∗\(grad\ (g\ h)\) ∗\(div(g\ \bv{A})\) ∗\(rot(g\ \bv{A})\) ∗\(grad\ (\bv{A}\cdot \bv{B})\) ∗\(div\ (\bv{A}\x \bv{B})\) ∗\(rot (\bv{A}\x \bv{B})\) ∗\(rot\ rot \bv{A}\) ∗\(rot\ grad f=0\) ∗\(div\ rot \bv{A}=0\) ∗\(rot\ rot\ rot \bv{A}\)
♦応用例と例題\(grad\ f(\b{r})=∇\ f\) \(= ( \pder{f(x,y,z)}{x},\pder{f(x,y,z)}{y},\pder{f(x,y,z)}{z} )\) \(\ :❶\)
\(\bv{f(r)}=f(x,y,z)\)\(=( f_x ( x,y,z ) ,f_y (x,y,z),f_z(x,y,z) )\)に対して
\( div \bv{f}\)\(=∇\cdot \bv{f}\) \(=\pder{f_x}{x}+\pder{f_y}{y}+\pder{f_z}{z}\) \(\ :❷\)
\(\bv{f(r)}=f(x,y,z)\)\(=( f_x ( x,y,z ) ,f_y (x,y,z),f_z(x,y,z) )\)に対して
\( rot\ \bv{f}\) \(=∇\x \bv{f}\) \(= (\pder{f_z}{y}-\pder{f_y}{z},\ \)\(\pder{f_x}{z}-\pder{f_z}{x},\ \)\( \pder{f_y}{x}-\pder{f_x}{y}\ )\) \(\ :❸\)
スカラー関数\(f\)の勾配は
\(grad\ f=∇ｆ\) \(=(\pder{f}{x},\pder{f}{y},\pder{f}{z})\)
\(div\ (grad\ f\ ) \)\(=∇\cdot (grad\ f)\)\(=∇\cdot\ (∇f) \)\(=∇\cdot ∇\ f \)
\(=(\pder{}{x},\pder{}{y},\pder{}{z})\)\(\cdot\) \( (\pder{f}{x},\pder{f}{y},\pder{f}{z})\)
\(=(\pdera{}{x}+\pdera{}{y}+ \pdera{}{z})\ f\)
\(=∆ \ f\)
\(∆f=∇\cdot ∇f=∇^2f\) \(= (\pdera{}{x}+\pdera{}{y}+ \pdera{}{z})f \)
\(\ :❹\)
\(f,g,h\):スカラー値関数
\(\bv{f},\bv{A},\bv{B}\):ベクトル値関数
\(\bv{r}=(x,y,z)\):位置ベクトル
\(=\left( \begin{array}{c} \pder{}{x}\\ \pder{}{y}\\ \pder{}{z} \end{array} \right) \) \( \x \) \(\left( \begin{array}{c} gA_x \\ gA_y \\ gA_z \end{array} \right) \) \(= \left( \begin{array}{c} \ul{\pder{gA_z}{y}-\pder{gA_y}{z}}\\ \pder{gA_x}{z}-\pder{gA_z}{x}\\ \pder{gAy}{x}-\pder{gA_x}{y} \end{array} \right) \)
上の下線部が回転(外積)のx成分です。
\(\color{bl(ue}{\pder{gA_z}{y}-\pder{gA_y}{z}}\)
\(=\pder{g}{y}A_z+g\pder{A_z}{y}-\pder{g}{z}A_y-g\pder{A_y}{z}\)
\(=\underbrace{\pder{g}{y}A_z-\pder{g}{z}A_y}_{Q=(∇g)\x\bv{A}}\)
\(+ \underbrace{g(\pder{A_z}{y}-\pder{A_y}{z})}_{R=∇\x \bv{A}}\)
\(\small{Q}\) \(=\left( \begin{array}{c} \pder{g}{x}\\ \color{blue}{\pder{g}{y}}\\ \color{blue}{\pder{g}{z}} \end{array} \right) \) \( \x \) \(\left( \begin{array}{c} A_x \\ \color{blue}{A_y} \\ \color{blue}{A_z} \end{array} \right)\) \(\small{R}\) \(=\left( \begin{array}{c} \pder{}{x}\\ \color{blue}{\pder{}{y}}\\ \color{blue}{\pder{}{z}} \end{array} \right) \) \( \x \) \(\left( \begin{array}{c} A_x \\ \color{blue}{A_y} \\ \color{blue}{A_z} \end{array} \right) \)
これより
4) \(grad\ (\bv{A}\cdot \bv{B})\)
\(=(\bv{B}\cdot ∇)\bv{A} + (\bv{A}\cdot ∇)\bv{B}\)
\(+ \bv{A}\x (∇\x \bv{B}) + \bv{B}\x (∇\x \bv{A})\)
ベクトル関数の"内積\(\bv{A}\cdot \bv{A}\)" の勾配
•ｘ成分について証明します。
•右辺から式展開して下記の式ⓐと同じになることが証明です。
左辺： \(∇(\bv{A}\cdot \bv{B})_x\)
\(=\pder{}{x}(A_x B_x+A_y B_y + A_z B_z)\)
\(=(A_x B_x+A_y B_y + A_z B_z)_x\)
\(=(A_xB_x)_x+ (A_yB_y)_x+ (A_zB_z)_x\)
\(=\color{red}{\pder{A_x}{x}}B_x +A_x\pder{B_x}{x}\) \(+\pder{A_y}{x}B_y + A_y\pder{B_y}{x}\)
\(+\pder{A_z}{x}B_z +A_z\pder{B_z}{x}\)
次のように簡略表示する
\(=\ul{ \color{red}{(A_x)_x}B_x +A_x (B_x)_x}\) \(\ul{+(A_y)_x B_y + A_y(B_y)_x}\)
\(\ul{+(A_z)_x B_z +A_z (B_z)_x}\)
\(ⓐ\)
右辺：
\((\underbrace{\bv{B}\cdot ∇)\bv{A}}_{①} + \underbrace{(\bv{A}\cdot ∇)\bv{B}}_{➁}\)
\(+ \underbrace{\bv{A}\x (∇\x \bv{B})}_{③} + \underbrace{\bv{B}\x (∇\x \bv{A}}_{④})\)
\(①_x=[B_x\pder{}{x}+ B_y\pder{}{y}\)\(+ B_z\pder{}{z}]A_x\) \(=B_x\pder{A_x}{x}+ B_y\pder{A_x}{y}\)\(+ B_z\pder{A_x}{z} \)
\(①_x=\ul{B_x(A_x)_x+ B_y(A_x)_y}\)\(\ul{+ B_z(A_x)_z}\)
\(=\left(
\begin{array}{c}
A_x \\
\color{blue}{A_y} \\
\color{blue}{A_z}
\end{array}
\right)\)
\( \x \)
\(\left(
\begin{array}{c}
\pder{}{x}\\
\color{blue}{\pder{}{y}}\\
\color{blue}{\pder{}{z}}
\end{array}
\right) \)
\( \x \)
\(\left(
\begin{array}{c}
B_x \\
\color{blue}{B_y} \\
\color{blue}{B_z}
\end{array}
\right) \)
\(=\left(
\begin{array}{c}
A_x \\
\color{blue}{A_y} \\
\color{blue}{A_z}
\end{array}
\right) \)
\( \x \)
\(\left(
\begin{array}{c}
\pder{B_z}{y}-\pder{B_y}{z}\\
\color{blue}{\pder{B_x}{z}-\pder{B_z}{x}}\\
\color{blue}{\pder{B_y}{x}-\pder{B_x}{y}}
\end{array}
\right)\)
\(=\left(
\begin{array}{c}
A_x \\
\color{blue}{A_y} \\
\color{blue}{A_z}
\end{array}
\right)\)
\( \x \)
\(\left(
\begin{array}{c}
(B_z)_y-(B_y)_z\\
\color{blue}{(B_x)_z-(B_z)_x}\\
\color{blue}{(B_y)_x-(B_x)_y}
\end{array}
\right)\)
③のｘ成分は：
\(③_ｘ=\ul{A_y[\color{blue}{(B_y)_x-(B_x)_y}]}\)\(\ul{-A_z[\color{blue}{(B_x)_z-(B_z)_x}]}\)
上記③と同様な計算をして
\(④_ｘ=\ul{B_y[\color{blue}{(A_y)_x-(A_x)_y}]}\)\(\ul{-B_z[\color{blue}{(A_x)_z-(A_z)_x}]}\)
\(①_x=\ul{B_x(A_x)_x+ B_y(A_x)_y+ B_z(A_x)_z}\)
\(②_x=\ul{A_x(B_x)_x+ A_y(B_x)_y+ A_z(B_x)_z}\)
\(③_ｘ=\ul{A_y[\color{blue}{(B_y)_x-(B_x)_y}]}\)\(\ul{-A_z[\color{blue}{(B_x)_z-(B_z)_x}]}\)
\(④_ｘ=\ul{B_y[\color{blue}{(A_y)_x-(A_x)_y}]}\)\(\ul{-B_z[\color{blue}{(A_x)_z-(A_z)_x}]}\)
\(①_x=B_x(A_x)_x+\cancel{B_y(A_x)_y}+ \bcancel{B_z(A_x)_z}\)
\(②_x=A_x(B_x)_x+ \xcancel{A_y(B_x)_y}+ \cancel{A_z(B_x)_z}\)
\(③_ｘ=A_y(B_y)_x-\xcancel{A_y(B_x)_y}\)\(-\cancel{A_z(B_x)_z}+A_z(B_z)_x\)
\(④_ｘ=B_y(A_y)_x-\cancel{B_y(A_x)_y}\)\(-\bcancel{B_z(A_x)_z}+B_z(A_z)_x\)
\(=\pder{}{x}(A_ｙ B_z- A_z B_y)\) \(+ \pder{}{y}(A_z B_x- A_x B_z)\)\(+ \pder{}{z}(A_x B_y- A_y B_x)\)
上式を展開して、次式にまとめることができる。
\(=B_x(\pder{A_z}{x}-\pder{A_y}{z})\) \(+ B_y(\pder{A_x}{z}-\pder{A_z}{x})\)
\(+ B_z(\pder{A_y}{x}-\pder{A_x}{y})\)
\(-A_x(\pder{B_z}{x}-\pder{B_y}{z})\) \(- A_y(\pder{B_x}{z}-\pder{B_z}{x})\)
\(- A_z(\pder{B_y}{x}-\pder{B_x}{y})\)
上式の括弧内は回転rot を示している。
\(=\ul{B_x(∇\x A)_x+ B_y(∇\x A)_y}\)\(\ul{+ B_z(∇\x A)_zul}\)
\(-\ul{A_x(∇\x B)_x- A_y(∇\x B)_y}\)\(\ul{- A_z(∇\x B)_z}\)
上式は下式に書ける。(内積の功により3項が1項にまとまる)
内積の【参考先】
\(左辺=rot \left([A_yB_z-A_zB_y],[A_zB_x-A_xB_z] \right.\) \(\left. ,[A_xB_y-A_yB_x]\right) \)
参考(回転のx成分は右辺に上段)
\(\left(
\begin{array}{c}
\pder{}{x}\\
\pder{}{y}\\
\pder{}{z}
\end{array}
\right)\)
\( \x \)
\(\left(
\begin{array}{c}
A \\
B \\
C
\end{array}
\right)\)
\(=\left(
\begin{array}{c}
\pder{C}{y}- \pder{B}{z}\\
\pder{A}{z}- \pder{C}{x}\\
\pder{B}{x}- \pder{A}{y}
\end{array}
\right)
\)
\(左辺_x\) (上の回転のx成分を参考にして)
\(=rot (\bv{A}\x \bv{B})_x\)
\(=\pder{}{y}[A_xB_y-A_yB_x]-\pder{}{z}[A_zB_x-A_xB_z]\)
\(=[A_xB_y-A_yB_x]_y-[A_zB_x-A_xB_z]_z\)
\(=\ul{[B_y(A_x)_y+A_x(B_y)_y-B_x(A_y)_y-A_y(B_x)_y]}\)\(\ul{-[B_x(A_z)_z+A_z(B_x)_z-B_z(A_x)_z-A_x(B_z)_z]}\)
\(右辺=(B_x\pder{}{x}+B_y\pder{}{y}+B_z\pder{}{z})[A_x,A_y,A_z]\) \(-(A_x\pder{}{x}+A_y\pder{}{y}+A_z\pder{}{z})[B_x,B_y,B_z]\) \(+((B_x)_x+(B_y)_y+(B_z)_z)[A_x,A_y,A_z]\) \(-((A_x)_x+(A_y)_y+(A_z)_z)[B_x,B_y,B_z]\)
\(右辺_x\) (上式のx成分を抽出)
\(=(B_x\pder{}{x}+B_y\pder{}{y}+B_z\pder{}{z})[A_x]\)
\(-(A_x\pder{}{x}+A_y\pder{}{y}+A_z\pder{}{z})[B_x]\)
\(+((B_x)_x+(B_y)_y+(B_z)_z)[A_x]\)
\(-((A_x)_x+(A_y)_y+(A_z)_z)[B_x]\)
\(=(\cancel{B_x(A_x)_x}+B_y(A_x)_y+B_z(A_x)_z)\)
\(-(\bcancel{A_x(B_x)_x}+A_y(B_x)_y+A_z(B_x)_z)\)
\(+(\bcancel{A_x(B_x)_x}+A_x(B_y)_y+A_x(B_z)_z)\)
\(-(\cancel{B_x(A_x)_x}+B_x(A_y)_y+B_x(A_z)_z)\)
\(=\ul{[B_y(A_x)_y+A_x(B_y)_y-B_x(A_y)_y -A_y(B_x)_y]}\)
\(\ul{-[B_x(A_z)_z+A_z(B_x)_z-B_z(A_x)_z-A_x(B_z)_z]}\)
x成分について解き、証明する
今までの外積演算のx成分を参考にして
\([rot\ (rot \bv{A})]_x\)\(=[∇\x ∇\x\bv{A}]_x\)\(=\pder{}{y}(∇\x\bv{A})_z- \pder{}{z}(∇\x\bv{A})\)
外積演算のｘ成分は上記の4)の式③ を参考にして
\(=\pder{}{y}(\pder{A_y}{x}-\pder{A_x}{y})\) \(- \pder{}{z}(\pder{A_x}{z}-\pder{A_z}{x})\)
\(=\pderc{A_y}{y}{x}- \pdera{A_x}{y}-\pdera{A_x}{z}+\pderc{A_z}{z}{x}\)
\(=\ul{\pdera{A_x}{x}}+\pderc{A_y}{y}{x}+ \pderc{A_z}{z}{x}\ul{-\pdera{A_x}{x}}-\pdera{A_y}{y}-\pdera{A_x}{z}\)
\(=\pder{}{x}(\pder{A_x}{x}+\pder{A_y}{y}+\pder{A_z}{z})\)
\(-(\ul{\pdera{}{x}+\pdera{}{y}+\pdera{}{z}})A_x\)
\(=\pder{}{x}(∇ \cdot \bv{A})- ∇^2 A_x\)
\(=∇(∇ \cdot \bv{A})- \color{red}{∆}\ A_x\)
\(=[∇\x (∇f)]_x\)\(=\pder{}{y}(∇f)_z-\pder{}{z}{(∇f)_y}\)
\(=\pder{}{y}(\pder{f}{z})-\pder{}{z}(\pder{f}{y})\)
シュワルツの定理\(f_{yz}=f_{zy}\)を使い
\(=\pderc{f}{y}{z}-\pderc{f}{z}{y}=0\)
\(=\pder{}{x}(∇\x \bv{A})_x+\pder{}{y}+(∇\x \bv{A})_y+\pder{}{z}(∇\x \bv{A})_y\)
\(=\pder{}{x}(\pder{A_z}{y}-\pder{A_y}{z})\)\(+\pder{}{y}(\pder{A_x}{z}-\pder{A_z}{x})\)
\(+\pder{}{z}(\pder{A_y}{x}-\pder{A_x}{y})\)
シュワルツの定理\(f_{yz}=f_{zy}\)を使い
\(=\cancel{\pderc{A_z}{x}{y}}-\bcancel{\pderc{A_y}{x}{z}}\)
\(+\pderc{A_x}{y}{z}-\cancel{\pderc{A_z}{y}{x}}\)
\(+\bcancel{\pderc{A_y}{z}{x}}-\pderc{A_x}{z}{y}\)
\(=0\)
\(\bv{F}=∇\x \bv{A}=rot\bv{A}\)とおき、上記の式7)を使うと:
\(rot\ rot\ \ul{rot \bv{A}}=rot\ rot\ \ul{\bv{F}}\)
\(=∇(∇\cdot \bv{F})-\color{red}{∆}\bv{F}\)
\(=∇(\ul{∇\cdot rot\bv{A}})-\color{red}{∆} rot\bv{A}\)
式9)より \(∇\cdot rot\bv{A}=0\) です。
\(=-\color{red}{∆} rot\bv{A}\)
