Timeline for Is this function positive?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 7, 2017 at 21:40 | comment | added | user268193 | I posted a question related to the local existence and continuity of my solution here mathoverflow.net/questions/287975/… If you could give a look it would be great! | |
| Dec 7, 2017 at 17:52 | comment | added | user268193 | Ok! Thanks for your answer! To be honest I don't know what an essential infimum is, and I don't understand well which are these measurability issues you are talking about.... anyway I'm going to study something about these essential infimum, maybe it will be more clear... Thanks a lot for your help! | |
| Dec 7, 2017 at 16:58 | comment | added | fedja | @user268193 Yes, of course, provided that you can still justify the integral representation and understand $\inf$ as the essential infimum, ignoring sets of measure $0$ (otherwise you may have measurability issues). | |
| Dec 7, 2017 at 16:23 | comment | added | user268193 | In that case, instead of the $\min_r u_1(t, r)$ I would take the $\inf_r u_1(t, r)$. | |
| Dec 7, 2017 at 16:19 | comment | added | user268193 | Ok I am sorry, I understand my mistake now. Thank you! I have another question/curiosity, since I see that there are also versions of Gronwall's inequality that do not need the continuity of the function, could I repeat all the procedure if I had the functions $u_0(\cdot, \cdot)$ and $u_1(\cdot, \cdot)$ in $L^\infty([0, \tilde t]\times [0,1])$? In other words my question is, can I relax the hypothesis over the continuity of $u_0$ and $u_1$ and ask for them to be just bounded in the infinity norm? | |
| Dec 7, 2017 at 15:27 | comment | added | fedja | @user268193 $-U(t)\le [-U(0)]e^{Ct}$, not just $e^{Ct}$, and $U(0)=0$. Looks like you are quite proficient in real analysis and, probably, in PDE as well, but have never used Gronwall before. I naturally wonder where they teach students like that. | |
| Dec 7, 2017 at 11:42 | comment | added | user268193 | If I change sign in both sides of the inequality I get $-U(t)\leq C\int_0^t-U(s)ds$ and then $-U(t)\leq e^{Ct}$,but this shouldn't help me... Or not? | |
| Dec 7, 2017 at 11:25 | comment | added | user268193 | Thanks a lot for your answer. I have a problem in understanding one step... I agree with the inequality you wrote but I don't understand how to apply the Gronwell inequality... To apply the Gronwell inequality I should have the inequality you wrote in the opposite direction... Am I wrong? | |
| Dec 7, 2017 at 4:30 | history | answered | fedja | CC BY-SA 3.0 |