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Eliminate $\dot{v}$ from the first equation before linearization:All partial derivatives below are evaluated at $(\mu ,g(\mu ))$. The Jacobian matrix isIts trace and determinant areBy linear stability of a planar equilibrium, a positive trace makes both real eigenvalues positive or gives a complex pair with positive real part. The equilibrium is consequently unstable precisely whenA negative trace gives asymptotic stability; equality leaves a purely imaginary pair and requires a separate nonlinear bifurcation calculation. In particular, the unique positive equilibrium is never a saddle equilibrium.

Introduce the velocity variable $v=\dot{x}$. The second-order differential equation becomes the autonomous planar systemA critical point of a dynamical system is a point where both components of the vector field vanish: here $v=0$ and $f(x,0)=0$. It is a stationary solution in the phase plane. The Jacobian matrix at each such point gives the local linear stability of a planar equilibrium. For the equations below, a conserved energy also determines the nonlinear phase portrait; this is important because purely imaginary eigenvalues alone do not generally prove that an equilibrium is a center equilibrium.

At a fixed point, the first equation gives $y=-x$. Substitution into the second gives $2x(x^2-1)=0$, so the equilibrium points are $(0,0),(1,-1),(-1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1\pm i$, so the origin is an unstable spiral. The local rotation is clockwise: on the positive $x$-axis the velocity points downward. At either nonzero equilibrium point,Thus both nonzero equilibria are saddles, with eigenvalues $-2,2$. The stable manifold is tangent to $(1,-3)$, and the unstable manifold to $(1,1)$, at each saddle equilibrium. Since all these equilibrium points are hyperbolic, linear stability of a planar equilibrium gives their actual local types.

For the phase portrait, the nullclines are $y=-x$ and $y=x/(1-2x^2)$, the latter defined away from $x=\pm 1/\sqrt{2}$. The horizontal direction has the sign of $x+y$; the vertical direction has the sign of $(1-2x^2)y-x$. The system is invariant under $(x,y)\mapsto (-x,-y)$, so trajectories occur in centrally symmetric pairs. The arrows below show these directions; the highlighted stable manifolds enter the saddle equilibria, whereas the highlighted unstable manifolds leave them.

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.

At a nontrivial equilibrium $I^{\ast }>0$, the equation $I^{\prime }=0$ givesThe equation $R^{\prime }=0$ gives $R^{\ast }=\nu I^{\ast }/f$, and conservation of population then yieldsThese values are positive precisely in the epidemic regime $\beta N>\nu$.

Eliminate $R=N-S-I$. The two-dimensional system isSet $a=\beta I^{\ast }=f(\beta N-\nu )/(f+\nu )>0$. The Jacobian matrix at the endemic equilibrium isso its eigenvalues satisfyTheir sum is $-(f+a)<0$ and their product is $a(f+\nu )>0$, proving local asymptotic stability.

The discriminant isWriting $r=\beta N-\nu >0$, we have $a=fr/(f+\nu )$. As $f\to 0^{+}$,whereas as $f\to \infty$,Hence sufficiently slow immunity loss gives a stable focus: $I(t)$ approaches $I^{\ast }$ through damped oscillations, corresponding to successively smaller epidemic waves. Sufficiently rapid immunity loss gives a stable node: small disturbances are sums of two decaying real modes and show no forced oscillation. These conclusions and the equilibrium formulas are collected in the Endemic equilibrium of the SIR model with waning immunity.

Stability theory studies the response of a dynamical system to small changes of its state. Linear stability of a planar equilibrium describes the response near equilibria using the linearized evolution.