1

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?

Community
  • 1
0xbadf00d
  • 14,208

1 Answers1

1

$u$ 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 $L^2(0,T)$.

mathematician
  • 2,529
  • 14
  • 13
  • The crucial part is $$\tilde u(t)\in V;;;\text{for all }t\in(0,T);.\tag 1$$ If $\partial\Lambda$ is Lipschitz, then $$V=\left{v\in H_0^1(\Lambda,\mathbb R^d):\nabla\cdot v=0\right};.\tag 2$$ Since $$\nabla\cdot\tilde u(t)=0;;;\text{for all }t\in(0,T)$$ (where $\nabla\cdot\tilde u(t)$ is the classical divergence), the only thing which isn't clear to me is that $$\tilde u(t)\in H_0^1(\Lambda,\mathbb R^d);;;\text{for all }t\in(0,T);.\tag 3$$ In light of an other discussion, it's not obvious to me that $(3)$ holds. – 0xbadf00d Sep 04 '16 at 08:50
  • It's $C^2$ on a bounded domain. The derivative is bounded, so it's square integrable. – mathematician Sep 04 '16 at 09:19
  • Did you take a look at the comments under the answer to the linked question? Why do you think that the partial derivatives $\frac{\partial u(t,;\cdot;)}{\partial x_i}$ are bounded? By definition, the $\frac{\partial u(t,;\cdot;)}{\partial x_i}$ are continuous in $\Lambda$ and I don't see that the other properties of $u$ yield that the $\frac{\partial u(t,;\cdot;)}{\partial x_i}$ can be continuously extended to $\overline\Lambda$ (which is a compact set and I assume that you're using that a continuous function is bounded on a compact set). – 0xbadf00d Sep 04 '16 at 09:51
  • I assume $C^2(\bar{Q})$ means the 2nd partial derivatives are continuous on $\bar{Q}$. – mathematician Sep 05 '16 at 00:08
  • Now, it's still not clear why $\tilde u(t)\in H_{\color{red}0}^1(\Lambda,\mathbb R^d)$. See my other question. – 0xbadf00d Sep 06 '16 at 09:03
  • Of course, $u$ is only 0 on the boundary if it is assumed so. The boundary conditions are a part of the PDE that $u$ is assumed to satisfy. – mathematician Sep 06 '16 at 11:54
  • Why do we need to consider the calculation in your answer? In light of our discussion, shouldn't $u,\nabla u\in L^2$ simply follow from boundedness of $u,\nabla u$ in $\Lambda$? – 0xbadf00d Sep 06 '16 at 12:06
  • Still interested in an answer to the last question. – 0xbadf00d Sep 08 '16 at 10:28
  • yup you're right – mathematician Sep 09 '16 at 17:03