# Can solution to differential equation be continuous extended?

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!

• You can have a look at these notes. The main theoretical tool that underpins continuation is Zorn's lemma. – Pantelis Sopasakis May 16 at 0:26
• $\gamma$ bounded implies it never leaves a compact set $K$. On $K$, $v$ is automatically uniformly continuous – Calvin Khor May 16 at 0:37
• @PantelisSopasakis Thank you so much! I think I get it! – Fluffy Skye May 16 at 2:18

## 1 Answer

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 $$\gamma=\left.\widehat\gamma\right\rvert_{[0,T)}$$.

• By Lagrange, you mean the mean value theorem right? – Fluffy Skye May 16 at 6:08
• Yes${}{}{}{}{}$. – Saucy O'Path May 16 at 6:08
• @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 $\overline f:\overline S\to Y$. – Saucy O'Path May 16 at 6:21
• I see, so essentially we're showing $\gamma$ is uniformly continuous and then show uniformly continuous function can be continuously extended. Thank you! – Fluffy Skye May 16 at 6:23