An ODE with smooth, bounded solutions that computes a discontinous function in finite time

I'd like to share an Ordinary Differential Equation (ODE) that I find entertaining.

For initial condition $x(0)=x_0\in \mathbb{R}$, the following ODE converges to $\mathrm{sign}(x_0)$ in finite time: $$\begin{array}{}\text{(1)}& \begin{array}{rl}\tau \dot{x}& =\mathrm{sq}\{x-x^3\},\end{array}\end{array}$$ where $\tau$ is a time constant controlling how quickly the ODE evolves and $\mathrm{sq}(\cdot )$ is the signed square root function $\mathrm{sq}(a)=\mathrm{sign}(\mathrm{a})\cdot \sqrt{|a|}$ . If we set the time constant to $\tau =\sqrt{8\pi }/\mathrm{\Gamma }(\frac{1}{4}{)}^2$, we find that all $x_0$ reach $\mathrm{sign}(x_0)$ by time $t=1$.

Figure 1a shows the vector field for Equation $(1)$. Note that it's slope is vertical (infinite) at $x\in \{-1,0,1\}$ and so the vector field is not Lipschitz continuous. This is required for finite-time convergence, but removes the uniqueness of solutions. In particular, you cannot continue the final-value problem back in time from $x_f\in \{-1,0,1\}$.

Derivation from polar form

Equation $(1)$ can be derived from the following ODE on a circular variable $\theta$ $$\begin{array}{}\text{(2)}& \begin{array}{rl}\dot{\theta }& =\frac{1}{2}\mathrm{sq}\{\sin (2\theta )\},\end{array}\end{array}$$ where $\theta \in [0,\pi )$ and $x\in \mathbb{R}$. Equation $(2)$ will pull $\theta$ toward $\theta \mathrm{mod}\pi =\pm \pi /2$, or leave it stationary at the unstable fixed points $\theta \mathrm{mod}\pi =0$. We apply the change of variables $x=\tan (\frac{\theta }{2})$ to Equation $(2)$. Note that $\theta =2{\tan }^{-1}(x)$ and $\frac{dx}{d\theta }=\frac{1}{2}(1+x^2)$. This yields: $$\dot{x}=\frac{1}{4}(1+x^2)\mathrm{sq}\{4\frac{x-x^3}{(1+x^2{)}^2}\}.$$ Most terms cancel if we expand the signed square root: $$\dot{x}=\frac{1}{4}(1+x^2)\sqrt{\left|4\frac{x-x^3}{(1+x^2{)}^2}\right|}\cdot \mathrm{sign}\{4\frac{x-x^3}{(1+x^2{)}^2}\},$$ and we recover Equation $(1)$ if we cancel the $1+x^2$ term and drop $4/(1+x^2{)}^2$ from inside the $\mathrm{sign}(\cdot )$ term (since it is always positive): $$\begin{array}{}\text{(1)}& \begin{array}{rl}\dot{x}& =\sqrt{\left|x-x^3\right|}\cdot \mathrm{sign}\{x-x^3\}=\mathrm{sq}\{x-x^3\}.\end{array}\end{array}$$

The peculiar time constant $\tau =\sqrt{8\pi }/\mathrm{\Gamma }(\frac{1}{4}{)}^2$ can be solved as the maximum convergence time for the polar form $(2)$ using Elliptic integrals.

Remarks

This ODE is notable in that the mapping $x_0\mapsto x(1)=\mathrm{sign}(x_0)$ computes a discontinuous function in finite time, with all solutions remaining smooth. It also misbehaves in some other interesting ways and is a nice pathological example of what can happen if your vector field isn't Lipschitz. My main motivation for deriving this was to find an ODE whose solutions are smooth and bounded, but for which backpropagation through time fails due to gradients vanishing.