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grad/div/curl of product of two vector fields

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hbghlyj 发表于 2022-4-20 07:14 |阅读模式

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 楼主| hbghlyj 发表于 2022-4-20 07:48
\begin{aligned}
&\underline{\nabla}(\underline{A} \cdot \underline{B})=(\underline{A} \cdot \underline{\nabla}) \underline{B}+(\underline{B} \cdot \underline{\nabla}) \underline{A}+\underline{A} \times(\underline{\nabla} \times \underline{B})+\underline{B} \times(\underline{\nabla} \times \underline{A}) \\
&\underline{\nabla} \cdot(\underline{A} \times \underline{B})=\underline{B} \cdot(\underline{\nabla} \times \underline{A})-\underline{A} \cdot(\underline{\nabla} \times \underline{B}) \\
&\underline{\nabla} \times(\underline{A} \times \underline{B})=\underline{A}(\underline{\nabla} \cdot \underline{B})-\underline{B}(\underline{\nabla} \cdot \underline{A})+(\underline{B} \cdot \underline{\nabla}) \underline{A}-(\underline{A} \cdot \underline{\nabla}) \underline{B}
\end{aligned}
证明
\begin{aligned}
\underline{\nabla}(\underline{A} \cdot \underline{B})&=\underline{e}_i\frac{\partial}{\partial x_i}(A_jB_j)
\\&=\underline{e}_iA_j\frac{\partial B_j}{\partial x_i}+\underline{e}_iB_j\frac{\partial A_j}{\partial x_i}
\\&=\underline{e}_iA_j\frac{\partial B_i}{\partial x_j}+\underline{e}_iB_j\frac{\partial A_i}{\partial x_j}+\underline{e}_i(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})A_j\frac{\partial B_m}{\partial x_l}+\underline{e}_i(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})B_j\frac{\partial A_m}{\partial x_l}
\\&=\underline{e}_iA_j\frac{\partial B_i}{\partial x_j}+\underline{e}_iB_j\frac{\partial A_i}{\partial x_j}+\underline{e}_i\epsilon_{ijk}\epsilon_{klm}A_j\frac{\partial B_m}{\partial x_l}+\underline{e}_i\epsilon_{ijk}\epsilon_{klm}B_j\frac{\partial A_m}{\partial x_l}
\\&=(\underline{A} \cdot \underline{\nabla}) \underline{B}+(\underline{B} \cdot \underline{\nabla}) \underline{A}+\underline{A} \times(\underline{\nabla} \times \underline{B})+\underline{B} \times(\underline{\nabla} \times \underline{A})
\end{aligned}


\begin{aligned}
\underline{\nabla} \cdot(\underline{A} \times \underline{B}) &=\frac{\partial}{\partial x_{i}} \epsilon_{i j k} A_{j} B_{k} \\
&=\epsilon_{i j k}\left(\frac{\partial A_{j}}{\partial x_{i}}\right) B_{k}+\epsilon_{i j k} A_{j}\left(\frac{\partial B_{k}}{\partial x_{i}}\right) \\
&=B_{k} \epsilon_{k i j} \frac{\partial A_{j}}{\partial x_{i}}-A_{j} \epsilon_{j i k} \frac{\partial B_{k}}{\partial x_{i}}\\&=\underline{B} \cdot(\underline{\nabla} \times \underline{A})-\underline{A} \cdot(\underline{\nabla} \times \underline{B})
\end{aligned}


\begin{aligned}\underline{\nabla} \times(\underline{A} \times \underline{B})
&=\underline e_i\epsilon_{ijk}\frac{\partial}{\partial x_j}\epsilon_{klm}A_lB_m
\\&=\underline e_i(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})B_m\frac{\partial A_l}{\partial x_j}+\underline e_i(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})A_l\frac{\partial B_m}{\partial x_j}
\\&=\underline e_iA_i\frac{\partial B_j}{\partial x_j}-\underline e_iB_i\frac{\partial A_j}{\partial x_j}+\underline e_iB_j\frac{\partial A_i}{\partial x_j}-\underline e_iA_j\frac{\partial B_i}{\partial x_j}
\\&=\underline{A}(\underline{\nabla} \cdot \underline{B})-\underline{B}(\underline{\nabla} \cdot \underline{A})+(\underline{B} \cdot \underline{\nabla}) \underline{A}-(\underline{A} \cdot \underline{\nabla}) \underline{B}
\end{aligned}

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 楼主| hbghlyj 发表于 2022-4-20 09:39
本楼删除common.css中的分隔线样式(不知道怎么搞的,需要删两次,只删一次的话布局会有一个小变化,不懂...)

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