5
$\begingroup$

Consider the problem $$(\star) \quad \begin{cases} \frac{d}{dt} X(t,x) = \sqrt{X(t,x)}, &t \in [0,T],\\ X(0,x) = x, &x \in \mathbb R \end{cases} $$ This is the prototype of non-uniqueness for ODEs in the classical sense. I'm tempted to say, however, that the Lagrangian flow in the sense of Ambrosio exists and should be made of trajectories that don't "stay" at zero. Is this intuition correct? How can the regular Lagrangian flow be computed explicitly in this example?

For the existence and uniqueness of RLF for this example (which is not covered by the classical DiPerna-Lions theory), see this paper.

$\endgroup$

1 Answer 1

4
$\begingroup$

Your intuition is right. The key is in the paper you cite, in that they consider uniqueness in the class $L^1_{\text{loc}}$, which does not allow for concentrations. If you add to this, that the Lagrangian flow conserves mass and that the the classical ODE has uniqueness away from zero, you thus get that the mass transported upwards to 0 from the negative real axis has to immediately passed on to the positive real axis.

So denote by $\Phi(t,x)$ the flow of the ODE with $\Phi(0,x)=x$ and $\partial_t \Phi(t,x) = \sqrt{|\Phi(t,x)|}$ (I assume the missing absolute values were a typo) as well as the added condition that it is strictly monotone, i.e. at 0 you always pick the solution that immediately continues onwards. This you can write down explicitly, with some tedious case distinctions, depending on if you switch sign or not. If I didn't miscalculate, it should be something like $$ \Phi(t,x) := \begin{cases} \frac{1}{4}(t+2\sqrt{x})^2 & \text{for } x\ge 0 \\ \frac{1}{4}(t-2\sqrt{|x|})^2 &\text{for } x<0, 2\sqrt{|x|} \le t \\ -\frac{1}{4}(2\sqrt{|x|}-t)^2 &\text{for }x<0,2\sqrt{|x|} > t. \end{cases} $$

Then the solution to the corresponding continuity equation should be given by $$u(t,\cdot) := u_0 \circ \Phi(t,\cdot)^{-1} \partial_x\Phi(t,\cdot)^{-1},$$ which again you can construct explicitly.

Let me however stress again that this is not necessarily "the solution" to the problem. In many contexts, it is equally possible to say that the natural class for solutions is the class of measures. In that case you are allowed to accumulate mass at zero and as a result can construct infinitely many solutions by only passing on some of the mass. So in some sense the reformulation into a continuity equation still retains the non-uniqueness of the ODE and only using some selection principle can you expect uniqueness. It just so happens that in this class of ODEs, disallowing concentrations is precisely enough to obtain uniqueness while still guaranteeing existence.

$\endgroup$
2
  • $\begingroup$ Thank you. How does one actually write down (in compact form) the explicit solution? $\endgroup$
    – Riku
    Jul 25, 2022 at 8:47
  • $\begingroup$ @Riku I have edited what I think the solution should be but there may be some mistakes in there. $\endgroup$
    – mlk
    Jul 25, 2022 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.