the maneuvering problem.]{}

Pick an arbitrary $x_0 \in (0,\pi/2)$ and $t_0 \in \Re$, and consider the VHC generator $$\label{eq:constraint_generator:proof}
\dot x = - \tilde a(t) \cos x + \epsilon \tilde \mu(t),$$ where $\tilde \mu(\cdot)>0$. Let $\cS(\epsilon)$ denote the solution of  at time $t_0+T$ with initial condition $x(t_0) = x_0$. Since the right-hand side of  is $T$-periodic, the solution of  with initial condition $x(t_0) =x_0$ is $T$-periodic if and only if $\cS(\epsilon) = x_0$. For all $\epsilon > \epsilon^+$ we have $$\begin{aligned}
\dot x & > -\tilde a(t) \cos x + \tilde \mu(t) \cos(x_0) / K^-\\
 & > -\tilde a(t) \left[ \cos x - \cos x_0 (\tilde \mu(t) / \tilde
  a(t) ) / K^- \right]] \\
& > -\tilde a(t) (\cos x - \cos x_0).
\end{aligned}$$ The solution of $\dot x = -\tilde a(t) (\cos x - \cos x_0)$ with initial condition $x(t_0) = x_0$ is $x(t) \equiv x_0$. Therefore, by the comparison lemma ([@Kha02 Lemma 3.4]), for all $\epsilon>
\epsilon^+$ the solution of  with initial condition $x(t_0) = x_0$ satisfies $x(t)> x_0$ for all $t >
t_0$, so that $\cS(\epsilon)> x_0$ for all $\epsilon> \epsilon^+$. A similar argument can be used to show that for all $\epsilon<
\epsilon^-$ the solution 