Nablaの演算法則メモ

$ \pmb{\nabla}\phi=\sum_i\pmb{e}_i\frac{\partial \phi}{\partial x_i}=\sum_i\frac{\partial \phi}{\partial x_i}\pmb{e}_i

$ \pmb{\nabla}(\phi+\psi)=\pmb{\nabla}\phi+\pmb{\nabla}\psi

$ \pmb{\nabla}(\phi\psi)=\sum\frac{\partial (\phi\psi)}{\partial x_i}\pmb{e}_i=\psi\pmb{\nabla}\phi+\phi\pmb{\nabla}\psi

$ \pmb{\nabla}\cdot\pmb{a}=\sum_i\pmb{e}_i\cdot\frac{\partial \pmb{a}}{\partial x_i}=\sum_i\frac{\partial [\pmb{a}]_i^\mathsf{E}}{\partial x_i}

$ \pmb{\nabla}\cdot(\pmb{a}+\pmb{b})=\pmb{\nabla}\cdot\pmb{a}+\pmb{\nabla}\cdot\pmb{b}

$ \pmb{\nabla}\pmb{a}=\sum\pmb{e}_i\frac{\partial \pmb{a}}{\partial x_i}=\sum\frac{\partial [\pmb{a}]_j^\mathsf{E}}{\partial x_i}\pmb{e}_i\pmb{e}_j

$ \mathrm{tr}(\pmb{\nabla}\pmb{a})=(\pmb{\nabla}\pmb{a}):\pmb{I}=\pmb{\nabla}\cdot\pmb{a}

$ \pmb{\nabla}(\pmb{a}+\pmb{b})=\sum\left(\pmb{e}_i\frac{\partial (\pmb{a}+\pmb{b})}{\partial x_i}\right)

$ =\pmb{\nabla} \pmb{a}+\pmb{\nabla} \pmb{b}

$ \pmb{\nabla} (\pmb{\nabla}\phi)=\sum\pmb{e}_i \frac{\partial \pmb{\nabla}\phi}{\partial x_i}

$ =\sum_{i,j}\frac{\partial^2 \phi}{\partial x_i\partial x_j}\pmb{e}_i \pmb{e}_j

曲線座標の場合、この展開が怪しくなるので注意

内側のnablaの基底に、外側のnablaの偏微分演算子がかかってくる

$ \because \mathrm{tr}(\pmb{\nabla} \pmb{a})=\pmb{\nabla}\cdot\pmb{a}

$ \pmb{\nabla} (\pmb{\nabla}\phi)=(\pmb{\nabla} \pmb{\nabla})\phi

と書ける(結合法則)

$ \mathrm{tr}(\pmb{\nabla} (\pmb{\nabla}\phi))=\mathrm{tr}(\pmb{\nabla} \pmb{\nabla})\phi=\pmb{\nabla}^2\phi

$ \pmb{\nabla}\cdot(\pmb{\nabla}\phi)=\sum\frac{\partial^2 \phi}{{\partial x_i}^2}

$ \pmb{\nabla}\cdot(\pmb{\nabla}\phi)= (\pmb{\nabla}\cdot\pmb{\nabla})\phi

$ = |\pmb{\nabla}|^2\phi

$ =: \Delta\phi

$ \pmb{\nabla} \pmb{T}=\sum\pmb{e}_i \frac{\partial \pmb{T}}{\partial x_i}

$ =\sum\frac{\partial [\pmb{T}]_{jk}^\mathsf{EE}}{\partial x_i}\pmb{e}_i \pmb{e}_j \pmb{e}_k

$ \pmb{\nabla} (\pmb{\nabla} \pmb{a})=\sum\frac{\partial^2 [\pmb{a}]_k^\mathcal{S}}{\partial x_i\partial x_j}\pmb{e}_i \pmb{e}_j \pmb{e}_k

$ =(\pmb{\nabla} \pmb{\nabla}) \pmb{a}

$ \pmb{\nabla}(\pmb{\nabla}\cdot\pmb{a})=\sum\frac{\partial^2 [\pmb{a}]_j^\mathsf{E}}{\partial x_i\partial x_j}\pmb{e}_i

$ \pmb{\nabla}\wedge\pmb{a}=\pmb{\nabla} \pmb{a}-\pmb{a} \pmb{\nabla}

$ =\sum_{i\neq j}\left(\frac{\partial [\pmb{a}]_j^\mathsf{E}}{\partial x_i}-\frac{\partial [\pmb{a}]_i^\mathsf{E}}{\partial x_j}\right)\pmb{e}_i \pmb{e}_j

$ \mathrm{div}(\pmb{T})=(\pmb{\nabla}\otimes\pmb{T}):\pmb{I}

$ =\sum\frac{\partial [\pmb{T}]_{jj}^\mathsf{EE}}{\partial x_i}\pmb{e}_i

$ =\pmb{\nabla}\mathrm{tr}(\pmb{T})

わかった。↑での勾配の定義が違ったんだ

左発散の場合

$ \mathrm{div}(\pmb{a} \pmb{b})=\sum_{i,j}\frac{\partial [\pmb{a}]_i^\mathsf{E}[\pmb{b}]_j^\mathsf{E}}{\partial x_j}\pmb{e}_i

$ =\sum_{i,j}[\pmb{a}]_i^\mathsf{E}\frac{\partial[\pmb{b}]_j^\mathsf{E}}{\partial x_j}\pmb{e}_i+\sum_{i,j}[\pmb{b}]_j^\mathsf{E}\frac{\partial[\pmb{a}]_i^\mathsf{E}}{\partial x_j}\pmb{e}_i

$ =\pmb a(\pmb\nabla\cdot\pmb b)+\pmb b\cdot\pmb\nabla\pmb a

$ =(\pmb{\nabla}\cdot\pmb{b})\pmb{a}+(\pmb{b}\cdot\pmb{\nabla})\pmb{a}

別解

$ \mathrm{div}(\pmb{a} \pmb{b})=\sum_i(\pmb{\nabla}\cdot([\pmb{a}]^\mathsf{E}_i\pmb{b}))\pmb{e}_i

$ =\sum_i\left((\pmb{\nabla}[\pmb{a}]^\mathsf{E}_i)\cdot\pmb{b}+[\pmb{a}]^\mathsf{E}_i\pmb{\nabla}\cdot\pmb{b}\right)\pmb{e}_i

$ =\sum_i(\pmb{e}_i \pmb{\nabla}[\pmb{a}]^\mathsf{E}_i)\pmb{b}+(\pmb{\nabla}\cdot\pmb{b})\pmb{a}

$ =(\pmb{\nabla}\cdot\pmb{b})\pmb{a}+(\pmb{b}\cdot\pmb{\nabla})\pmb{a}

別解

$ \mathrm{div}(\pmb{a} \pmb{b})=\sum_i\frac{\partial \pmb{a} \pmb{b}}{\partial x_i}\pmb{e}_i

$ =\pmb{a} \sum_i\frac{\partial\pmb{b}}{\partial x_i}\pmb{e}_i+\sum_i\left(\frac{\partial\pmb{a}}{\partial x_i} \pmb{b}\right)\pmb{e}_i

$ =\left(\sum_i\pmb{e}_i\cdot\frac{\partial\pmb{b}}{\partial x_i}\right)\pmb{a}+\sum_i(\pmb{b}\cdot\pmb{e}_i)\frac{\partial\pmb{a}}{\partial x_i}

$ =(\pmb{\nabla}\cdot\pmb{b})\pmb{a}+(\pmb{b}\cdot\pmb{\nabla})\pmb{a}

右発散の場合

$ \mathrm{div}_R(\pmb{a}\otimes\pmb{b})=\mathrm{div}(\pmb{b}\otimes\pmb{a})

$ =(\pmb{\nabla}\cdot\pmb{a})\pmb{b}+(\pmb{a}\cdot\pmb{\nabla})\pmb{b}

$ =\sum_{i,j}\frac{\partial}{\partial x_j}\left(\frac{\partial\pmb{a}}{\partial x_i}\cdot\pmb{e}_j\right)\pmb{e}_i

$ =\sum_{i,j}\frac{\partial^2[\pmb{a}]^\mathsf{E}_j}{\partial x_j\partial x_i}\pmb{e}_i

もし分母の偏微分を交換できるなら(普通は交換できる)、

$ =\pmb{\nabla}(\pmb{\nabla}\cdot\pmb{a})

$ =\sum_{i,j}\frac{\partial}{\partial x_j}\frac{\partial (\pmb{e}_i\cdot\pmb{e}_j)\pmb{a}}{\partial x_i}

$ =\sum_i\frac{\partial^2\pmb{a}}{{\partial x_j}^2}

$ =\pmb{\nabla}^2\pmb{a}

$ \bm\nabla*(\bm a*\bm b)=\sum_i\bm e_i*\frac{\partial (\bm a*\bm b)}{\partial x_i}

$ = (\bm\nabla*\bm a)*\bm b+\sum_i\bm e_i*\bm a*\frac{\partial\bm b}{\partial x_i}

$ \bm{\nabla}\cdot(\phi\bm{a})=(\bm{\nabla}\phi)\cdot\bm{a}+\phi\bm{\nabla}\cdot\bm{a}

一般式から求まらない形

$ \bm{\nabla} (\phi\bm{a})=(\bm{\nabla}\phi) \bm{a}+\phi(\bm{\nabla} \bm{a})

$ \bm\nabla(\bm a\cdot\bm b)=(\bm\nabla \bm a)\cdot\bm b+(\bm\nabla \bm b)\cdot\bm a

$ =(\bm\nabla \bm a)\cdot\bm b+\bm a\cdot(\bm b\overleftarrow{\bm\nabla})

$ \bm{\nabla} (\bm{a} \bm{b})=(\bm{\nabla} \bm{a}) \bm{b}+\sum_i\bm e_i\bm a\frac{\partial\bm b}{\partial x_i}

第2項はこれ以上簡略化できない

$ \bm{\nabla}\cdot(\bm{T}\bm{a})=(\bm{\nabla}\cdot\bm{T})\bm{a}+\bm{T}^\top\cdot(\bm{\nabla}\bm{a})

$ \bm{\nabla}\cdot(\bm{T}\cdot\bm{A})=(\bm{\nabla}\cdot\bm{T})\cdot\bm{A}+\bm{T}:\bm\nabla\bm A

$ \bm{\nabla}\cdot(\bm a\cdot\bm T)=\sum_i\bm e_i\cdot\frac{\partial (\bm a\cdot\bm T)}{\partial x_i}

$ = \bm a\overleftarrow{\bm\nabla}:\bm T+\bm a\cdot\bm T\cdot \overleftarrow{\bm\nabla}

これも一般式では表現できない

一般化すると

References