 Solution

The specified coordinate change gives the exact equationsAt 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.