I am trying to mark some homework solutions. The question is: suppose $\dot{x}=f(x)$, with $x \in \mathbb{S}^{1}$, and $f(x) \neq 0$. Assume existence and uniqueness theorem is satisfied. Is it possible for periodic solutions to exist?
Of course, the answer is that all solutions are periodic since $f(x) \neq 0$ which means $x(t)$ is monotone, and we have $x(t)=x(t) \mbox{mod} 1$. In other words, there exists time period $T>0$ s.t. $x(t,t_{0},x_{0})=x(t+T, t_{0}, x_{0})$, for any initial condition $x(t_{0})=x_{0}$.
However, apparently somebody told the students that we have $x(t,t_{0},x_{0})= x(t+2\pi, t_{0}, x_{0})$.
I am wondering why the apparent statement that $T=2\pi$. To me, the time period $T$ is dependent on the specific form of the solution (or equivalently, the form of $f(x)$).Hence we can't assume a priori that $T=2\pi$. So whoever said that, is making a mistake.
Am I right?