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.

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

  • $\begingroup$ What it the definition of regular Lagrangian flow? I have not found it in the linked paper: there is only definition of $\mathscr L_b$-Langrangian flow (Definition 13). $\endgroup$ – Skeeve Dec 10 '19 at 5:05
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    $\begingroup$ How specifically do you use the reference you are citing? Usually results on Lagrangian flows require some regularity of the divergence of the vector field. Does this really apply to your case? $\endgroup$ – Michael Renardy Dec 10 '19 at 12:37
  • $\begingroup$ @Skeeve Thank you for your comment. You can use definition 4 (page 5-6) of this paper: cvgmt.sns.it/media/doc/paper/1612/crippa_bari.pdf $\endgroup$ – Riku Dec 10 '19 at 12:40
  • $\begingroup$ @MichaelRenardy I think it does apply to my example. $\endgroup$ – Riku Dec 10 '19 at 12:41
  • $\begingroup$ @Riku then $X(t,x) = x + t \cdot \chi_{\{x>0\}}(x)$ seems to be a regular Lagrangian flow in the sense of definition 4 from the second paper. But why do you think it is unique? If it is indeed unique, I don't see the point in approximating it numerically. $\endgroup$ – Skeeve Dec 10 '19 at 14:22

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 $L^\infty([0,T]\times \mathbb{R}^d,\mathbb{R}^d)$.

  • $\begingroup$ Hi. Thanks for your answer. I don't quite understand if $X$ is also a classical solution of the ODE. $\endgroup$ – Riku Jan 22 at 10:56
  • $\begingroup$ Note that I've also asked a new related question: mathoverflow.net/questions/350934/… $\endgroup$ – Riku Jan 22 at 11:05
  • $\begingroup$ If classical solution means satisfying the equation point-wise, that is, the LHS has a derivative for all $t\in[0,T]$ which is equal to the RHS, yes. $\endgroup$ – Pietro Majer Jan 22 at 11:28
  • $\begingroup$ Thanks for the clarification. Then my additional question linked in the comment above becomes even more pressing to me: that is, what concrete (computable) example makes Ambrosio theory of regular Lagrangian flow relevant? $\endgroup$ – Riku Jan 22 at 11:47
  • $\begingroup$ Note that in the preceding comment I took X to be the one of the question; in general one needs a milder notion of solution for the problem to be well-posed. $\endgroup$ – Pietro Majer Jan 22 at 13:00

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.

  • $\begingroup$ Thanks for your answer. This makes me wonder another question, which I've just asked separately: Note that I've also asked a new related question: mathoverflow.net/questions/350934/… $\endgroup$ – Riku Jan 22 at 11:05

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