Local bifurcation normal forms and numerical certificates — Examples

Its Jacobian is $\left[\begin{array}{cc}\mu & -1\\ 1& \mu \end{array}\right]$, with eigenvalues $\mu \pm i$, so ${\beta }^{\mathrm{\prime }}(0)=1$. The vector field is linear; therefore $B=0$ and $C=0$.

By [F1], $G_{21}=0$ and ${\mathrm{\ell }}_1=0$. The nondegeneracy hypothesis of the generic Hopf theorem is therefore absent even though the eigenvalue crossing is transverse.

In polar coordinates $\dot{R}=\mu R$ and $\dot{\theta }=1$. If $\mu \mathrm{\ne }0$, every nonzero radius changes strictly exponentially and cannot be periodic; if $\mu =0$, every circle $R>0$ is periodic, so there is no isolated branch selected by crossing.

The Jacobian has eigenvalues $\mu \pm i$, so ${\beta }^{\mathrm{\prime }}=1$; the normalized eigenvectors are $q=p=(1,-i{)}^T\mathrm{/}\sqrt{2}$. The quadratic derivative vanishes. The cubic coefficient has $c=1\mathrm{/}200$, so $C(q,q,\bar{q})=4cq$, $G_{21}=4c=1\mathrm{/}50$, and ${\mathrm{\ell }}_1=1\mathrm{/}100$, which lies in $[-1\mathrm{/}10,1\mathrm{/}10]$.

Every other field in the input record is valid and strictly separated as assumed, but the interval $[-1\mathrm{/}10,1\mathrm{/}10]$ contains zero. Hence the interval for ${\mathrm{\ell }}_1$ fails the strict nonzero test, and [F2] returns inconclusive rather than asserting a Hopf branch.

We have $F(0,\mu ,\epsilon )=\epsilon \mathrm{\ne }0$, and $F(-x,\mu ,\epsilon )+F(x,\mu ,\epsilon )=2\epsilon \mathrm{\ne }0$. Thus both the trivial-branch and odd-symmetry hypotheses in [F1] fail at the former pitchfork point.

At $\mu =0$, $x_{\ast }=-{\epsilon }^{1\mathrm{/}3}$ is an equilibrium and $F_x(x_{\ast },0,\epsilon )=3{\epsilon }^{2\mathrm{/}3}>0$. By [F2] it continues as a unique smooth simple equilibrium branch near $(x_{\ast },0)$; the equilibrium at the origin has disappeared.

The origin is an equilibrium for all $\mu$, and its Jacobian is $A(\mu )=\left[\begin{array}{cc}\mu & -1\\ 1& \mu \end{array}\right]$ with eigenvalues $\mu \pm i$. Thus the critical pair is simple, $\beta (\mu )=\mu$, ${\beta }^{\mathrm{\prime }}(0)=1$, and there are no other eigenvalues.

At $\mu =0$, take $q=p=(1,-i{)}^T\mathrm{/}\sqrt{2}$. The quadratic derivative $B$ is zero. For the cubic term $-\mathrm{\parallel }(x,y){\mathrm{\parallel }}^2(x,y)$, its symmetric third derivative is $C(u,v,w)=-2((u\cdot v)w+(u\cdot w)v+(v\cdot w)u)$, using the complex-bilinear extension of the real dot product. Since $q\cdot q=0$ and $q\cdot \bar{q}=1$, $C(q,q,\bar{q})=-4q$, so $G_{21}=p^{\ast }C(q,q,\bar{q})=-4$.

Therefore ${\mathrm{\ell }}_1=-2$. The hypotheses of [F2] hold, and $\beta {\mathrm{\ell }}_1<0$ exactly when $\mu >0$; since there are no transverse modes, the small cycle on that side is orbitally stable.

In polar coordinates $R=\sqrt{x^2+y^2}$ the equations give $\dot{R}=\mu R-R^3$ and $\dot{\theta }=1$. For $\mu >0$ the nonzero radial equilibrium is $R=\sqrt{\mu }$, and its radial derivative is $\mu -3R^2=-2\mu <0$, verifying the stable cycle and period $2\pi$.

The origin is an equilibrium and $A(\mu )=\left[\begin{array}{cc}\mu & -1\\ 1& \mu \end{array}\right]$ has eigenvalues $\mu \pm i$. Hence ${\beta }^{\mathrm{\prime }}(0)=1$ and there is no transverse spectrum.

With $q=p=(1,-i{)}^T\mathrm{/}\sqrt{2}$, $B=0$. The cubic vector field is $\mathrm{\parallel }(x,y){\mathrm{\parallel }}^2(x,y)$, so $C(u,v,w)=2((u\cdot v)w+(u\cdot w)v+(v\cdot w)u)$. Since $q\cdot q=0$ and $q\cdot \bar{q}=1$, $C(q,q,\bar{q})=4q$ and $G_{21}=4$.

Thus ${\mathrm{\ell }}_1=2>0$, and [F2] gives a small branch only for $\mu <0$; its radial multiplier is unstable because $\beta <0$.

Directly, polar coordinates give $\dot{R}=\mu R+R^3$ and $\dot{\theta }=1$. At $\mu <0$, $R=\sqrt{-\mu }$ is a radial equilibrium with derivative $\mu +3R^2=-2\mu >0$, so the periodic orbit is unstable.