# ODE with a measurable vector field

Suppose we have a bounded Borel measurable vector field $$F:\mathbb{R}^n\to\mathbb{R}^n$$. To make the question non-trivial, assume that $$F\neq 0$$ eveywhere.

Question. Does there exist at least one Lipschitz integral curve? That is a Lipschitz function $$\varphi:(a,b)\to\mathbb{R}^n$$ such that $$\varphi'(t)=F(\varphi(t))$$ for almost all $$t\in (a,b)$$.

This question is related to: Set of integral curves of a vector field.

• If $A \subseteq \mathbb{R}$ is a measurable set such that $|(a, b) \cap A| > 0$ and $|(a, b) \setminus A| > 0$ whenever $a < b$, and $F(x) = -1 + 2 \times \mathbb{1}_A(x)$, then, if I am not mistaken, there is no solution $\varphi$. – Mateusz Kwaśnicki Feb 12 at 6:16
• @MateuszKwaśnicki I think this example can be easily generalized to higher dimensions by taking $F$ on each component with "separated variables". Please change your comment into an answer including higher dimensions. This is very nice. – Piotr Hajlasz Feb 12 at 12:24
• The example mentioned by @MateuszKwaśnicki is also mentioned in the paper "A Necessary and Sufficient Condition for Existence of Measurable Flow of a Bounded Borel Vector Field" by Nikolay A. Gusev, Moscow Mathematical Journal 18(1):85-92, 2018. – Skeeve Feb 12 at 19:06
• @Skeeve Thank for the reference! I will certainly read it. – Piotr Hajlasz Feb 12 at 19:21
• @Skeeve: If you like, feel free to write a more detailed answer based on Gusev's article, I'll be happy to read it. And I am too busy at the moment to turn my comment into an answer, sorry. – Mateusz Kwaśnicki Feb 12 at 21:05

For completeness let me add some details (as suggested by @MateuszKwaśnicki).

Let $$A\subseteq \mathbb R$$ be a Borel set of positive but not full measure in each interval, set $$F(x) = -1 + 2 \cdot 1_{A}(x)$$. Let $$\varphi\colon (a,b) \to \mathbb R$$ be a Lipschitz integral curve of $$F$$.

The set $$R = \varphi((a,b))$$ is an interval (by the intermediate value property) with positive Lebesgue measure (because $$\varphi'(t)\ne 0$$ for a.e. $$t\in(a,b)$$). Consider the set $$\tilde R = \{x \in R : \forall t \in \varphi^{-1}(\{x\}) \; \exists \varphi'(t) = F(\varphi(t)) \}$$ The set $$R\setminus \tilde R$$ is Lebesgue negligible, being the image of a Lebesgue negligible set $$\{t \in (a,b) : \not\exists \varphi'(t) \text{ or } \varphi'(t) \ne F(\varphi(t))\}$$ under Lipschitz map $$\varphi$$.

Note that for any $$x\in \tilde R$$ there exists just one $$t\in (a,b)$$ such that $$\varphi(t)=x$$. Indeed, otherwise one could find distinct points $$t_1, t_2 \in (a,b)$$ such that $$\varphi(t_1) = \varphi(t_2) = x \in \tilde R$$ and $$\varphi(t) - x$$ has constant sign on $$(t_1, t_2)$$. Then $$\varphi'(t_1)$$ and $$\varphi'(t_2)$$ have to have different signs, a contradiction with the definition of $$\tilde R$$. Therefore $$\varphi$$ is injective on $$\varphi^{-1}(\tilde R)$$. Then it is easy to show that $$\varphi$$ is injective on the whole $$(a,b)$$, using the fact that $$\varphi'(t)\ne 0$$ for a.e. $$t\in(a,b)$$.

By continuity and injectivity the function $$\varphi$$ is strictly monotone, hence one of the sets $$R \cap A$$ and $$R \setminus A$$ has measure zero (because $$\varphi$$ cannot map a negligible set to a set with positive measure). This contradicts the definition of $$A$$.

In higher dimensions one could take $$F(x) = (-1 + 2 \cdot 1_{A}(x_1), 1, \ldots, 1)$$.

• In higher dimensions you need to modify your construction. The set $A$ is one dimensional so $1_A(x)$ makes no sense. Did you mean $1_A(x_1)$? – Piotr Hajlasz Feb 17 at 17:36
• @PiotrHajlasz fixed, thanks! – Skeeve Feb 17 at 18:30