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Let $v:\mathbb{R}^n\to\mathbb{R}^n$ be continuous. Let $\gamma:[0,T)\to\mathbb{R}^n$ be the unique solution to $\frac{d\gamma}{dt}=v(\gamma),\gamma(0)=p$. Suppose $\gamma$ remains bounded. Show that $\lim_{t\to T}\gamma(t)$ exists so that $\gamma$ admits a continuous extension to $[0,T]$.

I don't have much idea. I'm thinking about Picard-Lindelof Theorem. Does the converse work? The solution is unique so $v$ is lipschitz? If so, uniform continuity can somehow get to what we want? And how does $\gamma$ being bounded come into play? The class I'm taking is not mainly about differential equations and I don't have much knowledge in it as well. This is like a question for us to play around ourselves. Any help is appreciated!

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  • $\begingroup$ You can have a look at these notes. The main theoretical tool that underpins continuation is Zorn's lemma. $\endgroup$ – Pantelis Sopasakis May 16 at 0:26
  • $\begingroup$ $\gamma$ bounded implies it never leaves a compact set $K$. On $K$, $v$ is automatically uniformly continuous $\endgroup$ – Calvin Khor May 16 at 0:37
  • $\begingroup$ @PantelisSopasakis Thank you so much! I think I get it! $\endgroup$ – Fluffy Skye May 16 at 2:18
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The Cauchy problem $\begin{cases}y'=1+\lvert y\rvert^{1/3}\\ y(0)=0\end{cases}$ has $v$ non-Lipschitz in any neighbourhood of the initial value, yet it has unique solution on $\Bbb R$: namely, $y$ is the inverse function of $\int_0^x(1+\lvert t\rvert^{1/3})^{-1}\,dt$.

Yes, $\gamma$ being bounded does come into play; see, for instance, the Cauchy problem $\begin{cases} y'=1+y^2\\ y(0)=0\end{cases}$, which has solution $\tan x$, and we all know what happens for $T=\frac\pi2$.

As for the problem at hand, let's call fix the notation that $g^i:\Bbb R^n\to\Bbb R$ is the $i$-th component of the function $g:\Bbb R^n\to\Bbb R^n$. The same notation will be adopted for vectors. Recall that:

  • given a function $f:[0,T)\to\Bbb R^n$ and some $c\in\Bbb R^n$, then $\lim_{t\to T^-} f(t)=c$ if and only if for all sequences $t_k\nearrow T$ there is a subsequence $\lbrace t_{k_h}\rbrace_{h\in\Bbb N}$ such that $\lim_{h\to\infty} f\left(t_{k_h}\right)=c$.

  • a continuous function $f:\Bbb R^n\to\Bbb R^n$ maps bounded sets to bounded sets, because the closure of a bounded set is compact.

Consider some sequence $s_k\nearrow T$. Since $\gamma(s_k)$ is a bounded sequence in $\Bbb R^n$, there is a convergent subsequence $\gamma\left(s_{k_h}\right)\to c\in \Bbb R^n$. We need to prove that $c=\lim_{t\to T^-} \gamma(t)$. Let $t_n\nearrow T$ be any sequence and let's select a subsequence $t_{n_h}$ such that $\gamma\left(t_{n_h}\right)\to b\in\Bbb R^n$. Assume as a contradiction that $c^i\ne b^i$ for some $i\in\{1,\cdots,n\}$ (notice that this is possibile only if $s_{k_h}\ne t_{n_h}$ eventually). Then, let's consider $u_h=\left\lvert \frac{\gamma^i\left(s_{k_h}\right)-\gamma^i\left(t_{n_h}\right)}{s_{k_h}-t_{n_h}}\right\rvert$. By Lagrange, for all $h$ there is some $\tau_h\in\left(\min\left\{s_{k_h},t_{n_h}\right\},\max\left\{s_{k_h},t_{n_h}\right\}\right)$ such that $u_h=\lvert \gamma'^i(\tau_h)\rvert=\lvert v^i(\gamma(\tau_h))\rvert$. Since $u_k$ is in the image of the bounded set $\gamma[0,T)$ by the (globally) continuous function $\lvert v^i\rvert$, we know that $u_k$ must be bounded.

Yet, since $b^i\ne c^i$, $$\lim_{h\to\infty}\left\lvert \frac{\gamma^i\left(s_{k_h}\right)-\gamma^i\left(t_{n_h}\right)}{s_{k_h}-t_{n_h}}\right\rvert=\left[\left\lvert\frac{c^i-b^i}{T-T}\right\rvert\right]=\left[\left\lvert\frac{c^i-b^i}{0}\right\rvert\right]=\infty$$ Absurd. Therefore $c^i=b^i$ for all $i$ and $c=\lim_{t\to T^-} \gamma(t)$.

So $\widehat\gamma(t)=\begin{cases}\gamma(t)&\text{if }t\in[0,T)\\ c&\text{if }t=T\end{cases}$ extends continuously $\gamma$ to $[0,T]$. Since $\lim_{t\to T^-}\widehat\gamma'(t)=v\left(\widehat\gamma(t)\right)$ exists, $\widehat\gamma$ is also differentiable on the left, and it solves the Cauchy problem. It is also completely determined by $\gamma=\left.\widehat\gamma\right\rvert_{[0,T)}$.

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  • $\begingroup$ By Lagrange, you mean the mean value theorem right? $\endgroup$ – Fluffy Skye May 16 at 6:08
  • $\begingroup$ Yes${}{}{}{}{}$. $\endgroup$ – Saucy O'Path May 16 at 6:08
  • $\begingroup$ @FluffySkye I skimmed them after posting the answer. It's certainly true, now that I think about it. And in fact the techinque I used is reminescent of the one that's used to prove that given a metric space $(X,d)$, a complete metric space $(Y,d')$, a subset $S\subseteq X$ and an uniformly continuous function $f:S\to Y$, there is exactly one extension $\overline f:\overline S\to Y$. $\endgroup$ – Saucy O'Path May 16 at 6:21
  • $\begingroup$ I see, so essentially we're showing $\gamma$ is uniformly continuous and then show uniformly continuous function can be continuously extended. Thank you! $\endgroup$ – Fluffy Skye May 16 at 6:23

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