In article Finite-Time Stability of Continuous Autonomous Systems i found this [page 4].

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That's what I don't understand:

  1. Can (2.7) $\dot{y}(t)=-k \cdot {\rm sign}(y(t)) \cdot \lvert y(t) \rvert^{\alpha}$ be reformulated as a condition for convergence in a finite-time, i.e. $\dot{y}(t) + k \cdot {\rm sign}(y(t)) \cdot \lvert y(t) \rvert^{\alpha}=0$ ?
  2. What if I have an equation of the form $\dot{y}=\nabla_y f$ and want $\nabla_y f \rightarrow 0$ in finite-time. Does this mean that I must meet the condition $\dot{\nabla_y f} + k \cdot {\rm sign}(\nabla_y f) \cdot \lvert \nabla_y f \rvert^{\alpha}=0$

Please help me figure it out. I would be grateful for help.

  • 1
    $\begingroup$ What do you mean with "condition for convergence in a finite-time"? Obviously you need that at the stationary point the ODE function is not Lipschitz, else a branching-in can not happen. Your second question makes little sense. $\endgroup$ Apr 24, 2021 at 7:11
  • $\begingroup$ @LutzLehmann Only unimodal functions are used as function $f$ in my task. An equation of the form $\dot{y}=\nabla_y f$ is always stable and from any starting point $y(0)$ converges to an equilibrium point. The problem is that this convergence occurs in a finite-time, and it seems to me that it can be solved if conditions are imposed on the motion of the state variables. $\endgroup$
    – dtn
    Apr 24, 2021 at 7:20
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    $\begingroup$ For the first question it is right. SInce all solutions of that equation converges to $0$ infinite time. For second one I think you can find $\nabla_y f$ not solution to the equation and still converging in finite time so it would be a sufficient but not neccesary condition (to be confirmed). $\endgroup$
    – nicomezi
    Apr 24, 2021 at 15:03
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    $\begingroup$ I have some similar questions on the finite-duration. In the paper by V. T. Haimo (cited your paper) are shown some examples and conditions to figure out if a 1st or 2nd order autonomous ODE stands finite duration solutions, like the example $\dot{x}=-\text{sgn}(x)\sqrt{|x|},\,x(0)=1$ which has the finite duration solution $x(t)=\frac{1}{4}\left(2-t\right)^2\theta(2-t) \cong \frac{1}{4} \left(1-\frac{t}{2}+\left|1-\frac{t}{2}\right|\right)^2$, even so, keeping the solution on the reals it also solves $\dot{x}=-\sqrt{x},\,x(0)=1$. Hope this helps. $\endgroup$
    – Joako
    Mar 27, 2022 at 0:24
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    $\begingroup$ I am not completely sure but maybe it could help: If I am not mistaken, $$\dot{y}=-\text{sgn}(y)|y|^{\frac{1}{n}}$$ with $n>1$ have as solution $$y(t) =\left[\frac{n-1}{n}(T-t)\right]^{\frac{n}{n-1}}\theta(T-t)$$ with $T>0$ a finite extinction time determined by initial conditions and $\theta(t)$ the Heaviside step function (I took it from here but is slightly different equation, but since are no sign changes I think is valid at least for $y(0)>0$). $\endgroup$
    – Joako
    May 5, 2023 at 2:04


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