Question Number 186077 by Mastermind last updated on 31/Jan/23
$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\bigtriangledown\bullet\left(\varnothing\overset{−} {\mathrm{A}}\right)=\left(\bigtriangledown\varnothing\right)\bullet\mathrm{A}+\varnothing\left(\bigtriangledown\bullet\overset{−} {\mathrm{A}}\right). \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Answered by aleks041103 last updated on 30/Apr/23
$${Interesting}\:{choice}\:{of}\:{notation}... \\ $$$${what}\:{you}\:{want}\:{is} \\ $$$${div}\left({a}\overset{\rightarrow} {{v}}\right)={grad}\left({a}\right).\overset{\rightarrow} {{v}}\:+\:{a}\:{div}\left(\overset{\rightarrow} {{v}}\right) \\ $$$$ \\ $$$${First}\:{way}: \\ $$$${div}\left({av}\right)=\frac{\partial\left({av}_{{x}} \right)}{\partial{x}}+\frac{\partial\left({av}_{{y}} \right)}{\partial{y}}+\frac{\partial\left({av}_{{z}} \right)}{\partial{z}}= \\ $$$$={v}_{{x}} \frac{\partial{a}}{\partial{x}}+{v}_{{y}} \frac{\partial{a}}{\partial{y}}+{v}_{{z}} \frac{\partial{a}}{\partial{z}}+{a}\left(\frac{\partial{v}_{{x}} }{\partial{x}}+\frac{\partial{v}_{{y}} }{\partial{y}}+\frac{\partial{v}_{{z}} }{\partial{z}}\right)= \\ $$$$=\begin{pmatrix}{{v}_{{x}} }\\{{v}_{{y}} }\\{{v}_{{z}} }\end{pmatrix}\:\:.\begin{pmatrix}{\partial{a}/\partial{x}}\\{\partial{a}/\partial{y}}\\{\partial{a}/\partial{z}}\end{pmatrix}\:\:+\:{a}\:{div}\left({v}\right)= \\ $$$$={v}.{grad}\left({a}\right)+{a}\:{div}\left({v}\right)={div}\left({av}\right) \\ $$$$ \\ $$$${Second}\:{way}:\:{using}\:{einstein}\:{notation} \\ $$$${div}\left({av}\right)=\partial_{{i}} \left({av}^{{i}} \right)={v}^{{i}} \left(\partial_{{i}} {a}\right)+{a}\left(\partial_{{i}} {v}^{{i}} \right)= \\ $$$$={v}.{grad}\left({a}\right)+{a}\:{div}\left({v}\right)={div}\left({av}\right) \\ $$$$ \\ $$$${Note}: \\ $$$${I}\:{dropped}\:{the}\:\overset{\rightarrow} {\:}\:{on}\:{top}\:{of}\:{the}\:{v}\:{for}\:{easy} \\ $$$${typesetting}. \\ $$