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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.

Lutz Lehmann
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dtn
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    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. – Lutz Lehmann Apr 24 '21 at 07:11
  • @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.

     – dtn Apr 24 '21 at 07:20
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    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). – nicomezi Apr 24 '21 at 15:03
  • Well, yes, I tried to conduct some experiments. Anyway, this is a "not very good" solution for my tasks. I would even say very weak. I am not satisfied with the results of "improving" the quality of convergence. – dtn Apr 25 '21 at 04:57
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    I have some similar questions on the [tag: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. – Joako Mar 27 '22 at 00:24
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    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 $y(0)>0$). – Joako May 05 '23 at 02:04

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