= Solution
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 $f_3+Ah-Dh\,A(x,y)^T$. Thus the cubic coefficients in the first equation are
$$
a_1+\alpha_2,\quad b_1+\beta_2-3\alpha_1,\quad c_1+\gamma_2-2\beta_1,\quad d_1+\delta_2-\gamma_1,
$$
while those in the second equation are
$$
a_2,\quad b_2-3\alpha_2,\quad c_2-2\beta_2,\quad d_2-\gamma_2.
$$
Consequently one explicit <near-identity transformation> is
$$
\alpha_2=-a_1,\quad\beta_2=c_2/2,\quad\gamma_2=d_2,\quad\delta_2=0,\qquad
\alpha_1={b_1+c_2/2\over3},\quad\beta_1={c_1+d_2\over2},\quad\gamma_1=d_1,\quad\delta_1=0.
$$
The transformed equations have no cubic correction in $\dot x$, and only $x^3,x^2y$ survive in $\dot y$. \b[The cubic coefficients are]
$$
\boxed{P=a_2,\qquad Q=b_2+3a_1.}
$$
This 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.
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