Consider the following ODE initial value problem 
\begin{align\*}
&\frac{d}{dt}\Phi(t,x) = \boldsymbol{F}(t,\Phi(t,x)), & t \in [0,T], \ \ x \in \mathbb{R}^N,\\
&\Phi(0,x) = x, & x \in \mathbb{R}^N.
\end{align\*}


We say that $\Phi: [0,T] \times \mathbb{R}^N \to \mathbb{R}^N$ is the flow of the ODE. 


We assume that the vector field $\boldsymbol{F}:[0,T]\times \mathbb{R}^N \to \mathbb{R}^N$ is such that that $$\frac{|\boldsymbol{F}|}{1+|x|} \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right) + L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right),$$
that is, there exist 
\begin{align\*}
&\boldsymbol{F}\_1 \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right)\\
&\boldsymbol{F}\_2 \in L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right)
\end{align\*}
 such that $$\frac{\boldsymbol{F}}{1+|x|} = \boldsymbol{F}\_1 + \boldsymbol{F}\_2.$$




---


If $x \in B\_{R}(0)$, what is the [truncated cone with base $B\_R(0)$](http://web.tecnico.ulisboa.pt/~mcasquilho/compute/misc/,cone_vol/truncConeVert.jpeg), which we shall call $C(T)$, such that  

$$\Phi(t,x) \in C(T) $$
for all $t \in [0,T]$.



Quantitative finite speed of propagation property for ODE (cone of dependence)

> 
> **Edited according to Martin Hairer's comment:** the flow $\Phi(t,0)$ can blow up in finite (and arbitrarily small) time if the
>  $L^1(0,T;L^1)$ component $F\_1\neq 0$ in the Minkowski sum $\frac{F}{(1+|x|)}= F\_1+F\_2\in L^1(0,T;L^1) + L^1(0,T;L^\infty)$. So with the OP's assumption there
>  is no hope for a reasonable answer, hence from now on we simply assume
>  that $$ \frac{F}{1+|x|}\in L^1(0,T;L^\infty). $$
> 
> 
> 


For simplicity let me define $\beta(s):=\|F(s,\cdot)/(1+|.|)\|\_{\infty}\in L^1(0,T)$ and $B\_T:=\int\_0^T\beta(s)ds=\|F/(1+|x|)\|\_{L^1L^\infty}$.
Writing
\begin{multline\*}
|\Phi(t,x)-x|\leq \int\_0^t |F(s,\Phi(s,x))|ds\\
\leq \int\_0^t\frac{|F(s,\Phi(s,x))|}{1+|\Phi(s,x)|}(1+|\Phi(s,x)|)ds
\leq
\int\_0^t \beta(s) (1+|\Phi(s,x)|)ds
\end{multline\*}
we get, with $|\Phi(s,x)|\leq |x|+|\Phi(s,x)-x|\leq R+|\Phi(s,x)-x|$,
$$
|\Phi(t,x)-x|\leq (1+R)\int\_0^t\beta(s)ds +\int\_0^t\beta(s)|\Phi(s,x)-x|ds.
$$
Applying [Grönwall's inequality](https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality#Integral_form_for_continuous_functions) in its integral form (and observing that $t\mapsto\int\_0^t\beta(s)ds$ is continuous nondecreasing), we can conclude that
$$
|\Phi(t,x)-x|\leq (1+R)\int\_0^t\beta(s)ds \exp\left(\int\_0^t\beta(s)ds\right))\leq (1+R)B\_T\exp(B\_T)
$$
and this gives the "truncated cone" $C(T)$.


