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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!

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  • $\begingroup$ Take $n=1$ and $f(x)=x^2+1,\ t_0=x_0=0$. One has $x(t)=tan(t)=\phi(t)$. On the other hand, $\phi(t+\pi)=\phi(t)$ (in particular for $t=t_0$), then contradicting "the solution is defined for all $t\in \mathbb{R}$". $\endgroup$ Commented Nov 21, 2016 at 21:35
  • $\begingroup$ The unique solution is $\phi:(-\pi/2,\pi/2)\rightarrow R$, $\phi(t)=\tan t$. It is not possible to compute $\phi(\pi)$, because the maximum interval of definition is $(-\pi/2,\pi/2)$. $\endgroup$
    – user355857
    Commented Nov 21, 2016 at 22:56

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

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  • $\begingroup$ Hi @user39756! What do you mean with $\varphi(\cdot + T)$? $\endgroup$
    – Relure
    Commented Nov 21, 2016 at 23:37
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    $\begingroup$ @Relure It is the function $t\mapsto \varphi(t+T)$. It is also a solution in $I-T$ because $\varphi(t_0+T)=\varphi(t_0)=x_0$ and $\varphi'(t+T)=f(\varphi(t+T))$ for all $t\in I-T$. $\endgroup$
    – user39756
    Commented Nov 22, 2016 at 9:40

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