与えられたベクトル場 $A$ と $B$ に対して、以下の2つの等式を証明する。 (1) $\text{div} (A \times B) = B \cdot \text{rot} A - A \cdot \text{rot} B$ (2) $\text{rot} \ \text{rot} A = \text{grad} \ \text{div} A - \nabla^2 A$

1. 問題の内容

与えられたベクトル場 $A$ と $B$ に対して、以下の2つの等式を証明する。

(1) $\text{div} (A \times B) = B \cdot \text{rot} A - A \cdot \text{rot} B$

(2) $\text{rot} \ \text{rot} A = \text{grad} \ \text{div} A - \nabla^2 A$

2. 解き方の手順

(1) $\text{div} (A \times B) = B \cdot \text{rot} A - A \cdot \text{rot} B$ の証明

$A = A_x i + A_y j + A_z k$、 $B = B_x i + B_y j + B_z k$ とする。

まず、$A \times B$ を計算する。

$A \times B = (A_y B_z - A_z B_y) i + (A_z B_x - A_x B_z) j + (A_x B_y - A_y B_x) k$

次に、$\text{div} (A \times B)$ を計算する。

$\text{div} (A \times B) = \frac{\partial}{\partial x} (A_y B_z - A_z B_y) + \frac{\partial}{\partial y} (A_z B_x - A_x B_z) + \frac{\partial}{\partial z} (A_x B_y - A_y B_x)$

上記の式を展開する。

$\text{div} (A \times B) = \frac{\partial A_y}{\partial x} B_z + A_y \frac{\partial B_z}{\partial x} - \frac{\partial A_z}{\partial x} B_y - A_z \frac{\partial B_y}{\partial x} + \frac{\partial A_z}{\partial y} B_x + A_z \frac{\partial B_x}{\partial y} - \frac{\partial A_x}{\partial y} B_z - A_x \frac{\partial B_z}{\partial y} + \frac{\partial A_x}{\partial z} B_y + A_x \frac{\partial B_y}{\partial z} - \frac{\partial A_y}{\partial z} B_x - A_y \frac{\partial B_x}{\partial z}$

次に、$B \cdot \text{rot} A - A \cdot \text{rot} B$ を計算する。

$\text{rot} A = (\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}) i + (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}) j + (\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}) k$

$\text{rot} B = (\frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z}) i + (\frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x}) j + (\frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y}) k$

$B \cdot \text{rot} A = B_x (\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}) + B_y (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}) + B_z (\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y})$

$A \cdot \text{rot} B = A_x (\frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z}) + A_y (\frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x}) + A_z (\frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y})$

$B \cdot \text{rot} A - A \cdot \text{rot} B = B_x (\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}) + B_y (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}) + B_z (\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}) - A_x (\frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z}) - A_y (\frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x}) - A_z (\frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y})$

上記の式を展開し整理すると、$\text{div} (A \times B)$ と等しくなる。したがって、$\text{div} (A \times B) = B \cdot \text{rot} A - A \cdot \text{rot} B$ が証明された。

(2) $\text{rot} \ \text{rot} A = \text{grad} \ \text{div} A - \nabla^2 A$ の証明

$\text{rot} A = (\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}) i + (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}) j + (\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}) k$

$\text{rot} \ \text{rot} A = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} & \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} & \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \end{vmatrix}$

$\text{rot} \ \text{rot} A = [\frac{\partial}{\partial y} (\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}) - \frac{\partial}{\partial z} (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x})] i + [\frac{\partial}{\partial z} (\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}) - \frac{\partial}{\partial x} (\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y})] j + [\frac{\partial}{\partial x} (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}) - \frac{\partial}{\partial y} (\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z})] k$

$\text{rot} \ \text{rot} A = (\frac{\partial^2 A_y}{\partial y \partial x} - \frac{\partial^2 A_x}{\partial y^2} - \frac{\partial^2 A_x}{\partial z^2} + \frac{\partial^2 A_z}{\partial z \partial x}) i + (\frac{\partial^2 A_z}{\partial z \partial y} - \frac{\partial^2 A_y}{\partial z^2} - \frac{\partial^2 A_y}{\partial x^2} + \frac{\partial^2 A_x}{\partial x \partial y}) j + (\frac{\partial^2 A_x}{\partial x \partial z} - \frac{\partial^2 A_z}{\partial x^2} - \frac{\partial^2 A_z}{\partial y^2} + \frac{\partial^2 A_y}{\partial y \partial z}) k$

$\text{div} A = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}$

$\text{grad} \ \text{div} A = (\frac{\partial}{\partial x} (\frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z})) i + (\frac{\partial}{\partial y} (\frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z})) j + (\frac{\partial}{\partial z} (\frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z})) k$

$\text{grad} \ \text{div} A = (\frac{\partial^2 A_x}{\partial x^2} + \frac{\partial^2 A_y}{\partial x \partial y} + \frac{\partial^2 A_z}{\partial x \partial z}) i + (\frac{\partial^2 A_x}{\partial y \partial x} + \frac{\partial^2 A_y}{\partial y^2} + \frac{\partial^2 A_z}{\partial y \partial z}) j + (\frac{\partial^2 A_x}{\partial z \partial x} + \frac{\partial^2 A_y}{\partial z \partial y} + \frac{\partial^2 A_z}{\partial z^2}) k$

$\nabla^2 A = (\frac{\partial^2 A_x}{\partial x^2} + \frac{\partial^2 A_x}{\partial y^2} + \frac{\partial^2 A_x}{\partial z^2}) i + (\frac{\partial^2 A_y}{\partial x^2} + \frac{\partial^2 A_y}{\partial y^2} + \frac{\partial^2 A_y}{\partial z^2}) j + (\frac{\partial^2 A_z}{\partial x^2} + \frac{\partial^2 A_z}{\partial y^2} + \frac{\partial^2 A_z}{\partial z^2}) k$

$\text{grad} \ \text{div} A - \nabla^2 A = (\frac{\partial^2 A_y}{\partial x \partial y} + \frac{\partial^2 A_z}{\partial x \partial z} - \frac{\partial^2 A_x}{\partial y^2} - \frac{\partial^2 A_x}{\partial z^2}) i + (\frac{\partial^2 A_x}{\partial y \partial x} + \frac{\partial^2 A_z}{\partial y \partial z} - \frac{\partial^2 A_y}{\partial x^2} - \frac{\partial^2 A_y}{\partial z^2}) j + (\frac{\partial^2 A_x}{\partial z \partial x} + \frac{\partial^2 A_y}{\partial z \partial y} - \frac{\partial^2 A_z}{\partial x^2} - \frac{\partial^2 A_z}{\partial y^2}) k$

上記の式は $\text{rot} \ \text{rot} A$ と等しい。

したがって、$\text{rot} \ \text{rot} A = \text{grad} \ \text{div} A - \nabla^2 A$ が証明された。

3. 最終的な答え

(1) $\text{div} (A \times B) = B \cdot \text{rot} A - A \cdot \text{rot} B$

(2) $\text{rot} \ \text{rot} A = \text{grad} \ \text{div} A - \nabla^2 A$