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Write $h=(h_1,h_2)$ for the homogeneous cubic correction and $A(x,y{)}^T=(y,0{)}^T$ for the nilpotent linear part. The chain rule gives the transformed cubic vector field as $f_3+Ah-DhA(x,y{)}^T$. Thus the cubic coefficients in the first equation arewhile those in the second equation areConsequently one explicit near-identity transformation isThe transformed equations have no cubic correction in $\dot{x}$, and only $x^3,x^{2}y$ survive in $\dot{y}$. The cubic coefficients areThis is the cubic elimination at a nilpotent double-zero point; the extra $3a_1$ is the nilpotent cubic damping invariant contribution from the first equation. Higher-order terms generated by the substitution are outside the cubic truncation.

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.