Let $d\in\mathbb N$, $\Omega\subseteq\mathbb R^d$ be bounded and open, $V:=H_0^1(\Omega)$, $H:=L^2(\Omega)$ and $a\in C^1([0,\infty)\times\overline\Omega)$ such that $\alpha:=\inf_{(0,\:\infty)\times\Omega)}a>0$. Moreover, let $$\mathfrak a_t(u,v):=\langle a(t)\nabla u,\nabla v\rangle_{L^2}\;\;\;\text{for }u,v\in V$$ and $$\mathcal A(t):V\to V'\;,\;\;\;v\mapsto-\mathfrak a_t(\;\cdot\;,v)$$ for $t\ge0$. Let $u_0\in L^\infty(\Omega)$.
We say that $u$ is a weak solution of $$\left\{\begin{align}\frac{\partial u}{\partial t}&=\nabla_x\cdot a\nabla_xu&\text{in }(0,\infty)\times\Omega\\u&=0&\text{on }(0,\infty)\times\partial\Omega\\u(0,\;\cdot\;)&=u_0\end{align}\right.\tag1$$ if $u\in L^2((0,T),H)$ has a weak derivative in $L^2((0,T),V')$ and $$\left\{\begin{align}u'&=\mathcal Au\\u(0)&=u_0\end{align}\right.\tag2$$ for all $T>0$.
Now assume $t_0>0$ and $v$ is a weak solution of $(1)$ with $u_0$ replaced by $v_0:=u(t_0)$. Can we somehow identify $u$ and $v$?
We may note that, for all $T>0$, $u$ has a modification$^1$ $\tilde u\in C([0,T],V')$ and $$\tilde u(t)=\tilde u(s)+\int_s^tu'(r)\:{\rm d}r\tag3$$ for all $s,t\in[0,T]$.
$^1$ That is, $$\forall\varphi\in V:\langle\varphi,u(t)\rangle_H=\langle\varphi,\tilde u(t)\rangle_{V,\:V'}\tag4$$ for almost all $t\in(0,T)$.