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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 Poincaré-Bendixson theorem then supplies a periodic orbit.

At an equilibrium $(v_{\ast },k{v}_{\ast })$, set $p=f^{\prime }(v_{\ast })$. The Jacobian matrix, its trace and its determinant areThe eigenvalues and their discriminant areThus the complete local linear stability test isIn the last two cases, $\mathcal{D}>0$ gives a node, while $\mathcal{D}<0$ gives a focus; the sign of $T$ distinguishes inward and outward trajectories. At $\mathcal{D}=0$, a nonzero repeated eigenvalue gives a degenerate node. At $p=\gamma <k$ the eigenvalues are purely imaginary, so linear stability analysis alone does not decide asymptotic stability; with the required crossing and nondegeneracy this is a Hopf bifurcation. At $p=k$ there is a zero eigenvalue, requiring a nonlinear test and generically marking the saddle-node bifurcation in the preceding answer.

For a unique simple equilibrium, $p<k$, so it can be a sink equilibrium or a source equilibrium. The number of equilibria alone does not decide stability. If $k\le \gamma$, the unique simple equilibrium is always a sink equilibrium. If $\gamma <k$, it can become a source equilibrium where $\gamma <p<k$. The same conclusion follows particularly simply if $\gamma >M$: the trace is negative at every nonsaddle equilibrium.

For three simple equilibria ordered by their $v$ coordinates, the middle one has $p>k$ and is a saddle equilibrium; the two outer ones have $p<k$. When both outer equilibria are sink equilibria, each has an attracting neighborhood, giving bistability. In the usual two-basin portrait, the stable manifold of the saddle equilibrium separates their basins. Each branch of its unstable manifold initially leaves towards a different side; its eventual destination requires a global argument. If $k\le \gamma$, both outer equilibria must be sink equilibria. If $\gamma <k$, either or both outer equilibria can instead be source equilibria; the same test $p\lessgtr \gamma$ applies separately to each. The three lower panels illustrate all three possibilities up to exchanging left and right.

The arrows in the phase plane follow directly from the nullclines: below $w=f(v)+I$ the horizontal component points right, and above it left; below $w=kv$ the vertical component points up, and above it down. Near a stable focus trajectories spiral inward; near an unstable focus they spiral outward. Near a saddle equilibrium they approach along its stable manifold and leave along its unstable manifold. These directions, together with the local tests, determine the sketches.

There is also a useful global conclusion when the unique equilibrium is a source equilibrium. Withthe leading term in $v$ is $-\beta v^4$, so sufficiently large ellipses are trapping regions. A small repelling neighborhood of the hyperbolic source equilibrium can be removed to form an invariant annulus. The Poincaré-Bendixson theorem then guarantees at least one periodic orbit in that annulus. The numerical sketch illustrates the resulting sustained oscillation; this argument does not assert uniqueness of a limit cycle. A stable equilibrium can instead represent a resting membrane potential, with perturbations producing damped relaxation; near a threshold, a sufficiently large perturbation can produce a large excursion before recovery. Global basin boundaries and coexistence of periodic orbits need more information than the number and local stability of equilibria.

The voltage nullcline is $g(v,n)=I^{\ast }$ and the recovery nullcline is $n=n_{\infty }(v)$. In the relevant region $g_n>0$, so below the voltage curve $\dot{v}>0$ and above it $\dot{v}<0$. Below the recovery curve $\dot{n}>0$, while above it $\dot{n}<0$.

For small $ϵ$, the outer voltage branches with $g_v>0$ attract fast voltage motion; the middle branch with $g_v<0$ repels it. Arrange the intersections so the equilibrium is on that middle branch, with recovery increasing along the high-voltage attracting branch and decreasing along the low-voltage attracting branch. The trajectory then drifts towards a fold, jumps rapidly to the other attracting branch at nearly constant $n$, and repeats. This is a relaxation oscillation.

The local linear stability calculation supports the construction. At an equilibrium,A sufficiently steep recovery curve gives positive determinant, while $g_v<-ϵ$ gives positive trace. Hence the equilibrium is a repeller. The bounded physical rectangle can be chosen forward invariant, and a small neighborhood of the repeller removed to give a trapping annulus. The Poincaré-Bendixson theorem then supplies a periodic orbit. These conditions can be realized because the recovery function and the applied current control the intersection and its slope; oscillation is possible, not forced for every choice.

For the illustration, ${\gamma }_K=0.1$, $H=0.8$, $v_K=0.2$, $ϵ=0.01$, $I^{\ast }=0$, $M(v)=(1+e^{-25(v-0.2)}{)}^{-1}$ and $n_{\infty }(v)=0.8/(1+e^{-20v})$. The rectangle $[-0.2,1]\times [0,0.8]$ has inward-pointing flow, and the displayed numerical trajectory realizes the folded-branch mechanism.

The nullclines are $w=g(v)$ and $v=0$. The two outer pieces of $g$ have slope $-1$; the middle piece has positive slopeThe curve is continuous, with folds $(\alpha ,-\alpha )$ and $(\beta ,1-\beta )$. The unique equilibrium is $(0,-\alpha /(\beta -\alpha ))$. Its Jacobian matrix has trace $\mu /ϵ>0$ and determinant $1/ϵ>0$, so it is unstable. The outer branches attract the fast voltage motion.

Let $w_L=-\alpha$ and $w_R=1-\beta$. The hypothesis $\beta -\alpha <1$ gives $0<w_L<w_R<1$. The singular relaxation oscillation follows this route:

The jumps are almost horizontal in the phase plane, because $w$ changes negligibly on the fast $O(ϵ)$ time. Below $w=g(v)$, voltage increases; above it, voltage decreases; $w$ increases on the right of $v=0$ and decreases on the left. These signs provide a closed trapping region around the indicated route for sufficiently small positive $ϵ$, with the unstable equilibrium excluded, giving a periodic orbit by the planar return-map or Poincaré-Bendixson theorem argument. The continuous piecewise linear field is locally Lipschitz, so solution uniqueness is available at the corners.

On the descending part of the graph, mark the two tangency points whereChoose the crossings that enclose the portion with $g^{\prime }(u)<-1/\gamma$. For fixed $\mu$ strictly between these values and sufficiently small $\epsilon$, the unique equilibrium is a repelling equilibrium by part (e). The endpoints here are leading singular limit values; at finite $\epsilon$ the boundaries instead solve $f_v-\gamma f_u=\epsilon$ and approach these tangencies when the crossings are nondegenerate.

There is a necessary qualification: such an interval exists only when $\gamma$ is large enough that the descending graph actually attains slopes below $-1/\gamma$. If $M={\max }_{u>0}[-g^{\prime }(u)]$, one needs $\gamma M>1$. The supplied graph has a very steep descending section, but an arbitrarily small positive $\gamma$ need not satisfy this condition. For $\delta =0.001$, numerical evaluation gives $M\simeq 511.49$, so the leading threshold is $\gamma \simeq 0.0019551$. A graph alone does not remove this parameter restriction.

The fast motion is especially informative. In fast time $s=\tau /\epsilon$,At leading order $z=u+\gamma v$ is constant, so fast paths are straight lines with slope $dv/du=-1/\gamma$. Along such a line, the fast eigenvalue at the nullcline isIt is negative on the two outer attracting segments and positive on the intervening repelling segment. The tangencies defining ${\mu }_{\pm }$ are consequently the fold points of a critical manifold when it is expressed in the coordinates $(z,v)$. They are not generally the maximum and minimum of $g$ itself.

For ${\mu }_{-}<\mu <{\mu }_{+}$, slow motion on an attracting segment followsOn the left attracting segment $u<{\mu }_{-}$ this moves towards increasing $u$; on the right attracting segment $u>{\mu }_{+}$ it moves towards decreasing $u$. Each segment ends at a tangency and is followed by a fast jump along a line of slope $-1/\gamma$ to the other attracting segment. This constructs the limiting relaxation oscillation. The figure uses $\gamma =0.01$, for which ${\mu }_{-}\simeq 0.62970$ and ${\mu }_{+}\simeq 1.01157$; the marked endpoints change with $\gamma$.

To justify existence of a periodic orbit of the full planar system, rather than just draw a formal cycle, first establish a compact trapping region. The positive quadrant is forward invariant: at $u=0$, $\dot{u}=\mu +\gamma \delta v/\epsilon >0$, and at $v=0$, $\dot{v}=u^2/[\epsilon (1+u^2)]\ge 0$. MoreoverChoose $V>1/\delta$ above the initial store value. Since $f(u,v)\ge -1-\delta V$ when $0\le v\le V$,Thus the rectangle $0\le u\le U$, $0\le v\le V$ is a positively invariant set if $U>\mu +\gamma (1+\delta V)/\epsilon$ and the initial point lies inside it. Every forward solution is bounded. There are no equilibria on its axes for $\mu >0$, and its sole interior equilibrium is a hyperbolic equilibrium point with both eigenvalues having positive real parts in the strict unstable range. A sufficiently small neighborhood of this repelling equilibrium has outward flow; deleting it leaves a compact trapping region with no equilibria. The Poincaré-Bendixson theorem then supplies a periodic orbit.

Hence periodic solutions exist throughout the strict unstable interval, interpreted with the finite-$\epsilon$ correction above. This argument establishes existence, without asserting uniqueness, the criticality of possible endpoint Hopf bifurcations, or absence of coexisting periodic orbits outside the interval.

For Hopf thresholds in a two-pool calcium model, with the unique equilibrium parametrized by $u=\mu$, locate changes of stability by solving the two equationsIf there are two simple roots ${\mu }_{-}<{\mu }_{+}$ with positive trace between them, the equilibrium is a repeller throughout that interval, since its determinant stays positive. At a nondegenerate crossing a Hopf bifurcation is possible, with linear frequency $[-f_v/ϵ{]}^{1/2}$. A complete classification of criticality requires a nonlinear coefficient, not merely the trace calculation. At leading singular order the thresholds approach the folds determined by $f=0$ and $f_v-\gamma f_u=0$, equivalently $V^{\prime }=-1/\gamma$.

Periodic solutions are also supported globally in the positive quadrant. The axes point inward for positive input, $v^{\prime }\le (\beta -\delta v)/ϵ$ bounds $v$, and the $u$ equation then bounds $u$ by a linear restoring term plus a bounded source. A large rectangle can thus be chosen forward invariant. For an initial point distinct from the repelling unique equilibrium, its omega-limit set cannot contain that equilibrium or a boundary equilibrium. The Poincaré-Bendixson theorem supplies a periodic orbit. In the strong separation regime it follows the two attracting branches and jumps between them, a relaxation oscillation.

The interval requires the maximum of $f_v-\gamma f_u$ along the critical manifold to exceed $ϵ$; the listed scale hierarchy alone is not a substitute for that test for arbitrary remaining constants. As an explicit example, $\beta =0.1$, $\delta =10^{-5}$, $ϵ=5\times 10^{-4}$ and $\gamma =0.1$ give approximately ${\mu }_{-}=0.22618$ and ${\mu }_{+}=0.37350$. At $\mu =0.3$ the trace is positive, whereas at $0.03$ and $1$ it is negative. The figures use this example and label their singular-limit portions accordingly.

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.

The Bendixson-Dulac criterion says that if a $C^1$ planar field admits a $C^1$ multiplier $B$ on a simply connected region such that $\nabla \cdot (Bf)$ has one strict sign there, no nonconstant periodic orbit lies wholly in that region. The Poincaré-Bendixson theorem says that a nonempty compact omega-limit set of a planar $C^1$ flow which contains no equilibria is a periodic orbit.

For $\mu <0$, the divergence of the given field is $\mu -2x^2-3y^2<0$ throughout the plane. Taking $B=1$ proves there are no periodic orbits.

For $\mu >0$, use $E=(x^2+y^2)/2$. Along trajectories,On a circle $r=r_{-}$ with $2r_{-}^2<\mu$ the radial flow is outward or tangent, and on a circle $r=r_{+}$ with $r_{+}^2>\mu$ it is inward or tangent. Their intervening closed annulus is positively invariant. At the tangent points $y=0$, the field has $\dot{y}=-x\ne 0$, so no equilibrium sits on its boundaries. Indeed the only equilibrium is the origin, excluded from the annulus. Every forward trajectory in the annulus has a nonempty compact omega-limit set without equilibria. The Poincaré-Bendixson theorem consequently proves at least one periodic orbit exists for every $\mu >0$. No claim of uniqueness follows from this trapping argument.

A form of the Poincaré-Bendixson theorem states that a nonempty omega-limit set of a forward orbit of a $C^1$ planar autonomous vector field, contained in a compact region and containing no equilibrium, is a periodic orbit.

Set $\rho =x^2+y^2$. Direct calculation givesbecause $0\le x^2{y}^2\le {\rho }^2/4$. At the inner boundary $\rho =2/(2+k^2)$ the flow points into or along the annulus, and at the outer boundary $\rho =1$ it does likewise. Smoothness and these inequalities make the compact annulus forward invariant.

For nonzero $(x,y)$, write $x=\sqrt{\rho }\cos \theta$, $y=\sqrt{\rho }\sin \theta$. Thenthroughout the annulus. Hence it contains no equilibrium. Any orbit starting there remains compactly trapped, has a nonempty omega-limit set, and the Poincaré-Bendixson theorem supplies the required periodic orbit in the stated annulus. For $k=0$ the annulus reduces to $\rho =1$ itself, and there $\dot{\theta }=1$, giving the periodic orbit directly.

The nullclines are $y=\mu x$ and $y=x^2/[\nu (1+x^2)]$. At a nonzero fixed point, cancellation of $x$ yields $\mu \nu =x/(1+x^2)$. The latter function attains its maximum $1/2$ at $x=1$. HenceFor $0<\mu <{\mu }_c$, $x_{-}<1<x_{+}$ and $x_{-}{x}_{+}=1$.

The Jacobian matrix isAt $O$ the distinct eigenvalues are $-\mu ,-\nu$, so the origin is a stable node. At a nonzero fixed point its trace is $-(\mu +\nu )<0$ andThe discriminant is $(\mu -\nu {)}^2+8x/(1+x^2{)}^2>0$. Thus $P_{-}$ is a saddle and $P_{+}$ is a stable node, never a focus.

Above the straight nullcline $\dot{x}>0$; below it $\dot{x}<0$. Below the curved nullcline $\dot{y}>0$; above it $\dot{y}<0$. At the saddle, an eigenvector can be chosen as $(1,\mu +\lambda )$: the stable eigenvalue gives negative slope, and the unstable eigenvalue gives positive slope. The stable manifold is the basin boundary. The two branches of the unstable manifold lead respectively to the origin and the positive stable node, as illustrated for representative parameters.

The nonnegative quadrant is forward invariant. Since $\dot{y}\le 1-\nu y$, $y$ stays bounded, and then $\dot{x}=-\mu x+y$ also bounds $x$. Moreover the divergence is the negative constant $-(\mu +\nu )$, excluding periodic or homoclinic loops by the Bendixson-Dulac criterion. The Poincaré-Bendixson theorem then leaves convergence to a fixed point. Initial data on the saddle's stable manifold approach the saddle; data in the two open basins approach extinction at $O$ or persistence at $P_{+}$, respectively.

A useful version of the Poincaré-Bendixson theorem says that a forward orbit of a continuously differentiable planar autonomous vector field which stays in a compact set has a nonempty compact omega-limit set; if that omega-limit set contains no equilibrium, it is a periodic orbit.

Let $r^2=x^2+y^2$. Direct calculation givesFor $r>0$, $|xy|\le r^2/2$, so $-9/8\le \dot{\theta }\le -7/8$. In particular there is no equilibrium away from the origin.

On $r=1/2$, both coefficients in the radial expression are strictly positive, so the flow crosses outward. On $r=2$, both are strictly negative, so the flow crosses inward. Consequently the compact annulusis forward invariant. Any orbit starting there stays there, and its omega-limit set avoids the only equilibrium, the origin. The Poincaré-Bendixson theorem therefore supplies at least one periodic orbit in this annulus.

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 Poincaré-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 Poincaré-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 Poincaré-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 Poincaré-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 Poincaré-Bendixson theorem therefore supplies a periodic orbit.

The Poincaré-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 Poincaré-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 Poincaré-Bendixson theorem gives a periodic orbit. HenceThis is the Brusselator periodic-orbit criterion.

The Poincaré-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.