$$\bex \n\cdot{\bf b}=0\ra \n\times [(\n\times {\bf b})\times {\bf b}]=\n\times [\n\cdot ({\bf b}\otimes {\bf b})]. \eex$$
证明: 右端第一个分量为 $$\beex \bea &\quad \sum_i \p_2(\p_i(b_ib_3))-\p_3(\p_i(b_ib_2))\\ &=\sum_i \p_2(b_i\p_ib_3)-\p_3(b_i\p_ib_2)\\ &=\sum_i \p_2b_i\p_ib_3-\p_3b_i\p_ib_2\\ &\quad +\sum_ib_i\p_i(\p_2b_3-\p_3b_2)\\ &=\p_2b_1\p_1b_3+\p_2b_2\p_2b_3+\p_2b_3\p_3b_3\\ &\quad-\p_3b_1\p_1b_2-\p_3b_2\p_2b_2-\p_3b_3\p_3b_2\\ &\quad+({\bf b}\cdot\n)j_1\\ &=\p_2b_1\p_1b_3-\p_2b_3\p_1b_1\\ &\quad -\p_3b_1\p_1b_2+\p_3b_2\p_1b_1\\ &\quad+({\bf b}\cdot\n)j_1\\ &=\p_2b_1\p_1b_3 -\p_3b_1\p_1b_2 -j_1\p_1b_1 +({\bf b}\cdot\n)j_1\\ &=-(\p_3b_1-\p_1b_3)\p_2b_1 -(\p_1b_2-\p_2b_1)\p_3b_1 -j_1\p_1b_1+({\bf b}\cdot\n)j_1\\ &=-j_2\p_2b_1-j_3\p_3b_1-j_1\p_1b_1 +({\bf b}\cdot\n)j_1\\ &=-({\bf j}\cdot\n)b_1+({\bf b}\cdot\n)j_1. \eea \eeex$$ 利用公式 (link) $$\bex \n\times({\bf a}\times{\bf b})=({\bf b}\cdot\n){\bf a} -({\bf a}\cdot\n){\bf b}+{\bf a}(\n\cdot{\bf b})-{\bf b}(\n\cdot{\bf a}), \eex$$ 我们知 $$\bex \n({\bf j}\times {\bf b})=({\bf b}\cdot\n){\bf j}-({\bf j}\cdot\n){\bf b}. \eex$$ 而左端的第一项也为 $-({\bf j}\cdot\n)b_1+({\bf b}\cdot\n)j_1$. 故有结论.
see [D. Chae, M. Schonbek, On the temporal decay for the Hall-magnetohydrodynamic equations, J. Differential Equations, 255 (2013), 3971--3982].
