勾配作用素
出典: フリー百科事典『ウィキペディア（Wikipedia）』 (2017/12/02 20:00 UTC 版)
勾配 grad ： ( grad f ) ( x , y , z ) = ( ( ∂ f / ∂ x ) ( x , y , z ) ( ∂ f / ∂ y ) ( x , y , z ) ( ∂ f / ∂ z ) ( x , y , z ) ) = ( ( ∂ f ∂ x ) ( x , y , z ) ) ( 1 0 0 ) + ( ( ∂ f ∂ y ) ( x , y , z ) ) ( 0 1 0 ) + ( ( ∂ f ∂ z ) ( x , y , z ) ) ( 0 0 1 ) {\displaystyle {\begin{aligned}(\operatorname {grad} f)(x,y,z)&=\left({\begin{matrix}(\partial f/\partial x)(x,y,z)\\(\partial f/\partial y)(x,y,z)\\(\partial f/\partial z)(x,y,z)\\\end{matrix}}\right)\\&=\left(\left({\frac {\partial f}{\partial x}}\right)(x,y,z)\right)\left({\begin{matrix}1\\0\\0\\\end{matrix}}\right)+\left(\left({\frac {\partial f}{\partial y}}\right)(x,y,z)\right)\left({\begin{matrix}0\\1\\0\\\end{matrix}}\right)+\left(\left({\frac {\partial f}{\partial z}}\right)(x,y,z)\right)\left({\begin{matrix}0\\0\\1\\\end{matrix}}\right)\end{aligned}}} (6-2-1) を円柱座標変換すると ．．． 以下、証明を行う。(grad f )(Φ(r , θ, ζ)) と、Nr (r , θ, ζ) の内積を取ると、 [ ( grad f ) ( Φ ( r , θ , ζ ) ) ] ⋅ ( N r ( r , θ , ζ ) ) = ( ( ∂ f ∂ x ) ( Φ ( r , θ , ζ ) ) , ( ∂ f ∂ y ) ( Φ ( r , θ , ζ ) ) , ( ∂ f ∂ z ) ( Φ ( r , θ , ζ ) ) ) ( cos θ sin θ 0 ) = cos θ ( ( ∂ f ∂ x ) ( Φ ( r , θ , ζ ) ) ) + sin θ ( ( ∂ f ∂ y ) ( Φ ( r , θ , ζ ) ) ) = cos θ [ cos θ ( ( ∂ ( f ∘ Φ ) ∂ r ) ( r , θ , z ) ) − sin θ r ( ( ∂ ( f ∘ Φ ) ∂ θ ) ( r , θ , z ) ) ] + sin θ [ sin θ ( ( ∂ ( f ∘ Φ ) ∂ r ) ( r , θ , z ) ) + cos θ r ( ( ∂ ( f ∘ Φ ) ∂ θ ) ( r , θ , z ) ) ] = ( ∂ ( f ∘ Φ ) ∂ r ) ( r , θ , z ) {\displaystyle {\begin{aligned}\left[\left(\operatorname {grad} f\right)(\Phi (r,\theta ,\zeta ))\right]\cdot \left({{\mathbf {N} }_{r}}(r,\theta ,\zeta )\right)=&\left(\left({\frac {\partial f}{\partial x}}\right)(\Phi (r,\theta ,\zeta )),\left({\frac {\partial f}{\partial y}}\right)(\Phi (r,\theta ,\zeta )),\left({\frac {\partial f}{\partial z}}\right)(\Phi (r,\theta ,\zeta ))\right)\left({\begin{matrix}\cos \theta \\\sin \theta \\0\\\end{matrix}}\right)\\=&\cos \theta \left(\left({\frac {\partial f}{\partial x}}\right)(\Phi (r,\theta ,\zeta ))\right)+\sin \theta \left(\left({\frac {\partial f}{\partial y}}\right)(\Phi (r,\theta ,\zeta ))\right)\\=&\cos \theta \left[\cos \theta \left(\left({\frac {\partial (f\circ \Phi )}{\partial r}}\right)(r,\theta ,z)\right)-{\frac {\sin \theta }{r}}\left(\left({\frac {\partial (f\circ \Phi )}{\partial \theta }}\right)(r,\theta ,z)\right)\right]\\&+\sin \theta \left[\sin \theta \left(\left({\frac {\partial (f\circ \Phi )}{\partial r}}\right)(r,\theta ,z)\right)+{\frac {\cos \theta }{r}}\left(\left({\frac {\partial (f\circ \Phi )}{\partial \theta }}\right)(r,\theta ,z)\right)\right]\\=&\left({\frac {\partial (f\circ \Phi )}{\partial r}}\right)(r,\theta ,z)\end{aligned}}} (6-2-2) [ ( grad f ) ( Φ ( r , θ , ζ ) ) ] ⋅ ( N θ ( r , θ , ζ ) ) = 1 r ( ∂ ( f ∘ Φ ) ∂ θ ) ( r , θ , z ) {\displaystyle \left[\left(\operatorname {grad} f\right)(\Phi (r,\theta ,\zeta ))\right]\cdot \left({{\mathbf {N} }_{\theta }}(r,\theta ,\zeta )\right)={\frac {1}{r}}\left({\frac {\partial (f\circ \Phi )}{\partial \theta }}\right)(r,\theta ,z)} (6-2-3) [ ( grad f ) ( Φ ( r , θ , ζ ) ) ] ⋅ ( N ζ ( r , θ , ζ ) ) = ( ∂ ( f ∘ Φ ) ∂ ζ ) ( r , θ , ζ ) {\displaystyle \left[\left(\operatorname {grad} f\right)(\Phi (r,\theta ,\zeta ))\right]\cdot \left({{\mathbf {N} }_{\zeta }}(r,\theta ,\zeta )\right)=\left({\frac {\partial (f\circ \Phi )}{\partial \zeta }}\right)(r,\theta ,\zeta )} (6-2-3) が成り立つ。従って「ベクトル場の円柱座標表示」と同様にして、上の等式が示せる。
※この「勾配作用素」の解説は、「円柱座標変換」の解説の一部です。
「勾配作用素」を含む「円柱座標変換」の記事については、「円柱座標変換」の概要を参照ください。
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