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Let $u:\mathbb R_+ \times \Omega \subset \mathbb R^N \to \mathbb R$ (sufficiently smooth). Are the following statements are equivalent?

  1. For every $\tau >0$ and level surface $S$ of $u(\tau,\cdot)$, it holds that $u|_{S \times [\tau, \infty)}$ depends only on $t$
  2. For every $\tau >0$, there exists $\mu^\tau: [\tau, \infty) \times \mathbb R \to \mathbb R$ such that $u(t,x) = \mu^{\tau}(t,u(\tau,x))$
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  • $\begingroup$ Is there some derivative missing in condition 2? As it is stated now, I could just write $\mu^\tau(t,y):=y$ to make it trivially true, even if condition 1 is not. $\endgroup$
    – mlk
    Commented Apr 12, 2021 at 6:39
  • $\begingroup$ @mlk There was a typo. The line is $u(t,x) = \mu^{\tau}(t,u(\color{red}{\tau},x))$ $\endgroup$
    – Riku
    Commented Apr 12, 2021 at 10:13
  • $\begingroup$ I assume that in 1. you also mean level sets of $u(\tau,.)$, (otherwise the "depends only on $t$" is a bit circular). With that correction, the statement looks true on the level of sets (no need for any smoothness), as 1. implies that $u(t,x)$ is independent of $x$ on any given level set which is enough to define $\mu^\tau$ and prove 2. and conversely on any level set $S$, the function $\mu^\tau$ only depends on $t$ which implies 1. $\endgroup$
    – mlk
    Commented Apr 12, 2021 at 11:23
  • $\begingroup$ @mlk Thank you. Yes, there was a typo in the first line too: it was $u(\color{red}{\tau}, \cdot)$. Could you expand on your comment with more detail in an answer? $\endgroup$
    – Riku
    Commented Apr 12, 2021 at 12:39

1 Answer 1

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If I am not missing something then the corrected statements look equivalent to me on the level of sets, no need for smoothness.

Fix $\tau > 0$ and assume that 1. holds. For any $t\geq \tau$ and $y \in \mathbb{R}$ we can set $\mu^\tau(t,y) := u(t,x)$ for some $x \in S:=(u(\tau,.))^{-1}(y)$. By 1. $u(t,x)$ is independent of our choice of $x$ and thus this is well defined. (if $S$ is empty, the value of $\mu^\tau$ does not matter as it is not used.) By the same argument $\mu^\tau$ has the required property.

Similarly, assume 2. holds. Let $S:=(u(\tau,.))^{-1}(y)$ be a level set for some $y \in \mathbb{R}$. Then $u(t,x) = \mu^\tau(t,y)$ for all $x\in S$ by 2.. But as $y$ is fixed, this expression only depends on $t$, as required.

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