# Regular Lagrangian flow for explicit ODE with discontinuous right-hand side

Consider the problem $$(\star) \quad \begin{cases} \frac{d}{dt} X(t,x) = \begin{cases} - 1 & \text{ if } X(t,x) >0, \\ 1 & \text{ if } X(t,x) < 0 \end{cases}, &t \in [0,T],\\ X(0,x) = x, &x \in \mathbb R \end{cases}$$

This is a typical example of ODE that does not have a solution in the classical sense. The right-hand side is not continuous. However, it is BV, so there exists a regular Lagrangian flow in the sense of Ambrosio. How can it be computed explicitly?

• Where is it proved that, if the right-hand side is BV, then there exists a regular Lagrangian flow? Jul 21, 2022 at 14:16
• The solution reaches $X=0$ in finite time $t=|x|$, and then the RHS is undefined at $X=0$. Jul 21, 2022 at 17:48
• @IosifPinelis L. Ambrosio: Transport equation and Cauchy problem for BV vector fields. Invent. Math., 158 (2004), 227–26
– Riku
Jul 21, 2022 at 20:28
• In Theorem 3.1 of that paper about the existence (and uniqueness) of a Lagrangian flow, there is the condition that $D\cdot b$ be absolutely continuous with respect to the Lebesgue measure, where $b$ is the right-hand side of the ODE. Here this condition obviously fails to hold, and I think it should not be hard to see that no Lagrangian flow solution exists here. Jul 22, 2022 at 1:53


Let us show that in your case there is in fact no regular Lagrangian flow. Indeed, suppose the contrary, that there is a regular Lagrangian flow $$X\colon[0,T]\times\R\to\R$$. Let $$\la$$ denote the Lebesgue measure over $$\R$$. Then, according to Definition 4 in the paper linked by the OP, there is a set $$\Om\subseteq\R$$ such that $$\la(\R\setminus\Om)=0$$ and the following two conditions hold:

(i) For each $$\om\in\Om$$ and all $$t\in[0,T]$$ $$\begin{equation*} X(t,\om)=\om+\int_0^t b(s,X(s,\om))\,ds, \tag{1}\label{1} \end{equation*}$$ where $$\begin{equation*} b(s,x):=1(X(s,x)<0)-1(X(s,x)>0). \end{equation*}$$

(ii) There is some real $$L$$ such that for all $$t\in[0,T]$$ $$\begin{equation*} \mu_t:=X(t,\cdot)_{\#}\la\le L\la, \end{equation*}$$ so that $$\mu_t$$ is the push-forward of the Lebesgue measure $$\la$$ via the map $$X(t,\cdot)$$.

Take any $$\om\in\Om\cap(0,T)$$. Let $$\begin{equation*} E_\om:=\{t\in[0,T]\colon X(s,\om)>0\ \forall s\in[0,t)\} \end{equation*}$$ and $$\begin{equation*} t_\om:=\sup E_\om=\max E_\om. \tag{2}\label{2} \end{equation*}$$ By \eqref{1}, $$\begin{equation*} X(t_\om,\om)=\om+\int_0^{t_\om} (-1)\,ds=\om-t_\om. \tag{3}\label{3} \end{equation*}$$ Again by \eqref{1}, $$X(t,\om)$$ is continuous in $$t$$. So, by \eqref{3} and \eqref{2}, $$\om-t_\om=X(t_\om,\om)\ge0$$. To obtain a contradiction, suppose that $$t_\om<\om$$, that is, $$X(t_\om,\om)>0$$. Then $$t_\om<\om and, again by the continuity of $$X(t,\om)$$ in $$t$$, we have $$X(s,\om)>0$$ for some $$u\in(t_\om,T)$$ and all $$s\in(t_\om,u)$$. So, $$u\in E_\om$$ and $$u>t_\om=\max E_\om$$, which gives the desired contradiction. So, $$$$\om-t_\om=X(t_\om,\om)=0. \tag{4}\label{4}$$$$

To obtain another contradiction, suppose that $$X(t_1,\om)>0$$ for some $$t_1\in(t_\om,T]$$. Again, because $$X(\cdot,\om)$$ is continuous, there is some $$t_2\in[t_\om,t_1]$$ such that $$X(s,\om)\le X(t_2,\om)$$ for all $$s\in[t_\om,t_1]$$. So, $$X(t_2,\om)\ge X(t_1,\om)>0$$. So, in view of \eqref{4}, $$t_2\ne t_\om$$ and hence $$t_2\in(t_\om,t_1]$$. Again by the continuity of $$X(\cdot,\om)$$, there is some $$t_3\in(t_\om,t_2)$$ such that $$X(s,\om)>0$$ for all $$s\in[t_3,t_2]$$. It follows by \eqref{1} that $$$$X(t_2,\om)=X(t_3,\om)+\int_{t_3}^{t_2}(-1)\,ds which contradicts the condition that $$X(s,\om)\le X(t_2,\om)$$ for all $$s\in[t_\om,t_1]$$. So, $$X(t,\om)\le0$$ for all $$t\in(t_\om,T]$$. Similarly, $$X(t,\om)\ge0$$ for all $$t\in(t_\om,T]$$.

So, $$X(t,\om)=0$$ for all $$t\in(t_\om,T]=(\om,T]$$ and all $$\om\in\Om\cap(0,T)$$. So, for each $$t\in(0,T)$$, $$$$\mu_t(\{0\})=\la(\{x\in\R\colon X(t,x)=0\}) \ge\la(\{\om\in\Om\cap(0,T)\colon \om0.$$$$ So, condition $$\mu_t\le L\la$$ in (ii) fails to hold. This final contradiction concludes the proof. $$\quad\Box$$

No. The theory of Regular Lagrangian Flows rests on two key assumptions: a Sobolev/BV regularity of the vector field and a bound from below on its divergence.

In your case you are solving $$X'=b(X)$$ with $$b(x)$$ equal to $$1$$ for $$x<0$$ and to $$-1$$ for $$x>0$$. It is true that this is BV, but its divergence is $$-2\delta_0$$ and thus the theory does not apply.

Notice that the bound on the divergence is related to the `bounded compressibility' that enters in the definition of Regular Lagrangian Flow.