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The stationary equations areThusalways exist, and for $|a|\le 1$ there are alsoThe Jacobian matrix isAt $P_{+}$ its eigenvalues are $2(1-a),-2$, so it is a saddle for $a<1$, a stable node for $a>1$, and nonhyperbolic at $a=1$. At $P_{-}$ they are $-2(1+a),2$, so it is a saddle for $a>-1$, an unstable node for $a<-1$, and nonhyperbolic at $a=-1$.

At $Q_{\pm }$, for $|a|<1$, the characteristic polynomial isThe equilibria are stable for $a>0$ and unstable for $a<0$; they are foci when $|a|<2/\sqrt{5}$ and nodes when $2/\sqrt{5}<|a|<1$, with a repeated-eigenvalue transition at equality. For $a=0$ the eigenvalues are purely imaginary. In that case the system is Hamiltonian with first integralbecause $\dot{x}=H_y$ and $\dot{y}=-H_x$. The Hessian of $H$ is positive definite at $(1,0)$ and negative definite at $(-1,0)$, so nearby regular level sets are closed curves. Hence both nonhyperbolic equilibria are genuine nonlinear centers.

Finally,For $a\ne 0$ this has a strict constant sign throughout the simply connected plane, so the Bendixson-Dulac criterion proves that there are no periodic orbits. These results comprise the phase portrait of x dot equals two x times y minus a.

Use the Dulac theorem on the simply connected positive quadrant with $\varphi (x,y)=1/(xy)$. If $F=(\dot{x},\dot{y})$, thenand henceFor $ab<0$, both terms have the same strict sign throughout $\Lambda$. The Bendixson-Dulac criterion therefore gives no periodic orbit in $\Lambda$. The same computation also excludes the case $ab=0$ unless $a=b=0$.

The Bendixson-Dulac criterion says the following. Let $D\subseteq {\mathbb{R}}^2$ be a simply connected domain, let $F=(f,g)$ be continuously differentiable, and suppose there is a continuously differentiable function $B$ such thathas one sign throughout $D$ and does not vanish identically on any open subset. Then $\dot{z}=F(z)$ has no periodic orbit contained in $D$.

Indeed, if a periodic orbit $\Gamma$ existed, simple connectivity would put its interior $U$ inside $D$. The vector field $F$, and hence $BF$, is tangent to $\Gamma$, so its outward flux is zero. The divergence theorem would givecontradicting the sign hypothesis.

The Poincare-Bendixson theorem states that a nonempty compact omega-limit set of a continuously differentiable planar flow that contains no equilibrium point is a periodic orbit. More generally, when the omega-limit set contains only finitely many equilibria, it is either an equilibrium, a periodic orbit, or a union of equilibria and connecting trajectories.

Take the Dulac function $B=1$. Fordirect differentiation givesThe quadratic form in parentheses has matrixwhose eigenvalues are $14$ and $40$, so it is positive definite. For $0<k\le 3$ the divergence is nonpositive and vanishes only at the origin when $k=3$. The Bendixson-Dulac criterion therefore gives

There is no stable positive coexistence state: the only one is a saddle for every $\rho ,\alpha >0$ when $\beta >1$. Moreover, the Bendixson-Dulac criterion with Dulac function $B(x,y)=1/(xy)$ applies in the positive quadrant, sinceThus no periodic orbit can support long-term coexistence. Except on the saddle's stable manifold, one species is excluded: species 2 becomes extinct for initial data in the basin of $(1,0)$, ecologically when species 1 is sufficiently abundant and species 2 sufficiently scarce. On the other side, species 1 becomes extinct and the unregulated second species grows.

Let $\dot{z}=F(z)$ be a $C^1$ planar system on a simply connected domain $D$. The Bendixson-Dulac criterion states that if some $B\in C^1(D)$ makeshave one sign throughout $D$ and not vanish identically on any open subset, then $D$ contains no periodic orbit.

Indeed, if a periodic orbit $\Gamma$ bounded a region $\Omega \subset D$, then $F$ and hence $BF$ would be tangent to $\Gamma$. Its normal flux would therefore vanish. The divergence theorem would givecontradicting the sign hypothesis. Taking $B=1$ gives the usual Bendixson divergence criterion.

The stability version of the divergence test concerns an existing periodic orbit $\gamma$ of period $T$. Liouville's formula for the variational equation shows that its nontrivial Floquet multiplier isThus the divergence test for a planar periodic orbit says that $\gamma$ is asymptotically stable if the integral is negative and unstable if it is positive.

This reflection-symmetric planar system has equilibria $(0,\pm 1)$ and, for $|a|\le 1$, $(\pm \sqrt{1-a^2},a)$. Its divergence is the constant $-2a$, excluding periodic orbits when $a\ne 0$ by the Bendixson-Dulac criterion. At $a=0$ it is Hamiltonian withand the nonzero equilibria are nonlinear centers. At $a=\pm 1$ the symmetric equilibria undergo pitchfork bifurcations.