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Consider the following nonliner system: \begin{align} \dot{x}=f(x) \end{align} where $x\in\mathbb{R}^n$ and $f(x)\in\mathbb{R}^n$ is sufficiently smooth and Lipschitz in $x$. Then the system is smooth and admits a unique solution. Suppose $x^*\in\mathbb{R}^n$ is an equilibrium of the system, i.e., $f(x^*)=0$. Is it possible that, for some initial condition $x(0)=x_0$, the solution of the system satisfies \begin{align} \lim_{t\to T}x(t)=x^{*}, \end{align} that is, the solution reaches the equilibrium $x^*$ in some finite time $T$. If it is not possible, is there a way to show that, the solution of a smooth system will take an infinite amount of time to converge to an equilibrium?

Update:Assume that $x_0\neq x^{*}$.

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  • $\begingroup$ Yes, use $x_0=x^*$. $\endgroup$
    – Michael
    Mar 23, 2020 at 4:54
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    $\begingroup$ I am sorry, what I meant is the nontrivial case where $x_0\neq x^{*}$. $\endgroup$
    – guluzhu
    Mar 23, 2020 at 5:06
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    $\begingroup$ I don't think it's possible to reach equilibrium in finite time. If it does, then this trajectory and the trivial trajectory starting from equilibrium intersects at the equilibrium. This is not possible because of uniqueness of solutions. $\endgroup$ Mar 23, 2020 at 5:17
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    $\begingroup$ @SudheeshSurendranath Thank you for the explanation. By your augment, I think we can show that for any $x_0\neq x^*$ and any finite positive constant $T$, $x(t)\neq x^{*}$ for all $t\in[0,T]$. $\endgroup$
    – guluzhu
    Mar 23, 2020 at 5:36
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    $\begingroup$ In the following papers "Finite Time Controllers" and "Finite Time Differential Equations" by V. T. Haimo are studied 1st and 2nd order scalar autonomous ODEs that achieve solutions of finite duration, where is explained uniqueness of solutions is not hold since is required for the ODE to have at least one point in time where is not-Lipschitz. $\endgroup$
    – Joako
    May 20, 2022 at 10:59

1 Answer 1

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Let $y(t) = x^*$ and note that $y$ is a solution to the system $\dot{y} = -f(y)$ with $y(0) = x^*$.

Suppose $x$ is a solution with $x(T) = x^*$ and $x(0) = x_0$. Then since $x(T)=y(T)$ we must have $x(t) = y(t)$ for $t <T$ (by uniqueness, running the system backwards, that is $\dot{z} = -f(z)$). Hence $x_0 = x^*$.

In particular, if $x_0$ is not an equilibrium and $x(t) \to y^*$ where $y^*$ is an equilibrium, then $x(t) \neq y^*$ for all $t$.

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  • $\begingroup$ I noticed your answer was similar to my comment, so I was about to delete my comment, then I noticed you seem to be making statements about the specific system $y' = -y$, I'm not sure why you focus on that system. $\endgroup$
    – Michael
    Mar 23, 2020 at 4:57
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    $\begingroup$ I am sorry. I didn't get it why $y(t)=x^*$ satisfies $\dot{y}=-y$? Did you mean $\dot{y}=-(y-x^*)$? $\endgroup$
    – guluzhu
    Mar 23, 2020 at 5:29
  • $\begingroup$ @Michael: The time reversed system is unnecessary since the solution is unique for all time, but the uniqueness is often presented as unique on some interval (as in the usual fixed point proof) with time increasing, so out of a pedantic abundance, I stuck with that model. But clearly my goal of clarity failed miserably. $\endgroup$
    – copper.hat
    Mar 23, 2020 at 5:31
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    $\begingroup$ @guluzhu: It is $\dot{z} = -f(z)$, not $-z$. I am just running the system 'backwards' from $T$ to $0$. In this case, if I start with $z(T) = y^*$ then $z(2T) = x(0) = x_0$ and $z(2T) = y(0) = y^*$. $\endgroup$
    – copper.hat
    Mar 23, 2020 at 5:35
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    $\begingroup$ @guluzhu: Oops, you are right, I had a typo. $\endgroup$
    – copper.hat
    Mar 23, 2020 at 5:39

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