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The spatially homogeneous Brusselator has its unique positive equilibrium at $(r,s/r)$. Its Jacobian there has determinant $r^2$ and trace $s-1-r^2$. If $s-1>r^2$, the equilibrium is a repeller inside the Brusselator trapping region; the Poincare-Bendixson theorem then supplies a periodic orbit.

This vector field is the negative of the one in part (a). For the same positive-definite quadratic form,The ellipse $V=1$ is invariant and has no stationary point. The angular variable increases around it, with a finite return time ${\int }_0^{2\pi }d\theta /(2{\cos }^2\theta +{\sin }^2\theta )=\sqrt{2}\pi$. It therefore explicitly supplies a periodic solution. Alternatively, the Poincare-Bendixson theorem applies to any compact invariant annulus $c\le V\le C$ with $0<c<1<C$, because its boundaries point inward and there is no equilibrium in the annulus.

If a periodic orbit had a point with $V<1$, then $V$ would strictly increase along it and could not return; a point with $V>1$ would make it strictly decrease. A periodic orbit meeting $V=1$ remains on that ellipse by uniqueness of solutions. Consequently the ellipse is the unique periodic orbit, up to time translation. It is attracting, and all nonzero trajectories approach it.

The Poincare-Bendixson theorem says that a nonempty compact omega-limit set of a smooth planar flow containing no equilibrium is a periodic orbit; use its allowed cylinder version here. For $F>1$, choose $p_{-}=(F-1)/(2k)$ and $p_{+}=2(F+1)/k$. At the lower boundary $\dot{p}>0$ and at the upper boundary $\dot{p}<0$, so the compact annulus $p_{-}\le p\le p_{+}$ is positively invariant. It has no equilibrium because $F-\sin \theta >0$. The theorem supplies a periodic orbit, and $\dot{\theta }=p>0$ makes it encircle the cylinder.

Every periodic orbit must lie in $p>0$: for $p\le 0$, $\dot{p}>0$, and a crossing of $p=0$ can only be upward. In $p>0$ a periodic orbit is a graph of a function satisfyingTwo distinct function graphs cannot cross by uniqueness of this scalar differential equation. If $p_1>p_2$, their difference obeysIt contracts by a factor strictly less than $1$ during one circuit, contradicting periodicity of both function graphs. The scalar return map is order preserving, so a multi-circuit periodic graph of a function also cannot have a period larger than one circuit. Therefore there is exactly one periodic orbit, and it encircles the cylinder. The standard auxiliary results used are existence/uniqueness of smooth differential equations and invariance at inward-pointing boundaries.

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.

The Jacobian matrix at the origin isIts eigenvalues areThus, when $k>3$, the origin is a hyperbolic repelling spiral. By part (ii), every forward trajectory is eventually trapped in a compact disc. For $k<10$ the hypothesis says that the origin is the only equilibrium; choose any nonstationary trajectory sufficiently close to it. Its omega-limit set is nonempty and compact, and it cannot contain the repelling origin. The Poincare-Bendixson theorem then says that this omega-limit set is a periodic orbit. Combining the restrictions gives

On each side of the stated parallelogram, substitution into the outward normal component ofshows that the leading $L^3$ terms point inward; the remaining terms are $O(L)$, so the region is positively invariant for sufficiently large $L$.

There is one equilibrium,Its Jacobian has determinant one and trace $\mu (1-F^2)$. If $F^2<1$ the equilibrium is unstable. A trajectory starting nearby remains in the compact trapping region, and its omega-limit set contains no stable equilibrium. The Poincare-Bendixson theorem therefore supplies a periodic orbit.

The Poincare-Bendixson theorem states that a nonempty compact $\omega$-limit set of a planar $C^1$ flow which contains no equilibrium is a periodic orbit.

The origin is the only equilibrium: at a nonzero equilibrium, putting $q=x^2+y^2-a$ would requirebut its determinant $q^2+3$ is positive. By local asymptotic stability, choose a small simple closed Lyapunov level curve on which the original vector field points inward. On a sufficiently large circle,so the original vector field points outward.

Reverse time. The annulus between these two curves is then a compact positively invariant trapping region: the reversed field points into it at both boundaries. It contains no equilibrium. The Poincaré--Bendixson theorem applied to any reversed trajectory in the annulus produces a periodic orbit. Reversing time does not change its image, so the original system also has a periodic orbit.

Dulac theorem states the following. Let $D$ be a simply connected region for the planar system $\dot{z}=f(z)$, and suppose there is a continuously differentiable function $B:D\to \mathbb{R}$ for whichhas one strict sign throughout $D$. Then there is no periodic orbit contained in $D$.

To prove it, suppose a periodic orbit $\Gamma$ existed and let $A$ be its interior. Simple connectedness ensures $A\subset D$. By the planar divergence theorem,The vector field $f$ is tangent to its trajectory $\Gamma$, so its scalar product with the outward normal $n$ vanishes. The right-hand side is therefore zero. The left-hand side cannot be zero because its integrand has one strict sign, a contradiction.

The Poincare-Bendixson theorem states that if a forward trajectory of a smooth planar system remains in a compact set and its nonempty omega-limit set contains no equilibrium point, then that omega-limit set is a periodic orbit.

At an equilibrium point of a dynamical system, the second equation givesThe choice $x=0$ is impossible because then $\dot{x}=r>0$. Hence $y=s/x$, and substitution into the first equation givesThus the unique fixed point is

Choosewhich satisfies $0<\alpha <r$. LetWe check the direction of the vector field on its four edges.

On $x=\alpha$,On $y=0$,On $y=s/\alpha$, where $x\ge \alpha$,Finally, throughout the system,On the sloping edge $x+y=r+s/\alpha$, the bound $y\le s/\alpha$ implies $x\ge r$, and hence $\dot{x}+\dot{y}\le 0$. Every boundary edge therefore points inward. Violations of the corresponding four inequalities are driven toward the boundary in the same order, so positive trajectories enter this compact trapping region and then remain there. This is the Brusselator trapping region.

The Jacobian matrix at the unique fixed point isIts determinant and trace areWhen $s-1>r^2$, the trace is positive, so linear stability of a planar equilibrium shows that the fixed point is a repeller. It lies in the interior of $R$.

Choose a trajectory in $R$ other than the equilibrium. Compactness of $R$ gives a nonempty compact omega-limit set, and the repelling equilibrium cannot belong to that set. Since there are no other equilibria, the Poincare-Bendixson theorem gives a periodic orbit. HenceThis is the Brusselator periodic-orbit criterion.

The Poincare-Bendixson theorem says that a nonempty compact $\omega$-limit set of a planar $C^1$ flow which contains only finitely many fixed points is either a fixed point, a periodic orbit, or a union of fixed points and connecting trajectories. In particular, if it contains no fixed point, it is a periodic orbit.

The region $D$ is compact and positively invariant and contains the unique fixed point. If that fixed point is unstable, choose an initial point in $D$ which is not on its stable set. Its nonempty compact $\omega$-limit set cannot be the fixed point. The Poincare-Bendixson theorem therefore supplies a periodic orbit. Hence a sufficient parameter condition, and the one selected by the preceding analysis, is

For $H=x^2+y^2$, $\dot{H}=2\mu x^2(1-a{x}^2-y^2)$. It points outward on a sufficiently small circle $r<1/\sqrt{a}$ and inward on a circle $r>1$. The intervening annulus is compact, forward invariant, and contains no equilibrium. Poincare-Bendixson theorem therefore gives a periodic orbit.