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The positive fixed point $r=(-1+\sqrt{1+4\mu })/2$ of $g$ has periodic-orbit multiplier $-2r$. At $\mu =3/4$, $r=1/2$ and the periodic-orbit multiplier crosses $-1$. Factoring the second-iterate fixed-point equation givesThe second factor yields a two-cycleIt exists for $\mu >3/4$ and has periodic-orbit multiplier $4pq=4(1-\mu )$. Its stability range is $3/4<\mu <5/4$, so $g$ undergoes a supercritical period-doubling bifurcation at $\mu =3/4$. At $\mu =1$ the two-cycle is $\{0,1\}$ and its periodic-orbit multiplier is zero, making it superstable.

For positive $\mu$ and $0<|x|<\sqrt{\mu }$, direct composition gives the local second iterateThe one-sided limits at zero are opposite, and agree with zero exactly when $\mu =1$ in this neighborhood. There $F_1(x)=\mathrm{sgn}(x)(2x^2-x^4)$ is continuous at zero and fixes it. Thus the local return branch has the same discontinuity-closing transition as the map in part (a), with effective parameter ${\mu }^2-\mu$ crossing zero.

For $3/4<\mu <1$, both $p,q$ are positive and the period transfer from a quadratic map to its Lorenz sign lift gives the two stable cyclesFor $1<\mu <5/4$, $p>0$ and $q<0$, so the periodic-orbit multiplier of the $g$ cycle is negative and the lifted orbit isIt is one stable four-cycle. Hence the gluing of cycles in an iterated Lorenz map givesEquivalently, $F_{\mu }$ changes from two small stable fixed points $\pm q$ to a stable small two-cycle $\pm (-q)$. At the critical value itself, the actual convention gives $f_L(0)=-1$, $f_L(-1)=0$, $f_L(1)=0$: the surviving critical two-cycle is $\{0,-1\}$, and $+1$ is preperiodic. This critical itinerary must be evaluated directly, not with part (b)'s noncritical sign product.

There is also a precise qualification to the printed use of “continuous”. The whole second iterate is not continuous everywhere at $\mu =1$: $F_1(1)=-1$, while ${\lim }_{x\to 1^{-}}{F}_1(x)=1$ and ${\lim }_{x\to 1^{+}}{F}_1(x)=-1$. Jumps persist at the nonzero preimages of the original discontinuity. Therefore the stated global-bifurcation criterion holds for the return map restricted to a neighborhood of zero, as in the sketches; if continuity on all of $\mathbb{R}$ is demanded literally, the claim about $f_L^2$ needs that restriction. The orbit creation and destruction above is unaffected.

Put $s=\sqrt{b}$. The two fixed points on the invariant line $x=0$ are $P_{+}=(0,s)$ and $P_{-}=(0,-s)$. There are also $Q_{\pm }=(\pm \sqrt{b-a^2/4},-a/2)$ when $a^2<4b$. At equality they coalesce with $P_{-}$.

The Jacobian matrix isAt $P_{+}$ its eigenvalues are $-a-2s$ and $2s$, so it is always a saddle equilibrium. At $P_{-}$ they are $2s-a$ and $-2s$, so it is a saddle equilibrium for $a<2s$ and a stable node for $a>2s$, with a zero eigenvalue at $a=2s$.

At $Q_{\pm }$ the characteristic equation is ${\lambda }^2+a\lambda +4b-a^2=0$, givingFor $0<a^2<16b/5$ these points are stable spirals; for $16b/5<a^2<4b$ they are stable nodes. At $a^2=16b/5$ the double negative eigenvalue has only one independent eigenvector, so these are improper stable nodes. When $a=0$ they are center equilibria, as established in part (iii). Thus the collision at $\boxed{a^2=4b>0}$ changes both the number and stability of fixed points; it is the bifurcation classified below.

The fixed points satisfy $x[rx(1-x)-1]=0$. There is always $x=0$, and the other branches areThey coincide at $(r,x)=(4,1/2)$. For $0<x<1$, $F(x)>0$ and its maximum occurs at $x=2/3$, with value $4r/27$. Consequently $\boxed{F((0,1))\subset (0,1)⟺0<r<27/4}$; equality at the upper endpoint fails because the maximum is attained inside the interval.

The fixed-point multiplier at zero is zero, so zero is attracting for every $r>0$. On either nonzero branch,The lower branch has multiplier greater than one for $r>4$ and is repelling. The upper branch is attracting for $4<r<16/3$, reaches multiplier $-1$ at $x=3/4$, and is repelling for $r>16/3$. At $r=4$ the multiplier is $+1$, and the two branches are created in a saddle-node bifurcation. The fixed point there is attracting from one side and repelling from the other: $F(x)-x=-4x(x-1/2{)}^2$.

The second change is a supercritical period-doubling bifurcation at $\boxed{(r,x)=(16/3,3/4)}$. To check its type, expand at that fixed point. The local map has quadratic coefficient $a=-20/3$ and cubic coefficient $b=-16/3$ when the multiplier is $-1$. Its second iterate has cubic coefficient $-2(a^2+b)=-704/9<0$. The multiplier decreases through $-1$ as $r$ increases, so a small attracting two-cycle is born on the $r>16/3$ side. These exhaust the fixed-point bifurcations, since their multipliers have no other crossings of $\pm 1$.

At a fixed point, the first equation gives $y=0$, the second then gives $x=0$, and the third gives $z=-\mu$. Thus there is exactly one equilibrium for every real $\mu$:The Jacobian matrix there isOne eigenvalue is $-1$. The other two have product one and sum $5ϵ\mu$. For $\mu <0$ their real parts are negative, so the equilibrium is asymptotically stable; for $\mu >0$ both real parts are positive, leaving two unstable directions and one stable direction. At $\mu =0$ the eigenvalues are $-1,i,-i$, withTherefore the only local bifurcation of this equilibrium is a Hopf bifurcation at $\mu =0$. A repeated real eigenvalue away from zero does not itself cause a bifurcation: the equilibrium remains hyperbolic there.

The cubic centre manifold calculation below gives the Hopf normal form radial coefficient $-9ϵ/8<0$, so this is a supercritical Hopf bifurcation. Small attracting periodic orbits occur on the $\mu >0$ side. In particular, at $\mu =0$ the leading cubic radial drift is inward, although linear stability alone cannot decide that boundary case.

The Jacobian matrix at the origin and its characteristic polynomial areThe fixed-point multipliers areBy the Jury stability criterion, both lie strictly inside the unit circle exactly whenThe last condition always holds, since ${\mu }^2+\mu +2=(\mu +1/2{)}^2+7/4$. Therefore the origin is asymptotically stable exactly for $-1<\mu <0$ among the hyperbolic parameter values.

For completeness, its linear stability throughout the parameter line is as follows. When $\mu <-1$, the complex-conjugate Floquet multipliers have modulus $|\mu |>1$, so the origin repels. When $-1<\mu <-1/3$, their modulus is $|\mu |<1$, so it attracts with oscillation. At $\mu =-1/3$ the repeated Floquet multiplier is $1/3$, still attracting. For $-1/3<\mu <0$, both Floquet multipliers are real and in $(0,1)$. For $0<\mu <1$, $p(1)<0$ and $p(0)>0$, so one Floquet multiplier lies in $(0,1)$ and the other exceeds one: the origin is a saddle fixed point of a map. For $\mu >1$, the complex-conjugate Floquet multipliers have modulus $\mu >1$, so it repels.

Possible changes at the unit circle are exhausted by $p(1)=0$, $p(-1)=0$, or a nonreal pair with product one. They giveAt $\mu =0$ the Floquet multipliers are $0,1$; the nonzero fixed point branches in part (a) meet the origin, and part (c) identifies a subcritical pitchfork bifurcation. Indeed at $\mu =0$ the exact restricted map is $y\mapsto y+y^3$, which makes the origin unstable despite its neutral linear direction.

At $\mu =-1$ the Floquet multipliers are $i,-i$: this is a 1:4 strong resonance of a planar map, coinciding with the change from repelling to attracting as $\mu$ increases. A generic nonresonant Neimark–Sacker bifurcation conclusion would be unjustified. At $\mu =1$ there is a double Floquet multiplier $+1$; the matrix has a nontrivial Jordan block, and the two further nonzero fixed point branches meet the origin. Linear stability alone is inconclusive at these unit-modulus boundary values. The repeated Floquet multiplier at $\mu =-1/3$ is strictly inside the unit circle and gives no bifurcation.

Adjoin ${\mu }_{n+1}={\mu }_n$. At the origin the stable direction is the $x$ axis, with Floquet multiplier zero, while the $y$ and parameter directions have Floquet multiplier one. By the centre manifold theorem for a discrete dynamical system, write the extended centre manifold asOn it the scalar map is $Y=(1+\mu )y-\mu h(y,\mu )+y^3$. Its discrete centre-manifold invariance equation isPut $h=h_2+O({\rho }^3)$ with $\rho =|y|+|\mu |$. Since $Y=y+O({\rho }^2)$, the quadratic equation is simply $h_2(y,\mu )=\mu y$. ThusIn particular, setting the stable variable identically to zero would lose the leading parameter correction. The reduced map through total degree three isIn the natural pitchfork bifurcation scaling $\mu =O(y^2)$ this has leading form $y_{n+1}=y_n+\mu y_n+y_n^3$. Its nonzero fixed points satisfy $y^2\sim -\mu$, and their reduced fixed-point multipliers satisfy $1-2\mu +O({\mu }^2)>1$ for small $\mu <0$. The transverse Floquet multiplier remains inside the unit circle. Therefore the bifurcation at $\mu =0$ is a subcritical pitchfork bifurcation: the origin attracts for small negative $\mu$, each nonzero branch is a saddle fixed point of a map, and the origin is a saddle fixed point of a map for small positive $\mu$.

An exact check uses the full Jacobian matrix on either nonzero branch. Its trace is $T_{\ast }=1-2\mu +3{\mu }^2$ and its determinant is $D_{\ast }={\mu }^2$. Thuswhenever a nonzero branch exists. Also $p_{\ast }(0)={\mu }^2>0$, so one real Floquet multiplier lies in $(0,1)$ and the other in $(1,\infty )$. This proves that each nonzero branch is a saddle fixed point of a map throughout $\mu <0$ and $\mu >1$, in particular near the requested bifurcation.

For the real logistic map, $f(x)=x$ factors asThere is always the fixed point $x_0=0$. If $\mu \ne 0$ there is also $x_{\ast }=1-1/\mu$, distinct from zero unless $\mu =1$. At $\mu =0$ the map is constant and zero is its only fixed point. Negative $\mu$ must be retained because the paper specifies $x\in \mathbb{R}$.

Since $f^{\prime }(x)=\mu (1-2x)$, the fixed-point multipliers areThe fixed point stability for an iteration criterion gives $x_0$ attracting for $-1<\mu <1$, and $x_{\ast }$ attracting for $1<\mu <3$. The respective fixed points are repelling when $|\mu |>1$ and $|2-\mu |>1$. At the unit-modulus boundaries the linear criterion needs a nonlinear supplement.

At $\mu =1$, the two fixed point branches meet and exchange stability: this is a transcritical bifurcation. Explicitly, with $\delta =\mu -1$,which has the required nonzero parameter and quadratic terms. At the critical value itself, $f(x)=x-x^2$ attracts small positive points and pushes negative points away, so the merged fixed point is not stable from both sides.

At $\mu =-1$ the origin’s fixed-point multiplier crosses $-1$, while at $\mu =3$ the nonzero fixed-point multiplier crosses $-1$. Both are period-doubling bifurcations. To check existence and stability of the new periodic orbit, factor the second iteration of a map:The two non-fixed roots areThey form a real period-two orbit for $\mu <-1$ or $\mu >3$. Its periodic-orbit multiplier isIt is less than one in absolute value just below $-1$ and just above $3$, respectively. Thus these are supercritical bifurcations in the direction where the parent fixed point loses stability. At the critical values the fixed points themselves are weakly attracting: at $\mu =-1$, $f^2(x)=x-2x^3+x^4$; at $\mu =3$, setting $u=x-2/3$ gives $u\mapsto u-18u^3-27u^4$ after two steps. Both shrink sufficiently small nonzero absolute values. The bifurcation values for finite fixed points are thereforeThe disappearance of the formula for $x_{\ast }$ at $\mu =0$ corresponds to that branch going to infinity, not a local bifurcation at a finite fixed point.

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 planform is the horizontal spatial structure of a pattern, often described by the Fourier modes active at a bifurcation. For example, one pair of opposite wavevectors gives parallel convection rolls, while two equal-amplitude perpendicular pairs give squares. A planform includes the relative amplitudes and phases needed to distinguish spatial symmetry types.