Timeline for Energy estimates for nonlinear wave type equation
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 1, 2021 at 11:06 | vote | accept | Student | ||
| Oct 28, 2021 at 16:54 | answer | added | Willie Wong | timeline score: 2 | |
| Oct 28, 2021 at 14:03 | comment | added | Student | @WillieWong thank you very much for your remarks. I apologize for the lack of clarity. Indeed my goal is to control the energy when $t=\epsilon.$ | |
| Oct 28, 2021 at 13:03 | comment | added | Willie Wong | In particular: if you are prescribing data at $t = \epsilon$ and solving FORWARD in time ($t$ increasing), then your equation is anti-damped and you cannot expect energy decay in any reasonable sense. Solving backwards in time your equation is damped and you may get some better control. You need to make your setup a lot clearer since the arrow of time matters in these situations. | |
| Oct 28, 2021 at 13:01 | comment | added | Willie Wong | I am still not sure about what your question is aiming at. You prescribe data to be identically vanishing at $t = \epsilon$. Are you trying to bound the energy at $t = 1$? Because that's the opposite of what I asked about in my previous comment which you seem to agree with. | |
| Oct 28, 2021 at 11:42 | comment | added | Student | @WillieWong not sure how you derived the lower bound. Could you please explain? | |
| Oct 28, 2021 at 3:10 | comment | added | Willie Wong | What do you exactly hope to accomplish? If you use $H(t) = \int |u_t|^2 + |\nabla u|^2$ the standard energy, you get $$ \frac{d}{dt} H(t) \geq - \frac{t}{4} \|g\|_{L^2}^2 $$ or that $$ H(1) + \int_\epsilon^1 \frac{t}{4} \|g\|_{L^2}^2 ~dt \geq H(\epsilon) $$ This gives boundedness of the standard energy and as $\epsilon \searrow 0$, and hence your weighted energy will decay like $\epsilon^2$. Are you looking somehow for more decay? | |
| Oct 28, 2021 at 2:38 | comment | added | Willie Wong | Then your energy inequality is going the wrong way. If you integrate from $t = \epsilon$ to $t = 1$, your inequality gives $E(1) \leq E(\epsilon) + \frac{6E}{t} + \ldots$ whereas a useful equality will allow you to bound $E(\epsilon)$ in terms of $E(1)$. | |
| Oct 27, 2021 at 20:51 | history | edited | Student | CC BY-SA 4.0 |
added 5 characters in body
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| Oct 27, 2021 at 20:24 | comment | added | Student | Yeah, that's right! | |
| Oct 27, 2021 at 19:12 | comment | added | Willie Wong | Hang on, are you trying to solve backwards in time (with data at $t = 1$ and trying to show decay as $t\to 0$)? | |
| Oct 27, 2021 at 18:37 | history | asked | Student | CC BY-SA 4.0 |