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](http://cvgmt.sns.it/media/doc/paper/1612/crippa_bari.pdf). How can it be computed explicitly?



Regular Lagrangian flow for explicit ODE with discontinuous right-hand side
$\newcommand{\Om}{\Omega}\newcommand{\om}{\omega}\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}$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 $t\_\om<\om<T$ 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<X(t\_3,\om),
\end{equation}
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 \om<t\})=t>0. 
\end{equation}
So, condition $\mu\_t\le L\la$ in (ii) fails to hold. This final contradiction concludes the proof. $\quad\Box$


