Choose $X_{\ast }=k_2{A}_0/(2k_1)$, $t_{\ast }=1/(k_2{A}_0)$, ${\ell }_{\ast }=(D_X/(k_2{A}_0){)}^{1/2}$, and an independently selected positive $Y_{\ast }$ representing its reference or supplied concentration. Set $X=X_{\ast }u$, $Y=Y_{\ast }v$, $t=t_{\ast }\tau$, $x={\ell }_{\ast }\xi$. Dividing the first equation by $X_{\ast }/t_{\ast }$ and the second by $Y_{\ast }/t_{\ast }$ yields the stated system withThe parameter $\beta$ can be changed through the chosen or physically supplied reference $Y$ concentration while these other combinations remain fixed. The combinations satisfy $\alpha \delta =\gamma$ for this reaction mechanism, so they need not be algebraically independent. The spatial scale makes the $u$ diffusion coefficient one, and the time and $X$ scales balance linear autocatalytic growth against the quadratic loss.

The law of mass action makes each elementary reaction rate proportional to the product of its reactant concentrations, with the stoichiometric change determining its contribution. Let $A=A_0$ be maintained constant, and let $D_X,D_Y>0$ be the diffusion coefficients. The first reaction consumes two $X$ molecules, the second creates one net $X$, the third consumes one each of $X,Y$, and the fourth consumes $Y$ while creating $X$. Fick's laws for diffusion giveHere the standard deterministic convention takes the event rate of the first reaction as $k_1{X}^2$, so its loss is $2k_1{X}^2$. A convention absorbing a combinatorial factor into $k_1$ leads to the same nondimensional model after consistently redefining that rate constant. The maintained reactant $A$ makes this an open chemical model; there is no conservation law requiring $A$ to decrease with reaction progress.

The fixed-point branches are $x=0$ and $x=\pm \sqrt{3-6/a}$ beginning at $a=2$. Zero is a saddle before that threshold and repelling afterwards. The nonzero branches are saddles until $a=3$, attracting for $3<a<7/2$, and repelling beyond their resonant unit-circle crossing. Solid portions in the diagram mark attracting states; dashed portions mark saddles or repellers.

At $a=3$, the two asymmetric two-cycles split from $x=\pm 1$ with components $[\pm \sqrt{3+3/a}\pm \sqrt{3-9/a}]/2$, pairing the two values with the same first sign. These branches are always unstable. In addition, the antipodal two-cycle occupies the constant component levels $x=\pm \sqrt{3}$ for every $a>0$, attracting only below $a=1/2$. It must be included for the actual printed map, even though the quadratic proposed in the previous part misses it. At a given parameter the two component levels represent one cycle, rather than two independent fixed points.

A genuine two-cycle must have the form $(x,y)\mapsto (y,x)$ with $x\ne y$. Hence $a(x-x^3/3)=x+y=a(y-y^3/3)$. Subtracting givesLet $S=x+y$ and $P=xy$. Using $x^3=3x-PS$ in either original equation gives $S(aP/3-1)=0$. The cubic swap-map two-cycle factorization exhibits two cases, so cancellation of $S$ would lose an actual branch.

If $S\ne 0$, then $P=3/a$ and $S^2=3+3/a$. Thus the source's quadratic holds for this family, withThere are two reflected genuine cycles for $a>3$, coalescing with the fixed points $p=\pm 1$ at $a=3$. For their two-step Jacobian matrix, $\mathrm{tr}D{F}^2=4-a$ and $\det D{F}^2=-2a^2+3a+9$. The unit-disk necessary inequality $1-\mathrm{tr}D{F}^2+\det D{F}^2>0$ fails, since it is $-2(a-3)(a+1)<0$. Therefore these quadratic-family cycles are unstable for every $a>3$. For these unstable asymmetric cycles of a cubic swap map, their periodic-orbit multipliers are real: one exceeds $1$, while the other crosses $-1$ at $a=(1+\sqrt{29})/2$. They are saddles before that secondary threshold and repelling beyond it. Their presence on the stable side of the fixed-point flip identifies the unstable branch of a subcritical flip.

If $S=0$, then $y=-x$ and the preceding quadratic identity gives $x^2=3$. Hence there is also the exact cycleThis antipodal cycle of a cubic swap map does not satisfy the printed $P=3/a$ condition: its product is $-3$. Along it the one-step Jacobian matrix is the same matrix $B=\left(\begin{array}{cc}0& 1\\ -2a& -1\end{array}\right)$ at both points, so the cycle periodic-orbit multipliers are the squares of the roots of ${\lambda }^2+\lambda +2a=0$. The Jury stability criterion conditions giveAt $a=1/2$ its periodic-orbit multipliers are on the unit circle and linearization is inconclusive. Its real-to-complex transition occurs at $a=1/8$. Thus the printed assertion about all period-two components is incomplete: it describes only $S\ne 0$, whereas the omitted antipodal cycle supplies the actual stable period-two interval.

A fixed point $z_{\ast }$ of a map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ satisfies $F(z_{\ast })=z_{\ast }$. Its linear perturbations evolve by the Jacobian matrix $DF(z_{\ast })$: both eigenvalues must lie strictly inside the unit disk for linear asymptotic stability. One inside and one outside gives a saddle; both outside give a repeller. Unit-modulus eigenvalues require nonlinear analysis, unlike the continuous-time test using signs of real parts.

Here a fixed point has $x=y=p$ and $2p=a(p-p^3/3)$. ThereforeAt zero the characteristic equation is ${\lambda }^2+\lambda -a=0$. Its negative root is less than $-1$ for every $a>0$, while its positive root crosses $+1$ at $a=2$. Thus zero is a saddle for $0<a<2$ and a real repeller for $a>2$, with a pitchfork of fixed-point branches at $a=2$.

At either nonzero point the characteristic equation is ${\lambda }^2+\lambda +2a-6=0$, whose discriminant is $25-8a$. For $2<a<3$ the fixed points are saddles. At $a=3$ their periodic-orbit multipliers are $0,-1$, giving a flip threshold. For $3<a<25/8$ they are attracting real nodes; at $25/8$ repeated-eigenvalue degenerate nodes; and for $25/8<a<7/2$ attracting foci. The Jury stability criterion conditions reduce to $a>3$ and $a<7/2$, confirming the full stable interval.

At $a=7/2$ the periodic-orbit multipliers are $e^{\pm 2\pi i/3}$. For larger $a$ they form a repelling focus equilibrium. This is a unit-circle crossing at a strong resonance of a planar map with ratio $1:3$; linearization alone does not justify asserting a generic nonresonant Neimark-Sacker bifurcation. The period-two branch at the flip threshold is derived next, including an additional family absent from the source's proposed quadratic.