## One regularity of NS 3-D

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

Thm: If ${u}$ solves NS 3-D and ${u\in C([0,T];L^3(\mathbb{R}^3))}$, then ${u}$ is regular.

Proof: Denote

$\displaystyle E=\int_{\mathbb{R}^3}|u_0|^2dx$

Note that ${\int_{\mathbb{R}^3}|u(x,t)|^2dx\leq E}$ for all ${t>0}$.

Fix ${M>0}$, which will be determined later, letting ${G_M=\{x|u(x,t)\leq M\}}$, by Chebyshev’s inequality

$\displaystyle M^2\left|G^c_M\right|\leq \int_{\mathbb{R}^3}|u(x,t)|^2dx\leq E$

$\displaystyle \left|G^c_M\right|\leq \frac{E}{M^2}.$

Since ${u\in C([0,T];L^3)}$, then ${u(\cdot,t)}$ is compact in ${L^3}$, in particular, ${u(\cdot,t)}$ is uniformly integrable, namely

$\displaystyle \sup_t\left|\int_B|u(x,t)|^3dx\right|\leq \varepsilon\text{ whenever }|B|\leq \delta(\varepsilon),t\in [0,T]$

Multiplying the NS equation by ${\Delta u}$, we get

$\displaystyle \frac{1}{2}\frac{d}{dt}\int |\nabla u|^2+\nu\int |\Delta u|^2dx\leq \int |u||\nabla u||\Delta u|$

$\displaystyle \leq \int_{G_M}|u||\nabla u||\Delta u|dx+\int_{G_M^c}|u||\nabla u||\Delta u|dx$

$\displaystyle \leq M\int |\nabla u||\Delta u|dx+\left(\int_{G_M^c} u^3dx\right)^{1/3}||\nabla u||_{L^6}||\Delta u||_{L^2}$

Choose ${M}$ such that ${\int_{G_M^c} u^3dx}$ small enough and use the Sobolev inequality ${||\nabla u||_{L^6}||\leq C||\Delta u||_{L^2}}$,

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

By Young’s inequality,

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

So ${\nabla u\in L^\infty([0,T]; L^2)}$. From the regularity theory, ${u}$ must be a regular solution. $\Box$

Remark: Follows from my note of lectures given by peter constantin.