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I am reading about the uniqueness problem for the continuity equation $\partial_t \mu_t + div_x (b \mu_t)=0$ in the lecture notes by Ambrosio (here: https://warwick.ac.uk/fac/sci/maths/research/events/2016-17/symposium/workshops/ambrosio.pdf) and the article by Di Perna-Lions "Ordinary differential equations, transport theory and Sobolev spaces". Now, the lecture notes start by reviewing the classical theory and showing that if $b \in L^1 _t (W^{1, \infty} _x)$ then it's all good, you have existence and uniqueness. There's a remark later on (beginning of page 9) where they point out that you can actually consider $b \in L^1 _t (W^{1, \infty} _{loc})$ and $\frac{|b|}{1+ |x|} \in L^1 _t (L^\infty _x)$ and the theory still works.

Then they continue the exposition ALWAYS requiring some conditions on $|b|/(1+|x|)$. Now the question is: why do you need that $1+|x|$? What goes wrong if you don't consider that second condition? Apparently everyone from the paper of Di Perna and Lions is using this assumptions, but I can't see where you're using it.Apparently it has to do with the fact that, under that assumption, the ODE

$$ \dot{\gamma} (t)= b(t, \gamma(t)) ; \qquad \gamma(0)=x$$

has a unique solution in the pointwise sense.

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A heuristic explanation:

If $b$ grows faster than $1+|x|$ (for example, is of size $(1+ |x|)^{1+\epsilon}$), then the trajectories can escape to infinite (or come in from infinity) in finite time.

Consider the ODE $$ \dot{x} = |x|^{1+\epsilon} $$ For positive initial data, we can write this as $$ \frac{d}{dt} x^{-\epsilon} = - \epsilon $$ So an initial value of $x = 1$ will escape to $\infty$ in time $t = 1/\epsilon$.

Thinking of the continuity equation as modeling some sort of transport phenomenon, having particle trajectories that can move from finite regions to infinity (and vice versa) in finite time obviously poses an issue for uniqueness, since your prescription of initial data only tells you about what happens on every finite region, but not about what happens at infinity, where there is a potential pool of additional particles.

I suspect a lot of the theory can be reproduced with slightly weaker weights that are still non-integrable (maybe something like $|b| / (1 + |x| \ln(1+|x|))$ being bounded). And $1+|x|$ is a reasonably broad condition that allows some growth at infinity for which you don't get too bogged down in detailed worries about the asymptotics.

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  • $\begingroup$ Thanks! Now that I think about it, there's another (smaller) issue I have. They say that whenever $b \in L^1 (0,T; W^{1, \infty})$, you have pointwise uniqueness for the ODE. Now, to me pointwise uniqueness means classical solution, but for that you'd need $b \in C([0,T];W^{1,\infty})$, don't you? I guess they mean they are pointwise unique in the sense that if you approximate $b$ with, say, smooth functions, the flows converge in $C([0,T]; \mathbb{R}^d)$ $\endgroup$
    – tommy1996q
    Commented Feb 8, 2023 at 18:53
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    $\begingroup$ I have not thought too much about these kinds of regularity issues, but if you take a simple example of $x' = f$ where $f\in L^1([a,b])$, then you still have pointwise uniqueness of the solution even when $f$ is not continuous (the integral of $f$ is absolutely continuous). So I don't really see the continuity of $b$ as a must. $\endgroup$
    – Willie Wong
    Commented Feb 8, 2023 at 20:03
  • $\begingroup$ I thought about that, but the problem is that if it was the case, why should the classic Cauchy-Lipshitz theorem for $y'= f(t,y)$ be taught requiring "continuity in the independent variable and lipshitzianity in the other one"? $\endgroup$
    – tommy1996q
    Commented Feb 8, 2023 at 20:28
  • $\begingroup$ Ok, probably both for historical reasons and because (at least for me) when they taught me about it I still had no idea what a Lebesgue integral was $\endgroup$
    – tommy1996q
    Commented Feb 8, 2023 at 20:29
  • $\begingroup$ For the classic result you also want $\frac{d}{dt} x$ exists at all $t$ in its domain. This is obtained by some versions of fundamental theorem of calculus. For the existence and uniqueness of the equivalent integral equations, you can have weaker regularity assumptions. $\endgroup$
    – Willie Wong
    Commented Feb 8, 2023 at 20:56

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