(1) $\nabla h = \text{grad } h$ (2) $\nabla \cdot (\nabla h) = \text{div grad } h$ (3) $\nabla \times (\nabla h) = \text{rot grad } h$ (4) $\nabla \cdot \mathbf{F} = \text{div } \mathbf{F}$ (5) $\nabla (\nabla \cdot \mathbf{F}) = \text{grad div } \mathbf{F}$ (6) $\nabla \times \mathbf{F} = \text{rot } \mathbf{F}$ (7) $\nabla^2 \mathbf{F} = (\nabla \cdot \nabla) \mathbf{F}$ (8) $\nabla \cdot (\nabla \times \mathbf{F}) = \text{div rot } \mathbf{F}$ (9) $\nabla \times (\nabla \times \mathbf{F}) = \text{rot rot } \mathbf{F}$ (10) $\nabla \cdot (h\mathbf{F}) = \text{div } (h\mathbf{F})$

## 問題の解答

以下に、与えられた問題の解答を示します。

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1. 問題の内容

1. ベクトル関数 $\mathbf{F}(x,y,z) = (3y+2z)^2 \mathbf{j}$ とスカラー関数 $h(x,y,z) = (3y+2z)^2$ について、以下の演算を行う。

(1) $\nabla h = \text{grad } h$

(2) $\nabla \cdot (\nabla h) = \text{div grad } h$

(3) $\nabla \times (\nabla h) = \text{rot grad } h$

(4) $\nabla \cdot \mathbf{F} = \text{div } \mathbf{F}$

(5) $\nabla (\nabla \cdot \mathbf{F}) = \text{grad div } \mathbf{F}$

(6) $\nabla \times \mathbf{F} = \text{rot } \mathbf{F}$

(7) $\nabla^2 \mathbf{F} = (\nabla \cdot \nabla) \mathbf{F}$

(8) $\nabla \cdot (\nabla \times \mathbf{F}) = \text{div rot } \mathbf{F}$

(9) $\nabla \times (\nabla \times \mathbf{F}) = \text{rot rot } \mathbf{F}$

(10) $\nabla \cdot (h\mathbf{F}) = \text{div } (h\mathbf{F})$

2. ベクトル関数 $\mathbf{A}(x,y,z) = y\mathbf{j} + z\mathbf{k}$ と $\mathbf{B}(x,y,z) = (2y-z)\mathbf{j} + (2z-y)\mathbf{k}$ について、以下の演算を行う。

(1) $\nabla (\mathbf{A} \cdot \mathbf{B}) = \text{grad } (\mathbf{A} \cdot \mathbf{B})$

(2) $\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \text{div } (\mathbf{A} \times \mathbf{B})$

(3) $\nabla \times (\mathbf{A} \times \mathbf{B}) = \text{rot } (\mathbf{A} \times \mathbf{B})$

(4) $(\mathbf{A} \cdot \nabla) \mathbf{B}$

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2. 解き方の手順

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1. ベクトル関数 $\mathbf{F}$ とスカラー関数 $h$ の計算

(1) $\nabla h = \text{grad } h = \left(\frac{\partial h}{\partial x}, \frac{\partial h}{\partial y}, \frac{\partial h}{\partial z}\right) = (0, 2(3y+2z) \cdot 3, 2(3y+2z) \cdot 2) = (0, 6(3y+2z), 4(3y+2z))$

(2) $\nabla \cdot (\nabla h) = \text{div grad } h = \frac{\partial}{\partial x}(0) + \frac{\partial}{\partial y}(6(3y+2z)) + \frac{\partial}{\partial z}(4(3y+2z)) = 0 + 6 \cdot 3 + 4 \cdot 2 = 18 + 8 = 26$

(3) $\nabla \times (\nabla h) = \text{rot grad } h = (0, 0, 0)$ (任意のスカラ関数の勾配の回転はゼロ)

(4) $\nabla \cdot \mathbf{F} = \text{div } \mathbf{F} = \frac{\partial}{\partial x}(0) + \frac{\partial}{\partial y}((3y+2z)^2) + \frac{\partial}{\partial z}(0) = 2(3y+2z) \cdot 3 = 6(3y+2z)$

(5) $\nabla (\nabla \cdot \mathbf{F}) = \text{grad div } \mathbf{F} = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) (6(3y+2z)) = (0, 6 \cdot 3, 6 \cdot 2) = (0, 18, 12)$

(6) $\nabla \times \mathbf{F} = \text{rot } \mathbf{F} = \left(\frac{\partial}{\partial y}(0) - \frac{\partial}{\partial z}((3y+2z)^2), \frac{\partial}{\partial z}(0) - \frac{\partial}{\partial x}(0), \frac{\partial}{\partial x}((3y+2z)^2) - \frac{\partial}{\partial y}(0) \right) = (-2(3y+2z) \cdot 2, 0, 0) = (-4(3y+2z), 0, 0)$

(7) $\nabla^2 \mathbf{F} = (\nabla \cdot \nabla) \mathbf{F} = \left(0, \frac{\partial^2}{\partial x^2}(0) + \frac{\partial^2}{\partial y^2}((3y+2z)^2) + \frac{\partial^2}{\partial z^2}(0), 0\right) = \left(0, \frac{\partial}{\partial y}(6(3y+2z)), 0 \right) = (0, 18, 0)$

(8) $\nabla \cdot (\nabla \times \mathbf{F}) = \text{div rot } \mathbf{F} = \frac{\partial}{\partial x}(-4(3y+2z)) + \frac{\partial}{\partial y}(0) + \frac{\partial}{\partial z}(0) = 0 + 0 + 0 = 0$ (回転の発散は常にゼロ)

(9) $\nabla \times (\nabla \times \mathbf{F}) = \text{rot rot } \mathbf{F} = \nabla (\nabla \cdot \mathbf{F}) - \nabla^2 \mathbf{F}$

$\nabla (\nabla \cdot \mathbf{F}) = (0, 18, 12)$

$\nabla^2 \mathbf{F} = (0,18,0)$

よって、$\nabla \times (\nabla \times \mathbf{F}) = (0, 18, 12) - (0, 18, 0) = (0, 0, 12)$

(10) $\nabla \cdot (h\mathbf{F}) = \text{div } (h\mathbf{F}) = \nabla \cdot ((3y+2z)^4 \mathbf{j})$

$\nabla \cdot (h\mathbf{F}) = \frac{\partial}{\partial x}(0) + \frac{\partial}{\partial y}((3y+2z)^4) + \frac{\partial}{\partial z}(0) = 4(3y+2z)^3 \cdot 3 = 12(3y+2z)^3$

####

2. ベクトル関数 $\mathbf{A}$ と $\mathbf{B}$ の計算

(1) $\mathbf{A} \cdot \mathbf{B} = (y\mathbf{j} + z\mathbf{k}) \cdot ((2y-z)\mathbf{j} + (2z-y)\mathbf{k}) = y(2y-z) + z(2z-y) = 2y^2 - yz + 2z^2 - yz = 2y^2 + 2z^2 - 2yz$

$\nabla (\mathbf{A} \cdot \mathbf{B}) = \text{grad } (\mathbf{A} \cdot \mathbf{B}) = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) (2y^2 + 2z^2 - 2yz) = (0, 4y - 2z, 4z - 2y)$

(2) $\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & y & z \\ 0 & 2y-z & 2z-y \end{vmatrix} = (y(2z-y) - z(2y-z))\mathbf{i} = (2yz - y^2 - 2yz + z^2)\mathbf{i} = (z^2 - y^2)\mathbf{i}$

$\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \text{div } (\mathbf{A} \times \mathbf{B}) = \frac{\partial}{\partial x}(z^2 - y^2) + \frac{\partial}{\partial y}(0) + \frac{\partial}{\partial z}(0) = 0$

(3) $\nabla \times (\mathbf{A} \times \mathbf{B}) = \text{rot } (\mathbf{A} \times \mathbf{B}) = \nabla \times ((z^2 - y^2)\mathbf{i}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 - y^2 & 0 & 0 \end{vmatrix} = (0, -2z, 2y)$

(4) $(\mathbf{A} \cdot \nabla) \mathbf{B} = (y\frac{\partial}{\partial y} + z\frac{\partial}{\partial z})((2y-z)\mathbf{j} + (2z-y)\mathbf{k}) = y\frac{\partial}{\partial y}((2y-z)\mathbf{j} + (2z-y)\mathbf{k}) + z\frac{\partial}{\partial z}((2y-z)\mathbf{j} + (2z-y)\mathbf{k}) = y(2\mathbf{j} - \mathbf{k}) + z(-\mathbf{j} + 2\mathbf{k}) = (2y-z)\mathbf{j} + (-y+2z)\mathbf{k} = (2y-z)\mathbf{j} + (2z-y)\mathbf{k}$

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3. 最終的な答え

####

1. ベクトル関数 $\mathbf{F}$ とスカラー関数 $h$ について

(1) $\nabla h = (0, 6(3y+2z), 4(3y+2z))$

(2) $\nabla \cdot (\nabla h) = 26$

(3) $\nabla \times (\nabla h) = (0, 0, 0)$

(4) $\nabla \cdot \mathbf{F} = 6(3y+2z)$

(5) $\nabla (\nabla \cdot \mathbf{F}) = (0, 18, 12)$

(6) $\nabla \times \mathbf{F} = (-4(3y+2z), 0, 0)$

(7) $\nabla^2 \mathbf{F} = (0, 18, 0)$

(8) $\nabla \cdot (\nabla \times \mathbf{F}) = 0$

(9) $\nabla \times (\nabla \times \mathbf{F}) = (0, 0, 12)$

(10) $\nabla \cdot (h\mathbf{F}) = 12(3y+2z)^3$

####

2. ベクトル関数 $\mathbf{A}$ と $\mathbf{B}$ について

(1) $\nabla (\mathbf{A} \cdot \mathbf{B}) = (0, 4y - 2z, 4z - 2y)$

(2) $\nabla \cdot (\mathbf{A} \times \mathbf{B}) = 0$

(3) $\nabla \times (\mathbf{A} \times \mathbf{B}) = (0, -2z, 2y)$

(4) $(\mathbf{A} \cdot \nabla) \mathbf{B} = (2y-z)\mathbf{j} + (2z-y)\mathbf{k}$