta_{j}$ is zero, since for otherwise $f$ would
have at least four completely ramified rational functions with multiplicity 3,
which is impossible. Then the transformation $f\to 1/f$ leads (2.2) to the
case that $p=n=3$, $q=0$ when $\alpha\equiv 0$ or to the case that $p=q=n=3$
and $Q(z,f)$ has only one root when $\alpha\not\equiv 0$. Second, when
$p=q=n=2$, if $P(z,f)$ has only one root $\alpha$ and $Q(z,f)$ has two
distinct nonzero roots $\beta_{1}$ and $\beta_{2}$, then the transformation
$f\to 1/f$ leads (2.2) to the case that $p=n=2$, $q=0$ when $\alpha\equiv 0$
or to the case that $p=q=n=2$ and $Q(z,f)$ has only one root when
$\alpha\not\equiv 0$. On the other hand, if one of the two roots $\beta_{1}$
and $\beta_{2}$ is zero, then the transformation $f\to 1/f$ leads (2.2) to the
case that $p=1$, $q=2$ and $Q(z,f)$ has only one root. Therefore, by Lemmas
2.2, 2.3 and 2.5, we see that we only need to consider the following six cases
of (2.2):

  * (1)

$p=n=3$, $q=0$ and $P(z,f)$ has three distinct non-zero roots $\alpha_{1}$,
$\alpha_{2}$ and $\alpha_{3}$;

  * (2)

$p=q=n=3$, $P(z,f)$ has three distinct non-zero roots $\alpha_{1}$,
$\alpha_{2}$ and $\alpha_{3}$ and $Q(z,f)$ has only one root $\beta$;

  * (3)

$p=n=2$, $q=0$ and $P(z,f)$ has two distinct non-zero roots $\alpha_{1}$ and
$\alpha_{2}$;

  * (4)

$p=q=n=2$, $P(z,f)$ has two distinct non-zero roots $\alpha_{1}$ and
$\alpha_{2}$ and $Q(z,f)$ has only one root $\beta$;

  * (5)

$p=1$, $q=n=2$, $P(z,f)$ has only one non-zero root $\alpha$ and $Q(z,f)$ has
only one root $\beta$;

  * (6)

$p=q=n=2$, $P(z,f)$ has two distinct non-zero roots $\alpha_{1}$ and
$\alpha_{2}$ and $Q(z,f)$ has two distinct roots $\beta_{1}$ and $\beta_{2}$.

For each of the above six cases, we shall use Lemma 2.6 together with Theorem
2.1 or the inequality (2.1) to consider (2.8) and (2.9) as in [30] to
determine the coefficients of $R(z,f)$ of (2.2), which then yield the $11$
equations (1.11)–(1.21) after doing a bilinear transformation $f\to\alpha f$
with a suitable algebraic function $\alpha$. Below we apply this strategy to
each of the above six cases respectively.

Consider first case (1). By Lemma 2.4, for each $\alpha_{i}$,
$\omega\alpha_{i}$ is a completely ramified rational function of $f$ with
multiplicity $3$ for any $\omega$ such that $\omega^{3}=1$. By the inequality
(2.1) we may suppose that $\alpha_{2}=\omega\alpha_{1}$ and
$\alpha_{3}=\omega^{2}\alpha_{1}$ for an $\omega$ such that
$\omega^{2}+\omega+1=0$. Thus, by doing a linear transformation
$f\to\alpha_{1}f$, we may rewrite equation (1.8) as

(2.10) $\overline{f}^{3}=c(1-f^{3}),$

where $c=-a_{p}\alpha_{1}^{3}/\overline{\alpha}_{1}^{3}$ is a rational
function. By (2.10), we consider

(2.11) $\overline{f}^{3}-1=c(1-f^{3})-1=-c\left(f^{3}-\frac{c-1}{c}\right).$

Note that $f$ now has three distinct completely ramified rational functions
$1,\omega,\omega^{2}$ with multiplicity 3 and has no other completely ramified
rational functions. By Lemma 2.6, we must have $c-1=0$. This gives the
equation (1.20).

Consider next case (2). By the same arguments as in case (1), we may suppose
that $\alpha_{2}=\omega\alpha_{1}$ and $\alpha_{3}=\omega^{2}\alpha_{1}$ for
an $\omega$ such that $\omega^{2}+\omega+1=0$. Thus, by doing a linear
transformation $f\to\alpha_{1}f$, we may rewrite equation (1.8) as

(2.12) $\overline{f}^{3}=\frac{c(f^{3}-1)}{(f-\delta)^{3}},$

where $c=a_{p}/\overline{\alpha}_{1}^{3}$ and $\delta=\beta/\alpha_{1}$ are in
general algebraic functions. By (2.12), we consider

(2.13)
$\overline{f}^{3}-1=\frac{c(f^{3}-1)}{(f-\delta)^{3}}-1=\frac{c(f^{3}-1)-(f-\delta)^{3}}{(f-\delta)^{3}}.$

Note that $f$ now has three distinct completely ramified rational functions
$1,\omega,\omega^{2}$ with multiplicity 3 and has no other completely ramified
rational functions. If $c\not=1$, then by Lemma 2.6 we must have
$c(f^{3}-1)-(f-\delta)^{3}=(c-1)(f-\gamma)^{3}$ for some algebraic function
$\gamma$ distinct from $\delta$. However, a simple comparison on the terms of
degrees $1$ and $2$ on both sides of this equation yields $\delta=\gamma$, a
contradiction. If $c=1$, since the terms of degree $3$ cancel out, then by
Lemma 2.6 we must have $\delta=0$ so that $c(f^{3}-1)-(f-\delta)^{3}$ reduces
to be an algebraic function. This gives the equation (1.15).

Consider next case (3). We claim that $\alpha_{1}+\alpha_{2}=0$. Otherwise, by
Lemma 2.4, $f$ has four completely ramified rational functions with
multiplicities 2, namely $\pm\alpha_{1}$ and $\pm\alpha_{2}$. Now we consider

(2.14)
$\overline{f}^{2}-\overline{\alpha}_{1}^{2}=a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{1}^{2}.$

By Lemma 2.6, if some root of the polynomial
$a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{1}^{2}$ is not equal to
$-\alpha_{1}$ or $-\alpha_{2}$, then this root has order two. This implies
that $-\alpha_{1}$ and $-\alpha_{2}$ are either both simple roots of the
polynomial $a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{1}^{2}$, or
neither of them are. Note that
$a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{i}^{2}$ cannot be a
square of some polynomial in $f$ for both $i=1,2$. By considering
$\overline{f}^{2}-\overline{\alpha}_{2}^{2}$ again, then, in the first case we
conclude by Lemma 2.6 that the polynomial
$a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{2}^{2}$ is a square of
some polynomial in $f$ and in the latter case we conclude by Lemma 2.6 that
$-\alpha_{1}$ and $-\alpha_{2}$ are both simple roots of the polynomial
$a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{2}^{2}$. We only need to
consider the first case. Now, by doing a linear transformation
$f\to\alpha_{1}f$, we have the system of two equations:

(2.15) $\begin{split}c(f-1)(f-\kappa)-1&=c(f+1)(f+\kappa),\\\
c(f-1)(f-\kappa)-\overline{\kappa}^{2}&=c(f-\delta)^{2},\end{split}$

where $c=a_{p}\alpha_{1}^{2}/\overline{\alpha}_{1}^{2}$,
$\kappa=\alpha_{2}/\alpha_{1}$ and $\delta$ are in general algebraic
functions. However, by comparing the coefficients on both sides of the two
equations in (2.15), we deduce from the resulting coefficient relations that
$1+\kappa=0$, a contradiction. Therefore, $\alpha_{1}+\alpha_{2}=0$. Then, by
doing a linear transformation $f\to\alpha_{1}f$, we may rewrite equation (1.8)
as

(2.16) $\overline{f}^{2}=c(1-f^{2}),$

where $c=-a_{p}\alpha_{1}^{2}/\overline{\alpha}_{1}^{2}$ is a rational
function. By (2.16), we consider

(2.17) $\overline{f}^{2}-1=c(1-f^{2})-1=-c\left(f^{2}-\frac{c-1}{c}\right).$

Note that $f$ now has two distinct completely ramified rational functions $\pm
1$. If $c-1=0$, then $c=1$ and we get the equation (1.11). Otherwise,
$c-1\not=0$, then by Lemma 2.6 we see that $\pm\sqrt{(c-1)/c}$ are both
completely ramified rational functions of $f$. Again, we consider

(2.18)
$\overline{f}^{2}-\frac{\overline{c}-1}{\overline{c}}=c(1-f^{2})-\frac{\overline{c}-1}{\overline{c}}=-c\left(f^{2}-\frac{\overline{c}c-\overline{c}+1}{\overline{c}c}\right).$

Since now $f$ has four completely ramified rational functions $\pm
1,\pm\sqrt{(c-1/)c}$, then by Lemma 2.6 we must have
$\overline{c}c-\overline{c}+1=0$, i.e., $c=-\omega^{2}$, where $\omega$ is a
constant such that $\omega^{2}+\omega+1=0$. This gives the equation (1.16).

Consider next case (4). In this case, if $\alpha_{1}+\alpha_{2}\not=0$, then
by Lemma 2.4, $f$ has four completely ramified rational functions with
multiplicities 2, namely $\pm\alpha_{1}$ and $\pm\alpha_{2}$. Suppose first
that $a_{p}\not=\overline{\alpha}_{1}^{2},\overline{\alpha}_{2}^{2}$. We
consider

(2.19)
$\overline{f}^{2}-\overline{\alpha}_{1}^{2}=\frac{a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{1}^{2}(f-\beta)^{2}}{(f-\beta)^{2}}.$

By the same arguments as in case (3), $-\alpha_{1}$ and $-\alpha_{2}$ are
either both simple roots of the polynomial
$a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{1}^{2}(f-\beta)^{2}$, or
neither of them are. If
$a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{i}^{2}(f-\beta)^{2}$ is
a square of some polynomial in $f$ for both $i=1,2$, then by computing the two
discriminants
$\Delta_{i}:=[a_{p}(\alpha_{1}+\alpha_{2})-2\overline{\alpha}_{i}^{2}\beta]^{2}-4(a_{p}-\overline{\alpha}_{i}^{2})(a_{p}\alpha_{1}\alpha_{2}-\overline{\alpha}_{i}^{2}\beta^{2})=0$,
$i=1,2$, we deduce that $(\beta-\alpha_{1})(\beta-\alpha_{2})=0$, which is
impossible. By considering $\overline{f}^{2}-\overline{\alpha}_{2}^{2}$ again,
then, in the first case we conclude by Lemma 2.6 that the polynomial
$a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{2}^{2}(f-\beta)^{2}$ is
a square of some polynomial in $f$ and in the latter case we conclude by Lemma
2.6 that $-\alpha_{1}$ and $-\alpha_{2}$ are both simple roots of the
polynomial
$a_{p}(f-\alpha_{1})(f-\alpha_{2})-\overline{\alpha}_{2}^{2}(f-\beta)^{2}$. We
only need to consider the first case. Now, by doing a linear transformation
$f\to\alpha_{1}f$, we have the system of two equations:

(2.20) $\begin{split}c(f-1)(f-\kappa)-(f-\delta)^{2}&=(c-1)(f+1)(f+\kappa),\\\
c(f-1)(f-\kappa)-\overline{\kappa}^{2}(f-\delta)^{2}&=(c-\overline{\kappa}^{2})(f-\gamma)^{2},\end{split}$

where $c=a_{p}/\overline{\alpha}_{1}^{2}$, $\kappa=\alpha_{2}/\alpha_{1}$,
$\delta=\beta/\alpha_{1}$ and $\gamma$ are in general algebraic functions. By
comparing the coefficients on both sides of the two equations in (2.20), we
deduce from the resulting coefficient relations that $\kappa=\delta^{2}$,
$\gamma=-\delta$, $c=\frac{1}{2}\frac{(1+\delta)^{2}}{1+\delta^{2}}$ and
$\delta\not=0,\pm 1,\pm i$ satisfies
$8\overline{\delta}^{4}(\delta^{2}+1)\delta=(\delta+1)^{4}$. Note that
$\delta\equiv 1$ solves this equation. We see that $\delta$ is a constant and
thus $\alpha_{1}$ and $\alpha_{2}$ are both rational functions. This gives the
equation (1.21). Now, if $a_{p}=\overline{\alpha}_{1}^{2}$ or
$a_{p}=\overline{\alpha}_{2}^{2}$, then by similar discussions as above, we
have the system of two equations:

(2.21) $\begin{split}(f-1)(f-\kappa)-(f-\delta)^{2}&=c_{1},\\\
(f-1)(f-\kappa)-\overline{\kappa}^{2}(f-\delta)^{2}&=(1-\overline{\kappa}^{2})(f-\gamma)^{2},\end{split}$

or

(2.22)
$\begin{split}\overline{\kappa}^{2}(f-1)(f-\kappa)-(f-\delta)^{2}&=(\overline{\kappa}^{2}-1)(f+1)(f+\kappa),\\\
\overline{\kappa}^{2}(f-1)(f-\kappa)-\overline{\kappa}^{2}(f-\delta)^{2}&=c_{2},\end{split}$

where $\delta=\beta/\alpha_{1}$, $c_{1}$, $c_{2}$ and $\gamma$ are in general
algebraic functions. However, by comparing the coefficients on both sides, we
deduce from the resulting coefficient relations obtained from the system of
two equations in (2.21) that $\delta=\gamma$ and the resulting coefficient
relations obtained from the system of two equations in (2.22) that
$1+\kappa=0$, both of which are impossible. On the other hand, if
$\alpha_{1}+\alpha_{2}=0$, then by doing a linear transformation
$f\to\alpha_{1}f$, we may rewrite equation (1.8) as

(2.23) $\overline{f}^{2}=\frac{c(1-f^{2})}{(f-\delta)^{2}},$

where $c=-a_{p}/\overline{\alpha}_{1}^{2}$ and $\delta=\beta/\alpha_{1}$ are
in general algebraic functions. By (2.23), we consider

(2.24)
$\overline{f}^{2}-1=\frac{c(1-f^{2})}{(f-\delta)^{2}}-1=\frac{c(1-f^{2})-(f-\delta)^{2}}{(f-\delta)^{2}}.$

When $c=-1$, if $\delta\not=0$, then by Lemma 2.6 we see that
$2\delta/(\delta^{2}+1)$ and $\infty$ are both completely ramified rational
functions of $f$. However, by Lemma 2.4 $f$ would have $5$ completely ramified
rational functions, a contradiction to Theorem 2.1. Therefore, we must have
$\delta=0$ when $c=-1$. When $c\not=-1$, if $c(1-f^{2})-(f-\delta)^{2}$ has
only one root, i.e., the discriminant
$\Delta:=4\delta^{2}+4(c+1)(c-\delta^{2})=0$, then $c=\delta^{2}-1$. The above
two cases give the equation (1.12). Otherwise, we have $c\not=-1$ and
$c(1-f^{2})-(f-\delta)^{2}$ has two distinct roots, say $\delta_{1}$ and
$\delta_{2}$, which are both completely ramified rational functions of $f$ by
Lemma 2.6. By Lemma 2.4, $\pm\delta_{1}$ and $\pm\delta_{2}$ are all
completely ramified rational functions of $f$. By Theorem 2.1 we must have
$\delta_{1}+\delta_{2}=0$. It follows that $\delta=0$ and
$\delta_{1}^{2}=c/(c+1)$. Again, we consider

(2.25)
$\overline{f}^{2}-\frac{\overline{c}}{\overline{c}+1}=\frac{c(1-f^{2})}{f^{2}}-\frac{\overline{c}}{\overline{c}+1}=\frac{c-[(\overline{c}c+c+\overline{c})/(\overline{c}+1)]f^{2}}{f^{2}}.$

If $\overline{c}c+c+\overline{c}\not=0$, then by Lemma 2.6 we see that
$\pm\sqrt{c(\overline{c}+1)/(\overline{c}c+c+\overline{c})}$ are both
completely ramified rational functions of $f$, a contradiction to Theorem 2.1.
Therefore, we must have $\overline{c}c+c+\overline{c}=0$ and thus $c=-2$. This
gives the equation (1.17).

Consider next case (5). In this case, by doing a linear transformation
$f\to-\alpha f$, we may rewrite equation (1.8) as

(2.26) $\overline{f}^{2}=\frac{c(f+1)}{(f-\delta)^{2}},$

where $c=-a_{p}/\alpha\overline{\alpha}^{2}$ and $\delta=-\beta/\alpha$ are
rational functions. Then from previous discussions and Lemmas 2.2 and 2.4, we
see that $\infty$, $\pm 1$ and $\pm\delta$ are all completely ramified
rational functions of $f$. By Theorem 2.1, we must have $\delta=1$. By (2.26),
we consider

(2.27)
$\overline{f}^{2}-1=\frac{c(f+1)}{(f-1)^{2}}-1=\frac{c(f+1)-(f-1)^{2}}{(f-1)^{2}}.$

If $c(f+1)-(f-1)^{2}$ has two distinct roots, then by Lemma 2.6 these two
roots are both completely ramified rational functions, a contradiction to
Theorem 2.1. Thus $c(f+1)-(f-1)^{2}$ can have only one root, which implies
that the discriminant $\Delta:=(c+2)^{2}+4(c-1)=0$, i.e., $c=-8$. This gives
the equation (1.13).

Consider finally case (6). By Lemma 2.2, $\pm\alpha_{1}$, $\pm\alpha_{2}$,
$\pm\beta_{1}$ and $\pm\beta_{2}$ are all completely ramified rational
functions of $f$. By Theorem 2.1, we must have either
$\alpha_{1}+\alpha_{2}=0$ and $\beta_{1}+\beta_{2}=0$ or
$\alpha_{1}+\beta_{1}=0$ and $\alpha_{2}+\beta_{2}=0$. When
$\alpha_{1}+\alpha_{2}=0$ and $\beta_{1}+\beta_{2}=0$, by doing a linear
transformation $f\to\beta_{1}f$, we may rewrite equation (1.8) as

(2.28) $\overline{f}^{2}=\frac{c(f^{2}-\kappa^{2})}{f^{2}-1},$

where $c=a_{p}/\overline{\beta}_{1}^{2}$ and $\kappa=\alpha_{1}/\beta_{1}$ are
in general algebraic functions. We see that $\kappa^{2}\not=0,1$. By (2.26),
we consider

(2.29)
$\overline{f}^{2}-1=\frac{c(f^{2}-\kappa^{2})}{f^{2}-1}-1=\frac{(c-1)f^{2}-(c\kappa^{2}-1)}{f^{2}-1}.$

If $c\not=1$ and $c\not=1/\kappa^{2}$, then by Lemma 2.6,
$\pm\sqrt{(c\kappa^{2}-1)/(c-1)}$ are both completely ramified rational
functions of $f$ and thus $f$ would have 6 completely ramified rational
functions, a contradiction to Theorem 2.1. Therefore, $c=1$ or
$c=1/\kappa^{2}$. If $c=1$, then we get the equation (1.14); if
$c=1/\kappa^{2}$, then we consider the equation
$\overline{f}^{2}-\overline{\kappa}^{2}$ and by the same arguments as above to
obtain that $1/\kappa^{2}=\overline{\kappa}^{2}$, i.e., $\kappa^{2}=-1$ and
$c=-1$ and thus we get the equation (1.18). On the other hand, when
$\alpha_{1}+\beta_{1}=0$ and $\alpha_{2}+\beta_{2}=0$, by doing a linear
transformation $f\to\sqrt{\alpha_{1}\alpha_{2}}f$, we may rewrite equation
(1.8) as

(2.30)
$\overline{f}^{2}=c\frac{(f-\delta)(f-\delta^{-1})}{(f+\delta)(f+\delta^{-1})},$

where $c=a_{p}/\overline{\alpha}_{1}\overline{\alpha}_{2}$ and
$\delta=\sqrt{\alpha_{1}/\alpha_{2}}$ are in general algebraic functions. By
the same arguments as for the equation (2.19), we may consider
$\overline{f}^{2}-\overline{\delta}^{2}$ and conclude by Lemma 2.6 that the
polynomial
$c(f-\delta)(f-\delta^{-1})-\overline{\delta}^{2}(f+\delta)(f+\delta^{-1})$ is
a square of some polynomial in $f$ when $c\not=\overline{\delta}^{2}$ or
reduces to be an algebraic function $c_{1}$ when $c=\overline{\delta}^{2}$.
Since $\delta^{2}\not=0,\pm 1,\pm i$, a straightforward comparison shows that
the latter case is impossible. Similarly, we may consider
$\overline{f}^{2}-\overline{\delta}^{-2}$ and conclude that the polynomial
$c(f-\delta)(f-\delta^{-1})-\overline{\delta}^{-2}(f+\delta)(f+\delta^{-1})$
is a square of some polynomial in $f$. Thus we have the system of two
equations:

(2.31)
$\begin{split}c(f-\delta)(f-\delta^{-1})-\overline{\delta}^{2}(f+\delta)(f+\delta^{-1})&=(c-\overline{\delta}^{2})(f-\gamma_{1})^{2},\\\
c(f-\delta)(f-\delta^{-1})-\overline{\delta}^{-2}(f+\delta)(f+\delta^{-1})&=(c-\overline{\delta}^{-2})(f-\gamma_{2})^{2},\end{split}$

where $\gamma_{1}$ and $\gamma_{2}$ are in general algebraic functions. By
comparing the coefficients on both sides of the two equations in (2.31), we
deduce from the resulting coefficient relations that
$\gamma_{1}^{2}=\gamma_{2}^{2}=c^{2}=1$ and $d=\delta+\delta^{-1}$ satisfies
$\overline{d}^{2}(d^{2}-4)=2(1-c)d^{2}-8(1+c)$. We see that $d$ is a constant.
This gives the equation (1.19) and also completes the classification for
equation (2.2).

###  2.2. Growth of meromorphic solutions of equation (1.21)

We show that all transcendental meromorphic solutions $f$ of equation (1.21)
have hyper-order $\geq 1$. Note that $f$ is twofold ramified over each of $\pm
1,\pm\delta^{2}$. Then there exists an entire function $\varphi(z)$ such that
$f$ is written as $f(z)=\text{sn}(\varphi(z))$, where
$\text{sn}(\varphi)=\text{sn}(\varphi,1/\delta^{2})$ is the Jacobi elliptic
function with the modulus $1/\delta^{2}$ and satisfies the second order
differential equation
$\text{sn}^{\prime}(\varphi)^{2}=(1-\text{sn}(\varphi)^{2})(1-\text{sn}(\varphi)^{2}/\delta^{4})$.
Moreover, by the second main theorem we have $T(r,f)=N(r,1/(f-1))+O(\log r)$.
Let $z_{0}$ be a point such that $f(z_{0})=\text{sn}(\varphi(z_{0}))=1$. It
follows from (1.21) that $f(z_{0}+1)=\text{sn}(\varphi(z_{0}+1))=0$. Computing
the Maclaurin series for $\text{sn}(\varphi)$ and
$\text{sn}(\overline{\varphi})$ around the point $z_{0}$, respectively, we get

(2.32)
$\begin{split}\text{sn}(\varphi(z_{0}))=1-\frac{\delta^{4}-1}{\delta^{4}}(\varphi(z)-\varphi(z_{0}))+\cdots=1-\frac{\delta^{4}-1}{\delta^{4}}\varphi^{\prime}(z_{0})(z-z_{0})+\cdots,\end{split}$

and

(2.33)
$\begin{split}\text{sn}(\varphi(z_{0}+1))=\varphi(z+1)-\varphi(z_{0}+1)+\cdots=\varphi^{\prime}(z_{0}+1)(z-z_{0})+\cdots.\end{split}$

By substituting the above two expressions into (1.21) and then comparing the
second-degree terms on both sides of the resulting equation, we find

(2.34)
$\varphi^{\prime}(z_{0}+1)^{2}=\frac{1}{2}\frac{(1+\delta)^{4}}{\delta^{4}}\varphi^{\prime}(z_{0})^{2}.$

A simple computation together with equation (1.22) shows that
$(1+\delta)^{4}\not=2\delta^{4}$. Define
$G(z):=\varphi^{\prime}(z+1)^{2}-\frac{1}{2}\frac{(1+\delta)^{4}}{\delta^{4}}\varphi^{\prime}(z)^{2}$.
From the discussions in [15] we know that $T(r,\overline{f})=T(r,f)+O(\log
r)$. Since $\text{sn}(z)$ has positive order of growth, then by [7, p. 50] we
have $T(r,\varphi)=o(T(r,f))$ and
$T(r,\overline{\varphi})=o(T(r,\overline{f}))$, where $r\to\infty$. If
$G\not\equiv 0$, then $T(r,G)\leq o(T(r,f))$, $r\to\infty$, which is
impossible since $G$ has $T(r,f)+O(\log r)$ many zeros. Thus $G(z)\equiv 0$.
Now,
$\varphi^{\prime}(z+1)=\pm\frac{1}{\sqrt{2}}\frac{(1+\delta)^{2}}{\sigma^{2}}\varphi^{\prime}(z)$
and by integration we see that $\varphi$ is an entire function such that
$T(r,\varphi)\geq Kr$ for some $K>0$ and all $r\geq r_{0}$ with some
$r_{0}\geq 0$. Since $\text{sn}(z)$ has positive exponent of convergence of
zeros and $f(z)=\text{sn}(\varphi(z))$, then the fact that $f$ is of hyper-
order at least one is a consequence of Lemma 2.7 below, which is a slightly
modified version of [17, Lemma 5.20].

######  Lemma 2.7.

Let $g$ be a meromorphic function such that the exponent of convergence
$\lambda=\lambda(g)>0$, and let $\varphi=\varphi(z)$ be an entire function
such that $T(r,\varphi)\geq cr$ for some $c>0$ and all $r\geq r_{0}$ with some
$r_{0}\geq 0$. Then the hyper order of $g\circ\varphi$ is at least one.

###### Proof..

We consider the zeros of $g\circ\varphi$. Since $\varphi$ is an entire
function, then $\varphi$ has at most one finite Picard’s exceptional value.
Thus we may choose a constant $r_{1}\geq r_{0}$ such that $\varphi$ takes in
$|z|<t$ every value $w$ in the annulus $r_{1}<|w|<M(t,\varphi)$, provided that
$t$ is large enough. Let $g$ have $\mu(t)$ zeros in this annulus, counted
according to their multiplicity. Then by the definition of $\lambda$, we have

(2.35) $\limsup_{r\to\infty}\frac{\log n(r)}{\log
r}=\limsup_{t\to\infty}\frac{\log\mu(t)}{\log M(t,\varphi)}=\lambda>0.$

Hence, for some $\tau>0$, there exists a sequence $(t_{n})$ tending to
$+\infty$ such that

(2.36)
$\mu(t_{n})>\left(M(t_{n},\varphi)\right)^{\tau}\geq\left(e^{ct_{n}}\right)^{\tau},$

where $c$ is a positive constant. The second inequality above follows by our
assumption since $\log M(t,\varphi)\geq T(t,\varphi)$ for all large $t$. Now,
$g\circ\varphi$ has at least $\mu(t)$ zeros in $|z|<t$. Making using of
(2.36), we have

(2.37) $\limsup_{r\to\infty}\frac{\log\log n(r,1/g\circ\varphi)}{\log
r}\geq\limsup_{t_{n}\to\infty}\frac{\log\log\mu(t_{n},1/g\circ\varphi)}{\log
t_{n}}\geq 1.$

By the fact that $T(r,1/g\circ\varphi)\geq N(r,1/g\circ\varphi)$, we conclude
that the hyper order of $g\circ\varphi$ is at least one. Thus our assertion
follows. ∎

We also note that the fact that all meromorphic solutions of each equation in
the list (1.16)–(1.20) can be proved using the above method since all
solutions of them are elliptic functions composed with entire functions.

##  3\. Relations between equations (1.2) and (1.8) when $n=2$

In this section, we use the bilinear method and the continuum limit method to
study the relations between equations (1.2) and (1.8) for the case $n=2$. For
the description of the bilinear method, see [8] or [11]. Here we provide a
brief overview of the continuum limit method: Let $k$ be a positive integer,
and $\varepsilon$ be a complex number. We set a pair of relations:

(3.1)
$\begin{split}\mu(z,t,\varepsilon)=0,\quad\nu(f(z),w(t,\varepsilon),\varepsilon)=0.\end{split}$

According to this, we transform a difference equation

(3.2) $\Omega_{0}(z,f(z+1),\cdots,f(z+k))=0$

to a certain difference equation

(3.3) $\Omega_{1}(t,w(t,\varepsilon),\cdots,w(t+k\varepsilon,\varepsilon))=0.$

Letting $\varepsilon\to 0$, with some conditions on coefficients of
$\Omega_{1}$, we derive a differential equation:

(3.4) $\Omega_{1}(t,w^{\prime}(t,0),\cdots,w^{(k)}(t,0))=0.$

It is clear that the first order linear difference equation has a continuum
limit to the first order linear differential equation in the autonomous case.
In the two subsections below, we describe the relations between the difference
equations (1.11)–(1.14) and the differential equations (1.1), (1.3) or (1.4).
We also show how to take the continuum limit for the five equations
(1.16)–(1.19) and (1.21).

###  3.1. Relations between (1.12) and (1.13) and the Riccati equation (1.1)

In [12], Ishizaki discussed the relation between a differential Riccati
equation and a difference Riccati equation. We first recall Ishizaki’s results
below. For the differential Riccati equation (1.1), we assume that
$a_{2}\not\equiv 0$ from now on. It is elementary to show that a suitable
linear transformation on $f$ leads equation (1.1) to

(3.5) $f^{\prime}=f^{2}+A(z),$

where $A(z)$ is a rational function formulated in terms of $a_{j}$ and their
derivatives; see [17, chapter 9]. Ishizaki used the bilinear method to derive
a difference Riccati equation from (3.5) in the following way: Setting
$f(z)=u(z)/v(z)$, then equation (3.5) becomes

(3.6)
$\begin{split}u^{\prime}(z)v(z)-u(z)v^{\prime}(z)=u(z)^{2}+A(z)v(z)^{2},\end{split}$

which is gauge invariant. In other words, for any $h(z)$,
$\tilde{u}(z)=u(z)h(z)$ and $\tilde{v}(z)=v(z)h(z)$ also satisfy the
differential equation (3.6) in place of $u(z)$ and $v(z)$, respectively.
Corresponding to equation (3.6), we choose a difference equation

(3.7)
$\begin{split}u(z+1)v(z)-u(z)v(z+1)=u(z)u(z+1)+A(z)v(z)v(z+1),\end{split}$

having the property of being gauge invariant. Setting $f(z)=u(z)/v(z)$ in the
difference equation above, then we obtain

(3.8) $\begin{split}f(z+1)-f(z)=f(z+1)f(z)+A(z),\end{split}$

i.e.,

(3.9) $\begin{split}f(z+1)=\frac{f(z)+A(z)}{1-f(z)}.\end{split}$

On the other hand, for the difference Riccati equation (1.10), Ishizaki showed
that if $b_{1}(z)\not=-b_{3}(z+1)$, by doing the transformation
$f(z)\to[(-b_{3}-\underline{b}_{1})f+(-b_{3}+\underline{b}_{1})]/2$ we obtain
the difference equation (3.9) with

(3.10)
$A(z)=\frac{-4b_{2}-b_{1}\underline{b}_{1}+3b_{1}b_{3}-\underline{b}_{1}\overline{b}_{3}-b_{3}\overline{b}_{3}}{(b_{3}+\underline{b}_{1})(\overline{b}_{3}+b_{1})}.$

Set

(3.11) $\begin{split}t=\varepsilon z,\quad f=\varepsilon
w(t,\varepsilon),\end{split}$

with the condition $A(z)=\varepsilon^{2}\tilde{A}(t,\varepsilon)$ and
$\lim_{\varepsilon\to 0}\tilde{A}(t,\varepsilon)=\tilde{A}(t,0)$. Since
$f(z+1)=\varepsilon w(\varepsilon(z+1),\varepsilon)=\varepsilon
w(t+\varepsilon,\varepsilon)$, we have

(3.12) $w(t+\varepsilon,\varepsilon)-w(t,\varepsilon)=\varepsilon
w(t+\varepsilon,\varepsilon)w(t,\varepsilon)+\varepsilon\tilde{A}(t,\varepsilon).$

By letting $\varepsilon\to 0$, we have

(3.13) $w^{\prime}(t,0)=w(t,0)^{2}+\tilde{A}(t,0),$

which is equation (3.5). In particular, if $A$ is a constant, then we replace
$A$ with $\varepsilon^{2}\tilde{A}$ with a constant $\tilde{A}$.

With the introduction above, let’s look at the two equations (1.12) and
(1.13), respectively. For equation (1.12), if we put
$f=(\gamma+\gamma^{-1})/2$, then it follows that

$\overline{\gamma}^{2}\pm
2i\frac{\delta\gamma^{2}-2\gamma+\delta}{\gamma^{2}-2\delta\gamma+1}\overline{\gamma}-1=0.$

Solving the equation above, we get four different difference Riccati
equations:

$\overline{\gamma}=\left\\{-\theta\frac{(\pm
i\delta-\sqrt{1-\delta^{2}})\gamma\pm i}{\gamma-\delta\pm
i\sqrt{1-\delta^{2}}}\right\\}^{\theta},\quad\theta=\pm 1.$

It is easy to see that these four difference equations do not have any common
solutions. Take the following difference Riccati equation as an example:

(3.14)
$\overline{\gamma}=\frac{(-i\delta+\sqrt{1-\delta^{2}})\gamma-i}{\gamma+(-\delta+i\sqrt{1-\delta^{2}})},$

consider the case where $\delta$ is a constant. If $2\delta^{2}\not=1$, then
by doing the transformation
$\gamma\to\frac{(1+i)(\delta-\sqrt{1-\delta^{2}})}{2}\gamma+\frac{(1-i)(\delta+\sqrt{1-\delta^{2}})}{2}$,
we obtain from the equation above that

(3.15) $\overline{\gamma}-\gamma=\overline{\gamma}\gamma+A,$

where $A$ has the form in (3.10) with $b_{1}=(-i\delta+\sqrt{1-\delta^{2}})$,
$b_{2}=-i$ and $b_{3}=-\delta+i\sqrt{1-\delta^{2}}$. Therefore, for the
solutions of (1.12) such that (3.14) hold, we set

(3.16) $\begin{split}t&=\varepsilon z,\\\
f&=\frac{1}{2}\left(\gamma+\frac{1}{\gamma}\right),\\\
\gamma&=\frac{(1+i)(\delta-\sqrt{1-\delta^{2}})}{2}\varepsilon
w(t,\varepsilon)+\frac{(1-i)(\delta+\sqrt{1-\delta^{2}})}{2},\end{split}$

and replace $A$ with $\varepsilon^{2}A$. Then we have the equation in (3.12)
and, by letting $\varepsilon\to 0$, we finally obtain the equation in (3.13).
Equation (1.13) is dealt with in a similar way. From the results in [15], if
we put $f=\frac{1-u^{2}}{u^{2}}$, $\sqrt{2}u=\frac{1}{2}(\gamma+\gamma^{-1})$,
then we also get four different difference Riccati equations which do not have
any common solutions. Consider the following case:

$f=\frac{1-u^{2}}{u^{2}}=\frac{8\gamma^{2}-(\gamma^{2}+1)^{2}}{(\gamma^{2}+1)^{2}},\quad\overline{\gamma}=-\frac{-(1+\sqrt{2})\gamma+i}{\gamma-i+i\sqrt{2}}.$

By making similar substitutions as in (3.16), we may obtain the equation in
(3.12) and, by letting $\varepsilon\to 0$, we obtain the equation in (3.13).

###  3.2. Relations between (1.11), (1.14), (1.16)–(1.19) and (1.21) and the
differential equations (1.3) or (1.4)

The autonomous versions of the seven difference equations (1.11), (1.14),
(1.16)–(1.19) and (1.21) are closely related to the QRT map [20, 21], which is
defined by the system of two equations:

(3.17) $\displaystyle x_{n+1}$ $\displaystyle=$
$\displaystyle\frac{f_{1}(y_{n})-x_{n}f_{2}(y_{n})}{f_{2}(y_{n})-x_{n}f_{3}(y_{n})},$
(3.18) $\displaystyle y_{n+1}$ $\displaystyle=$
$\displaystyle\frac{g_{1}(x_{n+1})-y_{n}g_{2}(x_{n+1})}{g_{2}(x_{n+1})-y_{n}g_{3}(x_{n+1})},$

where

(3.34) $\displaystyle\left(\begin{array}[]{c}f_{1}(x)\\\ f_{2}(x)\\\
f_{3}(x)\\\ \end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccc}\alpha_{0}&\beta_{0}&\gamma_{0}\\\
\delta_{0}&\varepsilon_{0}&\zeta_{0}\\\ \kappa_{0}&\lambda_{0}&\mu_{0}\\\
\end{array}\right)\left(\begin{array}[]{c}x^{2}\\\ x\\\ 1\\\
\end{array}\right)\times\left(\begin{array}[]{ccc}\alpha_{1}&\beta_{1}&\gamma_{1}\\\
\delta_{1}&\varepsilon_{1}&\zeta_{1}\\\ \kappa_{1}&\lambda_{1}&\mu_{1}\\\
\end{array}\right)\left(\begin{array}[]{c}x^{2}\\\ x\\\ 1\\\
\end{array}\right),$ (3.50) $\displaystyle\left(\begin{array}[]{c}g_{1}(x)\\\
g_{2}(x)\\\ g_{3}(x)\\\ \end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccc}\alpha_{0}&\delta_{0}&\kappa_{0}\\\
\beta_{0}&\varepsilon_{0}&\lambda_{0}\\\ \gamma_{0}&\zeta_{0}&\mu_{0}\\\
\end{array}\right)\left(\begin{array}[]{c}x^{2}\\\ x\\\ 1\\\
\end{array}\right)\times\left(\begin{array}[]{ccc}\alpha_{1}&\delta_{1}&\kappa_{1}\\\
\beta_{1}&\varepsilon_{1}&\lambda_{1}\\\ \gamma_{1}&\zeta_{1}&\mu_{1}\\\
\end{array}\right)\left(\begin{array}[]{c}x^{2}\\\ x\\\ 1\\\
\end{array}\right),$

where ’$\times$’ denotes the cross product of two vectors. In the symmetric
case, i.e.,

(3.57)
$\displaystyle\left(\begin{array}[]{ccc}\alpha_{i}&\beta_{i}&\gamma_{i}\\\
\delta_{i}&\varepsilon_{i}&\zeta_{i}\\\ \kappa_{i}&\lambda_{i}&\mu_{i}\\\
\end{array}\right)=\left(\begin{array}[]{ccc}\alpha_{i}&\delta_{i}&\kappa_{i}\\\
\beta_{i}&\varepsilon_{i}&\lambda_{i}\\\ \gamma_{i}&\zeta_{i}&\mu_{i}\\\
\end{array}\right),\quad i=0,1,$

the QRT family reduces into a single equation

(3.58) $\displaystyle w_{n+1}$ $\displaystyle=$
$\displaystyle\frac{f_{1}(w_{n})-w_{n-1}f_{2}(w_{n})}{f_{2}(w_{n})-w_{n-1}f_{3}(w_{n})},$

by taking $x_{n}=w_{2n}$ and $y_{n}=w_{2n+1}$ in (3.17) and (3.18). The
symmetric QRT family possesses an invariant:

(3.59)
$\begin{split}(\alpha_{0}&+K\alpha_{1})x_{n+1}^{2}x_{n}^{2}+(\beta_{0}+K\beta_{1})(x_{n+1}^{2}x_{n}+x_{n+1}x_{n}^{2})+(\gamma_{0}+K\gamma_{1})(x_{n+1}^{2}+x_{n}^{2})\\\
&+(\varepsilon_{0}+K\varepsilon_{1})x_{n+1}x_{n}+(\zeta_{0}+K\zeta_{1})(x_{n+1}+x_{n})+(\mu_{0}+K\mu_{1})=0,\end{split}$

where $K$ is a constant. By doing a Möbius transformation
$x_{n}\to\frac{\alpha_{1}x_{n}+\beta_{1}}{\alpha_{2}x_{n}+\beta_{2}}$ with
suitable constants $\alpha_{i}$ and $\beta_{j}$, the symmetric QRT map in
(3.59) takes the form:

(3.60) $\begin{split}\alpha
x_{n+1}^{2}x_{n}^{2}+\beta(x_{n+1}^{2}+x_{n}^{2})+\gamma
x_{n+1}x_{n}+\delta=0,\end{split}$

where $\alpha$, $\beta$, $\gamma$ and $\delta$ are constants; see [22]. By
reinterpreting discrete equations as difference equations (see [1]), we see
later that (1.14) reduces to a special case of the symmetric QRT map in the
generic case (i.e., $\alpha\delta\not=0$) and equation (1.11) is the symmetric
QRT map in the degenerate case (i.e., $\alpha\delta=0$). Moreover, the five
equations (1.16)–(1.19) and (1.21) can also be mapped to the symmetric QRT
map, as is shown below.

Suppose that $a$ and $b$ in (1.3) and (1.4) are both constants. For
simplicity, we treat equation (1.3) as a special case of equation (1.4) with
$\tau_{1}=\tau_{3}$. By doing a Möbius transformation
$f\to\frac{\alpha_{1}f+\beta_{1}}{\alpha_{2}f+\beta_{2}}$ with suitable
constants $\alpha_{i}$ and $\beta_{j}$, if necessary, we may assume that
$\tau_{1}+\tau_{3}=0$ and $\tau_{2}+\tau_{4}=0$. Thus we may write (1.4) as

(3.61)
$\begin{split}(f^{\prime})^{2}=a(f^{2}-\tau_{1}^{2})(f^{2}-\tau_{2}^{2}),\end{split}$

where $\tau_{1}^{2}\not=\tau_{2}^{2}$ and it is possible that
$\tau_{1}^{2}=0$. Now the bilinear method applies. Following Ishizaki, we set
$f(z)=u(z)/v(z)$ and obtain from equation (3.61) that

(3.62)
$\begin{split}[u^{\prime}(z)v(z)-u(z)v^{\prime}(z)]^{2}=a(z)[u(z)^{2}-\tau_{1}^{2}v(z)^{2}][u(z)^{2}-\tau_{2}^{2}v(z)^{2}],\end{split}$

which is gauge invariant. Corresponding to this equation, we choose a
difference equation

(3.63) $\begin{split}&[u(z+1)v(z)-u(z)v(z+1)]^{2}\\\
=&\,a(z)[u(z)u(z+1)-\tau_{1}^{2}v(z)v(z+1)][u(z)u(z+1)-\tau_{2}^{2}v(z)v(z+1)],\end{split}$

having the property of being gauge invariant. Setting $f(z)=u(z)/v(z)$ in the
difference equation above, we have

(3.64)
$\begin{split}[f(z+1)-f(z)]^{2}=a[f(z+1)f(z)-\tau_{1}^{2}][f(z+1)f(z)-\tau_{2}^{2}].\end{split}$

which is a special case of the symmetric QRT map (3.59) after expansion.

When $\tau_{1}^{2}=0$, we do the transformation $f\to 1/f$ and obtain from
(3.64) that

(3.65)
$\begin{split}[f(z+1)-f(z)]^{2}=-a\tau_{2}^{2}\left(f(z+1)f(z)-1/\tau_{2}^{2}\right).\end{split}$

We see that equation (1.11) is included in (3.65). By setting $t=\varepsilon
z$, $f(z)=w(t,\varepsilon)$ and giving $\varepsilon^{2}(-a\tau_{2}^{2})$ in
place of $-a\tau_{2}^{2}$, then equation (3.65) has a continuum limit to the
equation $(f^{\prime})^{2}=-a(\tau_{2}^{2}f^{2}-1)$; see [13]. The equation
$(f^{\prime})^{2}=-a(\tau_{2}^{2}f^{2}-1)$ can be obtained from equation
(3.61) by doing the transformation $f\to 1/f$.

When $\tau_{1}^{2}\not=0$, we re-scale $f$ by $f\to(\tau_{1}\tau_{2})^{1/2}f$
and then expand (3.64) to obtain the canonical form of the symmetric QRT map:

(3.66)
$\begin{split}\overline{f}^{2}f^{2}+A(\overline{f}^{2}+f^{2})+2B\overline{f}f+1=0,\end{split}$

where $A=-1/(a\tau_{1}\tau_{2})$ and
$B=[2-a(\tau_{1}^{2}+\tau_{2}^{2})]/(2a\tau_{1}\tau_{2})$. In particular, for
equation (1.14), we may re-scale $f$ by $f\to f/\kappa_{1}$ with a constant
$\kappa_{1}$ first to obtain the equation

(3.67)
$\begin{split}\overline{f}^{2}f^{2}-\kappa_{1}^{2}(\overline{f}^{2}+f^{2})+\kappa_{2}^{2}=0,\end{split}$

where $\kappa_{2}^{2}=\kappa_{1}^{4}\kappa^{2}$. By doing the transformation
$f\to\alpha\frac{f-\beta}{f+\beta}$ with constants $\alpha$, $\beta$
satisfying $\alpha^{4}=\kappa_{2}^{2}$, we obtain from (3.67) the canonical
form in (3.66) and the corresponding coefficients $A$ and $B$ in (3.66) are

$\begin{split}A&=\beta^{2},\\\
B&=2\frac{\alpha^{4}+2\kappa_{1}^{2}\alpha^{2}+\kappa_{2}^{2}}{\alpha^{4}-2\kappa_{1}^{2}\alpha^{2}+\kappa_{2}^{2}}\beta^{2},\end{split}$

respectively. It is well-known that equation (3.66) is parameterized by
elliptic functions; see [3, 22] or [5]. Here we incorporate the process of
parametrization from [5]. Define the parameters $k$ and $\varepsilon$ such
that

(3.68) $\begin{split}A&=-\frac{1}{k\text{sn}^{2}\ \varepsilon},\\\
B&=\frac{\text{cn}\ \varepsilon\ \text{dn}\ \varepsilon}{k\text{sn}^{2}\
\varepsilon},\end{split}$

respectively. These choices of $A$ and $B$ imply that

(3.69) $\begin{split}k+k^{-1}=(B^{2}-A^{2}-1)A^{-1}.\end{split}$

Therefore, considering equation (3.66) as a quadratic equation for
$\overline{f}$, and using the transformation $f=k^{1/2}\text{sn}\ u$, where
$\text{sn}\ u$ denotes the Jacobi elliptic sn function with argument $u$ and
modulus $k$, we have

(3.70) $\begin{split}\text{sn}\ \overline{u}=\frac{\text{cn}\ \varepsilon\
\text{dn}\ \varepsilon\ \text{sn}\ u\pm\text{sn}\ \varepsilon\ \text{cn}\ u\
\text{dn}\ u}{1-k^{2}\text{sn}^{2}\ \varepsilon\ \text{sn}^{2}\
u}.\end{split}$

This is solved by $u=\varepsilon z+C$, where $C$ is a free parameter. Using
the expressions of $A$ and $B$ in (3.68), we rewrite equation (3.66) as

(3.71) $\begin{split}(\overline{f}-f)^{2}=(k\text{sn}^{2}\
\varepsilon)\overline{f}^{2}f^{2}+2\left(\text{cn}\ \varepsilon\ \text{dn}\
\varepsilon-1\right)\overline{f}f+k\text{sn}^{2}\ \varepsilon.\end{split}$

By the above process, if we set

(3.72) $\begin{split}t=\varepsilon z,\quad
f=k^{1/2}w(t,\varepsilon),\end{split}$

then, since $f(z+1)=k^{1/2}w(t+\varepsilon,\varepsilon)$, by dividing
$k\text{sn}^{2}\ \varepsilon$ on both sides of equation (3.71) we get

(3.73)
$\begin{split}&\frac{[w(t+\varepsilon,\varepsilon)-w(t,\varepsilon)]^{2}}{\text{sn}^{2}\
\varepsilon}\\\
=&\,k^{2}w(t+\varepsilon,\varepsilon)^{2}w(t,\varepsilon)^{2}+\left(\frac{2\text{cn}\
\varepsilon\ \text{dn}\ \varepsilon-2}{\text{sn}^{2}\
\varepsilon}\right)w(t+\varepsilon,\varepsilon)w(t,\varepsilon)+1.\end{split}$

Recall the Maclaurin series for $\text{sn}\ \varepsilon$, $\text{cn}\
\varepsilon$ and $\text{dn}\ \varepsilon$, respectively:

(3.74) $\begin{split}\text{sn}\
\varepsilon&=\varepsilon-(1+k^{2})\frac{\varepsilon^{3}}{3!}+(1+14k^{2}+k^{4})\frac{\varepsilon^{5}}{5!}+\cdots,\\\
\text{cn}\
\varepsilon&=1-\frac{\varepsilon^{2}}{2!}+(1+k^{4})\frac{\varepsilon^{4}}{4!}+\cdots,\\\
\text{dn}\
\varepsilon&=1-k^{2}\frac{\varepsilon^{2}}{2!}+k^{2}(4+k^{2})\frac{\varepsilon^{4}}{4!}+\cdots.\end{split}$

By substituting the above series into (3.73) and then letting $\varepsilon\to
0$, we obtain the following differential equation:

(3.75)
$\begin{split}[w^{\prime}(t,0)]^{2}=(k^{2}w(t,0)^{2}-1)(w(t,0)^{2}-1),\end{split}$

which is equation (3.61). In particular, for equation (3.67), we see that this
process yields
$2\frac{\alpha^{4}+2\kappa_{1}^{2}\alpha^{2}+\kappa_{2}^{2}}{\alpha^{4}-2\kappa_{1}^{2}\alpha^{2}+\kappa_{2}^{2}}\to-1$
as $\varepsilon\to 0$. Recalling that $\alpha^{4}=\kappa_{2}^{2}$, this
implies that $\kappa_{2}\to\pm\frac{\kappa_{1}}{3}$ as $\varepsilon\to 0$. By
combining the results above together, we conclude that (1.14) has a continuum
limit to the differential equation (1.4).

We now consider the five equations (1.16)–(1.19) and (1.21). We take the
equation (1.16) as an example to show that this equation can be transformed
into the symmetric form, which is included in the QRT family, and have a
continuum limit to (3.61). It is easy to see that solutions of equation (1.16)
also satisfy the following two equations:

(3.76) $\begin{split}\overline{f}^{2}+\eta^{2}&=\eta^{2}f^{2},\\\
\overline{f}^{2}-1&=\eta^{2}(f^{2}+\eta^{2}).\end{split}$

Since all roots of $f(z)\pm i\eta=0$ have even multiplicities, we see that
$\frac{f+i\eta}{f-i\eta}=h^{2}$ for some meromorphic function $h$. It follows
that $f=i\eta\frac{h^{2}+1}{h^{2}-1}$. Denote $H=\frac{h^{2}+1}{2h}$. By
dividing the first equation in (3.76) by the second equation in (3.76) on both
sides, we obtain

(3.77)
$\begin{split}\frac{\overline{f}^{2}+\eta^{2}}{\overline{f}^{2}-1}&=\frac{f^{2}}{f^{2}+\eta^{2}}=\left(\frac{h^{2}+1}{2h}\right)^{2}=H^{2},\end{split}$

i.e.,

(3.78)
$\begin{split}\overline{f}^{2}H^{2}-(\overline{f}^{2}+H^{2})-\eta^{2}=0,\end{split}$

which is a biquadratic equation with respect to $\overline{f}$ and $H$.
Instead of considering equation (3.78) directly, we may first re-scale $f$ and
$H$ simultaneously by $f\to f/\kappa_{1}$ and $H\to H/\kappa_{1}$ with a
constant $\kappa_{1}$ to obtain the equation
$\overline{f}^{2}H^{2}-\kappa_{1}^{2}(\overline{f}^{2}+H^{2})-\eta^{2}\kappa_{1}^{4}=0$,
where $\kappa_{1}$ is a constant. Then we do the Möbius transformations:

(3.79)
$\begin{split}\overline{f}\to\alpha\frac{\overline{f}-\beta}{\overline{f}+\beta},\quad
H\to\alpha\frac{H-\beta}{H+\beta},\end{split}$

with suitable constants $\alpha$ and $\beta$, and obtain the canonical form of
the symmetric QRT map:

(3.80)
$\begin{split}\overline{f}^{2}H^{2}+A(\overline{f}^{2}+H^{2})+2B\overline{f}H+1=0,\end{split}$

where $A$ and $B$ are both nonzero constants dependent on $\alpha$, $\beta$
and $\kappa_{1}$. The process of solving (3.66) shows that equation (3.80) is
parameterized by elliptic functions and $\overline{f}=H(\overline{\varphi})$
for an entire function $\varphi$. In fact, if we define the parameters $k$ and
$\varepsilon$ as in (3.68) and consider equation (3.80) as a quadratic
equation for $\overline{f}$ with respect to $H$, then using the transformation
$H=k^{1/2}\text{sn}\ \varphi$ and $\overline{f}=k^{1/2}\text{sn}\
\overline{\varphi}$, where $\text{sn}\ \varphi$ denotes the Jacobi elliptic sn
function with argument $\varphi$ and modulus $k$, we have

(3.81) $\begin{split}\text{sn}\ \overline{\varphi}=\frac{\text{cn}\
\varepsilon\ \text{dn}\ \varepsilon\ \text{sn}\ \varphi\pm\text{sn}\
\varepsilon\ \text{cn}\ \varphi\ \text{dn}\ \varphi}{1-k^{2}\text{sn}^{2}\
\varepsilon\ \text{sn}^{2}\ \varphi},\end{split}$

which is solved by $\varphi=\varepsilon\phi+C$ such that $\varphi=\varphi(z)$
is an entire function satisfying $\varphi(z+1)=\varphi(z)+\varepsilon$, where
$C$ is a free parameter. It follows that $\phi=\phi(z)$ is an entire function
satisfying $\phi(z+1)=\phi(z)+1$. Thus $\phi(z)=\pi(z)+z$, where $\pi(z)$ is
an arbitrary non-constant periodic function of period 1. We may suppose that
$\pi(z)$ has a zero, say $\pi(z_{0})=0$. Then $z_{m}=z_{0}+m$ is a zero of
$\pi(z)$ for all integers $m\geq 0$. It follows that the infinite sequence
$\\{z_{m}\\}$ satisfies $z_{m}\to\infty$ as $m\to\infty$ and
$\phi_{m}=\phi(z_{m})=z_{m}$ for all $m$. Therefore, if we set

(3.82) $\begin{split}t=\varepsilon_{m}z_{m},\quad
H=k^{1/2}w(t,\varepsilon_{m}),\quad\overline{f}=k^{1/2}w(t+\varepsilon_{m},\varepsilon_{m}),\end{split}$

then we have from (3.80) that

(3.83)
$\begin{split}&\frac{[w(t+\varepsilon_{m},\varepsilon_{m})-w(t,\varepsilon_{m})]^{2}}{\text{sn}^{2}\
\varepsilon_{m}}\\\
=&\,k^{2}w(t+\varepsilon_{m},\varepsilon_{m})^{2}w(t,\varepsilon_{m})^{2}+\left(\frac{2\text{cn}\
\varepsilon_{m}\ \text{dn}\ \varepsilon_{m}-2}{\text{sn}^{2}\
\varepsilon_{m}}\right)w(t+\varepsilon_{m},\varepsilon_{m})w(t,\varepsilon_{m})+1.\end{split}$

For a fixed $t$, we choose $\varepsilon_{m}=\frac{t}{z_{m}}$. By using the
Maclaurin series for $\text{sn}\,\varepsilon_{m}$,
$\text{cn}\,\varepsilon_{m}$ and $\text{dn}\,\varepsilon_{m}$, respectively,
in (3.74), and then letting $\varepsilon_{m}\to 0$, we obtain exactly the
differential equation in (3.75). For each of the four equations (1.17), (1.18)
and (1.19) and (1.21), by using the same method as above we may obtain an
equation of the form in (3.78) with a certain meromorphic function $H$ and
then also obtain the differential equation in (3.75) after taking a continuum
limit. We omit those details.

##  4\. Concluding remarks

The Malmquist type difference equations (1.8) with $\deg_{f}(R(z,f))=n$ are
revisited in this paper. In section 2, we first complete the classification
for equation (1.8) with $\deg_{f}(R(z,f))=n$ by identifying one new equation
(1.21) left out in our previous work. We have actually derived the eleven
equations (1.11)–(1.21) using some recent observations on equation (1.8) in
[30]. In section 3, we study the relations between the Malmquist type
differential and difference equations in the case $n=2$. The seven equations
(1.9)–(1.15) singled out from (1.8) have finite order meromorphic solutions
and appear to be integrable from the viewpoint of the proposed difference
analogue of the Painlevé property suggested by Ablowitz, Halburd and Herbst
[1]. We point out that each of the equations (1.9)–(1.14) has a natural
continuum limit to equations (1.1), (1.3) or (1.4). The process of taking a
continuum limit from equation (1.14) to equation (1.4) also applies to some
more equations singled out from (1.8) in the case $n=2$, namely the five
equations (1.16)–(1.19) and (1.21). These equations only have infinite order
transcendental meromorphic solutions. However, they can also be mapped to the
symmetric QRT map with respect to $\overline{f}$ and a meromorphic function
$H$ dependent on $f$, so that $\overline{f}$ and $H$ are written in the form
$\overline{f}=H(\varphi+\varepsilon)$ and $H=H(\varphi)$ with an argument
$\varphi$ which is an entire function of $z$. In fact, by looking at the proof
of the main theorems in [30] and the discussions in the last section of [30],
we see that most equations singled out from (1.8) with $n=2$ in the autonomous
case in [30] can be written in the form

$\overline{f}^{2}+R_{1}^{2}=1,$

or the form

$\overline{f}^{2}R_{2}^{2}-(\overline{f}^{2}+R_{2})+\kappa^{2}=0,$

where $R_{1}$ and $R_{2}$ are rational functions in $f$ or in a certain
meromorphic function $g$ such that $f=g^{2}-1$ or $f=a\frac{g^{2}-b}{g^{2}-c}$
for some constants $a,b,c$ such that $abc\not=0$, and thus are included in the
QRT family defined in (3.17)–(3.58).

Recall from [22] that the QRT family defined in (3.17)–(3.58) possesses an
invariant which is biquadratic in $x_{n}$ and $y_{n}$:

(4.1)
$\begin{split}(\alpha_{0}&+K\alpha_{1})x_{n}^{2}y_{n}^{2}+(\beta_{0}+K\beta_{1})x_{n}^{2}y_{n}+(\gamma_{0}+K\gamma_{1})x_{n}^{2}+(\delta_{0}+K\delta_{1})x_{n}y_{n}^{2}\\\
&+(\varepsilon_{0}+K\varepsilon_{1})x_{n}y_{n}+(\zeta_{0}+K\zeta_{1})x_{n}+(\kappa_{0}+K\kappa_{1})y_{n}^{2}\\\
&+(\lambda_{0}+K\lambda_{1})y_{n}+(\mu_{0}+K\mu_{1})=0,\end{split}$

where $K$ plays the role of the integration constant. In the symmetric case,
the invariant in (4.1) becomes just (3.59). In the generic case, it is shown
in [22] that, by doing two Möbius transformations
$x_{n}\to\frac{\alpha_{1}x_{n}+\alpha_{2}}{\alpha_{3}x_{n}+\alpha_{4}}$ and
$y_{n}\to\frac{\beta_{1}y_{n}+\beta_{2}}{\beta_{3}y_{n}+\beta_{4}}$ with
suitable constants $\alpha_{i}$ and $\beta_{j}$, respectively, (4.1) can also
be mapped into the symmetric form:

(4.2)
$\begin{split}x_{n}^{2}y_{n}^{2}+A(x_{n}^{2}+y_{n}^{2})+2Bx_{n}y_{n}+1=0,\end{split}$

where $A$ and $B$ are constants. The process of solving (3.80) in section 3
shows that $x_{n}$ and $y_{n}$ in (4.2) are parameterized by elliptic
functions and $x_{n}=y_{n}(\overline{\varphi})$ with some entire function
$\varphi$. Combining this fact and the process of taking continuum limit from
equations (3.66) and (3.80) to the differential equation (1.4) in section 3,
we conclude that the QRT family defined in (3.17) and (3.18) always has a
continuum limit to the first order differential equation (1.4) in the generic
case.

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