Timeline for $L^{\infty}$ estimate for heat equation with $L^2$ initial data
Current License: CC BY-SA 4.0
15 events
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| Jan 13 at 12:33 | comment | added | Bogdan | Thank you very much. Now I understand! | |
| Jan 13 at 12:05 | comment | added | Giorgio Metafune | As explained by @MaoWao one has to apply Ouhabaz results to the form $a+1$ which gives global ultracontractivity to for the scaled semigroup and local for the original one. | |
| Jan 13 at 11:35 | vote | accept | Bogdan | ||
| Jan 13 at 9:54 | comment | added | MaoWao | The equivalent characterization in terms of a Sobolev inequality takes the form $\lVert u\rVert_{2d/(d-1)}^2\leq c(a(u,u)+\lVert u \rVert_2^2)$, which is true for bounded domains with Lipschitz boundary. | |
| Jan 13 at 9:53 | comment | added | MaoWao | One has to be a bit careful here, there a two (non-equivalent) forms of ultracontractivity. The one holds for the heat semigroup on full space and requires the $L^1$-$L^\infty$ bound from the question for all $t\geq 0$. As noted in Denis' answer, this cannot possible hold for the Neumann Laplacian on bounded domains. The second form of ultracontractivity requires the $L^1$-$L^\infty$ bound from the question only on bounded intervals with a constant $c$ depending on $T$. This form of ultracontractivity is true for the Neumann Laplacian on bounded domains with Lipschitz boundary. | |
| Jan 13 at 8:03 | answer | added | Denis Serre | timeline score: 4 | |
| Jan 13 at 6:52 | history | edited | Bogdan | CC BY-SA 4.0 |
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| Jan 13 at 5:40 | comment | added | Bogdan | Moreover, since Theorem 6.4 is an equivalence, we can say that the inequality from the statement is false. The same can be deduced from Theorem 2.4.6 in DAVIES - HEAT KERNELS AND SPECTRAL THEORY. But that contradicts other resources, such as Arendt-Semigroups and Evolution Equations: Functional Calculus, Regularity and Kernel Estimates, page 69 where it is stated that Neumann Laplacian on Lipschitz domains is ultracontractive (i.e. satisfies the inequality in the statement). It's very hard for me to decide where is the truth... | |
| Jan 13 at 4:40 | comment | added | Bogdan | In Theorem 6.3 and 6.4 this inequality in Ouhabaz (page 158) this inequality is equivalent with proving that $\Vert u\Vert^2_{\frac{2d}{d-2}}\leq c\cdot a(u,u),\ \forall u\in D(a)$. But this is not true in our case since $a(u,u)=\int_{\Omega}|\nabla u |^2\ dx$ and $D(a)=H^1(\Omega)$. For example $u=$constant do not satisfy such an inequality. | |
| Jan 12 at 22:14 | comment | added | Giorgio Metafune | This can be proved by Nash inequality, as explained in the book of Ouhabaz, Analysis of heat equations in domains | |
| Jan 12 at 20:50 | comment | added | Bogdan | Is there such a formula taking into account the Neumann bc? | |
| Jan 12 at 20:35 | comment | added | leo monsaingeon | Use the integral representation in terms of the fundamental solution? | |
| Jan 12 at 19:45 | history | edited | Bogdan | CC BY-SA 4.0 |
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| Jan 12 at 19:09 | history | edited | Bogdan | CC BY-SA 4.0 |
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| Jan 12 at 17:45 | history | asked | Bogdan | CC BY-SA 4.0 |