Let $X$ be a Banach space, we understand $L^1(0, T, X)$ is the space of strongly measurable functions from $[0, T]$ valued in $X$, that is integrable. Assume ${\bf u}\in L^1(0, T, X)$, we say ${\bf v}\in L^1(0, T, X)$ is the (weak time) derivative of ${\bf u}$ if for any $\varphi\in C_0^\infty(0, T)$ we have $$ \int_0^T {\bf u}(t) \varphi'(t)dt=-\int_0^T {\bf v}(t)\varphi(t)dt, $$ and we will write ${\bf v}={\bf u}'$.

Now in Evans' PDE book, he discussed a situation where ${\bf u}\in L^2(0, T, H_0^1(\Omega))$ while ${\bf u}'\in L^2(0, T, H^{-1}(\Omega))$. It seems there are two possible interpretations:

(A). The situation implies ${\bf u}\in L^1(0, T, H_0^1(\Omega))$, so we follow the DEFINITON of derivative I mentioned in the beginning and understand that ${\bf u}'\in L^1(0, T, H_0^1(\Omega))$. However in this situation is possible ${\bf u}'\notin L^2(0, T, H_0^1(\Omega))$, while from the continuous injection $H_0^1(\Omega)\to H^{-1}(\Omega)$ it happens that ${\bf u}'\in L^2(0, T, H^{-1}(\Omega))$. In conclusion, $$ {\bf u}'\in L^1(0, T, H_0^1(\Omega))\cap L^2(0, T, H^{-1}(\Omega)). $$

(B). It just means $$ \int_0^T {\bf u}(t) \varphi'(t)dt=-\int_0^T {\bf u}'(t)\varphi(t)dt $$ literally. This looks like a little miracle - the right hand side is supposed to be in $H^{-1}(\Omega)$, but the equation tells us it is indeed in $H_0^1(\Omega)$.

My question is, Assume one uses (B) as definition, is there a theorem which implies that indeed ${\bf u}'$ is also in $L^{1}(0, T, H_0^1(\Omega))$ ? Or counterexamples?

  • 1
    $\begingroup$ (B) is the correct definition, in particular, there is no requirement (or theorem) that $u'\in L^1(0,T,H^1_0)$. The "little miracle" comes from the smoothness of $\varphi$ which makes the integration by parts possible. $\endgroup$ – Michael Renardy Oct 3 at 2:54

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