# 2017-2018-2偏微分方程复习题解析10

Problem: Consider the three-dimensional Navier-Stokes equations $$\bee\tag{*} \seddm{ \p_tu+(u\cdot\n)u-\lap u+\n P=0,\\ \n\cdot u=0,\\ u|_{t=0}=u_0. } \eee$$ Let $u_0\in L^2(\bbR^3)$. Then by the Leray's famous work, there exists at least one weak solution of (*). Suppose that $u_0\in H^1(\bbR^3)$ and $$\bee\tag{**} u\in L^p(0,T;L^q(\bbR^3)),\quad \f{2}{p}+\f{3}{q}=1,\quad 3<q\leq\infty. \eee$$ Show that $u$ is a strong solution, i.e., $u\in L^\infty(0,T;L^2(\bbR^3))\cap L^2(0,T;H^1(\bbR^3))$. This is the classical Ladyzhenskaya-Prodi-Serrin condition.

Proof: Multiplying $(*)_1$ by $-\lap u$, integrating by parts gives $$\beex \bea &\quad\f{1}{2}\f{\rd }{\rd t} \sen{\n u}_{L^2}^2 +\sen{\lap u}_{L^2}^2\\ &=\int (u\cdot\n)u\cdot \lap u\rd x\\ &\leq \sen{u}_{L^q}\sen{\n u}_{L^a}\sen{\lap u}_{L^2}\qx{\f{1}{q}+\f{1}{a}+\f{1}{2}=1}\\ &\leq C\sen{u}_{L^q} \cdot\sen{\n u}_{L^2}^{1-\tt} \sen{\lap u}_{L^2}^\tt \cdot \sen{\lap u}_{L^2}\qx{\f{3}{a}=(1-\tt)\f{3}{2}+\tt\sex{-1+\f{3}{2}}}\\ &\leq C\sen{u}_{L^q}^\f{2}{1-\tt} \sen{\n u}_{L^2}^2 +\f{1}{2}\sen{\lap u}_{L^2}^2. \eea \eeex$$ Consequently, $$\bex \f{\rd }{\rd t} \sen{\n u}_{L^2}^2 +\sen{\lap u}_{L^2}^2\leq C\sen{u}_{L^q}^\f{2}{1-\tt} \sen{\n u}_{L^2}^2. \eex$$ Applying the Gronwall inequality yields $$\bex \sen{\n u(t)}_{L^2}^2 +\int_0^t \sen{\lap u(s)}_{L^2}^2\rd s \leq \sen{\n u_0}_{L^2}^2 \cdot \exp\sez{C\int_0^T \sen{u(s)}_{L^q}^\f{2}{1-\tt} \rd s}<\infty,\quad\forall\ 0\leq t\leq T. \eex$$ At last, you can directly calculate $$\bex \f{2}{2/(1-\tt)}+\f{3}{q}=1. \eex$$

posted @ 2018-05-15 15:08 张祖锦 阅读(...) 评论(...) 编辑 收藏