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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 $\dot{x}=2x-x^2-y$, $\dot{y}=y-kx$ 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.

Here a periodic solution must mean a nonconstant periodic orbit: constant equilibria are not excluded by the stated divergence hypothesis. Suppose such an orbit exists. Since the vector field is continuously differentiable, uniqueness for the autonomous differential equation makes its trajectory a simple closed curve $C$, bounding a region $D$ by the Jordan curve theorem. The Green theorem givesAlong the orbit, $dx=fdt$ and $dy=gdt$, so the line integral is identically zero. Reversing orientation if necessary does not change this conclusion. The area integral is strictly positive because its continuous integrand is positive everywhere and $D$ has positive area. This contradiction proves the Bendixson-Dulac criterion in this case.

For the second-order equation set $y=x^{\prime }$, giving $x^{\prime }=y$, $y^{\prime }=-y-\epsilon y^3-\sin x$. Choose the Dulac function $\rho =-1$. ThenThus there is no nonconstant periodic solution for $\epsilon >0$. The equilibria $x(t)=k\pi$ do remain solutions. As a direct check, the mechanical energy $H=y^2/2+1-\cos x$ obeys $H^{\prime }=-y^2-\epsilon y^4$, whose integral over a period could vanish only if $y\equiv 0$.

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.

Write $c=(1-a-b)/a$. The admissible equilibrium points are $O=(0,0)$, $E=(1-a,0)$ when $a\le 1$, and $P=(b,c)$ when $a+b\le 1$. The Jacobian matrix isAt $O$ the eigenvalues are $(1-a)/a$ and $-1$; at $E$ they are $-(1-a)/a$ and $(1-a-b)/b$. At a positive coexistence equilibrium point,It is therefore asymptotically stable. It is a node for $b^2>4a(1-a-b)$, a focus for the reverse inequality, and has repeated real eigenvalues at equality. The node–focus change does not alter the attracting equilibrium point's topology.

There are three open physical parameter regions:

The axes are invariant. On $x=0$, $y$ decays exponentially; on $y=0$, prey obey the logistic differential equation. Positivity follows because each right-hand side has its population as a factor. Also $x$ is bounded by logistic comparison, while $W=x+by$ satisfies $\dot{W}=(1-a)x/a-x^2/a-by$, so all forward trajectories are bounded. The Dulac function $1/(xy)$ has divergence $-1/(ay)<0$ in the interior, excluding periodic orbits by the Bendixson-Dulac criterion.

In the extinction region, $\dot{x}\le -(a-1)x/a$, and then $\dot{y}/y=x/b-1$ is eventually negative. In the prey-only region, the upper bound $\mathrm{limsup}x\le 1-a<b$ implies predator decay and convergence of interior prey to $1-a$. In the coexistence region the Lyapunov functionhas $\dot{\mathcal{V}}=-(x-b{)}^2/a\le 0$. Its positive-quadrant sublevel sets are compact, and the only invariant subset with $x=b$ also has $y=c$. The LaSalle invariance principle proves convergence of every interior trajectory to $P$. These facts justify the complete qualitative phase portraits, not just their local arrows.

At $a=1$, $O$ and $E$ meet in a transcritical bifurcation; positive prey decay algebraically. At $a+b=1$ with $a<1$, $E$ and $P$ exchange stability. On its centre manifold, $x=b-ay+O(y^2)$ gives $\dot{y}=-a{y}^2/b+O(y^3)$, so the physically positive side is still attracting. There is no interior Hopf bifurcation since the coexistence trace never vanishes.

First find the positive spatially homogeneous equilibrium. Setting the reactions to zero gives $v=u^2$ and $u/v=b$, and hence the unique positive equilibrium isThe zero pair is not an admissible positive state and is singular in the $u^2/v$ term. For homogeneous deviations, the reaction Jacobian matrix at this equilibrium isThus $\mathrm{tr}J=b-1$, $\det J=b$, and its eigenvalues areFor $0<b<1$, both real parts are negative: a real pair has positive product and negative sum, while a complex pair has real part $(b-1)/2$. The reaction equilibrium is therefore asymptotically stable, with no growing oscillatory mode. This does not exclude damped oscillations. Indeed, for $3-2\sqrt{2}<b<1$ the eigenvalues are complex; for example, $b=1/2$ gives $-1/4\pm i\sqrt{7}/4$, a stable focus with oscillatory transients. Thus a literal exclusion of every oscillatory solution is false; the intended exclusion of sustained or growing oscillations is the meaningful statement.

One can strengthen this to exclude nonlinear periodic oscillations in the positive quadrant without relying only on the local linear stability analysis. For the reactions $f=u^2/v-bu$, $g=u^2-v$, use the Dulac function $H=u^{-2}$. ThenWere there a periodic orbit wholly in the positive quadrant, the flux of $H(f,g)$ through it would vanish, since this vector is tangent to the orbit; Green's theorem would equate that flux to a strictly negative interior integral. This contradiction proves there is no sustained positive periodic oscillation when $0<b<1$, the positive-quadrant Dulac multiplier for quadratic activation argument. It leaves the damped transients entirely consistent with the conclusion.

Now introduce a perturbation $(\delta u,\delta v)=\mathbf{a}{e}^{\sigma t+ikx}$ in the quadratic activator-inhibitor model. The Fourier mode of squared wavenumber $q=k^2$ evolves underIts matrix trace and determinant areA Turing instability must start with reaction stability, requiring $0<b<1$. Then $\tau (q)<0$ for every $q\ge 0$, so a mode is unstable exactly when $D(q)<0$: its two real eigenvalues have opposite signs. This also proves the diffusion-driven onset is stationary rather than oscillatory.

The upward-opening quadratic $D(q)$ has its minimum at $q_m=(db-1)/(2d)$. To have a negative value at a physically allowed positive squared wavenumber, one needsWriting $r=db$, the second inequality is $r^2-6r+1>0$, with roots $3\pm 2\sqrt{2}$. The first inequality discards the lower branch, givingThese are the Turing threshold of the quadratic activator-inhibitor model conditions. Equivalently, in the $(b,d)$ plane the instability region lies strictly above $d=(3+2\sqrt{2})/b$ and strictly to the left of $b=1$, with $b>0$. The boundary diverges as $b\to 0^{+}$ and approaches $3+2\sqrt{2}$ as $b\to 1^{-}$.

At equality, the determinant has a double root and one growth rate is zero, while the other remains negative. Substituting $db=3+2\sqrt{2}$ into its minimum location givesBeyond onset, the unstable band isThe stated parameter region assumes continuously available wavenumbers, as on an infinite domain. With specified finite-domain boundary conditions, an allowed nonzero spatial eigenmode must additionally lie in this band; homogeneous stability alone does not ensure that a finite box admits the critical wavelength.

Assume first $\beta >0$ and $d\ge 0$. The nonnegative equilibrium points areand, when both coordinates are nonnegative,The Jacobian matrix at $O$ has eigenvalues $1,1-d$, so $O$ cannot attract an initially positive uninfected population. At $U$ the eigenvalues are $-1,\beta -d$. At $I$, for $d<1$, they are $d(1+\beta )-\beta$ and $-(1-d)$. At an interior $C$, the equilibrium equations reduce the Jacobian matrix toThus the coexistence equilibrium is asymptotically stable precisely when $\beta /(1+\beta )<d<\beta$.

For positive initial populations these local results identify the global outcome. The relation $\dot{N}=N(1-N)-dy$ bounds all trajectories. With the Dulac function $1/(xy)$, the divergence of the rescaled vector field is $-1/y-1/x<0$, excluding interior periodic orbits. Boundary saddles cannot attract an interior trajectory except at the merging thresholds. The resulting long-term outcomes areAt $d=\beta /(1+\beta )$, $C$ merges into $I$; at $d=\beta$, it merges into $U$. The merged equilibria still attract interior trajectories, but convergence along the zero-eigenvalue direction is algebraic. For example at $d=\beta$, put $x=1+u$, $y=v$. The local centre manifold has $u=-(1+\beta )v+O(v^2)$, and $\dot{v}=-{\beta }^2{v}^2+O(v^3)$. At the other threshold, using $y=(1-d)+v$, $x=u$, gives $v=(\beta -1)u+O(u^2)$ and $\dot{u}=-{\beta }^2{u}^2+O(u^3)$.

The invariant axes also matter: $y=0$ leads to $U$ whenever $x(0)>0$; $x=0$ leads to $I$ for $d<1$ and to extinction for $d\ge 1$. At $d=1$, $I$ coincides with $O$. If $\beta =0$, disease declines for $d>0$ and any initially positive $x$ tends to $U$. If also $d=0$, every point $x+y=1$ is an equilibrium and the initial population proportions are preserved.

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.

After an initial layer of duration $O(ϵ)$ in slow time, $v\simeq u/(u+K)$ while $u$ changes only slightly. The reduced substrate equation isBelow $u_e$ it decreases to zero; above $u_e$ it increases without bound. The exact threshold for initial $v=0$ is the intersection of the saddle equilibrium's stable manifold with that axis. For small $ϵ$ it isIn dimensional initial-substrate units the leading threshold is $s_0{u}_{\ast }\simeq k_2{e}_0/k_0-(k_{-1}+k_2)/k_1$. Comparing the actual initial value $u(0)=1$ to this threshold determines its fate; the reference scale can be held fixed when varying the initial concentration.

The initial sequestration produces an $O(ϵ)$ correction if desired. Write the stable manifold as $u=u_e+ϵ\varphi (v)+O({ϵ}^2)$. The invariance equation at leading order gives ${\varphi }^{\prime }(v)=F(u_e,v)/G(u_e,v)=-(1-\eta )$ and $\varphi (v_e)=0$. Hencefor fixed parameters away from degenerate limits. The positive correction reflects the substrate consumed in building complexes from the initially enzyme-free-complex state.

The physical strip $u\ge 0$, $0\le v\le 1$ is invariant, and the off-diagonal derivatives are nonnegative there, so ordered initial substrate values remain ordered. There are only the attracting origin and the saddle equilibrium. With Dulac function $1/u$, the divergence of the weighted vector field is $-(K-\lambda )v/u^2-(1+K/u)/ϵ<0$, excluding periodic orbits in its interior. The saddle equilibrium separatrix therefore gives the unique threshold on the initial axis: below it solutions tend to extinction, at it they tend to the saddle equilibrium, and above it they escape to infinity. Growth is at most exponential, so this is $u\to \infty$ as time tends to infinity, not a finite-time blowup. This is autocatalytic enzyme clearance with a saddle threshold.

For the diffusionless quadratic activator-inhibitor model, choose the Dulac function $u^{-2}$. Then $u^{-2}f=1/v-b/u$ and $u^{-2}g=1-v/u^2$, giving divergence $(b-1)/u^2$. For $0<b<1$ it is strictly negative throughout the simply connected positive quadrant. Green's theorem rules out a periodic orbit: its boundary flux would be zero because the rescaled vector field is tangent to the orbit, while its interior divergence integral would be negative.