Regular Lagrangian flow for the problem $\frac{d}{dt} X(t,x) = \chi_{\{x>0\}}(X(t,x))$

Question

Consider the problem $$(\star) \quad \begin{cases} \frac{d}{dt} X(t,x) = \chi_{\{x>0\}}(X(t,x)), &t \in [0,T],\\ X(0,x) = x, &x \in \mathbb R \end{cases} $$ where $\chi$ denotes the indicator function of a set.

• What is the unique regular Lagrangian flow of $(\star)$?
• How can it be approximated numerically?

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Questions related to the second point has been asked on Mathematica Stackexchange.

Answer 1

The flow $X:[0,T]\times \mathbb{R}\to\mathbb{R}$ defined by $X(t,x)= x+t\chi_{\mathbb{R}_+}(x)$, for all $(t,x)\in [0,T]\times \mathbb{R}$, is a regular Lagrangian flow solution to $(\star)$ in the sense of Definition (4) of the linked paper (for a.e. initial datum $x\in \mathbb{R}$, in fact for all, one has $(x+t\chi_{\mathbb{R}_+}(x))'=\chi_{\mathbb{R}_+}(x)=\chi_{\mathbb{R}_+}(x+t\chi_{\mathbb{R}_+}(x))$; moreover $X(t,\cdot)_\#\mathcal{L}^1\le\mathcal{L}^1$ for all $t\ge0$).

Note that according to the above definition, one can always modify a regular Lagrangian flow for a vector field $b$ at least on a countable number of flow lines (even to non-solution curves), always yielding to a regular Lagrangian solution; indeed the uniqueness has to be intended for the flow as an element of $L^\infty([0,T]\times \mathbb{R}^d,\mathbb{R}^d)$. The vector field $b$ itself is given as an element of $L^\infty([0,T]\times \mathbb{R}^d,\mathbb{R}^d)$.

Answer 2

Let $H$ be the Heaviside function (characteristic function of $(0,+\infty)$). The ODE $$ \dot x=H(x)\text{ on $t>0$}, \quad x(0)=0, $$ has solutions $ x_1(t) = 0 $ as well as $x_2(t)=t$. Thus non-uniqueness. Your example is one-dimensional: in that case, Lipschitz continuity of the flux is not required to get uniqueness. Take for instance $x_0\in \mathbb R, f\in C^0(\mathbb R, \mathbb R)$ such that $f(x_0)\not=0$. Then the ODE $$ \dot x= f(x), \quad x(0)=x_0, $$ has a unique (local) solution. Indeed, you can separate the variables and you get $$ \frac{dx}{f(x)} =dt. $$ Due to the assumption $f(x_0)\not=0$, you can consider an anti-derivative $G$ of $1/f$ near $x_0$: then the ODE is $\frac{d}{dt}\bigl(G(x(t))\bigr)=1,$ i.e. $$ G(x(t))=t+G(x_0). $$ It is easy to invert that relation, using the inverse function theorem since $G'(x_0)\not=0$ and you get eventually the unique solution, assuming as we may that $G(x_0)=0$, $$ x(t)=G^{-1}(t). $$ The difficulty with your example comes mainly from the fact that $H$ is vanishing at 0.