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

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 $$t \in [0,T]$$.

• Does your question simply mean: Given $R>0$ find upper estimates on the norm of solutions starting at $x$ with $\lvert x \rvert=R$? – user539887 Nov 24 '18 at 8:48
• @user539887: I guess it does... – leo monsaingeon Nov 24 '18 at 12:10
• what do you mean by $L^1L^1 + L^1L^\infty$? is it really the Minkowski sum, or (I as suspect) the intersection? – leo monsaingeon Nov 24 '18 at 13:01

## 1 Answer

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" $$C(T)$$.

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