0.
\end{aligned}$$ Similarly, letting $x^- = \cos^{-1}\big(\frac{K^+}{K^-} c_{x_0}\big)$, the subset $S^-= \{(x,t): x < x^-\}$ is positively invariant. Since $\{ (x,t): x= x_0\}$ has empty intersection with $S^+ \cup S^-$, and since $S^+ \cup S^-$ is positively invariant, it follows that the $T$-periodic solution $x(t)$ with initial condition $x(t_0)=x_0$ must be contained in the complement of $S^+ \cup S^-$, i.e., for all $t\in\Re$, $x^- \leq x(t) \leq x^+$. For, if for some $\bar t \in
\Re$, $x(\bar t) \in S^+ \cup S^-$, then the fact that $x(t) \in S^+
\cup S^-$ for all $t\geq \bar t$ would contradict the periodicity of $x(t)$.

\[ex:ellipse\]

Suppose $\C$ is an ellipse with major semiaxis $A$, minor semiaxis $B$, and $2 \pi$-periodic parametrization $\sigma(s) = (A
  \cos s, B \sin s)$, with $A=15$, $B=10$. The curvature is [$\kappa(s) = AB / ( A^2 \sin^2 s + B^2 \cos^2 s
    )^{3/2}$]{}. For the initial condition of the VHC generator , we pick $x(0) = \pi/8$. Following Proposition \[prop:generator:periodic\_solutions\], we need to choose a $2 \pi$-periodic function $\tilde \mu(t)>0$, set $\tilde \delta(t) = \epsilon \tilde \mu(t)$, and find the unique value of $\epsilon>0$ guaranteeing that the solution with initial condition $x(0) = \pi/8$ is $2\pi$-periodic. There 