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The Bendixson-Dulac theorem states that if $(P,Q)$ is a continuously differentiable planar vector field on a simply connected open region $D$, and a continuously differentiable Dulac function $B$ makeshave one weak sign throughout $D$, without vanishing identically on any nonempty open subset, then there is no nonconstant periodic orbit contained in $D$. To see why, a hypothetical simple periodic orbit bounds a region in $D$. The divergence theorem gives the integral of this expression as the outward flux of $B(P,Q)$ across the orbit. The flux is zero because the vector field is tangent to its orbit, whereas the one-sign area integral is nonzero, a contradiction. The usual stricter version assumes a strictly positive or strictly negative expression everywhere.

For the printed polynomial system, the origin is an equilibrium andThus its linearization is a centre equilibrium. For the full system take $B=1$. The divergence isOn the open disk $x^2+y^2<3$ this is nonpositive, and is negative wherever $x\ne 0$. Its zero set inside the disk is a line, so it cannot vanish on any open subset. The Bendixson-Dulac theorem therefore provesThis also exposes a terminology problem in the source: a nonlinear centre equilibrium has surrounding closed orbits, so the printed claim of a centre cannot hold in that sense. The imaginary eigenvalues establish only a centre of the linearized system. In fact the full equilibrium is an asymptotically stable weak focus, as the following direct check shows.

A quartic Lyapunov correction for a weak focus makes the nonlinear damping visible. Put $r^2=x^2+y^2$ and consider the Lyapunov functionIt is positive definite sufficiently close to zero. Differentiating along the polynomial vector field givesFor clarity, the quartic terms come from $-x^4$ in the derivative of $r^2/2$, and from applying $y{\partial }_x-x{\partial }_y$ to the quartic correction. Their sum is $-3x^4/8-3x^2{y}^2/4-3y^4/8$; all other terms have degree at least six. Therefore $\dot{V}<0$ in a sufficiently small punctured neighbourhood, proving asymptotic stability by Second Lyapunov theorem. Finallyso nonconstant nearby orbits turn clockwise as they converge. The equilibrium is a weak focus despite the centre in its linearization. This interpretation resolves both the local dynamics and the absence of closed orbits without changing the printed coefficients.

A Lyapunov function near an equilibrium is a continuously differentiable positive-definite function $V$, vanishing there, whose derivative along trajectories is nonpositive. The strict Second Lyapunov theorem says a negative-definite derivative implies local asymptotic stability.

For sharper bounds than the circular test $x^2+y^2$, put $z=(x,y{)}^T$, $r^2=z^{T}z$ andA direct multiplication gives $A^{T}P+PA=-2I$. Its eigenvalues are ${\lambda }_{\pm }=(13\pm \sqrt{13})/12>0$. Since the nonlinear vector field is $Az+r^{2}z$,For $0<V<1$, the derivative is strictly negative. The Lyapunov theorem proves asymptotic stability. In fact $V(0)<1$ gives $V(t)\le V(0)$ andso the solution stays bounded, exists for all positive time and converges exponentially to zero.

The ellipse $V=1$ is invariant. In polar coordinates, $\dot{\theta }=-(2{\cos }^2\theta +{\sin }^2\theta )<0$, so it is a nonconstant periodic orbit, not part of the basin. For $V(0)>1$, use $r^2\ge V/{\lambda }_{+}$ to obtain $\dot{V}\ge 2V(V-1)/{\lambda }_{+}$. This implies finite-time blow-up by comparison with the scalar equation. Thus the exact basin is $\{V<1\}$ and every point outside its closure escapes to unbounded norm within its maximal existence interval.

The largest open disc centred at zero lying in this exact basin has radiusThe smallest radius such that every point strictly outside its circle escapes is the largest radius on the invariant ellipse:These are optimal: a larger attraction disc contains a periodic-orbit point, and a smaller escape threshold leaves a periodic-orbit point in its exterior. This is an exact quadratic basin boundary.

The simpler radial derivative is $\frac{1}{2}(d/dt)r^2=r^2(r^2-1)-xy$. It only certifies attraction for $r<1/\sqrt{2}$ and strictly increasing norm for $r>\sqrt{3/2}$. Those are optimal thresholds for a sign test on that radial derivative, not for the actual attraction and escape regions. If “increases without bound” is read as requiring monotonic increase from the initial instant, the smallest such uniform radius is $\sqrt{3/2}$: at direction $x=y$, any smaller threshold permits an initially decreasing norm. The sharper $r_2$ above concerns eventual unbounded escape, which need not be monotone at every instant.

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.

Let $x_{\ast }$ be an equilibrium point of a dynamical system. A Lyapunov function on a neighbourhood of $x_{\ast }$ is a continuously differentiable function $V$ such thatand whose derivative along trajectories satisfies

The First Lyapunov theorem says that the existence of such a positive-definite $V$ proves Lyapunov stability of $x_{\ast }$. If $\dot{V}<0$ away from $x_{\ast }$, the Second Lyapunov theorem strengthens this to asymptotic stability.

LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set $\Omega$ on which $\dot{V}\le 0$, then it approaches the largest invariant subset ofIn particular, if that largest invariant subset consists only of $x_{\ast }$, every trajectory in $\Omega$ approaches $x_{\ast }$.

For $r>0$, differentiate $r^2=x^2+y^2$ and use the two differential equations:Substituting $x=r\cos \theta$, $y=r\sin \theta$ gives the required equations in polar coordinates. The origin is handled in Cartesian coordinates, where it is an equilibrium point; an angular coordinate is not defined there.

A Lyapunov function near an equilibrium point is a continuously differentiable positive-definite scalar function $V$ whose derivative along solutions is nonpositive. This implies local Lyapunov stability under the usual comparison bounds; if its derivative is negative definite, the Second Lyapunov theorem gives local asymptotic stability. Here takeIf $a<f(0,0)$, continuity gives a small ball and $\delta >0$ such that $f(x,y)-a\ge \delta$ inside it. Consequently $\dot{V}\le -2\delta V$. A sufficiently small initial sublevel set stays inside the ball, and $V(t)\le V(0)e^{-2\delta t}$. The origin is asymptotically stable, in fact locally exponentially stable. If $a>f(0,0)$, continuity instead gives $\dot{V}\ge 2\delta V$ in a small ball. Every nonzero initial point there exits that ball in forward time, proving instability of the origin. These inequalities hold along all local solutions; continuity of $f$ alone need not supply uniqueness for a general vector field.

If $f=g(r)$, a nonzero constant radius $r_0$ satisfying $a=g(r_0)$ gives a periodic orbit of period $2\pi$, since $\dot{\theta }=1$. Linearizing the scalar radial equation there givesThus the periodic orbit is orbitally asymptotically stable for $g^{\prime }(r_0)>0$ and unstable for $g^{\prime }(r_0)<0$. The direction along the periodic orbit is neutral, as expected for an autonomous periodic orbit; orbital asymptotic stability means attraction in a dynamical system to the orbit with a possible phase shift.

For the specified radial response,The origin is therefore stable for $0<a<3/2$ and unstable for $a>3/2$. The nonzero-radius equation has roots of an equationretaining only positive radii. For $1<a<3/2$, the square root exceeds one, so only $r_{+}>1$ exists and $g^{\prime }(r_{+})<0$: the periodic orbit is unstable. For $3/2<a<2$, both radii are positive; $0<r_{-}<1$ gives $g^{\prime }(r_{-})>0$ and a stable periodic orbit, while $r_{+}>1$ remains unstable.

As $a$ increases through $3/2$, the stable small-radius periodic orbit emerges from the origin while the origin loses Lyapunov stability. This is the supercritical Hopf-type transition with a nonsmooth radial response requested. It follows directly from the explicit radial dynamics. There is a regularity qualification: $g(r)=3/2+r/2+O(r^2)$ near zero, so the vector field in Cartesian coordinates contains $x\sqrt{x^2+y^2}$ and $y\sqrt{x^2+y^2}$ terms. It is continuously differentiable but not generally twice differentiable at the origin, and the periodic orbit radius grows as $r_{-}\sim 2(a-3/2)$ rather than the square-root law of a generic smooth Hopf normal form. One should therefore use the explicit computation here rather than invoke a smooth Hopf bifurcation theorem with unmet hypotheses. At $a=3/2$ itself the origin is still weakly attracting, because $a-g(r)<0$ for small positive $r$.

Consider a smooth planar system $\dot{x}=y+P_3(x,y)+O(r^5)$, $\dot{y}=-x+Q_3(x,y)+O(r^5)$, where $P_3,Q_3$ are homogeneous cubic polynomials and $r^2=x^2+y^2$. The leading derivative of $r^2/2$ is the quartic polynomial $H_4=x{P}_3+y{Q}_3$. Let $\kappa$ be the angular average of $H_4$ on the unit circle. A homogeneous quartic correction $V_4$ solvinggives $\dot{V}=\kappa r^4+O(r^6)$ for $V=r^2/2+V_4$. The right side has zero angular average. Since $y{\partial }_x-x{\partial }_y=-{\partial }_{\theta }$, its nonconstant angular Fourier modes can be integrated individually to obtain a periodic quartic correction; only modes of orders two and four can occur after removing the constant average.

The correction is $O(r^4)$, so $V$ remains positive definite close to zero. If $\kappa <0$, its orbital derivative is strictly negative there, and the Second Lyapunov theorem proves asymptotic stability. If $\kappa >0$, apply the same argument in reversed time to obtain repulsion. If $\kappa =0$, this quartic calculation alone does not decide stability. This method converts the averaged damping of a weak focus into a pointwise sign for a Lyapunov function.