Timeline for Regularity of $u$ in $u_t - \Delta \beta(t,u) = f$, can we get $u_t$ is a function?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2015 at 7:40 | comment | added | Pace | $\beta(t,\cdot)$ (fixed $t$) can be degenerate nonlinearity. For $t \mapsto \beta(t)$, as smooth as necessary. | |
| Oct 12, 2015 at 21:09 | comment | added | Deane Yang | What are you assuming about $\beta$? | |
| Oct 12, 2015 at 20:51 | comment | added | Pace | @DeaneYang Let us fix $\beta$ to be PME nonlinearity. One thing we would need is estimate of the form $|u(t+h)-u(t)|_{L^1} \leq C|h|$. This follows since: if $u$ is a (weak solution) solution with initial data $u_0$, then $v(t):= \lambda u(\lambda^{m-1}t)$ is solution with initial data $\lambda u_0$. We pick $\lambda^{m-1}t$ such that it equals $t+h$, and then a simple argument gives the result, using the $L^1$ cts. dependence. If eg. we have a time-dependence, the function $v$ is no longer solution to the same problem (think of the weak formulation), it would be a sort of "rescaled" problem. | |
| Oct 12, 2015 at 20:03 | comment | added | Deane Yang | What goes wrong if you try to use the same $L^1$ contraction argument used for $\beta$ independent of time? | |
| Oct 12, 2015 at 16:03 | history | edited | Pace | CC BY-SA 3.0 |
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| Oct 12, 2015 at 15:50 | review | First posts | |||
| Oct 12, 2015 at 19:15 | |||||
| Oct 12, 2015 at 15:47 | history | asked | Pace | CC BY-SA 3.0 |