Let $\Omega\subseteq\mathbb R^d$ be bounded and open, $V:=H_0^1(\Omega)$, $H:=L^2(\Omega)$, $u_0\in C(\overline\Omega)$, $T\in(0,\infty]$, $I:=(0,T)$ and $u\in\mathcal L^2(I,V)$ admit a weak derivative in $\mathcal L^2(I,V')$ and satisfy $$\langle\varphi,u(0)-u_0\rangle_H=0\tag1.$$ Note that, by the assumption that $u$ has a weak derivative in $\mathcal L^2(I,V')$, we know that $u$ has a modification in $C(\overline I,H)$ and hence $(1)$ is meaningful.
Question: Does $(1)$ necessary require that $\left.u\right|_{\partial\Omega}=0$?
A naive argumentation could be that $u_0$ and $u(0,\;\cdot\;)$ are both continuous on $\overline\Omega$ and hence, if $x\in\overline\Omega$ and $(x_n)_{n\in\mathbb N}\subseteq\Omega$ with $x_n\xrightarrow{n\to\infty}x$, we can infer that $u(0,x)=\lim_{n\to\infty}u(0,x_n)=u_0(x)$".
However, I wonder whether this argumentation is really correct and isn't missing some subtlety. In fact, while $u$ has a modification, say $\tilde u$, in $C(\overline I,H)$, are we able to conclude
- $\left.\tilde u(t,\;\cdot\;)\right|_{\partial\Omega}=0$ for all $t\in\overline I$
and/or
- $\tilde u(0,\;\cdot\;)=u_0$ (on $\overline\Omega$)
at all?