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Suppose I have a discrete dynamical system with a finite set X of states, and suppose I want to prove that every state of X ends up, sooner or later, in a subset Z under the dynamics of the system. Then a natural proof strategy is to break this "convergence" statement into two parts, by first showing that every state of X eventually ends up in a subset Y, and then showing that every state in Y ends up in Z (with Z $\subset$ Y $\subset$ X). This might simplify the problem quite a bit.

My question is,

Under what conditions will the above proof strategy work for a continuous dynamical system $\dot{x}=f(x)$ ($x \in R^n$)?

The main issue I have is that it might take infinite time to reach Y. To see an example where things go wrong if $f$ is not continuous, consider the example

• $\dot{x} = -1-x$, $\quad$ for $x<-1$,
• $\dot{x}=-x$, $\quad\quad$ for $-1 \le x \le 1$, $\quad$ and
• $\dot{x}=1-x$, $\quad$ for $x>1$.

In this case every $x$ in $[-1,1]$ converges to zero, and every point outside $[-1,1]$ converges to either $-1$ or $1$, but it takes infinite time to do so and it is not true that all points in $R$ converge to zero.

Smoothness of $f$ might be enough to rule out this behavior; I am unable to construct a counterexample with $f$ continuous. Is it actually enough, and if so, why?

Suppose I have a discrete dynamical system with a finite set X of states, and suppose I want to prove that every state of X ends up, sooner or later, in a subset Z under the dynamics of the system. Then a natural proof strategy is to break this "convergence" statement into two parts, by first showing that every state of X eventually ends up in a subset Y, and then showing that every state in Y ends up in Z (with Z $\subset$ Y $\subset$ X). This might simplify the problem quite a bit.

My question is,

Under what conditions will the above proof strategy work for a continuous dynamical system $\dot{x}=f(x)$ ($x \in R^n$)?

The main issue I have is that it might take infinite time to reach Y. To see how things may go wrong if $f$ is not continuous, consider the following example:

• $\dot{x} = -1-x$, $\quad$ for $x<-1$,
• $\dot{x}=-x$, $\quad\quad$ for $-1 \le x \le 1$, $\quad$ and
• $\dot{x}=1-x$, $\quad$ for $x>1$.

In this case every $x$ in $[-1,1]$ converges to zero, and every point outside $[-1,1]$ converges to either $-1$ or $1$, but it takes infinite time to do so and it is not true that all points in $R$ converge to zero.

Smoothness of $f$ might be enough to rule out this behavior; I am unable to construct a counterexample with $f$ continuous. Is it actually enough, and if so, why?

Suppose I have a discrete dynamical system with a finite set X of states, and suppose I want to prove that every state of X ends up, sooner or later, in a subset Z under the dynamics of the system. Then a natural proof strategy is to break this "convergence" statement into two parts, by first showing that every state of X eventually ends up in a subset Y, and then showing that every state in Y ends up in Z (with Z $\subset$ Y $\subset$ X). This might simplify the problem quite a bit.

My question is,

Under what conditions will the above proof strategy work for a continuous dynamical system $\dot{x}=f(x)$ ($x \in R^n$)?

The main issue I have is that it might take infinite time to reach Y. To see an example where things go wrong if $f$ is not continuous, consider the example

• $\dot{x} = -1-x$, $\quad$ for $x<-1$,
• $\dot{x}=-x$, $\quad\quad$ for $-1 \le x \le 1$, $\quad$ and
• $\dot{x}=1-x$, $\quad$ for $x>1$.

In this case every $x$ in $[-1,1]$ converges to zero, and every point outside $[-1,1]$ converges to either $-1$ or $1$, but it takes infinite time to do so and it is not true that all points in $R$ converge to zero.

Smoothness of $f$ might be enough to rule out this behavior; I am unable to construct a counterexample with $f$ continuous. Is it actually enough?

Suppose I have a discrete dynamical system with a finite set X of states, and suppose I want to prove that every state of X ends up, sooner or later, in a subset Z under the dynamics of the system. Then a natural proof strategy is to break this "convergence" statement into two parts, by first showing that every state of X eventually ends up in a subset Y, and then showing that every state in Y ends up in Z (with Z $\subset$ Y $\subset$ X). This might simplify the problem quite a bit.

My question is,

Under what conditions will the above proof strategy work for a continuous dynamical system $\dot{x}=f(x)$ ($x \in R^n$)?

The main issue I have is that it might take infinite time to reach Y. To see an example where things go wrong if $f$ is not continuous, consider the example

• $\dot{x} = -1-x$, $\quad$ for $x<-1$,
• $\dot{x}=-x$, $\quad\quad$ for $-1 \le x \le 1$, $\quad$ and
• $\dot{x}=1-x$, $\quad$ for $x>1$.

In this case every $x$ in $[-1,1]$ converges to zero, and every point outside $[-1,1]$ converges to either $-1$ or $1$, but it takes infinite time to do so and it is not true that all points in $R$ converge to zero.

Smoothness of $f$ might be enough to rule out this behavior; I am unable to construct a counterexample with $f$ continuous. Is it actually enough, and if so, why?

Suppose I have a discrete dynamical system with a finite set X of states, and suppose I want to prove that every state of X ends up, sooner or later, in a subset Z under the dynamics of the system. Then a natural proof strategy is to break this "convergence" statement into two parts, by first showing that every state of X eventually ends up in a subset Y, and then showing that every state in Y ends up in Z (with Z $\subset$ Y $\subset$ X). This might simplify the problem quite a bit.

My question is,

Under what conditions will the above proof strategy work for a continuous dynamical system $\dot{x}=f(x)$ ($x \in R^n$, $f$ smooth)?

Suppose I have a discrete dynamical system with a finite set X of states, and suppose I want to prove that every state of X ends up, sooner or later, in a subset Z under the dynamics of the system. Then a natural proof strategy is to break this "convergence" statement into two parts, by first showing that every state of X eventually ends up in a subset Y, and then showing that every state in Y ends up in Z (with Z $\subset$ Y $\subset$ X). This might simplify the problem quite a bit.

My question is,

Under what conditions will the above proof strategy work for a continuous dynamical system $\dot{x}=f(x)$ ($x \in R^n$)?

The main issue I have is that it might take infinite time to reach Y. To see an example where things go wrong if $f$ is not continuous, consider the example

• $\dot{x} = -1-x$, $\quad$ for $x<-1$,
• $\dot{x}=-x$, $\quad\quad$ for $-1 \le x \le 1$, $\quad$ and
• $\dot{x}=1-x$, $\quad$ for $x>1$.

In this case every $x$ in $[-1,1]$ converges to zero, and every point outside $[-1,1]$ converges to either $-1$ or $1$, but it takes infinite time to do so and it is not true that all points in $R$ converge to zero.

Smoothness of $f$ might be enough to rule out this behavior; I am unable to construct a counterexample with $f$ continuous. Is it actually enough?