 Solution

The cubic Taylor expansion at the double-zero point isThere are no quadratic or quartic terms because the local vector field is odd under $(q,p)\mapsto (-q,-p)$. Set $u=p$, $v=q-p$, and use $\tau =2t$. The linear transformation is invertible, and the positive time rescaling preserves orbit orientation. With primes denoting $d/d\tau$, the result isThis supplies the requested cubic functions, with a remainder stronger than $O(4)$.

There is an important normal form check. Setting $x=u$, $y=u^{\prime }$ straightens the first equation, and differentiating givesIn particular the nilpotent cubic damping invariant is $g_{21}+3f_{30}=3/2-3/2=0$. A nonsingular cubic near-identity transformation and rescaling cannot change this vanishing invariant into the nonzero $-x^{2}y$ coefficient in the supplied later normal form. The conclusions for that assumed system are therefore given separately from the independently derived conclusions for the printed original equations.