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Prove that the origin is not asymptotically stable for this system:

$$x'_1=f_1(x_1+x_2) \\ x'_2=f_2(x_1+x_2)$$

where $f$ is both continuous and derivable and $f_1(0)=0$, $f_2(0)=0$.

The second conditions to me only means that the origin is an equilibrium point, right?

Here’s what I thought: we can use the Lyapunov function $V(x_1,x_2)$, and then: $$\Delta V·f(x_1,x_2) = \frac{\delta V}{\delta x_1}·f_1(x_1+x_2)+\frac{\delta V}{\delta x_2}·f_2(x_1+x_2).$$

If we put $x_1=-x_2$ the value of $\Delta V·f(x_1,x_2)$ will be zero near an interval of zero $I_0\setminus \{0\}$, which means that the second Lyapunov theorem hypothesis are not satisfied. Do I have to check something else? Does that make sense?

Wrzlprmft
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Francesco
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  • This is likely insufficient. The existence of a single Lyapunov function that does not satisfy $\mathrm{D}V(x) < 0$ for $x\neq 0$ in a neighborhood of $0$ does not disprove asymptotic stability. You should show that for any $\delta > 0$ there is an $\lvert x_0\rvert < \delta$ such that $x(0) = x_0$ implies $x(t)\not\to 0$ as $t\to \infty$. – Michael L. Aug 25 '17 at 09:53
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    Consider initial points $(a,-a)$ and their dynamic and compare to the conditions on asymptotic stability. – Lutz Lehmann Aug 25 '17 at 10:39

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