## Smoothness for small initial data

$\displaystyle \begin{cases}u_t+u\cdot\nabla u-\nu\Delta u+\nabla p=0\\div u=0\\u(x,0)=u_0\end{cases}\text{ on }\mathbb{R}^3$

Thm(Kato): If ${u_0\in \dot{H}^{1/2}}$ is small, then the solution with initial data ${u_0}$ is global smooth.

Proof: ${\Lambda u=\sqrt{-\Delta}u}$ is a well defined operator on ${\dot{H}^{1/2}}$. Multiplying ${\Lambda u}$ on both sides, we get

$\displaystyle \frac{1}{2}\frac{d}{dt}||u||_{\dot{H}^{1/2}}^2+||u||_{\dot{H}^{3/2}}^2\leq \int |u||\nabla u||\Lambda u|\leq ||u||_{L^{3}}||\nabla u||_{L^3}||\Lambda u||_{L^3}$

By the Sobolev embedding,

$\displaystyle ||u||_{L^3}\leq C||u||_{\dot{H}^{1/2}}$

So we can get

$\displaystyle ||u||_{L^{3}}||\nabla u||_{L^3}||\Lambda u||_{L^3}\leq C||u||_{\dot{H}^{1/2}}||u||_{\dot{H}^{3/2}}^2$

Let ${G=||u||_{\dot{H}^{3/2}}^2}$, ${N=||u||_{\dot{H}^{1/2}}^2}$,

$\displaystyle \frac{1}{2}\frac{d}{dt}N+G(1-C\sqrt{N})\leq 0$

If ${N(0)<1/C^2}$ is small enough, then ${N'(t)}$ will be negative thus ${N(t)<1/C^2}$ and small enough. Consequently, ${||u||_{L^3}\leq C||u||_{\dot{H}^{1/2}}}$ will also be small. While from the NS equation,

$\displaystyle \frac{1}{2}\frac{d}{dt}|\nabla u|_{L^2}^2+|\Delta u|_{L^2}^2\leq\int|u||\nabla u||\Delta u|\leq C||u||_{L^3}||\Delta u||_{L^2}^2$

When ${||u||_{L^3}}$ small enough, the right hand side can be absorded by left hand side. Thus ${|\nabla u|_{L^2}^2}$ will be bounded. And from the standard regularity theory, ${u}$ must be smooth. $\Box$

Remark: Follows from my note of lectures given by Peter Constantin.