Let
- $T>0$
- $d\in\mathbb N$
- $\Omega\subseteq\mathbb R^d$ be bounded and open and $\partial\Omega$ be Lipschitz
- $Q:=(0,T)\times\Omega$
I'm reading Navier-Stokes Equations: Theory and Numerical Analysis by Roger Temam. The author states that if $u\in C^2(\overline Q,\mathbb R^d)$ is a classical solution of the Navier-Stokes equation, then it would be obvious that $$\tilde u(t):=u(t,\;\cdot\;)\;\;\;\text{for }t\in(0,T)$$ belongs to $L^2((0,T),V)$, where $V$ is a (special) closed subspace of $H_0^1(\Omega,\mathbb R^d)$.
Maybe it's cause I don't understand the definition of $C^2(\overline Q,\mathbb R^d)$, but I don't even understand why $\tilde u$ takes values in $V$.
So, how can we prove the claim?
$u$ is continuously differentiable on a bounded set, presumably required to zero on the boundary, by assumption, so it is obvious that $u$ is in $L^2$, its derivative is in $L^2$, and it is zero on the boundary. That the divergence of $u$ is zero is a part of the NSE. So $u(t)$ is in $V$ for all $t$. The fact that it is in $L^2(0,T;V)$ can be found with some simple calculus.
$$u_t + (u\cdot \nabla) u -\Delta u+\nabla p=0 $$
$$\frac{1}{2}\frac{d}{dt}\int |u|^2 \,dx + \int(u\cdot \nabla) u\cdot u\,dx +\int|\nabla u|^2dx=0 $$
Then integrating in time you get that $||\nabla u||_{L^2}$ is in $L^2(0,T)$.