1
$\begingroup$

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 $t \in [0,T]$.

$\endgroup$
  • 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 '18 at 8:48
  • $\begingroup$ @user539887: I guess it does... $\endgroup$ – leo monsaingeon Nov 24 '18 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? $\endgroup$ – leo monsaingeon Nov 24 '18 at 13:01
1
$\begingroup$

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" $C(T)$.

$\endgroup$
  • 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 $X = L^1$). $\endgroup$ – Martin Hairer Nov 25 '18 at 20:42
  • $\begingroup$ @MartinHairer I've a new follow up question here: mathoverflow.net/questions/328208/… $\endgroup$ – Riku Apr 16 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.