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Suppose a compact connected component $C$ of $\{x:V(x)\le c\}$ contains an equilibrium $x_{\ast }$ and $\dot{V}<0$ throughout $C\setminus \{x_{\ast }\}$. Then $C$ is an invariant sublevel set, and every trajectory in $C$ converges to $x_{\ast }$ by the LaSalle invariance principle. Thus the interior of $C$ lies in the basin of attraction of $x_{\ast }$. The largest such component supplies the strongest basin estimate available from that Lyapunov function alone.

For $k\ne 0$, put $t=k^2>0$ and use the positive-definite Lyapunov function $V=x^2+t{y}^2$. Its derivative isFor $V>0$ set $u=x^2/V\in [0,1]$. The brace divided by $V^2$ is $A{u}^2+2Bu(1-u)+C(1-u{)}^2$, where $A=1$, $B=\beta (1+t)/(2t)$ and $C=1/t$. Since $\beta >2$, $B>A,C$. Completing the square gives its maximumHence $\dot{V}\le -2V+2M{V}^2$. If $V(0)<1/M$, this sublevel region is positively invariant: $V$ decreases, and $\dot{V}\le -2(1-MV(0))V$. The exponential bound makes $V\to 0$. The trajectory is bounded in a compact ellipse, so smoothness of the vector field also guarantees existence for every positive time. ThereforeFor $k=0$ the displayed strict inequality has no solutions; negative $k$ gives the same ellipse as $-k$. Taking the union of the nonempty ellipses is valid because each point needs only one Lyapunov certificate.

At $k=1$ the certified disk has squared radius $2/(\beta +1)$. Along $y=0$, the limiting threshold as $k^2\to \infty$ is $4(\beta -1)/{\beta }^2$, which is strictly larger, since their difference is $2({\beta }^2-2)/({\beta }^2(\beta +1))>0$ for $\beta >2$. Choose $x^2$ strictly between these thresholds and then a sufficiently large finite $k$. This point is certified by that ellipse but lies outside the $k=1$ disk. Thus the union gives a strictly larger certified region. It need not equal the whole basin of attraction.

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

For $0<k\ll 1$, the same three fixed points remain. At the origin the Jacobian isIts eigenvalues have negative real part; for small $k$ they are complex, so the origin is a stable spiral. At $x=\pm 1$, $U^{\prime \prime }(\pm 1)<0$, making the Jacobian determinant negative, so both points remain saddles.

Every nonstationary trajectory moves across the former energy contours in the direction of decreasingThe unstable branches of the two saddles directed into the central region spiral into the origin, while their outward branches escape toward $x=\pm \infty$. The stable manifolds of $(\pm 1,0)$ are the separatrices that form the boundary of the basin of attraction of the origin. Initial points between these stable separatrices lose enough energy to remain between the barriers and spiral to the origin; points outside escape to one of the two infinities.

This gives the weakly damped damped rational double-barrier phase portrait and identifies the requested domain of stability as the open region bounded by those stable manifolds.

For $k=1$, the interval in part (b) contains one exactly when $a>1$. With $V=x^2+y^2=r^2$,because $2xy\le r^2$. Hence $\dot{V}<0$ throughoutEvery trajectory starting in the disc $x^2+y^2<a-1$ remains there and approaches the origin. Thus this disc is contained in the basin of attraction of the origin.

With $s=0$, the equilibria satisfyBesides $g=0$, the two positive equilibria arewhere $g_{-}{g}_{+}=1$. Sincethe nonzero equilibrium relation $k=(1+g^2)/g$ givesThe fixed point stability for an autonomous differential equation therefore shows that $g=0$ and $g_{+}$ are stable, while $g_{-}$ is unstable. The graph of $f(g,0)$ starts at zero with negative slope, crosses upward at $g_{-}$, crosses downward at $g_{+}$, and tends to $-\infty$ as $g\to \infty$.

For a constant input $s$, write the equilibrium equation asThe low stable equilibrium and the intervening unstable equilibrium coalesce in a saddle-node bifurcation. More precisely, let $g_c$ be the first positive solution ofand defineEquivalently, $f(g_c,s_c)=0$ and ${\partial }_{g}f(g_c,s_c)=0$, with $g_c$ on the low-concentration branch. If $s_1>s_c(k)$, that branch no longer exists. Holding the input long enough carries the trajectory into the basin of attraction of the high state. When the input returns to zero, the concentration converges toThis is the saturating autocatalytic switch.

For $k\gg 1$, the threshold occurs at $g=O(k^{-1})$, soThe approximate double-root conditions areThus $g_c\simeq 1/(2k)$ andHence the constant in $s_c(k)\simeq C{k}^{-1}$ is $\boxed{C=1/4}$, as recorded by the strong-autocatalysis switching threshold.