\[
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\]
はじめに
他の記事で参照するための補題をここに書いておく。
ベクトルのスカラー関数倍の回転
主張
$\bm{r}\in\realNumbers$とする。$f: \realNumbers^3\to\complexNumbers,\;A(\bm{r}): \realNumbers^3\to\realNumbers^3$はともに$\mathrm{C}^1$級とする。このとき次式が成り立つ。
\[ \nabla_\bm{r}\times f(\bm{r})\bm{A}(\bm{r}) = \bigl(\nabla_\bm{r}f(\bm{r})\bigr)\times \bm{A}(\bm{r}) + f(\bm{r})\nabla_\bm{r}\times\bm{A}(\bm{r}) \]導出
$i+j$を3で割った余りを$[i+j]$と表すことにする。
\begin{align*} \phantom{=} &\nabla_\bm{r}\times f(\bm{r})\bm{A}(\bm{r}) = \sum_{i=1}^3 \bm{i}_i\left(\partDeriv{f(\bm{r})A_{[i+2]}(\bm{r})}{r_{[i+1]}}{} – \partDeriv{f(\bm{r})A_{[i+1]}(\bm{r})}{r_{[i+2]}}{}\right) \\ = &\sum_{i=1}^3 \bm{i}_i \left[\left(\partDeriv{f(\bm{r})}{r_{[i+1]}}{}A_{[i+2]}(\bm{r}) – \partDeriv{f(\bm{r})}{r_{[i+2]}}{}A_{[i+3]}(\bm{r})\right) + f(\bm{r})\left(\partDeriv{A_{[i+2]}(\bm{r})}{r_{[i+1]}}{} – \partDeriv{A_{[i+1]}(\bm{r})}{r_{[i+2]}}{}\right)\right] \\ = &\bigl(\nabla_\bm{r}f(\bm{r})\bigr)\times \bm{A}(\bm{r}) + f(\bm{r})\nabla_\bm{r}\times\bm{A}(\bm{r}) \end{align*}$\square$
定数ベクトルとの外積の回転
主張
$\bm{r},\bm{C}\in\realNumbers$とする。$A: \realNumbers^3\to\realNumbers^3$は$\mathrm{C}^1$級とする。このとき次式が成り立つ。\[ \nabla_\bm{r}\times\bigl(\bm{C}\times\bm{A}(\bm{r})\bigr) = (\nabla\cdot\bm{A}(\bm{r}))\bm{C} – J_\bm{A}\bm{C} \]ここに$J_\bm{A}$は$A$のJacobi行列である。
導出
\begin{align*} &\phantom{=} \nabla_\bm{r}\times\bigl(\bm{C}\times\bm{A}(\bm{r})\bigr) \\ &= \bm{i}_1\left[C_1\partDerivLong{A_2}{r_2}{} – C_2\partDerivLong{A_1}{r_2}{} – C_3\partDerivLong{A_1}{r_3}{} + C_1\partDerivLong{A_3}{r_3}{}\right] \\ &\phantom{=} + \bm{i}_2\left[C_2\partDerivLong{A_3}{r_3}{} – C_3\partDerivLong{A_2}{r_3}{} – C_1\partDerivLong{A_2}{r_1}{} + C_2\partDerivLong{A_1}{r_1}{}\right] \\ &\phantom{=} + \bm{i}_3\left[C_3\partDerivLong{A_1}{r_1}{} – C_1\partDerivLong{A_3}{r_1}{} – C_2\partDerivLong{A_3}{r_2}{} + C_3\partDerivLong{A_2}{r_2}{}\right] \end{align*}$\bm{i}_1$の係数を変形して次式を得る。
\[ C_1\left(\partDerivLong{A_1}{r_1}{} + \partDerivLong{A_2}{r_2}{} + \partDerivLong{A_3}{r_3}{}\right) – C_1\partDerivLong{A_1}{r_1}{} – C_2\partDerivLong{A_1}{r_2}{} – C_3\partDerivLong{A_1}{r_3}{} = C_1\nabla\cdot\bm{A} – \bm{C}\cdot\nabla A_1 \]$\bm{i}_2, \bm{i}_3$についても同様にして、結局次式を得る。
\[ \nabla_\bm{r}\times\bigl(\bm{C}\times\bm{A}(\bm{r})\bigr) = (\nabla\cdot\bm{A})\bm{C} – (\bm{C}\cdot\nabla A_1)\bm{i}_1 – (\bm{C}\cdot\nabla A_2)\bm{i}_2 – (\bm{C}\cdot\nabla A_3)\bm{i}_3 = (\nabla\cdot\bm{A}(\bm{r}))\bm{C} – J_\bm{A}\bm{C} \]$\square$
ベクトルLaplacianの発散
主張
$\bm{r}\in\realNumbers$とする。$A: \realNumbers^3\to\realNumbers^3$は$\mathrm{C}^3$級とする。次式が成り立つ。
\[ \nabla\cdot(\nabla^2\bm{A}) = \Delta(\nabla\cdot\bm{A}) \]導出
\begin{align*} \nabla\cdot\nabla^2\bm{A} &= \nabla\cdot\left(\bm{i}_1 \Delta A_1 + \bm{i}_2 \Delta A_2 + \bm{i}_3 \Delta A_3\right) = \partDerivLong{\Delta A_1}{r_1}{} + \partDerivLong{\Delta A_2}{r_2}{} + \partDerivLong{\Delta A_3}{r_3}{} \\ &= \Delta\left(\partDerivLong{A_1}{r_1}{} + \partDerivLong{A_2}{r_2}{} + \partDerivLong{A_3}{r_3}{}\right) = \Delta(\nabla\cdot\bm{A}) \end{align*}$\square$
