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.
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
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
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 $f(z) = 2\pi\mathrm{i}\operatorname{E}_2(z)$, $D = \mathbb{H}$:
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.