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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 $\{x:\dot{V}(x)=0\}$.

The factor $x^2(2-x^2)$ is positive for $0<|x|<\sqrt{2}$. The product $y^3\sin 2y$ is positive for $0<|y|<\pi /2$. Therefore $\dot{V}<0$ away from the origin throughoutNeither side length can be enlarged while retaining a strict Lyapunov function on the whole rectangle: at $(x,0)$ with $|x|>\sqrt{2}$, the first term makes $\dot{V}>0$, while at $(0,y)$ immediately beyond $|y|=\pi /2$, the second term makes $\dot{V}>0$.

Strict decrease on this rectangle does not by itself put the whole rectangle in the domain of stability, because the rectangle has not been shown to be positively invariant; a trajectory may leave it before convergence.

On the two vertical sides of the square, the $x$ component of the vector field satisfiesOn the horizontal sides,Thus the vector field never points outward through the boundary, so the closed square is a positively invariant set and trajectories cannot leave $S$.

At a fixed point, $\dot{y}=0$ gives $x=-y^3$. Substitution into $\dot{x}=0$ givesFor $0<y<1$, both $y^3(2-y^6)$ and $\sin (2y)$ are positive; for $-1<y<0$, both are negative. Hence $y=0$, and then $x=0$, is the unique equilibrium point in $S$.

This alone does not prove that $S$ lies in the domain of stability. A compact trapping region containing a unique equilibrium can also contain a periodic orbit or another nontrivial omega-limit set.