= Solution
At $(\sigma,\mu)=(1,0)$ the origin's <matrix> is $\begin{pmatrix}0&0\\2&0\end{pmatrix}$: it has a double zero <eigenvalue> and one <eigenvector>. <Trace> and <determinant> vary independently with $\mu$ and $\sigma$ there, so two parameters are needed. The symmetry of an <odd function> removes quadratic terms, giving a <reflection-symmetric cubic double-zero normal form>. The quadratic nonlinearity required for a generic <Bogdanov–Takens bifurcation> vanishes.
Use a prime for <differentiation> with respect to the rescaled time. Direct substitution, with all rescaled variables written without tildes, gives
$$
\begin{aligned}
u'&=-sv+\tfrac12v^3+\varepsilon(\mu-v^2/2)u
+\tfrac12\varepsilon^2u^2v-\tfrac12\varepsilon^3u^3,\\
v'&=2u+\varepsilon(\mu-v^2/2)v
+\varepsilon^2(s-v^2/2)u-\tfrac12\varepsilon^3u^2v-\tfrac12\varepsilon^4u^3.
\end{aligned}
$$
At $\varepsilon=0$, the equations are $u'=-sv+v^3/2$, $v'=2u$. For
$$
\boxed{H=u^2+\frac s2v^2-\frac18v^4,}
$$
<differentiation> gives $H'=2u(-sv+v^3/2)+(sv-v^3/2)2u=0$. This is the <Hamiltonian blow-up of a reflection-symmetric double-zero point>.
For $s>0$, the <equilibrium points> are a <centre equilibrium> at $(0,0)$ and saddles at $(0,\pm\sqrt{2s})$. The <centre equilibrium> has <eigenvalues> $\pm i\sqrt{2s}$; at either <saddle equilibrium> they are $\pm2\sqrt s$. Levels $0<H<s^2/2$ contain closed inner ovals around the <centre equilibrium>. The <saddle equilibrium> level has
$$
\boxed{H_{\rm het}=\frac{s^2}{2},\qquad
u=\pm\frac{2s-v^2}{2\sqrt2},\quad|v|\le\sqrt{2s}.}
$$
These two branches form a <heteroclinic cycle> connecting the saddles. The branch with $u>0$ runs from the lower <saddle equilibrium> to the upper one because $v'=2u>0$; the other runs back. Outside the potential well, trajectories are open. The right panel of the preceding figure uses the coordinates $(v,u)$ to display the well and these directed connections.
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