[Mathematical Analysis for small readers] (vector formula: the hall term)

$ \ 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 $

Proof: the first component on the right side is $ \ 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_3 \ p_2b_3 \ products \ & \ quad-\ p_3b_1 \ p_1b_2-\ p_3b_2 \ p_2b_1 \ p_3b_3 \ p_3b_3 \\ & \ 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_1 + \ 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_1bb_1-\ 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 $ use formula (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 }), \ EEx $ we know $ \ Bex \ n ({\ bf j} \ times {\ BF B}) = ({\ BF B} \ cdot \ n) {\ bf j}-({\ bf j} \ cdot \ n) {\ BF B }. \ EEx $ and the first item on the left end is $-({\ bf j} \ cdot \ n) B _1 + ({\ BF B} \ cdot \ n) J_1 $. therefore, there is a conclusion.

See [D. Chae, M. schonbek, on the temporal decay for the hall-magnetohydrodynamic equations, J. differential equations, 255 (2013),].