# When are attracting sets invariant?

Consider a control system of the form $\dot{x}(t) = f(x(t), u(t))$ where $u(t)$ is the control input, $x \in \mathbb{R}^{n}$, $u \in \mathbb{R}^{m}$. Assume $f$ is Lipschitz continuous so that the existence and uniqueness of solutions holds.

Following are the definitions of attracting sets and invariant sets that I use:

• A set $S$ is (control) invariant if for any initial state $x(0) \in S$, there exists a control signal $u(\cdot)$ such that $x(t) \in S$ for all $t \geq 0$.
• A set $A$ is (weakly) attracting with basin of attraction $B$ if for any initial state $x(0) \in B$, there exists a control signal $u(\cdot)$ such that $x(t)$ converges to $A$ as $t \to +\infty$, that is $\lim_{t \to +\infty} \operatorname{dist} (x(t), A) = 0$.

It seems to me that attracting sets and invariant sets are closely related, in particular when $B = \mathbb{R}^{n}$ then in many cases, $A$ is also invariant. My questions are:

1. Is there any example in which an attracting set is not invariant?
2. Under what condition an attracting set is invariant?

I come from the control community and not many control textbooks mention these concepts. If there are any good (math) books that discuss these sets well, please let me know.

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This is an answer to 1. In two dimensions ($n=2$) consider the system $f(x,u)=(1,-x_2)$ (here $x=(x_1,x_2)$ and $f$ does not depend on $u$ at all. All solutions are of the form $x(t)=(t,ce^{-t})$. The curve $A=\{(x_1,x_2) : x_2=e^{-x_1}\sin x_1\}$ is an attracting set because $\mathrm{dist}(x(t),A)\to 0$ as $t\to\infty$ for every solution. Yet, the set is not invariant because a solution cannot stay within $A$: for one thing, the solutions have a constant sign of $x_2$ while $A$ wiggles between $x_2<0$ and $x_2>0$.
This example can also be modified to include nontrivial control: $f(x,u)=(2+\sin u,-x_2)$, where $u\in \mathbb R^1$ is our control. No matter what $u$ we use, two things happen: $x_1(t)$ increases to $+\infty$, and $x_2(t)$ tends to $0$ while keeping its sign. In particular, the solution cannot stay within $A$.