全体座標系 $(x, y, z)$ と局所座標系 $(\xi, \eta, \zeta)$ の間に以下の関係がある。 $x = \xi + 2\eta + 3\zeta$ $y = 2\xi - \eta - 2\zeta$ $z = -3\xi + 4\eta + 2\zeta$ (1) $(\frac{\partial \theta}{\partial \xi}, \frac{\partial \theta}{\partial \eta}, \frac{\partial \theta}{\partial \zeta})^T$ と $(\frac{\partial \theta}{\partial x}, \frac{\partial \theta}{\partial y}, \frac{\partial \theta}{\partial z})^T$ の関係を行列 $J$ を用いて表せ。 (2) 行列 $J$ の逆行列を求めよ。

1. 問題の内容

全体座標系 $(x, y, z)$ と局所座標系 $(\xi, \eta, \zeta)$ の間に以下の関係がある。

$x = \xi + 2\eta + 3\zeta$

$y = 2\xi - \eta - 2\zeta$

$z = -3\xi + 4\eta + 2\zeta$

(1) $(\frac{\partial \theta}{\partial \xi}, \frac{\partial \theta}{\partial \eta}, \frac{\partial \theta}{\partial \zeta})^T$ と $(\frac{\partial \theta}{\partial x}, \frac{\partial \theta}{\partial y}, \frac{\partial \theta}{\partial z})^T$ の関係を行列 $J$ を用いて表せ。

(2) 行列 $J$ の逆行列を求めよ。

2. 解き方の手順

(1) まずヤコビ行列 $J$ を計算する。

$J = \begin{pmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial x}{\partial \eta} & \frac{\partial x}{\partial \zeta} \\ \frac{\partial y}{\partial \xi} & \frac{\partial y}{\partial \eta} & \frac{\partial y}{\partial \zeta} \\ \frac{\partial z}{\partial \xi} & \frac{\partial z}{\partial \eta} & \frac{\partial z}{\partial \zeta} \end{pmatrix}$

与えられた式より、

$\frac{\partial x}{\partial \xi} = 1, \frac{\partial x}{\partial \eta} = 2, \frac{\partial x}{\partial \zeta} = 3$

$\frac{\partial y}{\partial \xi} = 2, \frac{\partial y}{\partial \eta} = -1, \frac{\partial y}{\partial \zeta} = -2$

$\frac{\partial z}{\partial \xi} = -3, \frac{\partial z}{\partial \eta} = 4, \frac{\partial z}{\partial \zeta} = 2$

したがって、

$J = \begin{pmatrix} 1 & 2 & 3 \\ 2 & -1 & -2 \\ -3 & 4 & 2 \end{pmatrix}$

連鎖律より、

$\begin{pmatrix} \frac{\partial \theta}{\partial \xi} \\ \frac{\partial \theta}{\partial \eta} \\ \frac{\partial \theta}{\partial \zeta} \end{pmatrix} = \begin{pmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi} & \frac{\partial z}{\partial \xi} \\ \frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta} & \frac{\partial z}{\partial \eta} \\ \frac{\partial x}{\partial \zeta} & \frac{\partial y}{\partial \zeta} & \frac{\partial z}{\partial \zeta} \end{pmatrix} \begin{pmatrix} \frac{\partial \theta}{\partial x} \\ \frac{\partial \theta}{\partial y} \\ \frac{\partial \theta}{\partial z} \end{pmatrix}^T$ではない。

$\begin{pmatrix} \frac{\partial \theta}{\partial \xi} \\ \frac{\partial \theta}{\partial \eta} \\ \frac{\partial \theta}{\partial \zeta} \end{pmatrix} = J^T \begin{pmatrix} \frac{\partial \theta}{\partial x} \\ \frac{\partial \theta}{\partial y} \\ \frac{\partial \theta}{\partial z} \end{pmatrix}$ と表現できる。

$(\frac{\partial \theta}{\partial \xi}, \frac{\partial \theta}{\partial \eta}, \frac{\partial \theta}{\partial \zeta})^T = J^T(\frac{\partial \theta}{\partial x}, \frac{\partial \theta}{\partial y}, \frac{\partial \theta}{\partial z})^T$

ここで

$J^T = \begin{pmatrix} 1 & 2 & -3 \\ 2 & -1 & 4 \\ 3 & -2 & 2 \end{pmatrix}$

(2) 行列 $J$ の逆行列を求める。

$J = \begin{pmatrix} 1 & 2 & 3 \\ 2 & -1 & -2 \\ -3 & 4 & 2 \end{pmatrix}$

$\det(J) = 1((-1)(2) - (-2)(4)) - 2((2)(2) - (-2)(-3)) + 3((2)(4) - (-1)(-3)) = 1(-2 + 8) - 2(4 - 6) + 3(8 - 3) = 6 + 4 + 15 = 25$

$J^{-1} = \frac{1}{\det(J)} \text{adj}(J)$

余因子行列を求める。

$C_{11} = (-1)(2) - (-2)(4) = -2 + 8 = 6$

$C_{12} = -(2(2) - (-2)(-3)) = -(4-6) = 2$

$C_{13} = 2(4) - (-1)(-3) = 8 - 3 = 5$

$C_{21} = -(2(2) - 3(4)) = -(4-12) = 8$

$C_{22} = 1(2) - 3(-3) = 2 + 9 = 11$

$C_{23} = -(1(4) - 2(-3)) = -(4+6) = -10$

$C_{31} = 2(-2) - 3(-1) = -4 + 3 = -1$

$C_{32} = -(1(-2) - 3(2)) = -(-2-6) = 8$

$C_{33} = 1(-1) - 2(2) = -1 - 4 = -5$

$\text{adj}(J) = \begin{pmatrix} 6 & 8 & -1 \\ 2 & 11 & 8 \\ 5 & -10 & -5 \end{pmatrix}$

$J^{-1} = \frac{1}{25} \begin{pmatrix} 6 & 8 & -1 \\ 2 & 11 & 8 \\ 5 & -10 & -5 \end{pmatrix}$

3. 最終的な答え

(1) $(\frac{\partial \theta}{\partial \xi}, \frac{\partial \theta}{\partial \eta}, \frac{\partial \theta}{\partial \zeta})^T = \begin{pmatrix} 1 & 2 & -3 \\ 2 & -1 & 4 \\ 3 & -2 & 2 \end{pmatrix} (\frac{\partial \theta}{\partial x}, \frac{\partial \theta}{\partial y}, \frac{\partial \theta}{\partial z})^T$

(2) $J^{-1} = \frac{1}{25} \begin{pmatrix} 6 & 8 & -1 \\ 2 & 11 & 8 \\ 5 & -10 & -5 \end{pmatrix}$