Flow of a vector field: how existence of a flow line implies existence of flow.

Question

I am unable to see why there exists $U$ such that $\phi_t(x)$ exists for all $t\in[0,T]$. Can you please help me to understand the argument above.

Thanks.

Answer 1

Let $D_X$ be the subset of $N\times \mathbb{R}$ on which the flow is defined. By using uniqueness of integral curves and local solvability of the corresponding differential equation, one shows that $D_X$ is open.

If for some $x\in N$ the integral curve through $x$ exists for $t\in[0,T]$, one has $\{x\}\times [0,T]\subset D_X$. For $t\in[0,T]$, let $N_t\times M_t\subset D_X$ be an open set, which contains $(x,t)$. Then one gets $$\{x\}\times [0,T]\subset\bigcup\limits_{t\in[0,T]}N_t\times M_t$$ and since $\{x\}\times [0,T]$ is compact, one finds finitely many $t_1,...,t_n$, such that the $N_{t_i}\times M_{t_i}$ cover $\{x\}\times [0,T]$. Then $N_{t_1}\cap...\cap N_{t_n}$ is an open neighbourhood of $x$ in $N$ on which the flow is defined for times $t\in[0,T]$.

If $X(x)=0$ one has $\{x\}\times\mathbb{R}\subset D_X$, so especially $\{x\}\times[0,T]\subset D_X$ for arbitrarily large $T$ and by the first observation, there is an open neighbourhood of $x$, on which the flow is defined for these times.

It's not the case that you can find a neighbourhood on which the flow is everywhere defined of every point, on which the vector field is zero. Consider the open 2-disk and take the radial vector field scaled by the square of the Euclidean norm. $(0,0)$ is a zero of this vector field, but the integral curves are not everywhere defined for all points except zero.