Compactly Supported Flow Consider the following initial value problem
$$ 
\begin{cases}
\frac{d}{dt} Y_t = \rho(Y_t)\\
Y_0 = 0
\end{cases}
$$
where $\rho(x)$ is a bump function supported near $0$ on $\mathbb{R}^1$.  Let $f:t\mapsto Y_t$.  
Why is $Im(f)\subseteq supp(\rho)$?
 A: Setup: I assume that by support you mean the set of points where $\rho\neq 0$ (as opposed to, say, the closure of this set) and  assume $0\in supp(\rho)$ i.e. $\rho(0)\neq 0$. Since the complement of $supp(\rho)$ is closed, its intersection with ray $x\geq 0$ being bounded below  must have minimum $b>0$ and and similarly intersection with $x\leq 0$ must have maximum $a<0$, so that on $(a,b)$ we have $\rho\neq 0$ (and $\rho(a)=\rho(b)=0$). Then without loss of generality $\rho(0)>0$ and so by continuity $\rho>0$ on $(a,b)$ (this is automatic if you assume $\rho \geq 0$ everywhere).  We will prove that if $\rho$ is differentiable at $b$ and at $a$ then $Im(f)=(a,b)$. 
As pointed out in an answer to An ODE involving bump functions we can explicitly solve the IVP; to do this we start by writing $g(t)=\int_0^t \frac{1}{\rho(s)}ds$. This is a continuous  monotone increasing function on $(a,b)$ and so it has range which is a (generalized) interval (which we will call $J$) and is invertible. The inverse $f:J\to (a,b)$ is a solution to the IVP on the time interval $J$. So all we need to show is that $J=(-\infty, \infty)$, that is to show that the improper integral $g(b)=\int_0^b \frac{1}{\rho(s)}ds$ diverges to $+\infty$ and improper integral $g(a)=\int_0^a \frac{1}{\rho(s)}ds$ diverges to $-\infty$. But this is not too bad. If $\rho$ is differentiable at $b$ then  there exists $c>0$ such that $\rho(x)\leq c (b-x)$ for all $x$ sufficiently close to $b$, say between $b_-$ and $b$. Then $\frac{1}{\rho(x)}\geq \frac{1}{c(b-x)}$ and 
$$\int_{b_-}^b \frac{1}{\rho(x)}dx \geq \int_{b_-}^b \frac{1}{c(b-x)} dx= \lim_{x\to b}\frac{1}{c}(\ln(b-b_-)-\ln(b-x))=\infty$$
Similar argument works "at the other end" near $a$, thus establishing $J=(-
\infty, \infty)$ and concluding the proof.
Of course when  $\rho$ is smoother the integral above diverges faster, and the flow approaches  $b$ and $a$ more slowly.
