Timeline for Is $(u\cdot\nabla)v\in H^1$, if $u,v\in H^2$?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Dec 17, 2016 at 14:53 | comment | added | Fan Zheng | Exactly. (Placeholder text) | |
| Dec 13, 2016 at 12:53 | comment | added | 0xbadf00d | @FanZheng I'm not sure, if I understand what you mean. We've got $W^{k,\:p}\subseteq C^0$, if $k>d/p$. Thus, $H^2\subseteq C^0$, if $d<4$. Is that what you mean? | |
| Dec 12, 2016 at 22:56 | comment | added | Fan Zheng | @0xbadf00d Beware that if $d\ge4$, then $H^2$ does not embed into $L^\infty$, let alone $C_0$, so there is some luck in that the NS equation is usually posted in 2 or 3 dimensions. | |
| Dec 12, 2016 at 20:26 | comment | added | Michael Renardy | Yes, that is what I mean. | |
| Dec 12, 2016 at 20:15 | comment | added | 0xbadf00d | @MichaelRenardy It's clear that $\overline u_n∈ C_c^∞(B_r(0))$ for all $n\inℕ$. Obviously, $(\overline u_n)_{n\inℕ}$ is convergent in $H^2(B_r(0))$. Call the limit $v$. Since $\partial B_r(0)\in C^1$, we can apply the argument from the question and find a continuous representative of $v$ which can be continuously extended to $\overline{B_r(0)}$. If we extend $u$ by zero to $B_r(0)$ and call the extension $\overline u$, it's easy to see that $\left\|\overline u-v\right\|_{H^2(B_r(0))}=0$, i.e. $\overline u=v$ a.e. and we're done. Is that what you mean? | |
| Dec 12, 2016 at 20:15 | comment | added | 0xbadf00d | @MichaelRenardy I'm a computer scientist, so, please be patient and let me try to understand this. We have $Λ⊆\mathbb R^d$, $d\inℕ$ arbitrary, being bounded and open (and I stress the fact that $Λ$ doesn't need to be connected, i.e. a domain). Now, if $u∈ H_0^2(Λ)$, we can find $(u_n)_{n\inℕ}⊆ C_c^∞(Λ)$ such that $\left\|u_n-u\right\|_{H^2(Λ)}\xrightarrow{n\to∞}0$ by definition of $H^2(Λ)$. Since $Λ$ is bounded, we can find a ball $B_r(0)$ around $0$ with radius $r>0$ such that $Λ⊆ B_r(0)$. Now, we can simply extend the $u_n$ to $B_r(0)$ by zero. Denote these extensions by $\overline u_n$. | |
| Dec 12, 2016 at 19:57 | comment | added | Fan Zheng | @0xbadf00d For the first term, $\nabla u$ and $\nabla v$ are in $H_0^1$, which embeds in $L^6$. For the second term, it is $L^\infty$ multiplied by $L^2$. | |
| Dec 12, 2016 at 19:42 | comment | added | Michael Renardy | If the functions are in $H^2_0$, you do not need the extension property because you can extend by zero. | |
| Dec 12, 2016 at 19:37 | comment | added | 0xbadf00d | @MichaelRenardy Yes, but how should that help? What I need is $W_k^p(\Lambda)\subseteq C^0(\overline\Lambda)$, if $k>d/p$, in the case where $\Lambda$ is just bounded and open. For $p=2$, I've found this statement in your book (Lemma 7.27) in the case where $\Lambda$ has the $k$-extension property. However, you deal with the stronger statement that the embedding is compact there. So, maybe we can always find a continuous representative in the bounded and open case, but I couldn't find a reference. | |
| Dec 12, 2016 at 19:25 | comment | added | Michael Renardy | If a function is in $H^2_0$, it can be extended by zero outside the domain, and the extended function is in $H^2$. | |
| Dec 12, 2016 at 19:04 | comment | added | 0xbadf00d | @FanZheng $H_0^2$ is the completion of $C_c^\infty$ with respect to the $H^2$-norm, sure, but how do you intend to conclude that $(1)$ is in $L^2$ from that? | |
| Dec 12, 2016 at 18:59 | comment | added | Fan Zheng | Well, $H_0^2$ by definition is the completion of $C_c^\infty$ in $H^2$ norm, so in this case the limiting sequence converges uniformly, and it has nothing to do with the regularity of the boundary. (Or could $H_0^2$ mean something else (like div/curl free vector fields) in the context of the NS equation?) | |
| Dec 12, 2016 at 18:42 | history | edited | 0xbadf00d | CC BY-SA 3.0 |
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| Dec 12, 2016 at 18:38 | comment | added | 0xbadf00d | @FanZheng Of course, you're right (the crucial point is that $d\le 3$). I didn't saw this at the first place, cause I got a domain in mind which is just bounded and open. I've modified the question accordingly. | |
| Dec 12, 2016 at 18:36 | history | edited | 0xbadf00d | CC BY-SA 3.0 |
added 111 characters in body
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| Dec 12, 2016 at 18:26 | comment | added | Fan Zheng | It also follows from Sobolev embedding. | |
| Dec 12, 2016 at 18:24 | history | asked | 0xbadf00d | CC BY-SA 3.0 |