Differential equation: autonomous system

This isn't homework. I have no idea what theorems I should be looking at to solve this. Guidance, partial and total solutions are all welcomed.

Let $f$ be a locally lipschitz function in an open set $G\subseteq \Bbb R^n$. Consider the autonomous system $y'=f(y)$.

Let $y(x,\xi)$ be the value at point $x$ of the maximal solution that satisfies the initial condition $y(0)=\xi$.

1. Prove that the domain of $y(\cdot, y(s, \xi))$ is $I-s$, where $I$ is the domain of $y(\cdot ,\xi)$.
2. Prove that for all $s, t$ such that $y(s, \xi)$ and $y(t+s, \xi)$ exist, then $y(t,y(s,\xi))$ also exists and $y(t,y(s,\xi))=y(t+s, \xi)$.
3. If $y$ is a maximal solution and there exists $T>0$ such that $y(0)=y(T)$ and $f(y(0))\neq 0$, then $y$ is a periodic solution and not constant.
4. If $y$ is a solution whose domain is $(a,+\infty)$, if $\eta:=\lim _{x\to +\infty}y(x)$ and $\eta \in G$, then $f(\eta)=0$.

EDIT

I found an alternative solution for 3. Please check my proof and give feedback in comments:

let $u$ be the restriction of $y$ to $[0,T]$, now let $\overline u$ be the periodic extension of $u$ to $\mathbb R$. It is easy to see that $\overline u$ is a solution to the given differential equation. But since $f$ is locally lipschitz, $\overline u$ must coincide with $y$ wherever they are both defined. Since $y$ is a maximal solution, it must be $\overline u$, so $y$ is defined on $\mathbb R$.

Solution 1:

These are the standard properties of the systems of autonomous equations. For example, property 1 follows from the fact that if $y(x)$ is a solution then $y(x+c)$ is also a solution to $$ y'=f(y) $$ for any constant $c$. To show that this is true plug in $y(x+c)$ into the system $$ \frac{dy(x+c)}{dx}=f(y(x+c)). $$ The left hand side can be rewritten, using the chain rule, as $$ \frac{dy(x+c)}{d(x+c)}, $$ which proves that $y(x+c)$ is a solution. Now this property allows you to consider solutions $y(0)=y_0$ and $y(x_0)=y_0$ as basically the same, the second is obtained from the first by the translation for $x_0$. Hence if $y(x;y(0;y_0))$ is the maximal solution on some interval, then $y(x;y(x_0;y_0))$ is the maximal solution on the translated interval.

Probably my explanation is too wordy, since this property (as well as others) are almost obvious. I will let you work on the rest of them.

Solution 2:

Regarding point 4., if $\lim_{x\to\infty} y(x) = \eta \in G$, then the fact that $y$ solves the differential equation together with the continuity of $f$ implies

$$f(\eta) = f\left(\lim_{x\to\infty} y(x)\right) = \lim_{x\to\infty} f(y(x)) = \lim_{x\to\infty} y'(x).$$

In particular, $\lim\limits_{x\to\infty} y'(x)$ exists. Now if $f(\eta)$ were $\neq 0$, then we would have $y_i'(x) > \delta > 0$ (or $y_i'(x) < -\delta < 0$) for all large enough $x$ and some component $y_i$ of $y$. But then we would have an inequality

$$y_i(x) \geqslant K + \delta\cdot x\qquad (\text{or } y_i(x) < K - \delta\cdot x)$$

for some $K\in\mathbb{R}$ and all $x \geqslant x_0$, and that implies that $y_i(x) \to \pm\infty$ for $x\to\infty$, in particular, $y(x)\not\to \eta$, contradicting the premise.

Solution 3:

For 3., suppose $I$, the maximal interval of existence, is bounded above, so $I=(a,b)$ with $b<\infty$. Then you can use the equation $y(x)=y(x-T)$ (a true fact for $a<x-T<x<b$) to define $y(x)$ for $x\in[b,b+T)$. Just check that this extended definition provides a solution on $(a,b+T)$, and you have a contradiction with $I$ being maximal. So $b=\infty$. Similar reasoning gives $a=-\infty$.