1.  引理 (极分解): 设 $|{\bf F}|\neq 0$, 则存在正交阵 ${\bf R}$ 及对称正定阵 ${\bf U},{\bf V}$ 使得 $$\bex {\bf F}={\bf R}{\bf U}={\bf V}{\bf R}. \eex$$ 此称为 ${\bf F}$ 的极分解.

证明:

(1)  先证明存在正交阵 ${\bf P},{\bf Q}$ 及对角阵 ${\bf D}$ 使得 $$\bex {\bf F}={\bf P}{\bf D}{\bf Q}. \eex$$ 事实上, 由 ${\bf F}$ 可逆知 ${\bf F}^T{\bf F}$ 正定, 而存在正交阵 ${\bf Q}$, 使得 $$\bex {\bf F}^T{\bf F}={\bf Q}^T\diag(\lm_1,\cdots,\lm_n){\bf Q},\quad(\lm_i>0). \eex$$ 取 $$\bex {\bf D}=\diag(\sqrt{\lm_1},\cdots,\sqrt{\lm_n}),\quad {\bf P}={\bf F}{\bf Q}^T{\bf D}^{-1}, \eex$$ 则可直接验证 ${\bf P},{\bf Q},{\bf D}$ 适合要求.

(2)  取 $$\bex {\bf R}={\bf P}{\bf Q},\quad {\bf U}={\bf Q}^T{\bf D}{\bf Q},\quad {\bf V}={\bf P}{\bf D}{\bf P}^T \eex$$ 即满足条件.

2.  由 $\rd {\bf y}={\bf F}\rd{\bf x}$, ${\bf F}={\bf R}{\bf U}$ 知 $$\bex {\bf y}={\bf R}\rd{\bf z},\quad\rd {\bf z}={\bf U}\rd {\bf x}, \eex$$ 而 $\rd {\bf x}\to\rd {\bf y}$ 是 ``在三个相互正交的方向上的伸长或压缩'' 与 ``刚体旋转'' 的复合.

3.  Cauchy - Green 应变张量

(1)  右: ${\bf C}={\bf F}^T{\bf F}={\bf U}^2$.

(2)  左: ${\bf B}={\bf F}{\bf F}^T={\bf V}^2$.

4.  稳态时, 已知 Cauchy - Green 应变张量求 ${\bf y}$ 的 PDE 组称为 Beltrami 方程组 (超定).

5.  总结: ${\bf B},{\bf C}$ 表示左、右 Cauchy - Green 应变张量, ${\bf F}$ 表示变形.