Dynamical system with vanishing perturbation

Question

Consider the following system:

$$ \dot{x} = f(x)+h(y),\\ \dot{y} = g(y), $$

with $x,y\in\mathbb{R}^n$, and where $f$, $g$, and $h$ are globally continuously differentiable.

We know that $y$ is asymptotically stable (i.e., for any $\epsilon>0$ there exists a $\delta>0$ such that, for any $||y(0)||\le\delta$, we have $||y(t)||\le\epsilon$ for all $t\le0$, and $y(t)\to0_n$ for $t\to\infty$).

Can we say that the solutions of $\dot{x}=f(x)+h(y)$ converge to the solutions of $\dot{x}=f(x)+h(0_n)$? Why?

Answer 1

I believe this is not the case.

Consider the following (linear) example $$ \dot x = 3 + y,$$ $$ \dot y = -3 y.$$ All functions are smooth.

However, solving you get that $y(t) = y_0\mathrm{e}^{-3t}$; and for $x$ you get that $$ \dot x = 3 + y_0\mathrm{e}^{-3t}$$ $$ x(t) = x_0 + 3t + y_0/3 - \mathrm{e}^{-3t}/3$$ whereas, solving $\dot x = 3$ you get $$ \tilde x(t) = x_0 + 3t$$ and $x(t) - \tilde x(t)$ will not vanish as $t\to \infty$.