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