If $u:[0,\infty)\times\mathcal V\to\mathbb R^d$ is divergence-free, then $\partial u/\partial t$ is divergence-free too

Question

Let

• $d\in\mathbb N$
• $\mathcal V\subseteq\mathbb R^d$ be open
• $u:[0,\infty)\times\mathcal V\to\mathbb R^d$ with
  • $u(\;\cdot\;,x)\in C^1(\mathbb R_{\ge 0},\mathbb R^d)$ for all $x\in\mathcal V$
  • $u(t,\;\cdot\;)\in H_0^1(\mathcal V,\mathbb R^d)$ for all $t\ge 0$

Suppose that $$\nabla\cdot u:=\sum_{i=1}^d\frac{\partial u_i}{\partial x_i}=0\;.\tag 1$$ I want to show that $$\nabla\cdot\frac{\partial u}{\partial t}=0\;.\tag 2$$

If $u(t,\;\cdot\;)\in C^1(\mathcal V,\mathbb R^d)$ for all $t\ge 0$, then $(1)$ is obvious by Schwarz' theorem. How can we show it in general?