设超弹性材料的贮能函数 $\hat W$ 满足 (4. 19) 式, 证明由它决定的 Cauchy 应力张量 ${\bf T}$ 满足各向同性假设 (4. 7) 式.
证明: 若贮能函数 $W$ 满足 ``$\hat W({\bf F}{\bf Q})=W({\bf F})$ 对任意正交阵 ${\bf Q}$'', 则 $$\beex \bea p_{ij}({\bf F})&=\cfrac{\p \hat W({\bf F})}{\p f_{ij}}\\ &=\cfrac{\p \hat W({\bf F}{\bf Q})}{\p f_{ij}}\\ &=\sum_{m,n}\cfrac{\p \hat W({\bf F}{\bf Q})}{\p z_{mn}}\cfrac{\p z_{mn}}{\p f_{ij}}\quad\sex{{\bf Z}={\bf F}{\bf Q}}\\ &=\sum_{m,n}p_{mn}({\bf F}{\bf Q})q_{jn}\delta_{mi}\\ &\quad\sex{z_{mn}=\sum_l f_{ml}q_{ln}\ra \cfrac{\p z_{mn}}{\p f_{ij}}=q_{jn}\delta_{mi}}\\ &=\sum_n p_{in}({\bf F}{\bf Q})q_{jn}. \eea \eeex$$ 于是 $$\bex {\bf P}({\bf F})={\bf P}({\bf F}{\bf Q}){\bf Q}^T. \eex$$ 又由 Piola 应力张量的定义 ${\bf P}=J\hat {\bf T}{\bf F}^{-T}$ 知 $$\beex \bea \hat {\bf T}({\bf F}){\bf F}^{-T}&=\hat{\bf T}({\bf F}{\bf Q})({\bf F}{\bf Q})^{-T}{\bf Q}^T\\ &=\hat{\bf F}({\bf F}{\bf Q}){\bf F}^{-T},\\ {\bf T}({\bf F})&=\hat{\bf T}({\bf F}{\bf Q}). \eea \eeex$$
