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$$\bex \bbu\in L^{p,r}(0,T;L^{q,\infty}(\bbR^3)),\quad\frac{2}{p}+\frac{3}{q}=1,\quad 3<q<\infty,\quad 2<p<r<\infty, \eex$$ or $$\bex \sen{\bbu}_{L^{p,\infty}(0,T;L^{q,\infty}(\bbR^3))}\leq \ve,\quad \frac{2}{p}+\frac{3}{q}=1,\quad 3<q<\infty,\quad 2<p<\infty, \eex$$
[Papers]NSE, $u$, Lorentz space [Sohr, JEE, 2001]
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原文地址:http://www.cnblogs.com/zhangzujin/p/4247901.html