Skip to main content
21 events
when toggle format what by license comment
Oct 24, 2023 at 15:40 comment added NancyBoy Hi @LorenzoPompili, you are right concerning the mistake (I updated the post). So now, I have $H(t)$ that tends to $\alpha x$ as $t$ tends to 0 if $0<x<1$. But I have a condition only on $x$, I don't have a condition on the ratio $x/t$. Am I missing something ? Thank you for your very helpful comments btw !
Oct 24, 2023 at 15:29 history edited NancyBoy CC BY-SA 4.0
edited body
Oct 24, 2023 at 15:24 history edited NancyBoy CC BY-SA 4.0
edited body
Oct 22, 2023 at 19:58 comment added Lorenzo Pompili I suspect the second term of $B$ should have $(x-1)$ instead of $(1-x)$ inside $\Phi$. It should converge to $\varepsilon x$ on $[0,1]$ by how yiu estimated $f$, so if it doesn’t there is surely a mistake there. So, what I meant is that at $t=0$, the term with $\Phi$ in $A$ is supported on $x\in(-\infty,0]$, the terms with $\Phi$ in $B$ are supported on $[0,1]$, the remaining terms should be zero.
Oct 22, 2023 at 18:02 comment added NancyBoy @LorenzoPompili But there is only one term that doesn't go to 0 for small $t$ (moreover only if x>1), am I wrong ? So I can't see what you mean by "leading part".
Oct 22, 2023 at 17:46 history edited NancyBoy CC BY-SA 4.0
added 218 characters in body
Oct 22, 2023 at 17:28 comment added NancyBoy Concerning the computation of the original post, thank you very much, I will think about this!
Oct 22, 2023 at 17:26 comment added NancyBoy Concerning (i), my feeling is that we have even more because if the initial condition crosses 0 with a positive slope, its convolution will also cross 0 with a positive slope. I am wondering how one could prove that if we fix $t>0$, then we have automatically $\partial_x u(t,x)>0$ but not only on a neighbourhood of 0.
Oct 22, 2023 at 17:22 comment added Lorenzo Pompili Hint: plug $t=0$ first in the estimate $u(t,x)\geq A+B$ and see what you get; then try to imagine what happens for very small $t$ instead. Remember that essentially you want to solve the equation $A+B=0$.
Oct 22, 2023 at 17:15 comment added Lorenzo Pompili Your estimate looks better now, it should suffice. To continue, note that the leading parts are the ones with $\Phi$, because the other terms go to zero for small $t$. Try to compare the terms with $\Phi$ first, and then try to add the other terms and see how to deal with them. It could be messy but probably not impossible, you should try and do some computation and estimates. There is no unique nice way of proving such statements, try and see what you get, and then if it is not enough come back and try to use better bounds
Oct 22, 2023 at 17:14 comment added Lorenzo Pompili (i) the point is that $x_t$ satisfies a differential equation, $\frac{d}{dt}x_t=\frac{u_{xx}(t,x)}{u_x(t,x)}$, so if the derivative vanishes at $t=0$, it seems that the derivative of $t\mapsto x_t$ at $t=0$ could be unbounded, or not defined. (ii) I guess it depends on whether $f'(0)>0$ or not. If yes, then it should be immediate because $\partial_x u(t,x)$ also satisfies the heat equation and it is smooth at $t=0$, so $\partial_xu(t,x)$ must be continuous, hence the derivative can not be zero in a neighbourhood of $x=0$, $t=0$. If $f'(0)=0$ I don't know
Oct 22, 2023 at 15:37 history edited NancyBoy CC BY-SA 4.0
added 3 characters in body
Oct 22, 2023 at 15:35 comment added NancyBoy Hi @LorenzoPompili, I have updated the post but I have to admit that I have a even more complicated expression to analyse (depending on $C$ and $\alpha$...). I am quite stuck here. Moreover, I have two questions: (i) Could you explain in what sense you think that the lower the derivative is, the worst it is for $x_t$ ? (ii) Secondly, do you think that it is hard to prove that $\partial_x u (t,x_t)>0$ Or even to bound it with a power of $t$ (e.g. $\partial_xu(t,x_t) > \lambda/t$?)? Thank you again!
Oct 22, 2023 at 15:31 history edited NancyBoy CC BY-SA 4.0
deleted 73 characters in body
Oct 22, 2023 at 15:06 history edited Michael Hardy CC BY-SA 4.0
added 2 characters in body
Oct 22, 2023 at 14:54 comment added NancyBoy Thank you for your answer, you are absolutely right for the $x$ multiplication. I will take into account your comments and will update the post today. Thank you again!
Oct 22, 2023 at 14:53 history edited NancyBoy CC BY-SA 4.0
edited body
Oct 22, 2023 at 12:35 comment added Lorenzo Pompili Admittedly, I have not much intuition on what happens when $f'(0)=0$, but I feel that the more derivatives vanish at $x=0$, the worse can be the behaviour of $x_t$ for small $t$. So it could be that you have to assume $f'(0)>0$ in the main statement to have a constant $K$ that works for all $t>0$.
Oct 22, 2023 at 12:24 comment added Lorenzo Pompili P.s. are you sure you computed $A$ correctly? I think there should be an $x$ multiplying $\Phi(-t^{-1/2}x)$, or something like that. If that is the case, then $A$ should be a good term, and you would have to compare it to the "new" $B$ that you obtain with the bound on $f$ that I said in the previous comment
Oct 22, 2023 at 12:10 comment added Lorenzo Pompili Hi again! It seems to me that the bound for positive $x$ is too weak, because for small $t$ the main contribution to $u(t,x)$ comes from values of $f$ close to $x=0$. I think the bound you want to use for positive $x$ is: $f(x)\geq \varepsilon x$ on (say) the interval $[0,1]$, with suitable $\varepsilon>0$. To have $\varepsilon>0$, of course you need to assume $f'(0)\neq 0$, and honestly I don't know what happens in the case $f'(0)=0$...
Oct 21, 2023 at 10:41 history asked NancyBoy CC BY-SA 4.0