3
$\begingroup$

In Chapter 5 of M. Spivak's A Comprehensive Introduction To Differential Geometry he states the existence and uniqueness of integral curves $\alpha_x:(-b,b)\rightarrow U $ of an ODE with initial condition $$\alpha_x'(t)=f(\alpha(t)) $$ $$\alpha_x(0)=x $$ This is proven for a map $f:U\rightarrow \mathbb{R}^n$ defined in an open set $U$ and every $x\in B_a(x_0)$ for some $a>0$ and $x_0\in U$ such that the closed ball $\overline{B}_{2a}(x_0)\subset U$, we also ask $f$ to be locally Lipschitz (there are as well some restriction on $b$). Next he introduces the flow $$\alpha:(-b,b)\times B_a(x_0)\rightarrow U\quad \;\text{as }\quad (t,x)\mapsto\alpha_x(t) $$ Ok! this is all great and fairly straightforward, but the next discussion has an important step that I do not follow, here it is:

He argues that $\alpha$ is continuous (proven, no problem!) and since every $\alpha_y$ satisfies the condition $\alpha_y(0)=y$, then we have $$\alpha:\{0\}\times \overline{B} _{a/2}(x_0)\rightarrow \overline{B} _{a/2}(x_0) $$ Then by continuity of $\alpha$ and compactness of $\{0\} \times \overline{B} _{a/2}(x_0)$, then there exists $\epsilon>0$ such that $$\alpha:(-\epsilon,\epsilon) \times B_{a/2}(x_0)\rightarrow B_a(x_0) $$ This last step I do not get, it does not seems trivial to me the existence of such $\epsilon$. My attempt of give a proof of such statement is the following:

First, since $B_a(x_0)$ is an open neighborhood of $x_0$ and $\alpha$ is continuous we have an open set $V=(-\epsilon,\epsilon)\times B_\delta(x_0)$ for some $\epsilon,\delta>0$, such that $$\alpha(V)\subset B_a(x_0)$$ Here compactness should play an important role to bound $\delta$, but I cannot see this. Can someone please give me a hint or full proof of this last statement?

Thanks in advance!

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.