Consider the ODE $$\frac{d}{dt}\Phi(t,x) = \boldsymbol{F}(t,\Phi(t,x))$$ $$\Phi(0,x) = x, \quad x \in \mathbb{R}^N.$$

$\Phi: [0,T] \times \mathbb{R}^N \to \mathbb{R}^N$ is the flow of this ODE. On the vector field $\boldsymbol{F}$ we assume that $$\frac{|\boldsymbol{F}|}{1+|x|} \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right) + L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right),$$ by which I mean that there exists $\boldsymbol{F}_1 \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right)$ and $\boldsymbol{F}_2 \in L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right)$ 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 $B_R(0)$, which we shall call $C(T)$, such that
$$\Phi(t,x) \in C(T) $$ for all $t \in [0,T]$.

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 $B_R(0)$, which we shall call $C(T)$, such that
$$\Phi(t,x) \in C(T) $$ for all $t \in [0,T]$.

Consider the ODE $$\frac{d}{dt}\Phi(t,x) = \boldsymbol{F}(t,\Phi(t,x))$$ $$\Phi(0,x) = x, \quad x \in \mathbb{R}^N.$$

$\Phi: [0,T] \times \mathbb{R}^N \to \mathbb{R}^N$ is the flow of this ODE. On the vector field $\boldsymbol{F}$ we assume that $$\frac{|\boldsymbol{F}|}{1+|x|} \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right) + L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right).$$

If $x \in B_{R}(0)$, what is the truncated cone with base $B_R(0)$, which we shall call $C(T)$, such that
$$\Phi(t,x) \in C(T) $$ for all $t \in [0,T]$.

Consider the ODE $$\frac{d}{dt}\Phi(t,x) = \boldsymbol{F}(t,\Phi(t,x))$$ $$\Phi(0,x) = x, \quad x \in \mathbb{R}^N.$$

$\Phi: [0,T] \times \mathbb{R}^N \to \mathbb{R}^N$ is the flow of this ODE. On the vector field $\boldsymbol{F}$ we assume that $$\frac{|\boldsymbol{F}|}{1+|x|} \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right) + L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right),$$ by which I mean that there exists $\boldsymbol{F}_1 \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right)$ and $\boldsymbol{F}_2 \in L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right)$ 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 $B_R(0)$, which we shall call $C(T)$, such that
$$\Phi(t,x) \in C(T) $$ for all $t \in [0,T]$.

Consider the ODE $$\frac{d}{dt}\Phi(t,x) = \boldsymbol{F}(t,\Phi(t,x))$$ $$\Phi(0,x) = x, \quad x \in \mathbb{R}^N.$$

Assume that $$\frac{|\boldsymbol{F}|}{1+|x|} \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right) + L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right).$$

If $x \in B_{R}(0)$, what is the truncated cone with base $B_R(0)$, which we shall call $C(t)$, such that
$$\Phi(t,x) \in C(t) \ ?$$

Consider the ODE $$\frac{d}{dt}\Phi(t,x) = \boldsymbol{F}(t,\Phi(t,x))$$ $$\Phi(0,x) = x, \quad x \in \mathbb{R}^N.$$

$\Phi: [0,T] \times \mathbb{R}^N \to \mathbb{R}^N$ is the flow of this ODE. On the vector field $\boldsymbol{F}$ we assume that $$\frac{|\boldsymbol{F}|}{1+|x|} \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right) + L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right).$$

If $x \in B_{R}(0)$, what is the truncated cone with base $B_R(0)$, which we shall call $C(T)$, such that
$$\Phi(t,x) \in C(T) $$ for all $t \in [0,T]$.