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Substrate replication competes with saturating enzyme clearance in this fast-slow model, where $K>\lambda >0$ and $ϵ\ll 1$. The physical strip $u\ge 0$, $0\le v\le 1$ is invariant. If $0<\eta <\lambda /K$, the origin is attracting and a positive saddle exists at $u_e=\lambda /\eta -K$, $v_e=1-\eta K/\lambda$. Eliminating the fast complex fraction gives $u^{\prime }\simeq \eta u(u-u_e)/(u+K)$, separating extinction from runaway. For initially zero complexes, the stable-manifold threshold is $u_{\ast }=u_e+ϵ(1-\eta )v_e+O({ϵ}^2)$, away from singular parameter limits. To derive the correction, write the stable manifold as $u=u_e+ϵ\varphi (v)$; its invariance equation gives ${\varphi }^{\prime }=-(1-\eta )$ and $\varphi (v_e)=0$. A Dulac factor $1/u$ has strictly negative weighted divergence, excluding a periodic alternative to the two basins. Saturated clearance and autocatalytic growth can therefore create an unstable concentration threshold rather than a stable nonzero balance.

For positive $\sigma ,b,r$, the origin of the Lorenz system loses stability at $r=1$ and two equilibria $(\pm \sqrt{b(r-1)},\pm \sqrt{b(r-1)},r-1)$ emerge stably. Center-manifold projection gives $\dot{q}=\sigma (r-1)q/(\sigma +1)-\sigma q^3/[b(\sigma +1)]+\cdots$, proving a supercritical pitchfork. The nonzero-branch characteristic polynomial is ${\lambda }^3+(\sigma +b+1){\lambda }^2+b(\sigma +r)\lambda +2\sigma b(r-1)$. If $\sigma \le b+1$ these equilibria remain stable for all $r>1$. Otherwise their transverse Hopf crossing occurs at the displayed parameter, with frequency squared $b(\sigma +r_H)$. Its cubic Hopf coefficient is positive, so this Hopf is subcritical, producing an unstable small periodic orbit on the stable-equilibrium side. A zero eigenvalue detects the pitchfork, while the Hopf condition uses the Routh-Hurwitz equality; neither alone determines nonlinear criticality without the corresponding normal-form calculation.

The positive fixed point exists when $b<1+a$ and equals $(1,b)/(1+a-b)$; the negative one exists when $b>1-a$ and equals $(1,b)/(1-a-b)$. Equality on either boundary does not give a branch, because its fixed-point equation becomes $0=1$. For a genuine two-cycle and $a^2\ne (1-b{)}^2$, subtracting its scalar equations rules out equal signs of the two first coordinates. Solving with $x_1>0>x_2$ gives the displayed coordinates and second coordinates $y_1=b{x}_2$, $y_2=b{x}_1$. Their signs require exactly $a>|1-b|$. The cycle is unique up to interchanging its two points. Degenerate boundary parameter lines can support additional period-two families and are outside this statement.

A parameter specifies a member of a family of mathematical models and is held fixed while a particular solution evolves. Varying it can change equilibrium existence, stability or long-term dynamics, producing a bifurcation. A model parameter differs from its evolving state variables and from an initial condition, even though all may be numerical inputs to a computation.

After an initial layer of duration $O(ϵ)$ in slow time, $v\simeq u/(u+K)$ while $u$ changes only slightly. The reduced substrate equation isBelow $u_e$ it decreases to zero; above $u_e$ it increases without bound. The exact threshold for initial $v=0$ is the intersection of the saddle equilibrium's stable manifold with that axis. For small $ϵ$ it isIn dimensional initial-substrate units the leading threshold is $s_0{u}_{\ast }\simeq k_2{e}_0/k_0-(k_{-1}+k_2)/k_1$. Comparing the actual initial value $u(0)=1$ to this threshold determines its fate; the reference scale can be held fixed when varying the initial concentration.

The initial sequestration produces an $O(ϵ)$ correction if desired. Write the stable manifold as $u=u_e+ϵ\varphi (v)+O({ϵ}^2)$. The invariance equation at leading order gives ${\varphi }^{\prime }(v)=F(u_e,v)/G(u_e,v)=-(1-\eta )$ and $\varphi (v_e)=0$. Hencefor fixed parameters away from degenerate limits. The positive correction reflects the substrate consumed in building complexes from the initially enzyme-free-complex state.

The physical strip $u\ge 0$, $0\le v\le 1$ is invariant, and the off-diagonal derivatives are nonnegative there, so ordered initial substrate values remain ordered. There are only the attracting origin and the saddle equilibrium. With Dulac function $1/u$, the divergence of the weighted vector field is $-(K-\lambda )v/u^2-(1+K/u)/ϵ<0$, excluding periodic orbits in its interior. The saddle equilibrium separatrix therefore gives the unique threshold on the initial axis: below it solutions tend to extinction, at it they tend to the saddle equilibrium, and above it they escape to infinity. Growth is at most exponential, so this is $u\to \infty$ as time tends to infinity, not a finite-time blowup. This is autocatalytic enzyme clearance with a saddle threshold.

A saddle-node bifurcation creates or destroys two equilibria when a real eigenvalue passes through zero with a nonzero quadratic normal-form term. With a parameter orientation chosen so the pair appears for $\mu >0$, useThere are no equilibria for $\mu <0$, a double nonhyperbolic equilibrium at zero, and equilibria $q=\pm \sqrt{\mu }$ for $\mu >0$. The positive branch is stable and the negative branch unstable in the active direction; additional hyperbolic directions can make them a node and a saddle equilibrium. “Supercritical” for a fold depends on this chosen parameter orientation; unlike the pitchfork and Hopf cases it is not an intrinsic distinction between the stability of two possible cubic branches.

The supercritical pitchfork normal form of a dynamical system isThe symmetric equilibrium $q=0$ is stable for $\mu <0$ and unstable for $\mu >0$. Two new stable equilibria $q=\pm \sqrt{\mu }$ emerge on the unstable side of the original branch. Reflection symmetry $q\mapsto -q$ preserves the pair. The new equilibria approach zero continuously as the parameter reaches the threshold.

The Lorenz equilibria satisfy $y=x$, $z=x^2/b$ and $x(r-1-z)=0$. ThusAt the origin the third eigenvalue is $-b$ and the other two solve ${\lambda }^2+(\sigma +1)\lambda +\sigma (1-r)=0$. Hence the origin is stable for $r<1$, has one zero eigenvalue at $r=1$, and becomes a saddle equilibrium with one unstable direction for $r>1$. The symmetry $(x,y,z)\mapsto (-x,-y,z)$ interchanges the new branches.

The nonzero-equilibrium characteristic polynomial isNear $r=1$ its Routh-Hurwitz inequalities are strict, so both new branches are stable. More explicitly, center-manifold projection at $r=1$ givesThe linear coefficient crosses positively and the cubic coefficient is negative. Therefore the equilibrium bifurcation is a supercritical pitchfork at $r=1$, not a saddle equilibrium-node.

For completeness, the later stability boundary satisfiesIf $\sigma \le b+1$, the nonzero branches remain linearly stable for all $r>1$. If $\sigma >b+1$, they lose stability atAt that point $P=(\lambda +\sigma +b+1)({\lambda }^2+{\omega }_H^2)$ andThis establishes a genuine transverse Hopf crossing. It is subcritical for these positive parameters, so it is not the supercritical Hopf normal form of a dynamical system from (c). One way to check its nonlinear sign is the cubic Hopf formula. At $C_{+}$ write $s=\sqrt{b(r_H-1)}$ andFor right and adjoint eigenvectors $Aq=i{\omega }_{H}q$, $A^{\ast }p=-i{\omega }_{H}p$, $⟨p,q⟩=1$, the cubic coefficient has the sign ofSubstitution and clearing positive factors reduces its sign to that of the following polynomial in $h=\sigma -b-1>0$:Every term is positive. The radial cubic is therefore positive, giving the unstable small cycle on the stable-equilibrium side. There is no saddle equilibrium-node among the equilibrium branches for $\sigma ,b,r>0$: the only zero-eigenvalue branch intersection is the symmetry-forced pitchfork. This does not rule out separate global bifurcations of periodic orbits.

A fixed point $X_{\ast }$ satisfies $F(X_{\ast })=X_{\ast }$. The Jacobian matrix is $DF=\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right)$; its eigenvalues are the linear fixed-point multipliers used in (i)–(iii).

For the Lozi map, $y=bx$ and $x=1+bx-a|x|$. At a positive point this gives $(1+a-b)x=1$, and at a negative point $(1-a-b)x=1$. ThereforeThere is no fixed point at $x=0$, because then $y=0$ and the first component would be one. On an equality boundary the corresponding scalar equation is $0=1$, so that branch is absent, not a line of fixed points.

For $x\ne 0$, $DF=\left(\begin{array}{cc}-a\mathrm{sgn}x& 1\\ b& 0\end{array}\right)$. If $a=0$, there is exactly one fixed point $(1,b)/(1-b)$ unless $b=1$, when there is none. Its eigenvalues satisfy ${\lambda }^2=b$. It is a sink fixed point of a map for $|b|<1$ and a source fixed point of a map for $|b|>1$. At $b=-1$, the affine map about its fixed point is a quarter turn, $D{F}^2=-I$ and $F^4=I$, so it is neutral, not attracting. At $b=1$ no fixed point exists. The positive/negative branch labels follow the sign of $1-b$.

If $b=0$, $X_{+}=(1/(1+a),0)$ exists for $a>-1$ with fixed-point multipliers $-a,0$. It is a sink fixed point of a map for $-1<a<1$ and a forward saddle fixed point of a map for $a>1$. At $a=1$ the fixed-point multiplier $-1$ is neutral; on the invariant axis the interval $0\le x\le 1$ is mapped by $x\mapsto 1-x$, giving two-cycles and the fixed midpoint. The negative point $X_{-}=(1/(1-a),0)$ exists only for $a>1$ and has fixed-point multipliers $a,0$, again a forward saddle fixed point of a map. For $a\le -1$ there is no fixed point. At $a=b=0$, both fixed-point multipliers vanish and two iterates send all initial points to the fixed point.

For a genuine two-cycle let the first coordinates be $x_1,x_2$ and put $h=1-b$. The two equations are $h{x}_2=1-a|x_1|$, $h{x}_1=1-a|x_2|$. If both first coordinates had the same sign $s$, subtracting would give $(h-as)(x_2-x_1)=0$. Since distinct cycle points must have different first coordinates, this contradicts $a^2\ne h^2$. If one coordinate were zero, the remaining equations would instead force $a=\pm h$, also excluded. Hence $\boxed{x_1{x}_2<0}$.

Choose $x_1>0>x_2$. Solving $a{x}_1+h{x}_2=1$, $h{x}_1-a{x}_2=1$ givesThe required signs hold exactly when $\boxed{a>|1-b|}$. This is the unique two-cycle up to interchanging its two points, under the stipulated nondegeneracy. Borderline parameter lines can carry different degenerate families and were explicitly excluded. These are Lozi fixed points and the alternating-sign two-cycle.

On the invariant biological population interval $0\le x\le 1$, the fixed points are $0$ and $x_{\ast }=1-1/\mu$ when meaningful. The zero fixed-point multiplier is $\mu$: extinction attracts for $0\le \mu <1$. At $\mu =1$, the nonzero branch crosses the origin in the full real-line family, giving a transcritical bifurcation. The nonzero point lies in the biological population interval for $\mu >1$ and has fixed-point multiplier $2-\mu$, so it is stable for $1<\mu <3$ and loses stability through $-1$ at $\mu =3$.

Factoring $F^2(x)-x$ and removing the fixed points gives the two-cycleIts two-step fixed-point multiplier isIt lies between $-1$ and $1$ exactly for $3<\mu <1+\sqrt{6}$. At the upper endpoint it crosses $-1$, creating the next period doubling. The two-cycle amplitude tends to zero like $\sqrt{\mu -3}$ at its birth, so the first flip is supercritical. At the equality endpoints linear fixed-point multipliers alone do not establish asymptotic stability.

Period-doubling cascade universality concerns the continuation of this period-doubling cascade in smooth unimodal maps with a quadratic maximum. After suitable rescalings their repeated two-step maps approach the same limiting shape; successive parameter spacings have the Feigenbaum ratio approximately $4.6692$ and the spatial rescaling has magnitude approximately $2.5029$. These are asymptotic statements involving later doublings, not numbers that can be established from the first two thresholds alone. The accumulation near $\mu \simeq 3.56995$ is outside the first interval considered here. One should not claim chaos throughout the stable two-cycle interval.

Four useful dynamical properties to distinguish are sensitivity to initial conditions, topological transitivity or mixing, dense periodic points, and aperiodic orbits with persistent complex evolution. They concern the invariant set, not an assertion that every orbit must be aperiodic. A positive Lyapunov exponent for typical initial data is an additional quantitative diagnostic.

At $\mu =4$, set $x_0={\sin }^2{\theta }_0$. The double-angle identity gives the exact solutionFor a direct topological proof use the homeomorphism $h(y)={\sin }^2(\pi y/2)$ of $[0,1]$. Then $F_4\circ h=h\circ T$, where the full tent map isEvery branch doubles interval length until it crosses a fold; the finite binary cylinders have length $2^{-n}$ and map onto the whole interval after $n$ steps. Thus every open interval eventually has an iterate covering the full interval: the map is topologically mixing and sensitive. Repeating a finite left/right branch word gives a periodic itinerary; arbitrarily long prescribed prefixes locate such periodic points in arbitrarily small cylinders, proving density. Nonperiodic binary itineraries yield uncountably many aperiodic orbits. Conjugacy transfers these properties to $F_4$, proving chaotic behaviour while retaining its dense special periodic orbits.

As a metric check, its invariant measure has density $1/[\pi \sqrt{x(1-x)}]$ and its typical Lyapunov exponent is $\log 2>0$. This is a typical-orbit assertion, not a claim about the derivative along every exceptional critical or preperiodic orbit. The explicit trigonometric solution and tent-map conjugacy explain why deterministic iteration of a map can nevertheless amplify uncertainty.