# 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

$$\newcommand{\Om}{\Omega}\newcommand{\om}{\omega}\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}$$As stated in my previous comment, in Theorem 3.1 of the paper linked by the OP 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.

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$$, which gives the desired contradiction. So, $$\begin{equation} \om-t_\om=X(t_\om,\om)=0. \tag{4}\label{4} \end{equation}$$

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 $$\begin{equation} X(t_2,\om)=X(t_3,\om)+\int_{t_3}^{t_2}(-1)\,ds which contradicts the condition that $$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 $$t\in(0,T)$$, $$\begin{equation} \mu_t(\{0\})=\la(\{x\in\R\colon X(t,x)=0\}) \ge\la(\{\om\in\Om\cap(0,T)\colon \om0. \end{equation}$$ 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.