Timeline for Does integration by parts formula hold in $H^1(0,T,L^2(\Omega))$?
Current License: CC BY-SA 4.0
18 events
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| Jan 15 at 13:16 | comment | added | Bogdan | @Hannes Thank you very, very much for your response! | |
| Jan 15 at 8:06 | comment | added | Hannes | See Proposition V.2.4.7 in Linear and Quasilinear Parabolic Problems Volume I: Abstract Linear Theory by Amann. (Choose $E_0 = E_1 = L^2(\Omega)$ and use the fundamental theorem of calculus in the resulting product rule; an interval is always "minimally smooth".) | |
| Jan 12 at 4:58 | comment | added | Bogdan | @JochenGlueck Can you detail a little bit, please? | |
| Jan 11 at 12:13 | comment | added | Bogdan | I'm sorry...my limited knowledge does't allow me to see how you can reduce the problem to the product rule for scalar Sobolev functions (who are these functions? How are they obtained from $u,v$?). Also, I don't understand exactly what you mean by "testing against $L^{\infty}(\Omega)$ - functions". | |
| Jan 11 at 11:50 | comment | added | Jochen Glueck | Thanks for the edit! Could you specific where precisely you see a problem now? The product rule seems to follow from simply combining the following three things (i) the definition of the vector-valued weak derivative; (ii) testing against $L^\infty(\Omega)$-functions; (iii) using the product rule for scalar-valued Sobolev functions. The integration by part formula then follows from the product rule and the fundamental theorem of calculus (i.e., its version for vector-valued Sobolev functions). | |
| Jan 11 at 11:30 | comment | added | Bogdan | Exactly. I corrected it. | |
| Jan 11 at 11:28 | history | edited | Bogdan | CC BY-SA 4.0 |
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| Jan 11 at 11:19 | comment | added | Jochen Glueck | For the edited question to make sense you first have to specify what you mean by $(uv)'$ since, as pointed out by others, $uv$ does not take values in $X$, in general. Do you mean the distributiinal derivative of functions that take values in $L^1(\Omega)$? | |
| Jan 11 at 11:11 | history | edited | Bogdan | CC BY-SA 4.0 |
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| Jan 11 at 11:10 | comment | added | Bogdan | I have edited my post. | |
| Jan 11 at 10:55 | comment | added | Bogdan | Very interesting...I was wrong then by saying that $u\cdot v\in H^1(0,T,X)$. But I see the integration by parts applied in this context. Even in your example the integration by parts formula holds. For example in this course of G. Leoni, at page 86 he uses that in proving the maximum principle> giovannileoni.weebly.com/uploads/3/1/0/5/31054371/… | |
| Jan 11 at 10:20 | comment | added | Keba | No, it is wrong for $\Omega = (0, 1)$ and $u(t,x) = v(t,x) = x^{-1/4}$. | |
| Jan 11 at 9:11 | comment | added | Bogdan | The statement is true and it is a generalization of Brezis, p. 215. | |
| Jan 11 at 8:52 | comment | added | Hannes | It is not, Brezis p. 215 is about Sobolev functions on an interval which are automatically bounded and this is exactly what you need. In the question, $(uv)(t)$ is generically only an $L^1(\Omega)$ function and there is no real way out without further assumptions. | |
| Jan 11 at 8:38 | history | edited | Bogdan | CC BY-SA 4.0 |
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| Jan 11 at 7:41 | comment | added | Bogdan | This is an exception. See, the book of Brezis, page 215. | |
| Jan 11 at 7:20 | comment | added | Willie Wong | You can't. Did you perhaps mean that $u\cdot v \in H^1(0,T,L^1(\Omega))$? The product of two $L^2$ functions is generally not in $L^2$. | |
| Jan 11 at 7:02 | history | asked | Bogdan | CC BY-SA 4.0 |