Suppose that exists $T>0$ such that $\varphi(t_0+T)=\varphi(t_0)$. Prove that the solution to the IVP is defined for all $t \in \mathbb{R}$.

Question

I'm doing an exercise of my Differential Equations' book. It says the following:

Let $f: \mathbb{R}^n \to \mathbb{R}^n$ such that the initial value problem

\begin{cases} x'(t)=f(x)\\ x(t_0)=x_0\\ \end{cases}

has an unique solution, let's say $\varphi(t)$. Suppose that exists $T>0$ such that $\varphi(t_0+T)=\varphi(t_0)$. Prove that the solution is defined for all $t \in \mathbb{R}$ and that $\varphi(t)$ is a periodic function.

I've been taught several theorems (Picard's, Peano's, etc...) that gives conditions on the initial value problems to determine if they have an unique solution.

Can we use conversely that information to say that $\varphi(t)$ is Lipschitz on a small interval $(t_0-\epsilon, t_0+\epsilon)$ and start working over here or I'm not choosing the right path to solve this exercise?

Thanks for your time!

Answer 1

Call $I=\,]a,b[$ the maximum interval on which $\varphi$ is defined, $-\infty\leq a<b\leq\infty$. Notice that $\varphi(\cdot +T)$ is also a solution with maximum interval $I-T=\,]a-T,b-T[$. By uniqueness,

1. $a-T=a$ and $b-T=b$, so $a=-\infty$ and $b=+\infty$.

2. For all $t\in \mathbb{R}=I=I-T$, $\varphi(t)=\varphi(t+T)$.