Simple heat equation, solution regularity

Question

I have a small problem with a regularity result for a simple parabolic heat equation: Given a $C^2$ open subset of $\mathbb R^n$ called $\Omega$, and a time $T > 0$, i have the following heat equation: $$ y_t - \Delta y = u,\quad Q := [0,T] \times \Omega \\ y = 0,\quad \Sigma := [0,T] \times \delta \Omega \\ y(0) = y_0,\quad \Omega $$ where $y_0 \in H^1(\Omega)$. Now a common result proves, that the equation has a unique solution $$ y \in L^\infty ( 0, T, H^1_0(\Omega)) \cap L^2(0,T,H^2(\Omega)), \\ y_t \in L^2(0,T,L^2(\Omega)). $$ My question now is, why the solution is also in $$ H_\Sigma^{2,1} := \{ y \in L^2(0,T,H^2(\Omega)); y_t \in L^2(Q); y=0 \text{ over } \Sigma \}. $$ The question seems pretty simple, but somehow I have a problem with the trace and the fact, that the function is only zero in the $H^1_0(\Omega)$ sense? Thank you very much :-)

Answer 1

The solution is obviously in the standard $H^1$ Sobolev space on the space-time domain $Q$: $u \in H^{1}(Q) = \{ u : \Vert u\Vert_{L^2(Q)}^2+ \Vert \nabla u\Vert_{L^2(Q)}^2+\Vert \partial_t u \Vert_{L^2(Q)}^2 < \infty \} $ and thus the trace operation $tr|_{\Sigma} u: H^1(Q) \rightarrow L^2(\Sigma)$ is continuous. Thus the trace on $\Sigma$ is well-defined (at least in the $L^2$-sense). As such you can write $$ \int_{\Sigma} (tr|_{\Sigma}u)^2 dx = \int_{0}^T \int_{\partial \Omega} \big(tr|_{\partial \Omega}u(t)\big)(x)^2 dx dt $$ where $u(t) \in H^1_0(\Omega)$ for almost every $t \in [0,T]$. For almost every $t \in [0,T]$ you can evaluate the trace as $u(t) \in H^1(\Omega)$ and hence has to be zero.