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For $A,B,\rho >0$, these Lotka-Volterra equations describe two populations limited by their own density and depressed by the other population. The nonnegative quadrant is invariant. Positive equilibria are intersections of the non-axis nullclines $u+Av=a$ and $Bu+v=b$; their existence and stability distinguish stable coexistence, competitive exclusion, and bistability. Positive solutions have no periodic orbits: multiplying the vector field by the Dulac function $1/(uv)$ makes its divergence $-1/v-\rho /u<0$, so the Bendixson-Dulac theorem applies. Boundedness, boundary equilibria, and their stable manifolds must still be checked before concluding convergence to one particular equilibrium.

For a planar periodic orbit $\gamma$ of period $T$, its nontrivial Floquet multiplier isThe orbit is asymptotically stable when the integral is negative and unstable when it is positive.

A multiplier $B(x,y)$ whose rescaled planar vector field has divergence of one strict sign can exclude periodic orbits in a simply connected region. For a system $\dot{x}=xF$, $\dot{y}=yG$ in the positive quadrant, $B=1/(xy)$ is often effective.

For $\dot{x}=2x-x^2-y$, $\dot{y}=y-kx$ with $k>2$, take the positive Dulac function in the title. More generally, $\psi =e^{a(x+y)}$ givesThe equal coefficients in the exponential cancel the $y$ term. Setting $a=-2/(k-2)$ also cancels the linear $x$ term, leaving $\psi (3+2x^2/(k-2))>0$. The Bendixson-Dulac theorem excludes periodic orbits in the open positive quadrant. The axes are not invariant: the vector field points out of the quadrant at every nonzero axis point. A closed orbit staying in the nonnegative quadrant therefore cannot touch an axis, and cannot pass through the equilibrium at the origin by uniqueness. This extends the exclusion to the closed quadrant. Choosing an exponential multiplier turns a differential search for a useful Dulac function into cancellation of polynomial coefficients.

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.

One useful form of the Bendixson-Dulac theorem is this: let $D\subseteq {\mathbb{R}}^2$ be open and simply connected, and let $X,Y,\psi$ be continuously differentiable on $D$. Ifhas one strict sign throughout $D$, then there is no nonconstant periodic orbit lying entirely in $D$. More generally the divergence may have a weak sign provided it is not identically zero on any nonempty open subset. The multiplier $\psi$ is a Dulac function. Green's theorem proves the criterion: the vector field is tangent to a hypothetical closed orbit, so its scaled flux across that boundary is zero, whereas the signed area integral of its divergence would be nonzero.

Let $k>2$ and choose the positive Dulac functionMore generally, for $\psi =e^{a(x+y)}$, direct differentiation givesTaking $a=-2/(k-2)$ cancels the linear term and makes the quadratic term nonnegative:The open positive quadrant is simply connected, so the Bendixson-Dulac theorem excludes every nonconstant periodic orbit entirely inside it. Such an orbit cannot touch $x=0,y>0$ while staying in the closed quadrant, since its differentiable coordinate $x$ would have a local minimum there but $\dot{x}=-y\ne 0$. The same argument applies on $y=0,x>0$. It cannot pass through the origin either, by local uniqueness of this smooth autonomous system. Therefore there is no nonconstant closed trajectory in the entire nonnegative quadrant. The calculation illustrates an exponential Dulac multiplier for a quadratic planar system.

The Bendixson-Dulac theorem says that a continuously differentiable planar vector field has no nonconstant periodic orbit in a simply connected region if a continuously differentiable Dulac function makes its rescaled divergence have one sign and not vanish identically on any open subset of that region. Isolated zeros of the divergence are allowed in this form.

Take the Dulac function to be one. For the exact vector field $(P,Q)$,This is positive except at the origin. The plane is simply connected, so there are no nonconstant periodic solutions. Equivalently, a hypothetical closed trajectory would have zero outward flux of its tangent vector field, while Green's theorem would make that flux the strictly positive area integral of the divergence.

The origin is the only equilibrium point: $Q=0$ forces $y=0$, and then $P=x^3$ forces $x=0$. Its Jacobian matrix is zero, so the linearized equations are $\dot{x}=\dot{y}=0$. They give no stability classification.

Keeping cubic terms gives the homogeneous approximationIn a sector where $xy\ne 0$, use $y$ as the independent variable and put $w=x^{-2}$. Dividing the two equations givesThe integrating factor $y$ gives $(yw{)}^{\prime }=-y^{-2}$, hence $yw=1/y+C_0$. Rearranging and renaming the arbitrary constant yields the implicit trajectoriesThis is the first integral of the cubic approximation, not an exact first integral of the higher-degree original system. The invariant axes, excluded by the divisions, must be added: on $y=0$, $\dot{x}=x^3$; on $x=0$, $\dot{y}=2y^3$. The diagonal trajectories $y=\pm x$ correspond to $C=0$. For fixed finite $C$, non-axis trajectories approaching the origin satisfy $y^2/x^2=1-Cy\to 1$, so their limiting tangent lines are the diagonals. This is the cubic planar flow with diagonal approach directions.

Every nonzero trajectory moves outward locally. For $V=x^2+y^2$, the exact system givesThis implies departure from any fixed small circle for every nonzero starting point inside it: while a trajectory stayed inside forever, $\frac{d}{dt}(1/V)\le -1$ would eventually force $1/V$ negative. Thus the origin is unstable and repelling for both the cubic approximation and the full system. The argument establishes nonlinear instability despite the zero linearization.

Putting $\epsilon =0$ gives the local regular perturbation result $y_0=(x^2+1)/2$. It cannot describe the whole half-line: its derivative grows like $x$, making $\epsilon y^{\prime }$ order one at $x=O({\epsilon }^{-1})$.

The useful exact change of variables isDirect substitution turns the equation into two autonomous differential equations,The initial values here are $z(0)=w(0)=\epsilon$. Near the origin, linearizing this system gives eigenvalues $1$ and $-1-\epsilon$. Equivalently, neglecting only the derivative-square nonlinearity in the original equation gives the two modes $x^2$ and $x^{-\epsilon }$. Their matching coefficients areThus the growing derivative has $w\sim Cx$. Keeping these unexpanded linear-mode coefficients is convenient for matching; it does not assert that the linearized solution is the exact nonlinear one.

At the distinguished limit $X=Cx=O(1)$, set $R=\log X$. To leading order the derivative equation becomes $w_R=w(1-w)$, the logistic differential equation. Matching $w\sim e^R$ as $R\to -\infty$ selectsThen $z_R+z=w$, and the growing-mode matching $z\sim X/2$ givesThis proves the logistic crossover in a weakly nonlinear Euler equation. In particular the nonlinear-region leading solution is $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-2}H(\epsilon x)$ with the constants taken at their lowest order.

A composite asymptotic expansion retaining the small-distance stable mode and satisfying the initial value exactly isSince $H(X)=X^2/2+O(X^3)$, this matches $A{x}^2+B{x}^{-\epsilon }$ when $Cx\ll 1$ and hence $(x^2+1)/2$ at fixed $x$ as $\epsilon \to 0$. Its derivative at one differs from the prescribed value only by $O(\epsilon )$, within leading-order accuracy. For $Cx\gg 1$ it gives $y\sim x/\epsilon$. Thus the composite covers the initial region, the nonlinear transition and the arbitrarily large-$x$ regime; uniform leading accuracy is understood relative to the growing solution, not as an absolute error independent of unbounded $x$.

There is no later hidden growth regime. The exact positive system keeps $z,w$ in $[0,1+\epsilon {]}^2$, because its vector field points inward at every boundary. Its positive equilibrium point is $z=w=1+\epsilon$, with Jacobian matrix having trace $-2-3\epsilon$ and determinant $1+\epsilon$. The origin has no attracting direction inside the positive quadrant, while the divergence $-\epsilon -2w<0$ excludes cycles by the Bendixson-Dulac theorem. The Poincaré-Bendixson theorem therefore gives convergence to the positive stable equilibrium. At fixed small positive $\epsilon$, the exact far-field slope is $y/x\to (1+\epsilon )/\epsilon$, whose difference from $1/\epsilon$ is only relative $O(\epsilon )$, consistent with the global leading composite.

Within the positive-amplitude wedge the reduced periodic orbit equilibrium has amplitude trace ${\mu }_1$ and positive determinant $4r_{\ast }^2$. At ${\mu }_1=0,{\mu }_2<0$ the transverse pair is imaginary, so the possible secondary event is a Hopf bifurcation of the amplitude equations, corresponding to an invariant torus in the full system.

It is degenerate. Set ${\mu }_2=-a^2$ and ${\mu }_1=0$. The first integral of the quadratic fold-Hopf amplitude flow isThe amplitude equations have the Hamiltonian system form $\dot{r}=F_x$, $\dot{x}=-F_r$, so $\dot{F}=0$. At $(r,x)=(a,0)$, $F$ has a strict local maximum $2a^3/3$ and negative-definite Hessian. Every nearby regular level is a closed amplitude periodic orbit, rather than one isolated orbit selected by nonlinear damping. Adding the rotating phase gives an entire family of invariant tori; the full motion is quasiperiodic or periodic according to the phase advance during one amplitude period.

For ${\mu }_1\ne 0$, the amplitude vector field has constant divergence ${\mu }_1$. The Bendixson-Dulac theorem rules out closed amplitude trajectories in the physical simply connected half-plane $r>0$. Hence there is no ordinary branch of isolated invariant tori on either side in this quadratic system. The secondary event has vanishing radial saturation coefficients; its imaginary linear pair alone does not establish a nondegenerate torus bifurcation.

Under $A>1$, $B<1$, $AB>1$, the thresholds are ordered $0<1/A<B<1$. As $\gamma$ decreases from one, $S_2$ first becomes stable atFor $1/A<\gamma <B$, both boundary equilibria are stable, and the positive coexistence point $S_4$ is a saddle point. Its stable manifold is the separatrix between their basins of attraction. For every fixed $u_1(0)>0$, a sufficiently small positive $u_2(0)$ lies in the red-survival basin. To justify the quantifier beyond a neighborhood of $S_2$, start on the red-only axis: its logistic equation approaches $S_2$. Continuous dependence over the finite time needed to enter a stable neighborhood shows that sufficiently small positive initial grey populations do so as well.

At the second threshold,$S_4$ meets $S_3$. For $0<\gamma <1/A$, $S_3$ is a saddle point, its stable manifold is the grey-only axis, and there is no positive coexistence equilibrium. Boundedness and the same Bendixson-Dulac theorem argument rule out an interior cycle. Consequently every initial state with $u_1(0)>0$ and $u_2(0)\ge 0$ approaches $S_2=(1,0)$. This is the removal regime that eliminates grey squirrels regardless of their initial abundance. At threshold equality linearization is nonhyperbolic; the strict parameter intervals above avoid inferring stability from a zero eigenvalue.

The non-axis nullclines are $u_1+A{u}_2=1$ and $B{u}_1+u_2=1$. For $A>1$, $B<1$, the first has intercepts $(1,0),(0,1/A)$ and the second has $(1/B,0),(0,1)$. Their axis intercept order rules out a positive intersection. Below the first line $u_1$ increases and above it decreases; below the second $u_2$ increases and above it decreases. Thus $S_2$ is a saddle point and $S_3=(0,1)$ is stable.

For strictly positive initial populations, both coordinates stay positive and are bounded above by their individual logistic equations. There are no interior periodic orbits: the Dulac function $1/(u_1{u}_2)$ makes the divergence of the multiplied vector field $-1/u_2-\rho /u_1<0$, so the Bendixson-Dulac theorem applies. The stable manifold of $S_2$ is the invariant red-only axis, and $S_1$ is repelling. The phase portrait, or the Poincaré-Bendixson theorem with these boundary directions, therefore givesThe qualification $u_2(0)>0$ matters: a grey population identically absent cannot drive extinction. In dimensional variables, positive grey presence leads to $N_1\to 0$, $N_2\to K_2$ in this no-hunting case.

The planar divergence is $\nabla \cdot F=-(\Omega +x^2)<0$ for $\Omega >0$. Since the plane is simply connected, the Bendixson-Dulac theorem excludes any nonconstant periodic orbit: the flux across a closed trajectory would be zero because the vector field is tangent, whereas Green's theorem makes it the strictly negative integral of the divergence over the enclosed region.

Alternatively define $E=y^2/2-x^2/2+x^4/4$. Direct differentiation givesA nonconstant periodic solution would require the integral of this strictly dissipative expression over a period to vanish. That forces $y\equiv 0$ and hence an equilibrium, a contradiction. This energy argument identifies the damping mechanism behind the geometric criterion.