利用者:知識熊/sandbox/自然法則/ベクトル解析の公式の証明

ここではベクトル解析の公式 ${\mathsf {rot}}({\mathsf {rot}}\,\mathbf {F} )={\mathsf {grad}}({\mathsf {div}}\,\mathbf {F} )-\Delta \,\mathbf {F}$ を証明する。

まず、ベクトル解析のナブラ ∇ を

$\nabla \equiv {\tfrac {\partial }{\partial x}}\,\mathbf {i} +{\tfrac {\partial }{\partial y}}\,\mathbf {j} +{\tfrac {\partial }{\partial z}}\,\mathbf {k} =\left({\tfrac {\partial }{\partial {x}}},{\tfrac {\partial }{\partial {y}}},{\tfrac {\partial }{\partial {z}}}\right)$

と定義すると、回転 ${\mathsf {rot}}$ は単位ベクトル i, j, k を用いて

${\begin{aligned}{\mathsf {rot}}\,\mathbf {F} &=\nabla \times \mathbf {F} ={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\{\tfrac {\partial }{\partial x}}&{\tfrac {\partial }{\partial y}}&{\tfrac {\partial }{\partial z}}\\F_{x}&F_{y}&F_{z}\end{vmatrix}}\\&=\left({\tfrac {\partial F_{z}}{\partial y}}-{\tfrac {\partial F_{y}}{\partial z}}\right)\mathbf {i} +\left({\tfrac {\partial F_{x}}{\partial z}}-{\tfrac {\partial F_{z}}{\partial x}}\right)\mathbf {j} +\left({\tfrac {\partial F_{y}}{\partial x}}-{\tfrac {\partial F_{x}}{\partial y}}\right)\mathbf {k} \\&=\left({\tfrac {\partial {F_{z}}}{\partial {y}}}-{\tfrac {\partial {F_{y}}}{\partial {z}}},{\tfrac {\partial {F_{x}}}{\partial {z}}}-{\tfrac {\partial {F_{z}}}{\partial {x}}},{\tfrac {\partial {F_{y}}}{\partial {x}}}-{\tfrac {\partial {F_{x}}}{\partial {y}}}\right)\end{aligned}}$

と表され、${\mathsf {rot}}({\mathsf {rot}}\,\mathbf {F} )$ は

${\begin{aligned}{\mathsf {rot}}({\mathsf {rot}}\,\mathbf {F} )&={\begin{pmatrix}{\tfrac {\partial }{\partial y}}\left({\tfrac {\partial {F_{y}}}{\partial {x}}}-{\tfrac {\partial {F_{x}}}{\partial {y}}}\right)-{\tfrac {\partial }{\partial z}}\left({\tfrac {\partial {F_{x}}}{\partial {z}}}-{\tfrac {\partial {F_{z}}}{\partial {x}}}\right)\\{\tfrac {\partial }{\partial z}}\left({\tfrac {\partial {F_{z}}}{\partial {y}}}-{\tfrac {\partial {F_{y}}}{\partial {z}}}\right)-{\tfrac {\partial }{\partial x}}\left({\tfrac {\partial {F_{y}}}{\partial {x}}}-{\tfrac {\partial {F_{x}}}{\partial {y}}}\right)\\{\tfrac {\partial }{\partial x}}\left({\tfrac {\partial {F_{x}}}{\partial {z}}}-{\tfrac {\partial {F_{z}}}{\partial {x}}}\right)-{\tfrac {\partial }{\partial y}}\left({\tfrac {\partial {F_{z}}}{\partial {y}}}-{\tfrac {\partial {F_{y}}}{\partial {z}}}\right)\end{pmatrix}}\\&={\begin{pmatrix}{\tfrac {\partial ^{2}F_{y}}{\partial x\partial y}}-{\tfrac {\partial ^{2}F_{x}}{\partial y^{2}}}-{\tfrac {\partial ^{2}F_{x}}{\partial z^{2}}}+{\tfrac {\partial ^{2}F_{z}}{\partial x\partial z}}\\{\tfrac {\partial ^{2}F_{z}}{\partial y\partial z}}-{\tfrac {\partial ^{2}F_{y}}{\partial z^{2}}}-{\tfrac {\partial ^{2}F_{y}}{\partial x^{2}}}+{\tfrac {\partial ^{2}F_{x}}{\partial x\partial y}}\\{\tfrac {\partial ^{2}F_{x}}{\partial x\partial z}}-{\tfrac {\partial ^{2}F_{z}}{\partial x^{2}}}-{\tfrac {\partial ^{2}F_{z}}{\partial y^{2}}}+{\tfrac {\partial ^{2}F_{y}}{\partial y\partial z}}\end{pmatrix}}\end{aligned}}$

(1)

となる。

また、ベクトル解析の勾配 ${\mathsf {grad}}$ と発散 ${\mathsf {div}}$ を単位ベクトル i, j, k を用いてそれぞれ

${\mathsf {grad}}\,\mathbf {F} =\nabla \mathbf {F} ={\tfrac {\partial \mathbf {F} }{\partial x}}\,\mathbf {i} +{\tfrac {\partial \mathbf {F} }{\partial y}}\,\mathbf {j} +{\tfrac {\partial \mathbf {F} }{\partial z}}\,\mathbf {k} =\left({\tfrac {\partial \mathbf {F} }{\partial {x}}},{\tfrac {\partial \mathbf {F} }{\partial {y}}},{\tfrac {\partial \mathbf {F} }{\partial {z}}}\right)$

${\mathsf {div}}\,\mathbf {F} =\nabla \cdot \mathbf {F} ={\tfrac {\partial {F_{x}}}{\partial {x}}}+{\tfrac {\partial {F_{y}}}{\partial {y}}}+{\tfrac {\partial {F_{z}}}{\partial {z}}}$

と表すと、${\mathsf {grad}}({\mathsf {div}}\,\mathbf {F} )$ は

${\mathsf {grad}}({\mathsf {div}}\mathbf {F} )=\left({\tfrac {\partial ^{2}F_{x}}{\partial x^{2}}}+{\tfrac {\partial ^{2}F_{y}}{\partial x\partial y}}+{\tfrac {\partial ^{2}F_{z}}{\partial x\partial z}},{\tfrac {\partial ^{2}F_{x}}{\partial x\partial y}}+{\tfrac {\partial ^{2}F_{y}}{\partial y^{2}}}+{\tfrac {\partial ^{2}F_{z}}{\partial y\partial z}},{\tfrac {\partial ^{2}F_{x}}{\partial x\partial z}}+{\tfrac {\partial ^{2}F_{y}}{\partial y\partial z}}+{\tfrac {\partial ^{2}F_{z}}{\partial z^{2}}}\right)$

と表される。したがって、${\mathsf {grad}}({\mathsf {div}}\,\mathbf {F} )-\Delta \,\mathbf {F}$ は

${\begin{aligned}{\mathsf {grad}}({\mathsf {div}}\,\mathbf {F} )-\Delta \,\mathbf {F} &={\begin{pmatrix}{\tfrac {\partial ^{2}F_{x}}{\partial x^{2}}}+{\tfrac {\partial ^{2}F_{y}}{\partial x\partial y}}+{\tfrac {\partial ^{2}F_{z}}{\partial x\partial z}}\\{\tfrac {\partial ^{2}F_{x}}{\partial x\partial y}}+{\tfrac {\partial ^{2}F_{y}}{\partial y^{2}}}+{\tfrac {\partial ^{2}F_{z}}{\partial y\partial z}}\\{\tfrac {\partial ^{2}F_{x}}{\partial x\partial z}}+{\tfrac {\partial ^{2}F_{y}}{\partial y\partial z}}+{\tfrac {\partial ^{2}F_{z}}{\partial z^{2}}}\end{pmatrix}}-{\begin{pmatrix}{\tfrac {\partial ^{2}F_{x}}{\partial x^{2}}}+{\tfrac {\partial ^{2}F_{x}}{\partial y^{2}}}+{\tfrac {\partial ^{2}F_{x}}{\partial z^{2}}}\\{\tfrac {\partial ^{2}F_{y}}{\partial x^{2}}}+{\tfrac {\partial ^{2}F_{y}}{\partial y^{2}}}+{\tfrac {\partial ^{2}F_{y}}{\partial z^{2}}}\\{\tfrac {\partial ^{2}F_{z}}{\partial x^{2}}}+{\tfrac {\partial ^{2}F_{z}}{\partial y^{2}}}+{\tfrac {\partial ^{2}F_{z}}{\partial z^{2}}}\end{pmatrix}}\\&={\begin{pmatrix}{\tfrac {\partial ^{2}F_{y}}{\partial x\partial y}}+{\tfrac {\partial ^{2}F_{z}}{\partial x\partial z}}-{\tfrac {\partial ^{2}F_{x}}{\partial y^{2}}}-{\tfrac {\partial ^{2}F_{x}}{\partial z^{2}}}\\{\tfrac {\partial ^{2}F_{x}}{\partial x\partial y}}+{\tfrac {\partial ^{2}F_{z}}{\partial y\partial z}}-{\tfrac {\partial ^{2}F_{y}}{\partial x^{2}}}-{\tfrac {\partial ^{2}F_{y}}{\partial z^{2}}}\\{\tfrac {\partial ^{2}F_{x}}{\partial x\partial z}}+{\tfrac {\partial ^{2}F_{y}}{\partial y\partial z}}-{\tfrac {\partial ^{2}F_{z}}{\partial x^{2}}}-{\tfrac {\partial ^{2}F_{z}}{\partial y^{2}}}\end{pmatrix}}\end{aligned}}$

(2)

となる。

ここで、(1)式と(2)式は合同なので、${\mathsf {rot}}({\mathsf {rot}}\,\mathbf {F} )={\mathsf {grad}}({\mathsf {div}}\,\mathbf {F} )-\Delta \,\mathbf {F}$ が証明された。

Category:ベクトル解析 Category:数学に関する記事