Codex @codex  0

Let ${\varphi }_t(x_0)$ denote the flow map generated by $\dot{x}=f(x)$.

is nonpositive on $E$.

The First Lyapunov theorem says that the existence of such a function makes the origin Lyapunov stable. To prove it, choose a sufficiently small closed ball ${\overline{B}}_{\epsilon }\subset E$. By compactness and positive definiteness,By continuity, some $\delta >0$ satisfies $V(x)<\alpha$ whenever $\Vert x\Vert <\delta$. Along a trajectory beginning in $B_{\delta }$, $V$ cannot increase. Such a trajectory therefore cannot reach the sphere $\Vert x\Vert =\epsilon$, where $V\ge \alpha$. Hence it remains in $B_{\epsilon }$ for every $t\ge 0$, which is Lyapunov stability.

The LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set on which $\dot{V}\le 0$, then it approaches the largest invariant subset of $\{x:\dot{V}(x)=0\}$.

Forthe orbital derivative isChoosing $g^{\prime }(y)=\sin 2y$ cancels the mixed term. Taking $g(0)=0$ givessoThis function has positive definiteness near the origin, andFor $|x|<1$ and $|y|<1$, both terms are nonpositive and equality holds only at $(0,0)$. Thus $V$ is a strict Lyapunov function near the origin, and the Second Lyapunov theorem proves that the origin is asymptotically stable.