Timeline for Euler-Lagrange Equation and "Eigen Value "
Current License: CC BY-SA 3.0
16 events
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| Mar 22, 2015 at 17:00 | comment | added | Hachino | @Learner : The Sobolev spaces of fractional order are usually defined as the interpolates of their counterparts with integer order. Thus, for instance, the (complex) interpolates of $L^2$ and $H^1$ are all $H^s$ spaces with $s$ between $0$ and $1$. (I said usually, because one also may use the Besov spaces to define in a few lines all Sobolev spaces, but then you would have to show that this definition is equivalent to the one using weak derivatives. Anyway, since you are seemingly not comfortable with the standard Sobolev spaces, looking at Besov spaces would probably be unwise.) | |
| Mar 22, 2015 at 15:31 | comment | added | Learner | Can you explain me the interpolation between $L^2(\Omega)$ and $H^1$ ? | |
| Mar 22, 2015 at 10:24 | comment | added | Hachino | @Learner : Nope, I meant homogeneous spaces. The difference is we remove the "zeroth order" term in the definition of the sapce, that is : $\dot{H}^1(\Omega) = \{u ; \nabla u \in L^2(\Omega) \}$, whereas $H^1(\Omega) = \{u; u , \nabla u \in L^2(\Omega) \}$. | |
| Mar 22, 2015 at 10:18 | comment | added | Learner | I am unaware of the interpolation between $L^2$ and $H^1$ as well as the continuity of trace operator from $\dot H^{\frac{3}{4}}(\Omega) $ and $\dot H^{\frac{1}{4}}(\partial \Omega)$ , I suppose by dot you meant functions with trace $0$ . | |
| Mar 19, 2015 at 16:46 | comment | added | Hachino | @Learner : I have just posted a more thorough explanation on the influence of the value of $q$ on existence (or nonexistence) of solutions to your problem. Please tell me if you find that something isn't quite clear. | |
| Mar 19, 2015 at 16:43 | history | edited | Hachino | CC BY-SA 3.0 |
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| Mar 18, 2015 at 7:41 | comment | added | Hachino | Here it is, though I'm not aware of any translation of this founding article in English, sorry. However, Terry Tao has written a blog post which is quite clear (as clear as one could expect from Terry, actually :) ). | |
| Mar 18, 2015 at 7:10 | comment | added | Learner | Could you let me know the source where i can get to read about it ? (defects of compactness of bounded sequences of $H^s$ in Sobolev-critical $L^p$ spaces ) | |
| Mar 16, 2015 at 18:00 | comment | added | Hachino | @Learner : See the comment right above yours. | |
| Mar 16, 2015 at 16:02 | comment | added | Learner | So you think there is a bound on $q$ ? Could you tell me how to go about to find the bound on $q$ ? | |
| Mar 16, 2015 at 16:02 | comment | added | Hachino | OTOH, if $q < q^*$ where $1 = d\left(\frac{1}{2} - \frac{1}{q^*} \right)$, then minimizing sequences will be compact in $L^q$, thus giving strong convergence of the denominator. The existence of a minimizer in $H^1$ follows readily by applying Fatou lemma. If $q = q^*$ I'm not quite sure right now, but profile decomposition could help you - this is precisely studying defects of compactness of bounded sequences of $H^s$ functions in Sobolev-critical $L^p$ spaces. | |
| Mar 16, 2015 at 15:44 | comment | added | Hachino | Not necessarily, because of Sobolev embeddings. Consider, for simplicity, functions belonging to $\mathcal{C}^{\infty}_c(\Omega)$, thus cancelling part of the numerator. If $q$ is too high (above the exponent given by the Sobolev embedding), then you can build up Dirac-like functions with $H^1$ norm equal to $1$ and big $L^q$ norm. If $d$ is the ambient dimension and $\rho$ some smooth function with compact support, consider $\rho_n(x) := n^{\frac{d}{2}-1} \rho(nx)$. | |
| Mar 16, 2015 at 15:27 | comment | added | Learner | But , does the solution(i.e. minimum) really exist for all $q \in \mathbb R$? I have been thinking for quite sometime but couldn't really prove . | |
| Mar 16, 2015 at 14:24 | history | edited | Hachino | CC BY-SA 3.0 |
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| Mar 15, 2015 at 11:09 | comment | added | Hachino | @Learner : regarding guidance, I strongly suggest you to use such Taylor-like expansions instead of directional derivatives. I believe they are less misleading, especially with intricate computations - think that this case was an introductory exercise. | |
| Mar 15, 2015 at 11:06 | history | answered | Hachino | CC BY-SA 3.0 |