= Solution
The specified coordinate change gives the exact equations
$$
\dot z=v/2,\qquad\dot v=-3v/2+3w,\qquad\dot w=-(z-v)^3/9.
$$
At linear order $v=2w$ on the <center manifold>. More precisely the stable coordinate $\eta=v-2w$ obeys $\dot\eta=-3\eta/2$ at linear order, while $\dot z=w$, $\dot w=0$. Thus $v$ relaxes towards its center-manifold value and can be eliminated by <center manifold> reduction or a properly ordered <adiabatic elimination>.
Write $v=2w+h_2(z,w)+h_3(z,w)+\cdots$, where $h_j$ is homogeneous of degree $j$. At degree two the graph invariance equation is $w\partial_z h_2=-3h_2/2$. On homogeneous quadratic polynomials the operator $w\partial_z$ is nilpotent, so adding $3/2$ makes it invertible. Hence $h_2=0$. Equivalently, the odd <vector field> permits a symmetry-compatible odd graph: <odd symmetry removes even center-manifold jets>. No quadratic forcing is present for <adiabatic elimination> either. However, setting $\dot v=0$ alone is insufficient at cubic order, because the evolving graph has a nonzero derivative.
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