Timeline for Chain rule for distributional derivative
Current License: CC BY-SA 3.0
18 events
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| Mar 26 at 4:40 | comment | added | Willie Wong | @No-one: are you sure the proof of the linked MO question applies here? At the level the OP is asking about I don't even believe $f(u)$ makes sense as an $L^2$ function.... | |
| Feb 25, 2023 at 15:19 | comment | added | k3thomps | $C^1$ functions with bounded first derivatives are dense in the space of differentiable functions with bounded derivative. You can approximate $f$ using $f_\eta \in C^1$ with $f_\eta \rightarrow f$, use the argument above to show that it's true for all $\eta > 0$, and take $\eta \rightarrow 0$ to re-coupe the above for all differentiable functions with bounded derivative. | |
| Dec 2, 2022 at 1:45 | comment | added | No-one | -1, the second addend at the RHS doesn't go to $0$ if $f$ is not $C^1$. The answer to OP's question is still positive though. See mathoverflow.net/questions/430332/chain-rule-in-sobolev-space | |
| Jun 14, 2014 at 13:28 | comment | added | k3thomps | I am actually using that $v$ is smooth with compact support. You can then use density of $C_0^\infty$ to extend the above identity to $L^2(0,T;H^1)$. | |
| Jun 14, 2014 at 12:28 | comment | added | mathias_l | Hmm. I don't understand $v'$ because $v$ is just in $L^2(0,T;H^1)$. | |
| Jun 13, 2014 at 18:57 | comment | added | k3thomps | Oh, the above argument shows that what you suggest is true. I showed that $f(u)' = f'(u) u'$. You can then plug this into the dual pairing and rearrange to get exactly the quantity you want. | |
| Jun 13, 2014 at 18:50 | comment | added | mathias_l | @k3thomps Thanks for answer. When I was trying to write with the duality pairing is this. If $(f(u))' \in L^2(0,T;H^{-1})$ is a weak derivative, then it seems natural to me that its action (recall that $(f(u))'$ is an element of the dual space to $L^2(0,T;H^1)$) should be $$\langle (f(u))', v\rangle_{L^2(0,T;H^{-1}), L^2(0,T;H^1)} = \langle u', f'(u)v \rangle_{L^2(0,T;H^{-1}), L^2(0,T;H^1)}$$ if $f'(u)v \in L^2(0,T;H^1)$. This is what I would expect because we know what $u'$ is. But whether this is how it should be is what I am not sure about. | |
| Jun 13, 2014 at 17:52 | comment | added | k3thomps | What is a vectoral $L^p$ space? I'm not familiar with this terminology. Do you mean functions whose image is in a vector space? | |
| Jun 13, 2014 at 17:36 | comment | added | Mark Peletier | Do you know of a good reference for smoothing properties of vectorial Lp spaces? | |
| Jun 13, 2014 at 14:49 | history | edited | k3thomps | CC BY-SA 3.0 |
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| Jun 13, 2014 at 14:48 | comment | added | k3thomps | OK, I now think that the angled brackets refer to space-time integration (i.e. $\langle u,v\rangle = \int_0^T \int_{\mathbb{R}} uv$). I think what I wrote above still works. You may just need to smooth everything out in space-time now instead of just space. | |
| Jun 13, 2014 at 14:33 | history | edited | k3thomps | CC BY-SA 3.0 |
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| Jun 13, 2014 at 14:33 | comment | added | k3thomps | Yes, I am missing a $'$ in the second bracket. That is a typo. Your questions are fair and now I am thoroughly confused about what I have. I am considering what I've written. | |
| Jun 13, 2014 at 14:30 | comment | added | Mark Peletier | Shouldn't the second bracket read $\langle v,f'(u_\epsilon)u'_\epsilon \rangle$? | |
| Jun 13, 2014 at 14:24 | comment | added | Mark Peletier | If the brackets are in space, then it seems the full time derivative is missing .. | |
| Jun 13, 2014 at 14:17 | comment | added | k3thomps | It should only be space, no? I meant it as mathias did in his question and I thought he had taken it to only be a space integral. The idea is to use a smoothing argument and take limits. | |
| Jun 13, 2014 at 13:58 | comment | added | Mark Peletier | Do you mean the brackets to be in space or space-time? I'm trying to understand the meaning of the displayed equation ... | |
| Jun 13, 2014 at 12:50 | history | answered | k3thomps | CC BY-SA 3.0 |