$x(t)$ is smooth because the right-hand side of  is smooth. Since $x(t)$ is $T$-periodic, $\Phi:=x \circ \pi^{-1}$ is a well-defined function $[\Re]_T \to \Re$, and by [@Lee13 Theorem 4.29] it is smooth. Since $x = \Phi \circ \pi$, we have $\dot x(t) =
\Phi'(\pi(t)) \dot \pi(t) = \Phi'(\pi(t))$, so that $\Phi'(s) =\dot
x(t)|_{t = \pi^{-1}(s)}$. Using , we get $$\begin{aligned}
\Phi'(s) & = -\tilde a(\pi^{-1}(s)) \cos \Phi(s) + \tilde
\delta(\pi^{-1}(s)) \nonumber \\
&=- a(s) \cos \Phi(s) +
\delta(s). \label{eq:Phiprime}\end{aligned}$$ Since $\delta(s) \neq 0$ for all $s \in [\Re]_T$, by Lemma \[lem:feasibility\] the relation $\varphi = \Phi(s)$ is a VHC for system .

In light of Lemma \[lem:vcg\], we call the differential equation  a [*VHC generator*]{}, for which one is to pick a $T$-periodic $\tilde \delta:\Re \to \Re
\backslash \{0\}$ yielding a $T$-periodic solution. The next proposition shows how to pick $\tilde \delta$.

\[prop:generator:periodic\_solutions\] Consider the smooth and $T$-periodic differential equation , and set $\tilde \delta(t) =
\epsilon \tilde \mu(t)$, where $\tilde \mu: \Re \to (0,\infty)$ is smooth and $T$-periodic. Let $K^+ = \max_{t \in [0, T]}[ \tilde
  \mu(t)/\tilde a(t)]$, $K^- = \min_{t \in [0,T]} [\tilde
  \mu(t)/\tilde a(t)]$. Then, for any $x_0 \in (0,\pi/2)$ and $t_0
\in \Re$, the following two 