Timeline for Trace theorem for $L^2([0,1]; H^k(S^2))$
Current License: CC BY-SA 4.0
9 events
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| Apr 26 at 19:49 | comment | added | Hannes | @Laithy Actually, I would say "trace theorem" is just fine and in fact regularly used. :-) | |
| Apr 26 at 15:59 | comment | added | Laithy | @Hannes Thank you! That was what I was looking for. I guess "trace theorem" is not the right phrase to use. | |
| Apr 26 at 7:59 | comment | added | Hannes | Yes there is, by interpolation theory, see e.g. here. This will give you $u(\tau) \in H^\ell(S^2)$ with $\ell = \frac12 k + \frac12 k'$ for every $\tau \in [0,1]$ with a uniform-in-$\tau$ norm dependence against $\|u\|_{L^2(H^k)} + \|u'\|_{L^2(H^{k'})}$. | |
| Apr 26 at 1:00 | history | edited | Laithy | CC BY-SA 4.0 |
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| Apr 26 at 0:43 | comment | added | Laithy | Oh I didn't mean to add the "$u_1$". Thanks for pointing that out. I edited the post. I see, so we cannot make sense of $u(r)$ for a given $r$. If we in addition know that $\partial_r u \in L^2([0,1];H^{k-1})$, then is there a trace theorem saying something about $u(0)$? | |
| Apr 26 at 0:39 | history | edited | Laithy | CC BY-SA 4.0 |
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| Apr 25 at 22:48 | comment | added | Willie Wong | For the question in your title: the constant function $1$ on $S^2$ is in any $H^k$. So any $L^2$ function $u$ can be realized as an $L^2([0,1]; H^k(S^2))$ function. But there is no "trace theorem" allow you to take pointwise values of an arbitrary $L^2$ function. // For the motivation: when $F \equiv 0$, your ODE is equivalent to $(e^r u(r))' = 0$. It is absolutely false that for arbitrary $u_0, u_1$ you can find a solution. Any solution must satisfy $u_0 = e \cdot u_1$. // Considering that you are studying a first order ODE, it doesn't make sense to give two boundary values. | |
| Apr 25 at 17:04 | history | edited | Laithy | CC BY-SA 4.0 |
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| Apr 25 at 16:33 | history | asked | Laithy | CC BY-SA 4.0 |