Chazy equation and movable singularity Given the non-linear ODE $$f'''-ff''+\frac{3}{2}(f')^2=0$$ has the Eisenstein series $E_2$ as a solution.
I want to know what is so special about this ODE. Wikipedia says that this is 

an example of a third-order differential equation with a movable singularity that is a natural boundary for its solutions

and 

Acting on this [$E_2$] solution by the group $\text{SL}_2(\mathbb{Z})$ gives a 3-parameter family of solutions.

I want to understand what this means in details. 
Is there somebody who can explain it to me?
Thank you.
 A: Caution: I have included animated plots. GIF is not particularly effective
at compressing those, so this page may take long to load.
Let us get rid of some inaccuracies first. The ODE
$$2f''' - 2f f'' + 3{f'}^2 = 0,\qquad
(\ )' = \frac{\mathrm{d}(\ )}{\mathrm{d}z}\tag{*}$$
is not fulfilled by $f(z) = \operatorname{E}_2(z)$ but by
$f(z) = 2\pi\mathrm{i}\operatorname{E}_2(z)
= \frac{\mathrm{d}}{\mathrm{d}z}\ln\Delta(z)$
where $\Delta(z)$ is the modular discriminant.
You can have that factor $2\pi\mathrm{i}$ absorbed
by rescaling the differentiation operator:
The ODE
$$2\dddot{u} - 2u \ddot{u} + 3\dot{u}^2 = 0,\qquad
\dot{(\ )} = \frac{1}{2\pi\mathrm{i}}\frac{\mathrm{d}(\ )}{\mathrm{d}z}
\tag{**}$$
is indeed fulfilled by $u(z) = \operatorname{E}_2(z)$.
Using $\dot{(\ )}$ instead of $(\ )'$ seems quite natural
when dealing with the associated $q$-series: In fact,
$$\dot{(\ )} = q\frac{\mathrm{d}(\ )}{\mathrm{d}q}
\quad\text{for}\quad q = \exp(2\pi\mathrm{i}z)$$
You can deduce $(**)$ from Ramanujan's system of differential equations by eliminating $\operatorname{E}_4$ and $\operatorname{E}_6$ via
differentiation.
In this post I will continue using $(\ )'$ instead of $\dot{(\ )}$
and focus on solutions of $(*)$.
Consider $a,b,c,d\in\mathbb{Z}$ with $ad-bc=1$.
(We will generalize that later.)
Then the modular symmetries of $\Delta$ imply
$$\Delta(z) = \frac{\Delta\!\left(\frac{az+b}{cz+d}\right)}{(cz+d)^{12}}$$
Taking the logarithmic derivative, we obtain
$$2\pi\mathrm{i}\operatorname{E}_2(z)
= 2\pi\mathrm{i}\frac{ad-bc}{(cz+d)^2}
\operatorname{E}_2\!\left(\frac{az+b}{cz+d}\right)
- \frac{12c}{cz+d}$$
Now suppose we relax the parameter constraints
to $a,b,c,d\in\mathbb{C}$ with $ad-bc\neq0$, but keep calling the
above right-hand side $f(z)$, now implicitly depending on $a,b,c,d$:
$$f(z) = 2\pi\mathrm{i}\frac{ad-bc}{(cz+d)^2}
\operatorname{E}_2\!\left(\frac{az+b}{cz+d}\right)
- \frac{12c}{cz+d}\tag{1}$$
This parameterization via $a,b,c,d$ has three complex degrees of freedom:
There are four parameters, but multiplying all of them with the same
nonzero factor does not change $f$ at all, therefore only three degrees of
freedom. This is the right amount of flexibility to be expected from solutions
to a third-order ODE, and so we might suspect that $(1)$ fulfills $(*)$
not only for $\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)
\in\operatorname{SL}_2(\mathbb{Z})$
but for the continuous parameter family
$\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)
\in\operatorname{PGL}_2(\mathbb{C})$.
This is indeed the case, as can be verified directly, although some background
theory e.g. about the
Schwarzian derivative
would certainly help. I will not cover that here.
This brings us to the second inaccuracy to be addressed:
The Wikipedia quote

Acting on this solution by the group $\operatorname{SL}_2$
  gives a 3-parameter family of solutions.

can be interpreted as meaning $\operatorname{SL}_2(\mathbb{C})$,
or rather $\operatorname{PGL}_2(\mathbb{C})$,
the group of complex fractional linear transforms.
In some applications, it might be reasonable to restrict the base
field to $\mathbb{R}$, so the Wikipedia snippet's author has chosen wisely
to omit such detail. The important point is that the group is
continuous here, not discrete like $\operatorname{SL}_2(\mathbb{Z})$.
Sidenote: $(1)$ is not the only solution to $(*)$; in particular,
replacing $\operatorname{E}_2(\cdots)$ with a constant in $(1)$
also yields a solution to $(*)$.
Now there is a caveat.
Let us write $\tau = \frac{az+b}{cz+d}$.
First thing to notice is that we need $z\neq-\frac{d}{c}$ if $c\neq0$,
so the domain of $f$ according to $(1)$ might be punctured. Moreover,
since $\operatorname{E}_2(\tau)$ is defined only for $\Im\tau>0$,
the variable $z = \frac{d\tau-b}{-c\tau+a}$ must be taken from
a certain fractional linear transformation of the upper half-plane.
If $a,b,c,d$ were real and $ad-bc>0$, that transformation would yield
the same upper half-plane, but for $ad-bc<0$ or generally complex $a,b,c,d$,
the domain of $f(z)$ may differ from the upper half-plane. It might be


*

*a translated and rotated half-plane (if $c=0$),

*the interior of a circle,

*the exterior of a circle,

*the entire complex plane if $\operatorname{E}_2$ is replaced with a constant,

*any of the above, without one point $-\frac{d}{c}$ (if $c\neq 0$).


So the domain of the solution depends on $a,b,c,d$.
Apart from the puncture, the domain's boundary is the fractional-linearly
transformed real line where $\operatorname{E}_2$ runs into singularities.
Thus, the ODE $(*)$ has solutions with movable singularities that form
a natural boundary.
Let us denote the domain of $f$ as
$$D = \left\{z\in\mathbb{C}\setminus\left\{-\frac{d}{c}\right\}
\colon\ \Im\frac{az+b}{cz+d} > 0\right\}
= \left\{\frac{d\tau-b}{-c\tau+a}
\middle|\tau\in\mathbb{H}\setminus\left\{\frac{a}{c}\right\}\right\}$$
In the remainder of this post, I want to show some plots of $f$ with varying
$a,b,c,d$, indicating the domain $D$.
The plots color-code the phase, i.e. $\arg f(z)$,
and do not contain coordinate annotations.
Ranges should be clear enough from the context,
e.g. plots for $D=\mathbb{H}$ have the real axis as lower boundary.
Basic things first:


*

*$f(z)$ for $\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)
  = \bigl(\begin{smallmatrix}1 & 0\\0 & 1\end{smallmatrix}\bigr)$,
so $f(z) = 2\pi\mathrm{i}\operatorname{E}_2(z)$, $D = \mathbb{H}$:
$\quad
  \begin{array}[b]{c}\text{Colormap}\\\text{(phase only):}\end{array}\quad
  $
You can see some of the symmetries of $\operatorname{E}_2$:
It is singly periodic and has a fractal structure.
The color vortex points in the interior of the half-plane are simple zeros
of $\operatorname{E}_2$.
The half-plane's boundary is crammed with singularities.
One of the easiest variations of $f$ that keeps fulfilling $(*)$
is achieved by translating $z$:


*

*$f(z)$ for $\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)
  = \bigl(\begin{smallmatrix}1 & -t\\0 & 1\end{smallmatrix}\bigr),\ 
  t\in\mathbb{R},\ D = \mathbb{H}$:

Now this is boring.
But compose this with a transformation that maps the unit disk to
the upper half-plane, and you get the following:


*

*$f(z)$ for $\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)
  = \bigl(\begin{smallmatrix}1 & -t\\0 & 1\end{smallmatrix}\bigr)
  \bigl(\begin{smallmatrix}1 & \mathrm{i}
  \\\mathrm{i} & 1\end{smallmatrix}\bigr),\ t\in\mathbb{R},
  \ D = \{z\in\mathbb{C}\colon\ \vert z\vert < 1\}$:

The north pole of that disk does not move because it corresponds to
the half-plane's idea of infinity.
However, the north pole can be moved as well:


*

*$f(z)$ for $\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)
  = \bigl(\begin{smallmatrix}1 & 0\\t & 1\end{smallmatrix}\bigr)
  \bigl(\begin{smallmatrix}1 & \mathrm{i}
  \\\mathrm{i} & 1\end{smallmatrix}\bigr),\ t\in\mathbb{R},
  \ D = \{z\in\mathbb{C}\colon\ \vert z\vert < 1\}$:

Back in the half-plane, this looks like the following:


*

*$f(z)$ for $\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)
  = \bigl(\begin{smallmatrix}1 & 0\\t & 1\end{smallmatrix}\bigr),
  \ t\in\mathbb{R},\ D = \mathbb{H}$:

This nicely demonstrates Farey-type relations among the rationals in the
boundary.
So far I have used variations of $a,b,c,d$ that leave the domain invariant.
The trick was to compose a variable transformation from
$\operatorname{SL}_2(\mathbb{R})$ with a fixed complex transformation
from $\operatorname{PGL}_2(\mathbb{C})$.
Varying the complex transformation can be used to animate domain changes,
e.g. like this:


*

*$f(z)$ for various
$\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)
  \in\operatorname{PGL}_2(\mathbb{C})$:

(There is an image size limit of 2MB here, so I cannot upload a larger
version.)
This summarizes the previous animations and directly demonstrates moving
boundaries.
Remember that each frame shows a valid solution to the Chazy ODE $(*)$.
And we have barely scratched the surface of the solution's
parameter space with its three complex dimensions.
