Timeline for Equivalence of statements about level sets: $u|_{S \times [\tau, \infty)}$ depends only on $t$ $\iff$ $u(t,x) = \mu^{\tau}(t,u(\tau,x))$
Current License: CC BY-SA 4.0
8 events
when toggle format | what | by | license | comment | |
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Apr 12, 2021 at 12:46 | answer | added | mlk | timeline score: 1 | |
Apr 12, 2021 at 12:39 | comment | added | Riku | @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? | |
Apr 12, 2021 at 12:36 | history | edited | Riku | CC BY-SA 4.0 |
added 3 characters in body
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Apr 12, 2021 at 11:23 | comment | added | mlk | 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. | |
Apr 12, 2021 at 10:13 | comment | added | Riku | @mlk There was a typo. The line is $u(t,x) = \mu^{\tau}(t,u(\color{red}{\tau},x))$ | |
Apr 12, 2021 at 10:13 | history | edited | Riku | CC BY-SA 4.0 |
added 3 characters in body
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Apr 12, 2021 at 6:39 | comment | added | mlk | 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. | |
Apr 11, 2021 at 23:19 | history | asked | Riku | CC BY-SA 4.0 |