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Purely imaginary nonzero eigenvalues make a real planar linearization a centre equilibrium. They do not control the nonlinear radial drift. For example,has the same linearization as a uniform rotation. In polar coordinates, however, $\dot{r}=-r^3$ and $\dot{\theta }=1$. Every sufficiently small nonzero orbit spirals towards the origin, so the origin is an asymptotically stable weak focus, and no nonzero orbit is periodic. Reversing the signs of the cubic terms gives a repelling weak focus. Removing them gives a centre. Thus a nonlinear centre equilibrium requires a further argument, such as a suitable first integral; the imaginary pair alone is insufficient.

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.

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.

This real planar dynamical system has a saddle point at $(-1,0)$ and a repelling weak focus at $(1,0)$. The latter has purely imaginary linearized eigenvalues $\pm 2i$, so its linearization has closed elliptical orbits. Nonlinear terms nevertheless produce outward drift.

Set $X=2(x-1)$, $Y=y$, $\tau =2t$ and $r^2=X^2+Y^2$. The cubic expansion isThe linear rotation acts on scalar functions by $L=Y{\partial }_X-X{\partial }_Y$. For $H_2=r^2/2$, its cubic derivative is $X{Y}^2-3X^{2}Y/4-Y^3/2$. The homogeneous correction $H_3=7X^3/12+X{Y}^2/2+Y^3/3$ satisfies $L{H}_3$ equal to the negative of that derivative. The remaining quartic derivative of $H_2+H_3$ is $3X^2{Y}^2/2-X^{3}Y$. Withdirect differentiation givesfor all sufficiently small nonzero $r$. Meanwhile $H_2+H_3+H_4=r^2/2+O(r^3)>0$. In reversed time it is a strict Lyapunov function, proving asymptotic stability of the reversed system and hence repulsion forward in time. The angular speed stays close to $-1$, so the nearby nonzero trajectories spiral rather than approaching along a fixed direction. This explicit calculation illustrates why a centre in the linearization need not be a nonlinear centre.