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.

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