 Solution

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.