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Let $f: E \rightarrow \mathbb{R}^n$ be a locally Lipschitz vector field for Initial Value Problem $\dot{x} = f(x)$ with $x(0)=x_0$, where $E \subset \mathbb{R}^n$ is open. Let $\bar{x}$ be an equilibrium of $f$ (and hence of $-f$).

First part: Prove that if $\bar{x}$ is asymptotically stable for $-f$ then it is unstable for $f$.

and

Second part: Is it true that if $\bar{x}$ is stable for $-f$ then it is unstable for $f$? Prove or provide counter example.

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  • $\begingroup$ A counterexample for the second question: $E=\mathbb{R}^1$, $f(x)\equiv0$, $\bar{x}=0$. $\endgroup$ – user539887 Apr 25 '19 at 4:34
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Indeed, it suffices to assume only that there is $\tilde{x} \in E$, $\tilde{x} \ne \bar{x}$, such that $\lim_{t\to\infty}\phi_t(\tilde{x})=\bar{x}$. In order to prove that $\bar{x}$ is unstable for $-f$ we need to show that there is $\delta > 0$ such that for each $\epsilon > 0$ there are $y \in E$ and $\tau > 0$ such that $\lVert y - \bar{x} \rVert < \epsilon$ and $\lVert \phi_{-\tau}(y) - \bar{x} \rVert \ge \delta$ (we use here the fact you asked in How to show flow of vector field (−f(y)) is ϕ−t(x) where ϕt(x) is the flow of (f(y)) ?).

Take $\delta = \lVert \tilde{x} - \bar{x}\rVert$. For $\epsilon > 0$ let $\tau > 0$ be so large that $\lVert \phi_\tau(\tilde{x}) - \bar{x} \rVert < \epsilon$. Put $y := \phi_\tau(\tilde{x})$. Then $\phi_{-\tau}(y)= \tilde{x}$ is at a distance $\delta$ of $\bar{x}$.

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For the first part use that If $x_\text{eq}$ is an asymptotically stable equilibrium point of $\dot{x}=-f(x)$ then we know by Lyapunov's converse theorem that there exists a Lyapunov function $V$ with $\dot{V}=\dfrac{\partial V}{\partial x}(-f(x))<0$. If we use $V$ for the system $\dot{x}=f(x)$ we will get

$$\dot{V}=\dfrac{\partial V}{\partial x}f(x)=-\dfrac{\partial V}{\partial x}(-f(x))>0.$$

The last condition implies instability.

This method will not work for the case in which we only have stability. You can use the counterexample form the comment of @user539887.

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