# Non-uniqueness of flow for divergence free vector fields

I am looking for a (possibly time-dependent) vector field $v: [0,1] \times \mathbb R_x^d \to \mathbb R^d$ such that

1. $\text{div}_x v = 0$ ;

2. $v$ has more than one (measure-preserving) flow, i.e. there exist two different maps $X_1,X_2$ such that $$X_i(t,x) = x + \int_0^t v(s,X_i(s,x))\,ds$$ for every $x \in \mathbb R^d$ and $t \in [0,1]$, for $i=1,2$;

3. $v$ is compactly supported (say in the unit ball) and bounded.

The paper by Aizenman, M. On Vector Fields as Generators of Flows: A Counterexample to Nelson's Conjecture (Annals of Mathematics, Second Series, 107, no. 2 (1978): 287-96. doi:10.2307/1971145) seems to address the points 1. & 2. but I have no idea if one can include also point 3. I believe that, given a vector field solving 1 & 2, it should be rather easy to cook up one satisfying 1,2 & 3 but I do not see how to handle it. There are also some related results in the periodic setting by Depauw.

• It seems that Aizenman's article deals with a non-uniqueness result and not with the existence of two different flows. The Depauw paper provides an Eulerian perspective, i.e. replaces the ODE by a first-order PDE. On the other hand, to get 3 in your question, it is quite likely that Aizenman's vector field $X$ is three-dimensional and thus $X=\text{curl} V$. To get 3, it should be enough to take $\tilde X=\text{curl}(\phi V)$ where $\phi$ is a cutoff function. May 21 '18 at 9:55
• @Bazin Thanks, this is indeed what I was looking for but I am not totally sure of how to do this. I agree that Aizenman's field is a curl and multiply times a test function seems a reasonable thing to do to have (3) satisfied. But how can we be sure that we have non-unique measure preserving flow? Do you see how to fix these details? I would be grateful if you could kindly expand your comment. Thanks. May 21 '18 at 18:57
• +1, nice question! I am interested into the Eulerian side and I know Depauw's example. Could one do something similar to what @Bazin suggests also in that case? May 22 '18 at 13:42
• @user111164 Well, I think that this requires some work: taking a close look at Depauw's example, you will see that his construction of a vector field $V$ so that $curl V$ does not have uniqueness is rather explicit (as well as Aizenman's). Then you have to check $curl(\phi V)$ where $\phi$ is a cutoff function and carefully follow what happens in the transition region where $0<\phi <1$. Some "good" choices of $\phi$ may simplify matters. May 22 '18 at 16:23
• @Bazin Mmm... what do you mean by "good"? I have read Depauw's example, I agree that the construction is rather explicit. The point is that I do not see how to "localize" (this is probably related also to what Uncle Sam was saying). Thanks for the comment. May 22 '18 at 17:06

The problems seems far from easy. This very recent paper gives an example: https://arxiv.org/pdf/1611.05928.pdf .

On $\mathbb{R}^2$, what about : $$v(x,y)= \begin{cases} e_y & \text{ if } y\geq |x| \\ -e_y & \text{ if } -y\geq |x| \\ -e_x & \text{ if } x\geq |y| \\ e_x & \text{ if } -x\geq |y| \end{cases} \\$$ ($e_x,e_y$: the canonical basis of $\mathbb{R}^2$)?

• Why does this have more than one measure preserving flow? May 18 '18 at 9:38
• This is valid only for $x=0$. Sorry because I missed the "all $x\in \mathbb{R}^d$": Maybe with $v(t,x,y)=v(x+P_t,y+P_t )$ with $v$ the function defined above and $P_t$ a Peano curne it could work. May 18 '18 at 9:52

Another recent paper that deals with this question is this one:

http://cvgmt.sns.it/paper/3700/

The spirit is Eulerian and they use the convex integration technique (so unluckily no explicit vector field): they show that basically they can reach whatever smooth measure you want (besides the lebesgue measure), with a sobolev vector field.

However it is set on the torus and so you don't have the compact support hypotesis; but the same technique has been used also with the euler equation, and in that case they could also have a counterexample to uniqueness with compact support, so I would expect this is also the case.

Hope this helps!