# Exact meromorphic solutions of Schwarzian differential equations

Liangwen Liao and Chengfa Wu

Mathematics Subject Classification (2020): Primary 34M05; Secondary 30D35.

Key words and phrases. differential equation, exact solution, Schwarzian
differential equation, elliptic function.

> Abstract: This paper studies exact meromorphic solutions of the autonomous
> Schwarzian differential equations. All transcendental meromorphic solutions
> of five canonical types (among six) of the autonomous Schwarzian
> differential equations are constructed explicitly. In particular, the
> solutions of four types are shown to be elliptic functions. Also, all
> transcendental meromorphic solutions that are locally injective or possess a
> Picard exceptional value are characterized for the remaining canonical type.

##  1 Introduction and Lemmas

The Schwarzian derivative of a meromorphic function $f$ is defined as

$S(f,z)=\left({{f^{\prime\prime}}\over{f^{\prime}}}\right)^{\prime}-{1\over
2}\left({{f^{\prime\prime}}\over{f^{\prime}}}\right)^{2}={{f^{\prime\prime\prime}}\over{f^{\prime}}}-{3\over
2}\left({{f^{\prime\prime}}\over{f^{\prime}}}\right)^{2}.$

It is well-known that $S(f,z)\equiv 0$ if and only if $f$ is a Möbius
transformation. This property reveals that the Schwarzian derivative $S(f,z)$
measures how much $f$ differs from being a Möbius transformation. Another
basic property of the Schwarzian derivative is that it is invariant under the
Möbius group in the sense that $S(f,z)=S(\gamma\circ f,z)$, where $\gamma$ can
be any Möbius transformation. The converse is also true, namely, if
$S(g,z)=S(f,z)$, where $f,g$ are meromorphic functions, then there exits a
Möbius transformation $\gamma$ such that $g=\gamma\circ f$.

The Schwarzian derivative plays an essential role in various branches of
complex analysis [5, 9, 12] including univalent functions and conformal
mappings. It has also been shown that the Schwarzian derivative has close
connections with second-order linear differential equations [8] and Lax pairs
of certain integrable partial differential equations [13]. In particular, it
appears in the differential equation

$S(f,z)^{p}=R(z,f)={{P(z,f)}\over{Q(z,f)}},$ (1)

where $p$ is a positive integer, and $R(z,f)$ is an irreducible rational
function in $f$ with meromorphic coefficients. The equation (1) is known as
the Schwarzian differential equation. Ishizaki [7] obtained some Malmquist-
type theorems of this equation and results concerning the deficiencies of its
meromorphic solutions. The growth of meromorphic solutions of the equation (1)
with polynomial coefficients has been studied by Liao and Ye [10]. A more
complicated Schwarzian type differential equation was considered by Hotzel and
Jank [6]. If we restrict ourselves to the autonomous Schwarzian differential
equation

$S(f,z)^{p}=R(f)={{P(f)}\over{Q(f)}},$ (2)

where $P,Q$ are co-prime polynomials with constant coefficients, Ishizaki [7]
obtained a Malmquist-Yosida-type result in which he gave a complete
classification of the equation (2) possessing transcendental meromorphic
solutions.

######  Theorem A.

Suppose that the autonomous Schwarzian differential equation (2) admits a
transcendental meromorphic solution. Then for some Möbius transformation
$u=(af+b)/(cf+d),ad-bc\not=0,$ (2) reduces into one of the following types

$\displaystyle S(u,z)$ $\displaystyle=$ $\displaystyle
c\frac{(u-\sigma_{1})(u-\sigma_{2})(u-\sigma_{3})(u-\sigma_{4})}{(u-\tau_{1})(u-\tau_{2})(u-\tau_{3})(u-\tau_{4})}$
(3) $\displaystyle S(u,z)^{3}$ $\displaystyle=$ $\displaystyle
c\frac{(u-\sigma_{1})^{3}(u-\sigma_{2})^{3}}{(u-\tau_{1})^{3}(u-\tau_{2})^{2}(u-\tau_{3})}$
(4) $\displaystyle S(u,z)^{3}$ $\displaystyle=$ $\displaystyle
c\frac{(u-\sigma_{1})^{3}(u-\sigma_{2})^{3}}{(u-\tau_{1})^{2}(u-\tau_{2})^{2}(u-\tau_{3})^{2}}$
(5) $\displaystyle S(u,z)^{2}$ $\displaystyle=$ $\displaystyle
c\frac{(u-\sigma_{1})^{2}(u-\sigma_{2})^{2}}{(u-\tau_{1})^{2}(u-\tau_{2})(u-\tau_{3})}$
(6) $\displaystyle S(u,z)$ $\displaystyle=$ $\displaystyle
c\frac{(u-\sigma_{1})(u-\sigma_{2})}{(u-\tau_{1})(u-\tau_{2})}$ (7)
$\displaystyle S(u,z)$ $\displaystyle=$ $\displaystyle c$ (8)

where $c\in\mathbb{C},\tau_{j}$ are distinct constants, and $\sigma_{j}$ are
constants, not necessarily distinct, $j=1,\dots,4$.

######  Remark 1.

We remark that the conclusion of Theorem A does not hold for rational
solutions of equation (2). For instance, the function

$f(z)=-\frac{3}{2(z+a)^{2}},$

where $a$ is an arbitrary constant, satisfies the equation $S(u,z)=u$ but it
cannot be transformed into any type of (3)-(8) via Möbius transformations. It
is also noted that $f$ can be viewed as a fixed point of the Schwarzian
operator and we refer the readers to the reference [15] for the details on
fixed points and $N$-cycles of the Schwarzian operator.

The above theorem intimates that to study the autonomous Schwarzian
differential equation (2), it suffices to consider the equations (3)–(8). We
will show that all transcendental meromorphic solutions of the equations
(3)-(6) are elliptic functions and can be explicitly constructed. It is also
shown that all transcendental meromorphic solutions of the equation (2) can be
characterized by imposing some conditions on them. The precise statements of
these results are as follows.

######  Theorem 1.

If the Schwarzian differential equation (2) admits a transcendental
meromorphic solution $f$ with a Picard exceptional value
$\xi\in\hat{\mathbb{C}}$, then by some M$\ddot{o}$bius transformation
$f=\gamma_{1}(u)$, (2) reduces into either

$S(u,z)=c\frac{(u-\sqrt{2}i)(u+\sqrt{2}i)}{(u-1)(u+1)},$

and the transcendental meromorphic solutions of (2) are
$f(z)=\gamma_{1}(\sin(\alpha z+\beta))$, where $\alpha=\sqrt{2c}$ and $\beta$
is a constant; or

$S(u,z)=c,$

and all solutions of (2) are $f(z)=\gamma_{2}(e^{\alpha z})$, where
$\alpha=\sqrt{-2c}$ and $\gamma_{2}$ is any Möbius transformation.

######  Remark 2.

Theorem 1 shows that any transcendental meromorphic solution of equations
(3)-(6) must have infinitely many poles.

The result below follows immediately from Theorem 1.

######  Corollary 1.

If the Schwarzian differential equation (2) admits a transcendental entire
solution $f$, then by some linear transformation $f=L_{1}(u)$, (2) reduces
into either

$S(u,z)=c\frac{(u-\sqrt{2}i)(u+\sqrt{2}i)}{(u-1)(u+1)},$

and the entire solutions of (2) are $f(z)=L_{1}(\sin(\alpha z+\beta))$, where
$\alpha=\sqrt{2c}$ and $\beta$ is a constant; or

$S(u,z)=c,$

and all entire solutions of (2) are $f(z)=L_{2}(e^{\pm\alpha z})$, where
$\alpha=\sqrt{-2c}$ and $L_{2}$ is any linear transformation.

######  Theorem 2.

If the Schwarzian differential equation (2) admits a locally injective
transcendental meromorphic solution, then by some Möbius transformation
$f=\gamma(u)$, (2) reduces into

$S(u,z)=c,$

and all solutions of (2) are $f(z)=\gamma(e^{\alpha z})$, where
$\alpha=\sqrt{-2c}$ and $\gamma$ is any Möbius transformation.

Rewrite the equation (3) as

$\displaystyle S(u,z)$ $\displaystyle=$ $\displaystyle
c\frac{(u-\sigma_{1})(u-\sigma_{2})(u-\sigma_{3})(u-\sigma_{4})}{(u-\tau_{1})(u-\tau_{2})(u-\tau_{3})(u-\tau_{4})}$
(9) $\displaystyle=$
$\displaystyle\frac{r_{4}u^{4}+r_{3}u^{3}+r_{2}u^{2}+r_{1}u+r_{0}u}{(u-\tau_{1})(u-\tau_{2})(u-\tau_{3})(u-\tau_{4})},$

and denote by

$\displaystyle e_{1}=\sum_{j=1}^{4}\tau_{j},\quad e_{2}=\sum_{1\leq j<k\leq
4}\tau_{j}\tau_{k},\quad e_{3}=\sum_{1\leq j<k<l\leq
4}\tau_{j}\tau_{k}\tau_{l},\quad e_{4}=\prod_{j=1}^{4}\tau_{j}.$ (10)

Then we can construct all transcendental meromorphic solutions to the equation
(3).

######  Theorem 3.

All transcendental meromorphic solutions of the equation (9) are elliptic
functions of the form

$\displaystyle u(z)$ $\displaystyle=$ $\displaystyle
a-\frac{b}{\wp(z-z_{0};g_{2},g_{3})-d},$ (11)

where $\wp(z;g_{2},g_{3})$ is the Weierstrass elliptic function,
$z_{0}\in\mathbb{C}$ is arbitrary, $a=\tau_{i}$ and $b,d,g_{2},g_{3}$ are
constants that depend on $c$, $\sigma_{i}$ and $\tau_{i},i=1,2,3,4$. Further,
with $e_{i}\,(i=1,2,3,4)$ defined in (10) and

$q_{i}=\prod_{\begin{subarray}{c}1\leq j\leq 4\\\
j\not=i\end{subarray}}(\tau_{i}-\tau_{j}),\quad i=1,2,3,4,$

the equation (9) admits solutions of the form (11) if and only if the
following parameter relations hold

$\displaystyle r_{0}$ $\displaystyle=$
$\displaystyle\frac{b}{2q_{i}}\left(3e_{3}^{2}-8e_{2}e_{4}\right),\quad
r_{1}=\frac{2b}{q_{i}}\left(6e_{1}e_{4}-e_{2}e_{3}\right),\quad
r_{2}=\frac{b}{q_{i}}\left(2e_{2}^{2}-3e_{1}e_{3}-24e_{4}\right),$ (12)
$\displaystyle r_{3}$ $\displaystyle=$
$\displaystyle\frac{2b}{q_{i}}\left(6e_{3}-e_{1}e_{2}\right),\quad
r_{4}=\frac{b}{2q_{i}}\left(3e_{1}^{2}-8e_{2}\right),$ (13) $\displaystyle d$
$\displaystyle=$
$\displaystyle\frac{b}{6q_{i}}\left[\sum_{\begin{subarray}{c}1\leq j<k\leq
4\\\
j,k\not=i\end{subarray}}(\tau_{j}-\tau_{k})^{2}-\sum_{\begin{subarray}{c}1\leq
j\leq 4\\\ j\not=i\end{subarray}}2(\tau_{i}-\tau_{j})^{2}\right],$ (14)
$\displaystyle g_{2}$ $\displaystyle=$
$\displaystyle\frac{4b^{2}}{3q_{i}^{2}}\left(e_{2}^{2}-3e_{1}e_{3}+12e_{4}\right),\quad
g_{3}=\frac{4b^{3}}{27q_{i}^{3}}\left(2e_{2}^{3}-9e_{1}e_{2}e_{3}-72e_{2}e_{4}+27e_{3}^{2}+27e_{1}^{2}e_{4}\right),$
(15) $\displaystyle\Delta$ $\displaystyle=$ $\displaystyle
g_{2}^{3}-27g_{3}^{2}=\frac{16b^{6}}{q_{i}^{6}}\prod_{1\leq j<k\leq
4}(\tau_{j}-\tau_{k})^{2}\not=0.$ (16)

######  Remark 3.

Theorem 3 indicates that the equation (9) has transcendental meromorphic
solutions only if the parameters $c,\sigma_{i},\tau_{i},i=1,2,3,4$ satisfy the
conditions (12) and (13). In addition, the solution (11) has just one free
parameter, which implies that the general solution of equation (9) should have
more complicated singularities other than poles.

In view of the invariance of Schwarzian derivatives under the Möbius group, we
may compose the solution $u$ of equations (4)-(6) with a Möbius transformation
such that $\tau_{1},\tau_{2}$ and $\tau_{3}$ can be any distinct desired
numbers, and this allows us to derive all transcendental meromorphic solutions
to the equation equations (4)-(6) explicitly.

######  Theorem 4.

Let $\tau_{1}=4,\tau_{2}=-3,\tau_{3}=0$, then all transcendental meromorphic
solutions to the equation (4) are elliptic functions. Moreover, these
solutions exist if and only if

$\displaystyle\\{\sigma_{1},\sigma_{2}\\}=\left\\{\sqrt{5}i,-\sqrt{5}i\right\\}.$

and in this case, all the transcendental meromorphic solutions to the equation
(6) are given by

$u(z)=-\frac{3c}{c-74088\wp\left(z-z_{0};g_{2},g_{3}\right)^{3}},$

where $\wp(z;g_{2},g_{3})$ is the Weierstrass elliptic function with
$g_{2}=0,g_{3}=c/10584$, and $z_{0}\in\mathbb{C}$ is arbitrary.

######  Theorem 5.

Let $\\{\tau_{1},\tau_{2},\tau_{3}\\}=\\{0,1,-1\\}$, then all transcendental
meromorphic solutions to the equation (5) are elliptic functions. Moreover,
these solutions exist if and only if

$\displaystyle\\{\sigma_{1},\sigma_{2}\\}=\left\\{\frac{i}{\sqrt{3}},-\frac{i}{\sqrt{3}}\right\\},$

and in this case, all the transcendental meromorphic solutions to the equation
(5) are given by

$u(z)=\frac{9\left[9\wp\left(z-z_{0};g_{2},g_{3}\right)+L^{2}\right]\wp^{\prime}\left(z-z_{0};g_{2},g_{3}\right)}{2L\left[81\wp\left(z-z_{0};g_{2},g_{3}\right)^{2}-9L^{2}\wp\left(z-z_{0};g_{2},g_{3}\right)+L^{4}\right]}$

where $L^{6}=-27c/64$, $\wp(z;g_{2},g_{3})$ is the Weierstrass elliptic
function with $g_{2}=0,g_{3}=c/432$, and $z_{0}\in\mathbb{C}$ is arbitrary.

######  Theorem 6.

Let $\tau_{1}=0,\tau_{2}=1,\tau_{3}=-1$, then all transcendental meromorphic
solutions to the equation (6) are elliptic functions. Moreover, these
solutions exist if and only if

$\displaystyle\\{\sigma_{1},\sigma_{2}\\}=\left\\{\frac{i}{2},-\frac{i}{2}\right\\},$

and in this case, all the transcendental meromorphic solutions to the equation
(6) are given by

$u(z)=-\frac{1}{2L}\frac{\left(8\wp\left(z-z_{0};g_{2},g_{3}\right)+L^{2}\right)^{2}\wp^{\prime}\left(z-z_{0};g_{2},g_{3}\right)}{\wp\left(z-z_{0};g_{2},g_{3}\right)\left(64\wp\left(z-z_{0};g_{2},g_{3}\right)^{2}+L^{4}\right)},$

where $c=9L^{4}/4$, $\wp(z;g_{2},g_{3})$ is the Weierstrass elliptic function
with $g_{2}=-c/36,g_{3}=0$, and $z_{0}\in\mathbb{C}$ is arbitrary.

######  Remark 4.

It follows from Theorems 1-6 that all transcendental meromorphic solutions of
the canonical Schwarzian differential equations (3)-(8) have been derived,
except the solutions of equation (7) that have no Picard exceptional values.
Although we are not able to prove that any transcendental meromorphic solution
of equation (7) must have Picard exceptional value(s), we conjecture this is
true.

##  2 Preliminaries

The important tools in our proof include Wiman-Valiron theorem and Wiman-
Valiron theory. Let $f$ be a transcendental entire function, and write

$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}.$

As usual, for $r>0$, we denote the maximum term by $\mu(r,f)$, the central
index by $\nu(r,f)$, and the maximum modulus by $M(r,f)$, i.e.,

$\mu(r,f)=\max_{|z|=r}|a_{n}z^{n}|,\ \
\nu(r,f)=\sup\\{n||a_{n}|r^{n}=\mu(r,f)\\},\ \ M(r,f)=\max_{|z|=r}|f(z)|.$

######  Lemma 1 (Wiman-Valiron Theorem [1]).

There exists a set $F\subset[1,+\infty)$ satisfying

$\int_{F}\frac{dt}{t}<\infty$

with the following property: if $(z_{k})$ is a sequence in $\mathbb{C}$ with
$|f(z_{k})|=M(|z_{k}|,f),|z_{k}|\not\in F$ and $z_{k}\to\infty$, and if
$\nu_{k}=\nu(|z_{k}|,f),$ then

$\frac{f\left(z_{k}+\dfrac{z_{k}}{\nu_{k}}z\right)}{f(z_{k})}\to e^{z}$

as $k\to\infty.$

######  Lemma 2 ([8]).

Let $f$ be a transcendental entire function, $0<\delta<{1\over 4}$ and $|z|=r$
such that

$|f(z)|>M(r,f){\nu(r,f)}^{-{1\over 4}+\delta}$ $None$

holds. Then there exists a set $F\subset(0,+\infty)$ of finite logarithmic
measure, i.e., $\int_{F}dt/t<+\infty$ such that

$f^{(m)}(z)={\left({{\nu(r,f)}\over z}\right)}^{m}(1+o(1))f(z)$ $None$

holds for all $m\geq 0$ and all $r\not\in F$.

The Schwarzian derivative has a fundamental relation with second-order linear
ordinary differential equations.

######  Lemma 3.

[8, p. 110] Let $A(z)$ be analytic in a simply connected domain $\Omega$.
Then, for any two linearly independent solutions $f_{1},f_{2}$ of

$f^{\prime\prime}(z)+A(z)f(z)=0,$ (17)

their quotient $g=f_{1}/f_{2}$ is locally injective and satisfies the
differential equation

$S(g,z)=2A(z).$ (18)

Conversely, let $g$ be a locally injective meromorphic function in $\Omega$
and define $A(z)$ by (18). Then $A(z)$ is analytic in $\Omega$ and the
differential equation (17) admits two linearly independent solutions
$f_{1},f_{2}$ such that $g=f_{1}/f_{2}$.

######  Remark 5.

The lemma above has crucial applications in differential equations. In
particular, it has been used by Bergweiler and Eremenko [2] to solve the Bank-
Laine conjecture, which concerns the zero distribution of solutions of
equation (17) where $A$ is an entire function of finite order.

Now we introduce some terminologies in Nevanlinna theory [8]. Let $f$ be a
meromorphic function on $\mathbb{C}$ and $n(r,f)$ denote the number of poles
of $f$ in the disk $\mathbb{D}(r)=\\{z\in\mathbb{C}||z|<r\\}$, counting
multiplicity. The Nevanlinna characteristic function of $f$ is defined as

$T(r,f)=m(r,f)+N(r,f),$

where

$\displaystyle m(r,f)$ $\displaystyle=$
$\displaystyle\int_{0}^{2\pi}\log^{+}|f\left(re^{i\theta}\right)|\frac{d\theta}{2\pi},$
$\displaystyle N(r,f)$ $\displaystyle=$ $\displaystyle n(0,f)\log
r+\int_{0}^{r}\left[n(t,f)-n(0,f)\right]\frac{dt}{t},$

with $\log^{+}x=\max\\{0,\log x\\}$. We note that $m(r,f)$ and $N(r,f)$ are
called the proximity function and integrated counting function, respectively.
Next, we define the order of $f$ by

$\rho(f)=\mathop{\overline{\rm lim}}_{r\rightarrow\infty}\frac{\log
T(r,f)}{\log r}.$

The following result of Liao and Ye [10] says that the order of meromorphic
solutions of equation (2) is bounded from above by $2$.

######  Lemma 4.

Let $f$ be a meromorphic solution of the autonomous Schwarzian differential
equation (2), then $\rho(f)\leq 2$.

##  3 Proof of main results

We first recall the definition of totally ramified values: we call a point
$a\in\overline{\mathbb{C}}$ a totally ramified value of a meromorphic function
$f$ if all $a$-points of $f$ are multiple. According to a classical result of
Nevanlinna, a non-constant function meromorphic in the plane can have at most
four totally ramified values while a non-constant entire function can have at
most two finite totally ramified values. We also need the following results.

######  Lemma 5 ([8]).

Let $f(z)$ be a nonconstant meromorphic function. Then

$m\left(r,\frac{f^{\prime}}{f}\right)=O(\log r),$

if $f$ is of finite order, and

$m\left(r,\frac{f^{\prime}}{f}\right)=O(\log(rT(r,f))),$

possibly outside a set $E$ of $r$ with finite linear measure, if $f(z)$ is of
infinite order.

######  Lemma 6 ([14]).

If the differential equation

$w^{2}+R(z)(w^{\prime})^{2}=Q(z),$ (19)

where $R,Q$ are nonzero rational functions, admits a transcendental
meromorphic solution $f$, then $Q\equiv A$ is a constant, the multiplicity of
zeros of $R(z)$ is no greater than 2 and $f(z)=\sqrt{A}\cos\alpha(z)$, where
$\alpha(z)$ is a primitive of $1/\sqrt{R(z)}$ such that
$\sqrt{A}\cos\alpha(z)$ is a transcendental meromorphic function.

###  3.1 Proof of Theorem 1

Let $f$ be a transcendental meromorphic solution with a Picard exceptional
value of the equation (2). It follows from Theorem A that by some Möbius
transformation

$u=\dfrac{af+b}{cf+d},\quad ad-bc\not=0,$

$u$ satisfies one of the equations (3)-(8).

If $u$ satisfies the equation (3), then $u$ has four totally ramified values
$\tau_{1},\tau_{2},\tau_{3},\tau_{4}$. This is impossible since $u$ has a
Picard exceptional value. If $u$ satisfies the equation (4), then $u$ has
three totally ramified values $\tau_{1},\tau_{2},\tau_{3}$. Thus, the Picard
exceptional value of $u$ must be one of them. Without loss of generality, we
may assume $\tau_{3}$ is a Picard exceptional value of $u$. Let

$v=\dfrac{1}{u-\tau_{3}},$

then $v$ has at most finitely many poles and satisfies the following
differential equation

$S(v,z)=c^{\prime}\frac{(v-\sigma^{\prime}_{1})^{3}(v-\sigma^{\prime}_{2})^{3}}{(v-\tau^{\prime}_{1})^{3}(v-\tau^{\prime}_{2})^{2}}$
(20)

Assume $\zeta_{1},\cdots,\zeta_{n}$ are the poles (counting multiplicities) of
$v$, then $v(z)=g(z)/P(z)$, where $g(z)$ is a transcendental entire function
and $P(z)=(z-\zeta_{1})\cdots(z-\zeta_{n}).$ We choose $z_{k}\to\infty$ such
that $|z_{k}|\not\in F$ and $|g(z_{k})|=M(|z_{k}|,g).$ Let

$h_{k}(z)=\frac{v(z_{k}+\rho_{k}z)}{v(z_{k})},$

where $\displaystyle\rho_{k}=\frac{z_{k}}{\nu_{k}},\nu_{k}=\nu(|z_{k}|,g)$,
then by Lemma 1, we have

$\lim_{k\to\infty}h_{k}(z)=\lim_{k\to\infty}\frac{v(z_{k}+\rho_{k}z)}{v(z_{k})}=\lim_{k\to\infty}\frac{g(z_{k}+\rho_{k}z)}{g(z_{k})}\frac{P(z_{k})}{P(z_{k}+\rho_{k}z)}=e^{z}.$

Thus

$\lim_{k\to\infty}\frac{\rho_{k}v^{\prime}(z_{k}+\rho_{k}z)}{v(z_{k})}=\lim_{k\to\infty}h_{k}^{\prime}(z)=e^{z},$

and

$\lim_{k\to\infty}\frac{\rho_{k}^{2}v^{\prime\prime}(z_{k}+\rho_{k}z)}{v(z_{k})}=\lim_{k\to\infty}h_{k}^{\prime\prime}(z)=e^{z}.$

It follows from (20) that

$\frac{1}{v(z_{k})}\left(\frac{1}{\rho_{k}}\right)^{2}\left(\frac{h_{k}^{\prime\prime\prime}(z)}{h_{k}^{\prime}(z)}-\frac{3}{2}\left(\frac{h_{k}^{\prime\prime}(z)}{h_{k}^{\prime}(z)}\right)^{2}\right)=c^{\prime}\frac{(h_{k}(z)-\sigma_{1}^{\prime}/v(z_{k}))^{3}(h_{k}(z)-\sigma_{2}^{\prime}/v(z_{k}))^{3}}{(h_{k}(z)-\tau_{1}^{\prime}/v(z_{k}))^{3}(h_{k}(z)-\tau_{2}^{\prime}/v(z_{k}))^{2}}.$
(21)

Noting the selection of $z_{k}$, we have

$\displaystyle\lim_{k\to\infty}\frac{\nu_{k}^{M}}{v(z_{k})}=0$

for any positive number $M$. Thus, the left side of the equation (21) tends to
zero while the right side of equation (21) tends tends to $c^{\prime}e^{z}$ as
$k\to\infty$, which is a contradiction. Thus $u$ cannot satisfy (4). With
similar arguments, we can prove that $u$ satisfies neither (5) nor (6).

If $u$ satisfies the equation (7), then $u$ has two totally ramified values
$\tau_{1},\tau_{2}$. Then we distinguish two cases.
Case 1: one of $\tau_{1}$ and $\tau_{2}$ is the Picard exceptional value of
$u$, by the same arguments as above, we get a contradiction.
Case 2: both of $\tau_{1}$ and $\tau_{2}$ are not the Picard exceptional value
of $u$. Without loss of generality, we may assume the Picard exceptional value
of $u$ is infinity. Otherwise, we may consider a composition of a Möbius
transformation and the function $u$. Thus we can express $u$ as

$u(z)=\frac{g(z)}{P(z)},$

where $g(z)$ is a transcendental entire function and $P(z)$ is a polynomial.
For any $r>0,$ let

$|g(z_{0})|=M(g,r),\quad|z_{0}|=r.$

Then, by Lemma 2, there exists a set $F\subseteq(0,+\infty)$ with a finite
logarithmic measure such that

$\frac{u^{\prime}(z_{0})}{u(z_{0})}=\frac{g^{\prime}(z_{0})}{g(z_{0})}-\frac{P^{\prime}(z_{0})}{P(z_{0})}=\frac{\nu(g,r)}{z_{0}}(1+o(1)),$
$\frac{u^{\prime\prime}(z_{0})}{u(z_{0})}=\frac{g^{\prime\prime}(z_{0})}{g(z_{0})}-\frac{P^{\prime\prime}(z_{0})}{P(z_{0})}-2\frac{u^{\prime}(z_{0})}{u(z_{0})}\frac{P^{\prime}(z_{0})}{P(z_{0})}=\left(\frac{\nu(g,r)}{z_{0}}\right)^{2}(1+o(1)),$

and

$\frac{u^{\prime\prime\prime}(z_{0})}{u(z_{0})}=\frac{g^{\prime\prime\prime}(z_{0})}{g(z_{0})}-\frac{P^{\prime\prime\prime}(z_{0})}{P(z_{0})}-3\frac{u^{\prime\prime}(z_{0})}{u(z_{0})}\frac{P^{\prime}(z_{0})}{P(z_{0})}-3\frac{u^{\prime}(z_{0})}{u(z_{0})}\frac{P^{\prime\prime}(z_{0})}{P(z_{0})}=\left(\frac{\nu(g,r)}{z_{0}}\right)^{3}(1+o(1)),$

for all sufficiently large $r\not\in F.$ Thus the equation (7) becomes

$\left(\frac{\nu(g,r)}{z_{0}}\right)^{2}(1+o(1))-\frac{3}{2}\left(\frac{\nu(g,r)}{z_{0}}(1+o(1))\right)^{2}=c^{\prime}(1+o(1)).$

This leads to

$\nu(r,g)\sim Ar\text{ and }\rho(g)=1.$

Hence $\rho(u)=1.$

By computing the Laurent expansions on both sides of (7), we may obtain

  * •

$u^{\prime}(z)=0$ if and only if $u(z)=\tau_{1}$ or $u(z)=\tau_{2}$.

  * •

all the zeros of $u^{\prime}$ are simple.

Without loss of generality, we may assume $\tau_{1}=1,\tau_{2}=-1.$ Thus,

$\displaystyle\frac{(u^{\prime})^{2}}{u^{2}-1}$

is a meromorphic function having only finitely many poles and no zeros. Noting
$\rho(u)=1$, we have

$\displaystyle Q^{2}(z)\frac{(u^{\prime})^{2}}{u^{2}-1}=e^{h(z)},$

where $Q(z)$ is a nonzero polynomial with simple zeros and $h(z)$ is an entire
function. Then, by Lemma 5, we have

$\displaystyle T(r,e^{h})$ $\displaystyle=$ $\displaystyle m(r,e^{h})$
$\displaystyle\leq$ $\displaystyle
2m\left(r,Q\right)+m\left(r,\frac{u^{\prime}}{u-1}\right)+m\left(r,\frac{u^{\prime}}{u+1}\right)$
$\displaystyle=$ $\displaystyle O(\log r).$

This implies $e^{h}$ is a polynomial and hence $h$ is a constant. Without loss
of generality, we may assume $h=1$, then $u$ satisfies the differential
equation

$u^{2}-Q(z)^{2}(u^{\prime})^{2}=1.$ (22)

If $\deg Q\geq 1$, then by the equation (22) and Lemma 2, we have

$\nu(r,u)\sim Ar^{1-\frac{\deg P}{2}},$

where $A$ is a positive number, but this contradicts with $\rho(u)=1.$ Hence
$Q(z)$ is a constant. It is easy to see that the solutions of (22) are of the
form

$u=\sin(\alpha z+\beta),$

where $\alpha,\beta$ are constants with $Q^{2}\alpha^{2}=-1.$ Substituting
$u=\sin(\alpha z+\beta)$ into (7) and noting $\tau_{1}=1,\tau_{2}=-1$, we
obtain that

$\alpha=\sqrt{2c},\quad\sigma_{1}=\sqrt{2}i,\quad\sigma_{2}=-\sqrt{2}i.$

Thus we get the conclusion.

Finally, if $u$ satisfies equation (8), then $R(f)$ must be a constant, say A,
and hence $c^{p}=A$. It is easy to check that $u(z)=e^{\alpha z}$ is a
solution of the equation (8), where $\alpha=\sqrt{-2c}$. Then it follows from
the invariance property of the Schwarzian derivative under Möbius
transformations that all the solutions of (8) are given by $u=\gamma(e^{\alpha
z})$, where $\gamma$ is a Möbius transformation and $\alpha=\sqrt{-2c}.$
Hence, in this case, all the solutions of the equation (2) are
$f(z)=\gamma(e^{\alpha z})$, where $\gamma$ is a Möbius transformation and
$\alpha=\sqrt{-2}A^{\frac{1}{2p}}.$

###  3.2 Proof of Theorem 2

Suppose $f$ is a locally injective transcendental meromorphic solution of the
equation (2). According to Theorem A, there exits a Möbius transformation
$\gamma_{1}$ such that $u=\gamma_{1}(f)$ is also a locally injective
transcendental meromorphic function and satisfies one of the equations
(3)-(8). Then it follows from Lemma 3 that $S(u,z)$ is entire. This implies
$u$ cannot satisfy any of the equations (3)-(7). Otherwise, $u$ has at least
one Picard exceptional value. By Theorem 1, it indicates that
$u=\gamma_{2}(\sin(\alpha z+\beta))$, where $\alpha,\beta$ are constants and
$\gamma_{2}$ is a Möbius transformation. Nevertheless, this contradicts with
the fact that $u$ is locally injective. As a consequence, $u$ can only satisfy
equation (8) and then the conclusion follows immediately from Theorem 1.

###  3.3 Proof of Theorem 3

Suppose $u$ is a transcendental meromorphic solution to the equation (9), then
Theorem 1 shows that $u$ must have infinitely many poles. By comparing the
Laurent expansions on both sides of the equation (9), we deduce that all the
poles of $u$ are simple and all the poles (if they exist) of $S(u,z)$ come
from the zeros of $u^{\prime}$. Since all the poles of $S(u,z)$ are double, it
follows that all zeros of $u^{\prime}$ should be simple, and at any zero of
$u^{\prime}$, $u(z)$ assumes one of the $\tau_{i},i=1,2,3,4$. This means any
zero of $u-\tau_{i}$ must be double. Therefore,

$G(z)=\frac{u^{\prime 2}}{(u-\tau_{1})(u-\tau_{2})(u-\tau_{3})(u-\tau_{4})}$
(23)

is a nonvanishing entire function, and there exists an entire function $g(z)$
such that $G=e^{g}$. According to Theorem 4, $u$ has finite order of growth.
Then we have

$\displaystyle T(r,e^{g})$ $\displaystyle=$ $\displaystyle m(r,e^{g})$
$\displaystyle\leq$ $\displaystyle
m\left(r,\frac{u^{\prime}}{(u-\tau_{1})(u-\tau_{2})}\right)+m\left(r,\frac{u^{\prime}}{(u-\tau_{3})(u-\tau_{4})}\right)$
$\displaystyle\leq$
$\displaystyle\sum_{i=1}^{4}m\left(r,\frac{u^{\prime}}{u-\tau_{i}}\right)+O(1)$
$\displaystyle=$ $\displaystyle O(\log r),$

where the last equality follows from Lemma 5. This implies $e^{g}$ is a
polynomial and hence $g=C$ is a constant. As a consequence, $u$ satisfies the
differential equation

$u^{\prime 2}=K(u-\tau_{1})(u-\tau_{2})(u-\tau_{3})(u-\tau_{4}),\quad K=e^{C}$

whose general solution is given by [3]

$\displaystyle u(z)$ $\displaystyle=$ $\displaystyle
K^{-1/2}\left(A-\frac{\wp^{\prime}(w;g_{2},g_{3})}{\wp(z-z_{0};g_{2},g_{3})-\wp(w;g_{2},g_{3})}\right)$
(24) $\displaystyle=$ $\displaystyle a-\frac{b}{\wp(z-z_{0};g_{2},g_{3})-d}$

where $\wp(z;g_{2},g_{3})$ is the Weierstrass elliptic function,
$z_{0}\in\mathbb{C}$ is arbitrary and $a,b,d,g_{2},g_{3}$ are constants that
depend on $K$ and $\tau_{i},i=1,2,3,4$. Finally, by substituting (24) into (9)
and applying the differential equation satisfied by $\wp(z;g_{2},g_{3})$

$\wp^{\prime 2}=4\wp^{3}-g_{2}\wp-g_{3},$

where $\Delta=g_{2}^{3}-27g_{3}^{2}\not=0$, it can be computed that $a$ should
be equal to one of the $\tau_{i},i=1,2,3,4$, and other parameters should
satisfy the relations (12)-(15).

###  3.4 Proof of Theorem 4

Let $u$ be a transcendental meromorphic solution to the equation (4), then
Theorem 1 implies that $u$ must have infinitely many poles. With similar
arguments as in Theorem 3, we find that

  * •

all the poles of $u$ are simple;

  * •

$u^{\prime}(z)=0$ if and only if $u(z)=\tau_{i}$ for some $i\in\\{1,2,3\\}$;

  * •

if $u(z)=\tau_{1}$, then $z$ is a simple zero of $u^{\prime}$;

  * •

if $u(z)=\tau_{2}$, then $z$ is a double zero of $u^{\prime}$;

  * •

if $u(z)=\tau_{3}$, then $z$ is a zero of $u^{\prime}$ of order $5$.

It follows that

$G(z)=\frac{u^{\prime 6}}{(u-\tau_{1})^{3}(u-\tau_{2})^{4}(u-\tau_{3})^{5}}$
(25)

is a nonvanishing entire function, and hence, there exists an entire function
$g(z)$ such that $G=e^{g}$. According to Theorem 4, $u$ has finite order of
growth. Then we have

$\displaystyle T(r,e^{g})$ $\displaystyle=$ $\displaystyle m(r,e^{g})$
$\displaystyle\leq$ $\displaystyle m\left(r,\frac{u^{\prime
3}}{(u-\tau_{1})^{3}(u-\tau_{2})^{3}(u-\tau_{3})^{3}}\right)+m\left(r,\frac{u^{\prime}}{u-\tau_{2}}\right)+m\left(r,\frac{u^{\prime
2}}{(u-\tau_{3})^{2}}\right)$ $\displaystyle\leq$ $\displaystyle
3m\left(r,\frac{u^{\prime}}{u-\tau_{1}}\right)+4m\left(r,\frac{u^{\prime}}{u-\tau_{2}}\right)+5m\left(r,\frac{u^{\prime}}{u-\tau_{3}}\right)+O(1)$
$\displaystyle=$ $\displaystyle O(\log r).$

This indicates that $g=C$ is a constant and hence $u$ satisfies the
differential equation

$u^{\prime 6}=K(u-\tau_{1})^{3}(u-\tau_{2})^{4}(u-\tau_{3})^{5},\quad
K=e^{C}.$ (26)

Since the elliptic curve parametrized by $u$ and $u^{\prime}$ has genus one,
the general solution of the above equation should be elliptic functions. Let

$u(z)=\frac{1}{v(z)}+\tau_{3},$ (27)

then the equation (26) reduces to

$v^{\prime 6}=K[(\tau_{1}-\tau_{3})v-1]^{3}[(\tau_{2}-\tau_{3})v-1]^{4}.$ (28)

By using the singularity methods (see [4, 11] and the references therein), we
find that the general solution to (28) reads

$v(z)=h-\frac{23328\left[6\wp(z-z_{0};g_{2},g_{3})^{3}+\wp^{\prime}(z-z_{0};g_{2},g_{3})^{2}\right]}{5K(\tau_{1}-\tau_{3})^{3}(\tau_{2}-\tau_{3})^{4}},$
(29)

where $\wp(z;g_{2},g_{3})$ is the Weierstrass elliptic function with
$g_{2}=0$, $z_{0}\in\mathbb{C}$ is arbitrary and $h,g_{3}$ are constants
depending on $K$ and $\tau_{i},i=1,2,3$. Finally, with
$\tau_{1}=4,\tau_{2}=-3,\tau_{3}=0$, substituting (27) and (29) into (4)
yields the solution of equation (4)

$u(z)=-\frac{3c}{c-74088\wp\left(z-z_{0};0,g_{3}\right)^{3}},$

where $g_{3}=c/10584$ and $z_{0}\in\mathbb{C}$ is arbitrary, provided that

$\displaystyle\\{\sigma_{1},\sigma_{2}\\}=\left\\{\sqrt{5}i,-\sqrt{5}i\right\\}.$

This completes the proof.

###  3.5 Proof of Theorem 5

Suppose $u$ is a transcendental meromorphic solution to the equation (5), then
Theorem 1 implies that $u$ must have infinitely many poles. Using similar
arguments as in Theorem 3, we can show that

  * •

all the poles of $u$ are simple;

  * •

$u^{\prime}(z)=0$ if and only if $u(z)=\tau_{i}$ for some $i\in\\{1,2,3\\}$;

  * •

all the zeros of $u^{\prime}$ are double.

It follows that

$G(z)=\frac{u^{\prime 3}}{(u-\tau_{1})^{2}(u-\tau_{2})^{2}(u-\tau_{3})^{2}}$
(30)

is a nonvanishing entire function, and hence, there exists an entire function
$g(z)$ such that $G=e^{g}$. Since the order of $u$ is finite, we have

$\displaystyle T(r,e^{g})$ $\displaystyle=$ $\displaystyle m(r,e^{g})$
$\displaystyle\leq$ $\displaystyle
m\left(r,\frac{u^{\prime}}{u-\tau_{1}}\right)+m\left(r,\frac{u^{\prime}}{(u-\tau_{2})(u-\tau_{3})}\right)+m\left(r,\frac{u^{\prime}}{(u-\tau_{1})(u-\tau_{2})(u-\tau_{3})}\right)$
$\displaystyle\leq$ $\displaystyle
2\left[\sum_{i=1}^{3}m\left(r,\frac{u^{\prime}}{u-\tau_{i}}\right)\right]+O(1)$
$\displaystyle=$ $\displaystyle O(\log r).$

This implies that $g=C$ is a constant and hence $u$ satisfies the differential
equation

$u^{\prime 3}=K(u-\tau_{1})^{2}(u-\tau_{2})^{2}(u-\tau_{3})^{2},\quad
K=e^{C}.$ (31)

Since the elliptic curve parametrized by $u$ and $u^{\prime}$ has genus one,
the general solution of the above equation should be elliptic functions. By
using the singularity methods, we find that the general solution to (34) can
be expressed as

$\displaystyle u(z)$ $\displaystyle=$
$\displaystyle\frac{1}{L}\left(\frac{\left(1+i\sqrt{3}\right)\left(\wp^{\prime}\left(z-z_{0};g_{2},g_{3}\right)-A_{1}\right)}{4\left(\wp\left(z-z_{0};g_{2},g_{3}\right)-B_{1}\right)}+\frac{\left(1-i\sqrt{3}\right)\left(\wp^{\prime}\left(z-z_{0};g_{2},g_{3}\right)-A_{2}\right)}{4\left(\wp\left(z-z_{0};g_{2},g_{3}\right)-B_{2}\right)}\right)$
(32) $\displaystyle+\frac{1}{3}(\tau_{1}+\tau_{2}+\tau_{3})$

where $L^{3}=K$, $\wp(z;g_{2},g_{3})$ is the Weierstrass elliptic function,
$z_{0}\in\mathbb{C}$ is arbitrary and $A_{1},A_{2},B_{1},B_{2},g_{2},g_{3}$
are constants depending on $K$ and $\tau_{i},i=1,2,3$. Finally, with
$\\{\tau_{1},\tau_{2},\tau_{3}\\}=\\{0,1,-1\\}$, substituting (32) into (5)
yields that

$\displaystyle\\{\sigma_{1},\sigma_{2}\\}=\left\\{\frac{i}{\sqrt{3}},-\frac{i}{\sqrt{3}}\right\\},\quad
A_{1}=A_{2}=g_{2}=0,\quad g_{3}=\frac{c}{432},$ $\displaystyle
B_{1}=\frac{1}{18}\left(1-i\sqrt{3}\right)L^{2},\quad
B_{2}=\frac{1}{18}\left(1+i\sqrt{3}\right)L^{2},\quad L^{6}=-\frac{27}{64}c.$

In this case, the equation (5) reduces to

$\displaystyle S(u,z)^{3}=c\frac{(u^{2}+1/3)^{3}}{u^{2}(u^{2}-1)^{2}},$

and the solution (32) becomes

$u(z)=\frac{9\left[9\wp\left(z;g_{2},g_{3}\right)+L^{2}\right]\wp^{\prime}\left(z;g_{2},g_{3}\right)}{2L\left[81\wp\left(z;g_{2},g_{3}\right)^{2}-9L^{2}\wp\left(z;g_{2},g_{3}\right)+L^{4}\right]}$

where $L^{6}=-27c/64,g_{2}=0,g_{3}=c/432$.

###  3.6 Proof of Theorem 6

Let $u$ be a transcendental meromorphic solution to the equation (6), then
Theorem 1 implies that $u$ must have infinitely many poles. With similar
arguments as in Theorem 3, we find that

  * •

all the poles of $u$ are simple;

  * •

$u^{\prime}(z)=0$ if and only if $u(z)=\tau_{i}$ for some $i\in\\{1,2,3\\}$;

  * •

if $u(z)=\tau_{1}$, then $z$ is a simple zero of $u^{\prime}$;

  * •

if $u(z)=\tau_{j},j=2,3$, then $z$ is a triple zero of $u^{\prime}$;

It follows that

$G(z)=\frac{u^{\prime 4}}{(u-\tau_{1})^{2}(u-\tau_{2})^{3}(u-\tau_{3})^{3}}$
(33)

is a nonvanishing entire function, and hence, there exists an entire function
$g(z)$ such that $G=e^{g}$. As the order of $u$ is finite, we have

$\displaystyle T(r,e^{g})$ $\displaystyle=$ $\displaystyle m(r,e^{g})$
$\displaystyle\leq$ $\displaystyle m\left(r,\frac{u^{\prime
2}}{(u-\tau_{1})^{2}(u-\tau_{2})^{2}(u-\tau_{3})^{2}}\right)+m\left(r,\frac{u^{\prime}}{u-\tau_{2}}\right)+m\left(r,\frac{u^{\prime}}{u-\tau_{3}}\right)$
$\displaystyle\leq$ $\displaystyle
2m\left(r,\frac{u^{\prime}}{u-\tau_{1}}\right)+3m\left(r,\frac{u^{\prime}}{u-\tau_{2}}\right)+3m\left(r,\frac{u^{\prime}}{u-\tau_{3}}\right)+O(1)$
$\displaystyle=$ $\displaystyle O(\log r).$

This indicates that $g=C$ is a constant and hence $u$ satisfies the
differential equation

$u^{\prime 4}=K(u-\tau_{1})^{2}(u-\tau_{2})^{3}(u-\tau_{3})^{3},\quad
K=e^{C}.$ (34)

Since the elliptic curve parametrized by $u$ and $u^{\prime}$ has genus one,
the general solution of the above equation should be elliptic functions. Then
the singularity methods indicate that the general solution to (34) can be
expressed as

$\displaystyle u(z)$ $\displaystyle=$ $\displaystyle
h+\frac{1}{2L}\frac{\wp^{\prime}\left(z-z_{0};g_{2},g_{3}\right)-A_{1}}{\wp\left(z-z_{0};g_{2},g_{3}\right)-B_{1}}+$
(35)
$\displaystyle\frac{i}{2L}\left(\frac{\wp^{\prime}\left(z-z_{0};g_{2},g_{3}\right)-A_{2}}{\wp\left(z-z_{0};g_{2},g_{3}\right)-B_{2}}-\frac{\wp^{\prime}\left(z-z_{0};g_{2},g_{3}\right)-A_{3}}{\wp\left(z-z_{0};g_{2},g_{3}\right)-B_{3}}\right)$

where $L^{4}=K$, $\wp(z;g_{2},g_{3})$ is the Weierstrass elliptic function,
$z_{0}\in\mathbb{C}$ is arbitrary and $A_{j},B_{j},g_{2},g_{3}$ are constants
depending on $K$ and $\tau_{j},j=1,2,3$. Finally, with
$\tau_{1}=0,\tau_{2}=1,\tau_{3}=-1$, substituting (35) into (6) yields that

$\displaystyle\\{\sigma_{1},\sigma_{2}\\}=\left\\{\frac{i}{2},-\frac{i}{2}\right\\},\quad
g_{2}=-\frac{c}{36},\quad B_{2}=-B_{3}=\frac{L^{2}}{8}i,$ $\displaystyle
c=\frac{9}{4}L^{4},\quad A_{1}=A_{2}=A_{3}=B_{1}=g_{3}=h=0.$

In this case, the equation (5) reduces to

$\displaystyle S(u,z)^{2}=c\frac{(u^{2}+1/4)^{2}}{u^{2}(u^{2}-1)},$

and the solution (35) becomes

$u(z)=-\frac{1}{2L}\frac{\left(8\wp\left(z-z_{0};g_{2},g_{3}\right)+L^{2}\right)^{2}\wp^{\prime}\left(z-z_{0};g_{2},g_{3}\right)}{\wp\left(z-z_{0};g_{2},g_{3}\right)\left(64\wp\left(z-z_{0};g_{2},g_{3}\right)^{2}+L^{4}\right)},$

where $g_{2}=-c/36,g_{3}=0$ and $c=9L^{4}/4$. This completes the proof.

######  Remark 6.

Since elliptic functions are of order $2$, Theorems 3-6 indicate that the
estimate on the growth of meromorphic solutions of the equation (2) given in
Lemma 4 is sharp.

###  3.7 Examples

We present some examples to illustrate all the possible configurations of the
transcendental meromorphic solutions given in Theorem 3.

######  Example 1.

The Schwarzian differential equation

$\displaystyle
S(u,z)=\frac{3\left(25u^{4}+20u^{3}+14u^{2}+4u+1\right)}{2u(u-1)(u+1)\left(3u+1\right)}$

has the solution

$u(z)=\frac{1}{\wp(z-z_{0};g_{2},g_{3})-1},$ (36)

where $z_{0}\in\mathbb{C}$ is arbitrary, $g_{2}=16$ and $g_{3}=0.$

######  Example 2.

The Schwarzian differential equation

$\displaystyle
S(u,z)=\frac{3\left(25u^{4}+20u^{3}+14u^{2}+4u+1\right)}{u(u-1)(u+1)\left(3u+1\right)}$

admits the solution

$u(z)=1-\frac{16}{\wp(z-z_{0};g_{2},g_{3})+12},$ (37)

where $z_{0}\in\mathbb{C}$ is arbitrary, $g_{2}=64$ and $g_{3}=0.$

######  Example 3.

The Schwarzian differential equation

$\displaystyle
S(u,z)=-\frac{3\left(25u^{4}+20u^{3}+14u^{2}+4u+1\right)}{u(u-1)(u+1)(3u+1)}$

has the solution

$u(z)=-1-\frac{8}{\wp(z-z_{0};g_{2},g_{3})-8},$ (38)

where $z_{0}\in\mathbb{C}$ is arbitrary, $g_{2}=64$ and $g_{3}=0.$

######  Example 4.

The Schwarzian differential equation

$\displaystyle
S(u,z)=\frac{3\left(225u^{4}+180u^{3}+126u^{2}+36u+9\right)}{u(u-1)(u+1)(3u+1)},$

admits the solution

$u(z)=-\frac{1}{3}-\frac{16}{\wp(z-z_{0};g_{2},g_{3})-12},$ (39)

where $z_{0}\in\mathbb{C}$ is arbitrary, $g_{2}=5184$ and $g_{3}=0.$

## Acknowledgement

We would like to thank Robert Conte for the helpful discussions.

## Funding

The first author was supported by the National Natural Science Foundation of
China (Grant No. 11671191). The second author was supported by the National
Natural Science Foundation of China (Grant Nos. 11701382 and 11971288).

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Department of Mathematics

Nanjing University

Nanjing, China

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Institute for Advanced Study

Shenzhen University

Shenzhen, China

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