If 0 is an equilibrium point and an attractor, then 0 is also stable?

Question

I need to show that, given $x'=f(x)$ with $f$ Lipschitz, $f(0)=0$, and that $\varphi(t,x_0) \rightarrow 0$ as $t \rightarrow \infty$ for all $x_0 \in \mathbb{R}$, then $0$ is stable. I feel like you need to relate $\varphi$ to $f$ somehow but I'm very hazy on how flows actually work, so I feel pretty stuck with this.

Answer 1

Well, think of the phase portrait. The only way for the origin to be unstable is if some solution moves away from it, and for systems in $\mathbb{R}$, such an outward-moving solution will be a barrier for other solutions to approach the origin from that direction.

Note, by the way, that this statement is false for systems in $\mathbb{R}^2$ (or higher dimension), as shown for example by the system $$ \dot x = \frac{x^2 (y-x)+y^5}{1+(x^2+y^2)^2} ,\qquad \dot y = \frac{y^2 (y-2x)}{1+(x^2+y^2)^2} ; $$ It can be shown that all of its solutions tend to $(0,0)$ as $t \to +\infty$, and that the origin is unstable anyway. (See phase portrait plotted on Wolfram Alpha to get an idea of how this is possible.)