Heteroclinic splitting of a fold-Hopf amplitude cycle (${\lambda }_1={\lambda }_2/4+O(\epsilon )$)

Perturb the first integral of the quadratic fold-Hopf amplitude flow by $u^{\prime }=-2uv+\epsilon u({\lambda }_1+v^2)$, keeping $v^{\prime }={\lambda }_2+v^2+u^2$. Then $F^{\prime }=-\epsilon u({\lambda }_1+v^2)({\lambda }_2+u^2+v^2)$. On the interior heteroclinic orbit for ${\lambda }_2=-a^2$, $u=\sqrt{3(a^2-v^2)}$ and $v^{\prime }=2(a^2-v^2)$, so the first-order change is $-\epsilon \sqrt{3}{\int }_{-a}^a({\lambda }_1+v^2)\sqrt{a^2-v^2}dv=-\epsilon \pi \sqrt{3}{a}^2({\lambda }_1+a^2/4)/2$. The boundary connection contributes zero. The simple zero gives the displayed leading persistence curve, rather than an exact all-orders parameter equality.