# How to show if $\bar{x}$ is asymptotically stable for $-f$ then it is unstable for $f$?

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.

• A counterexample for the second question: $E=\mathbb{R}^1$, $f(x)\equiv0$, $\bar{x}=0$. – user539887 Apr 25 '19 at 4:34

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