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An interior equilibrium point of a dynamical system must satisfyIf $ab\ne 1$, the unique candidate isIt belongs to $\Lambda$ precisely in either of the two casesAt this equilibrium the Jacobian matrix has determinantIn the first region the equilibrium has Poincare index $+1$, so index considerations alone do not rule out a periodic orbit. In the second region it is a saddle equilibrium of index $-1$. Everywhere else there is no interior equilibrium, except at $a=b=-1$, where the segment $x+y=1$ consists of equilibria.

Every simple closed curve in the convex set $\Lambda$ bounds a region contained in $\Lambda$. A periodic orbit has index $+1$, equal to the sum of the indices of the equilibria that it encloses. Thus a region with no equilibrium has total index zero, while the region with the single saddle has total index $-1$; both contradict the index of a planar periodic orbit. At $a=b=-1$, a periodic orbit cannot cross the equilibrium segment and must lie in one of its two complementary convex regions, where it encloses no equilibrium, giving the same contradiction. Therefore index considerations exclude periodic orbits except possibly when

Let $x_{\ast }$ be an equilibrium point of a dynamical system. A Lyapunov function on a neighbourhood of $x_{\ast }$ is a continuously differentiable function $V$ such thatand whose derivative along trajectories satisfies

The First Lyapunov theorem says that the existence of such a positive-definite $V$ proves Lyapunov stability of $x_{\ast }$. If $\dot{V}<0$ away from $x_{\ast }$, the Second Lyapunov theorem strengthens this to asymptotic stability.

LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set $\Omega$ on which $\dot{V}\le 0$, then it approaches the largest invariant subset ofIn particular, if that largest invariant subset consists only of $x_{\ast }$, every trajectory in $\Omega$ approaches $x_{\ast }$.

At an equilibrium, each product inmust vanish. This gives the two parameter-independent equilibriathe equilibria on $y=0$and the equilibria with both coordinates nonzerowhere matching signs are used.

The $y=0$ pair is born at $k=-9/4$, suggesting a saddle-node bifurcation. At $k=0$, one of these equilibria meets $(0,0)$, suggesting a transcritical bifurcation, while the two equilibria near $(1,2)$ are simultaneously born in a saddle-node. At $k=1$, the minus branch $(1-\sqrt{k},2-\sqrt{k})$ meets $(0,1)$, suggesting another transcritical bifurcation. At $k=4$, that branch meets the $y=0$ equilibrium $(-1,0)$, again suggesting a transcritical bifurcation. Thus the bifurcation parameter values are

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.

Put $y=\dot{x}$. The first-order system isAt an equilibrium point of a dynamical system, $y=0$ and $x(1-x^2{)}^2=0$, so the three fixed points are

DefineThen $U^{\prime }(x)=2x(1-x^2{)}^2=-F(x)$ andMoreover,so $E$ is positive definite at the origin. It is therefore a damped mechanical energy as a Lyapunov function, and the First Lyapunov theorem proves that the origin is Lyapunov stable.

Since $U(\pm 1)=1/3$, choose a compact energy sublevel $\{E\le c\}$ with $0<c<1/3$. In this set, $\dot{E}=0$ means $y=0$. A trajectory can remain in $y=0$ only when $F(x)=0$, and the chosen sublevel excludes $x=\pm 1$. Thus the largest invariant subset of $\{\dot{E}=0\}$ is the origin. The LaSalle invariance principle proves that the origin is asymptotically stable.

Finally, $E$ is radially unbounded because its leading terms are $x^6/3+y^2/2$. Every forward trajectory is therefore bounded. LaSalle's principle puts its omega-limit set insideAn omega-limit set of a bounded continuous trajectory is nonempty and connected. A connected subset of this three-point set is a singleton, so every trajectory has precisely one of the three fixed points as its omega-limit set.

At an equilibrium, the second equation gives $x=y=:s$. The first equation then givesHence the fixed-point branches in the $(a,x)$ plane areandThe three branches meet at $a=0$. The branch $x=a^2$ meets the positive square-root branch again when $a^2=\sqrt{a}$, so the second bifurcation value is

The Jacobian matrix at a general point isOn the branch $P_0=(a^2,a^2)$ it is triangular, with eigenvaluesThus $P_0$ is unstable for $a<0$, asymptotically stable for $0<a<1$, and unstable for $a>1$.

On a square-root branch write $s=\pm \sqrt{a}$. Thenwhose trace is $-1$ and determinant is $2s^2(s^3-1)$. The negative branch $s=-\sqrt{a}$ is therefore a saddle equilibrium for every $a>0$. The positive branch is a saddle for $0<a<1$ and asymptotically stable for $a>1$; close to $a=1$ it is a stable node.

To resolve the nonhyperbolic point at $a=0$, make the prescribed substitutionand append $\dot{a}=0$. The equations becomeThe centre subspace is $Y=X$. Seek the extended centre manifold for a parameter asThe centre-manifold invariance equation isTo second order, $\dot{X}=-aX+O(3)$, so comparison of coefficients givesand hence $A=C=0$, $B=1$. ThereforeSubstitution into the $X$ equation gives the reduced flowIts central branch $X=0$ is unstable for $a<0$ and stable for $a>0$, while the two nonzero branches $X\sim \pm \sqrt{a}$ for $a>0$ are unstable. Thus the bifurcation at zero is a subcritical pitchfork bifurcation with reversed normal-form parameter $\mu =-a$, exactly as recorded by the extended centre manifold of the 2023 Cambridge quadratic-product system.

The complete bifurcation diagram is therefore as follows. For $a<0$, only $P_0$ exists and is unstable. For $0<a<1$, $P_0$ is stable while both $P_{+}$ and $P_{-}$ are saddles. For $a>1$, $P_{+}$ is stable while $P_0$ and $P_{-}$ are saddles. At $a=1$, $P_0$ and $P_{+}$ cross and exchange stability, so the bifurcation is transcritical. This gives the bifurcation diagram of the 2023 Cambridge quadratic-product system.

Finally consider the phase plane near $(x,y)=(1,1)$ with $|a-1|\ll 1$. If $a<1$, the lower equilibrium $(a^2,a^2)$ is a stable node and the upper equilibrium $(\sqrt{a},\sqrt{a})$ is a saddle. If $a>1$, the lower equilibrium $(\sqrt{a},\sqrt{a})$ is the stable node and the upper equilibrium $(a^2,a^2)$ is the saddle. In each case the saddle has one-dimensional stable and unstable manifolds; one unstable separatrix runs toward the nearby stable node, while the other runs out of the local neighbourhood. The two equilibrium branches and their local invariant manifolds exchange roles as $a$ passes through one, which is the standard local phase portrait of a transcritical bifurcation.