Periodic orbit shooting collocation and Floquet error — Examples

The identities $\dot{r}=(x{x}^{\mathrm{\prime }}+y{y}^{\mathrm{\prime }})\mathrm{/}r$ and $\dot{\vartheta }=(x{y}^{\mathrm{\prime }}-y{x}^{\mathrm{\prime }})\mathrm{/}{r}^2$ give $\dot{r}=r(1-r)$ and $\dot{\vartheta }=\omega$. Hence $r=1$ is a periodic orbit with least period $2\pi \mathrm{/}\omega =3$. Differentiating $r(1-r)$ at $r=1$ gives $\dot{\eta }=-\eta$, whose exact solution is $\eta (3)=e^{-3}\eta (0)$; this transverse multiplier lies strictly between zero and one.

By the forward-Euler update specified in the Statement, with $a=-1$ and $h=3$ we have ${\eta }_1=(1-3){\eta }_0=-2{\eta }_0$. Thus its one-step amplification has modulus $2>1$, despite the exact transverse multiplier being stable. Since this calculation provides no enclosure of the truncation defect or exact variational flow, it cannot serve as a certified multiplier claim under [F1].

The identities $\dot{r}=(x{x}^{\mathrm{\prime }}+y{y}^{\mathrm{\prime }})\mathrm{/}r$ and $\dot{\vartheta }=(x{y}^{\mathrm{\prime }}-y{x}^{\mathrm{\prime }})\mathrm{/}{r}^2$ give $\dot{r}=r(1-r)$ and $\dot{\vartheta }=1$. Thus $r=1$ is invariant and $\vartheta (t)=\vartheta (0)+t$, so the least period is $2\pi$. Differentiating $r(1-r)$ at $r=1$ gives the radial variational equation $\dot{\rho }=-\rho$, whose multiplier is $e^{-2\pi }<1$; the angular variational equation is $\dot{\alpha }=0$, whose multiplier is $1$. Because the polar-coordinate Jacobian at times $0$ and $2\pi$ is the same invertible matrix, the Cartesian monodromy is conjugate to $\mathrm{diag}(e^{-2\pi },1)$. For $r<1$ the radial derivative is positive, and for $r>1$ it is negative, so each nearby radius moves monotonically toward $1$ and remains between its initial value and $1$. It has a limit $\bar{r}$; if $\bar{r}(1-\bar{r})\mathrm{\ne }0$, then $\dot{r}=r(1-r)$ is eventually bounded away from zero, contradicting convergence. Hence $\bar{r}$ is an equilibrium, and the interval between the initial radius and $1$ contains only the equilibrium $1$, so $r(t)\to 1$. These bounds also give Lyapunov stability of the unit circle. Thus the orbit is orbitally asymptotically stable and its neutral multiplier is simple.

At $(1,0)$ the vector field is $(0,1)$. For ${\psi }_b=x-1$, $c_b=(1,0)$ and $c_b^{\mathrm{\top }}f(x_{\ast })=0$, so [F2] gives a singular shooting Jacobian. For ${\psi }_g=y$, $c_g=(0,1)$ and $c_g^{\mathrm{\top }}f(x_{\ast })=1$; the phase multiplier is simple by step 1.1, so [F2] gives an invertible Jacobian.

Both $0.995$ and $1.005$ lie in $[0.99,1.01]$, but the first has modulus less than one and the second has modulus greater than one. Thus the interval alone admits opposite stability classifications, and it also intersects the unit circle at $1$.

A contour count can certify the number of roots enclosed without sharpening their location relative to the circle. Since the enclosure does not place every non-phase multiplier strictly inside the unit disk or strictly outside it, the hypotheses required for a stability verdict in [F2] are not met. The sound output is inconclusive.

Under the stated hypothetical corrector output, [F1] supplies the exact periodic orbit and a monodromy matrix in the enclosure family.

The assumed cluster certificates apply to that matrix by [F2]. The van der Pol field is polynomial and hence $C^2$. The neutral multiplier cannot be on a certified polygon boundary, so it is interior to its count-one polygon. The other filled polygon is compact and lies strictly inside the unit disk, so its maximum modulus $r$ satisfies $r<1$; take $\sigma =1-r>0$. The two counts sum to the state dimension $2$, giving the complete cluster hypotheses of [F3], so the orbit is orbitally asymptotically stable. No numerical conclusion is claimed without those certificates.