4

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 $\mathbb R$.

antifb
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3 Answers3

3

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.

Artem
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  • Thank you. Part of the reason I was having trouble is that I read $I-s$ as $I\setminus {s}$, I realised what it meant after reading your answer. Can you help me out with 3.? I proved that $y(x+T)=y(x)$ for all $x$ where $y(x), y(x+T)$ makes sense and I proved that $y$ isn't constant. I'm missing that $y$ is defined on $\Bbb R$. – antifb Oct 22 '13 at 18:14
  • This follows from 1, since if you have $y(0)=y(T)$ then you have it defined on $(T,2T)$ and so on. In general, any bounded solution is defined on $\bf R$ – Artem Oct 22 '13 at 21:49
  • I'm having trouble writing up the solution using 1. I do have a result I can use to show $y$ is defined on $\mathbb R$ if it is bounded, but I can´t prove it is bounded. How does that follow? Why can't $y$ to go $\pm \infty$ after $T$? – antifb Oct 23 '13 at 11:13
  • Because property 2 implies that $y(t;y_0)$ do not intersect. If you got beck to $y(0)$ does not matter at which time, you'll have to go along exactly the same way as before. – Artem Oct 23 '13 at 12:58
  • I'm having trouble both understanding and formalizing your reasoning. The only thing I'm missing in the whole problem is exactly that in 3., $y$ is defined on $\mathbb R$. Can you please add the details to the answer? Preferably an algebraic sort of answer. Anyway, thanks for the help. – antifb Oct 24 '13 at 09:12
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.

Daniel Fischer
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2

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 $a=-\infty$.

Bob Pego
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