Creating equilibria/stability by adding a time-periodic term to an ODE?

Question

Let $f\colon \Bbb R^N\to \Bbb R^N$, $g\colon[0,\infty)\to\Bbb R^N$ and consider the ODE $$ \dot x(t) = f(x(t)) + g(t), $$ where space- and time-dependence are additively separated.

Is it possible to choose $f$ such that the time-homogeneous equation with $g\equiv 0$ has no (stable) equilibria or periodic orbits, while the equation with some periodic $g\not\equiv 0$ does have (stable) equilibria or periodic orbits?

The idea is to interpret $g$ as an external input that induces (stable) periodic behavior in an otherwise "equilibrium-free"/unstable system.

Answer 1

It's relatively easy to give examples for your conjectures -- the examples can seem a bit trivial, but carry a lot of general insight.

First, to build a system without equilibria, we can just demand that $f_i(x) = 1$ for some $1 \leq i \leq N$. Then, choosing $g_i(x) = -1$ (which is constant, so in particular periodic) negates this obstacle to having equilibria.

Based on this idea, let's take the following one-dimensional example (which can be copied to more dimensions, obviously) $$ f(x) = 1,\quad g(t) = -1+\cos(t), $$ where the full system has solutions $x(t) = x_0 + \sin(t)$; moreover, every solution is stable (albeit not asymptotically stable).

NB. If you allow $g$ to be state-dependent, i.e. $g(x,t)$, it's quite easy to extend the above example to produce a unique, asymptotically stable periodic orbit by choosing $g(x,t) = -1 + \cos(t) - a x$, with any $a > 0$.

Answer 2

I have found a nice and very intuitive example which illustrates exactly what I had in mind (even though I have to admit that my question was too vague for someone else to grasp that).

Let $N=2$ and set

$$ f(x)=- \big(|x|^2-1\big) x, \quad x \in\Bbb R^2. $$

This refers to a particle that is subject to a mexican hat potential. The top of the hat is located at $x=0$ and the bottom of the hat is exactly $\Bbb S^1$, the boundary of the unit ball centered at zero. Obviously, the set of equilibria for the corresponding system with $g\equiv 0$ is exactly $\Bbb S^1 \cup \{0\}$, but none of these equilibria are asymptotically stable ($0$ is definitely not stable at all).

If we set

$$g(t)= \begin{pmatrix} \cos(t) \\ \sin(t) \end{pmatrix}, \quad t\in[0,\infty),$$

this signal "does nothing on average" in the sense that its integral over one period is zero. Intuitively speaking, it refers to rotating the mexican hat periodically in time. While the origin is no longer an equilibrium for the corresponding system with this choice of $g$ at all, the curve of equilibria $\Bbb S^1$ is turned into a periodic orbit which is in fact globally asymptotically stable. Checking this rigorously takes a little effort, but on an intuitive level, this is immediately plausible.