Timeline for Prove Liouville theorem without using mean value property
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 13, 2021 at 8:47 | comment | added | Lao | @RBega2 Thank you. I'm trying unsuccessfully to do this and posted it as a separate question here: mathoverflow.net/questions/381087/… | |
| Jan 12, 2021 at 0:11 | comment | added | RBega2 | @Lao That would be easiest if $u$ was in $L^2$. This is in general not true (e.g. a constant function) so the argument gets more involved. It's possible that playing around with the Poincare inequality would work. | |
| Jan 12, 2021 at 0:05 | comment | added | Lao | @RBega2 Thank you! What do you think about the discussion above regarding using $\int \phi_R |\nabla u|^2$ instead of the hessian and estimating the boundary term $\int_{\partial B} \partial_{\nu} u \phi_R u$? | |
| Jan 12, 2021 at 0:04 | history | edited | RBega2 | CC BY-SA 4.0 |
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| Jan 12, 2021 at 0:03 | comment | added | RBega2 | @Lao I used finite energy twice. First, for the final inequality in the string of inequalities and secondly to rule out affine functions. | |
| Jan 12, 2021 at 0:03 | comment | added | Lao | @leomonsaingeon The boundary term in this case is $\int_{\partial B} \partial_\nu u \phi_R u$ How can you show that this satisfies the decay? | |
| Jan 12, 2021 at 0:01 | history | edited | RBega2 | CC BY-SA 4.0 |
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| Jan 12, 2021 at 0:00 | comment | added | leo monsaingeon | Sure, but the problem remains that you will not get rid of the boundary term on $\partial B_R$. The whole point of the exercise is precisely showing that if $\nabla u\in L^2$ then this remainder decays to zero as $R\to\infty$. | |
| Jan 11, 2021 at 23:51 | comment | added | Lao | @leomonsaingeon And if we consider the test functions as in the answer above, could it be possible to consider the quantity $\int \phi_R |\nabla u|^2 $ instead of the one with the hessian? | |
| Jan 11, 2021 at 23:41 | comment | added | Lao | @leomonsaingeon I see, but I'm still a bit confused: isn't it very common in PDE books to integrate by parts in the whole space and neglect boundary terms when the involved functions are in $L^1$? | |
| Jan 11, 2021 at 23:37 | comment | added | leo monsaingeon | A priori you can only integrate by parts on bounded sets, say balls. When you do so you have a non-zero term on the boundary, and this term does not vanish when the radius of the ball $R\to+\infty$ unless $\nabla u$ (actually $\partial_r u$) decays at infinity (which is precisely what the "finite energy" assumption guarantees, in some weak sense) | |
| Jan 11, 2021 at 23:34 | comment | added | Lao | @leomonsaingeon Why can't you integrate by parts in the whole space in this case? | |
| Jan 11, 2021 at 23:33 | comment | added | leo monsaingeon | @Lao: well, for us to try pinpointing your mistake you must first give us your proof! But this is a classical mistake, in fact: you cannot integrate by parts in the whole space, that's all | |
| Jan 11, 2021 at 23:31 | comment | added | Lao | @leomonsaingeon (1) Thank you: I missed the last line. (2) Indeed it must be wrong, but I cannot pinpoint the flaw: do you see where the mistake is? | |
| Jan 11, 2021 at 23:28 | comment | added | leo monsaingeon | (1): Errr... finite energy was used when RBega writes "finite energy forces"! (2) try $u(x)=x$ in dimension 1. Surely your "proof" is wrong! | |
| Jan 11, 2021 at 23:19 | comment | added | Lao | Thank you. Two questions: (1) At which point was the assumption $\int_{\mathbb{R}^n}|\nabla u|^2 dx \leq C$ used? (2) Why the simpler observation $-\Delta u = 0 \implies \int_{\mathbb R^n} |\nabla u|^2 = 0$ does not work? | |
| Jan 11, 2021 at 23:10 | history | answered | RBega2 | CC BY-SA 4.0 |