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 $t \in [0,T]$.

  • 1
    $\begingroup$ Does your question simply mean: Given $R>0$ find upper estimates on the norm of solutions starting at $x$ with $\lvert x \rvert=R$? $\endgroup$
    – user539887
    Nov 24, 2018 at 8:48
  • $\begingroup$ @user539887: I guess it does... $\endgroup$ Nov 24, 2018 at 12:10
  • $\begingroup$ what do you mean by $L^1L^1 + L^1L^\infty$? is it really the Minkowski sum, or (I as suspect) the intersection? $\endgroup$ Nov 24, 2018 at 13:01

1 Answer 1


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 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" $C(T)$.

  • 2
    $\begingroup$ @Riku It's easy to find $F \in L^\infty([0,1],X)$ such that $\Phi(t,0)$ explodes in finite time for any function space $X$ that is translation invariant and contains some unbounded functions (in particular $X = L^1$). $\endgroup$ Nov 25, 2018 at 20:42
  • $\begingroup$ @MartinHairer I've a new follow up question here: mathoverflow.net/questions/328208/… $\endgroup$
    – Riku
    Apr 16, 2019 at 17:12

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