1
$\begingroup$

The usual version of Aubin-Lions lemma states that: if $X_0 \subset X \subset X_1$ where $X_0$, $X$ and $X_1$ are three Banach spaces and (the embedding $X_0$ into $X$ is compact and of $X$ into $X_1$ is continuous) then for some $p\geq 1$ and $q\geq 1$ the embedding of $$W=\{v\in L^p([0,T];X_0),~~\dot{v}\in L^q([0,T];X_1)\}$$ into $L^p([0,T];X)$ is compact. Usualy, the aforementioned apply to $X_0=H^1$ and $X=L^2$. The question is: can i apply Aubin-Lions lemma on the Sobolev spaces $X_0=H^{s+1}$, $X=H^{s}$ where $s>3/2$? In other words, is $H^{s+1}$ embedded compactly into $H^{s}$?

  • note that i am working on 3D domain.
$\endgroup$
  • $\begingroup$ Generally, compact embedding for scalar Sobolev functions + compact embedding between target spaces should imply compact embedding for vector-valued Sobolev functions; but finding a reference is tricky. $\endgroup$ – user357151 May 4 '18 at 23:08
  • $\begingroup$ The space dimension that i am working on is the 3D periodic vector valued space. I am asking about the Sobolev index which concerns the regularity $\endgroup$ – AlphaXY May 4 '18 at 23:21
1
$\begingroup$

Yes, $H^{s+1}$ is compactly embedded in $H^{s}$ for $s>3/2$. The restriction on $s$ is even unnecessary and this result holds in any dimension, we just need a bounded domain with sufficiently smooth boundary.

Indeed, let $s \in \mathbb{R}$. Then $H^{s}$ is compactly embedded in $H^{s-k}$ for any $k>0$, see 'Non-Homogeneous Boundary Value Problems Vol. I' by Lions & Magenes, Theorem 16.1.

Hence by Aubin-Lions it is true that $W$ is compactly embedded in $L^p(0,T;H^s)$ for the underlying Hilbert triple

$$H^{s+1} \hookrightarrow \hookrightarrow H^s \hookrightarrow (H^{s+1})'.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.