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.

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