设材料是超弹性的, 并设参考构形为自然状态, 证明由线性化得到的张量 ${\bf A}=(a_{ijkl})=\sex{2\cfrac{\p \bar p_{ij}}{c_{kl}}}$ 具有以下的对称性: $$\bex a_{ijkl}=a_{klij}. \eex$$
证明: 注意到 $$\beex \bea {\bf C}={\bf F}^T{\bf F}&\ra c_{mn}=\sum_t f_{tm}f_{tn}\\ &\ra \cfrac{\p c_{mn}}{\p f_{kl}}=\delta_{lm}f_{kn}+\delta_{ln}f_{km}, \eea \eeex$$ 我们有 $$\beex \bea &\quad\cfrac{\p^2\hat W}{\p f_{kl}\p f_{ij}} =\cfrac{\p^2\hat W}{\p f_{ij}\p f_{kl}} \ra \cfrac{\p p_{ij}}{\p f_{kl}}=\cfrac{\p p_{kl}}{\p f_{ij}}\\ &\ra \sum_s\cfrac{\p }{\p f_{kl}}(f_{is}\bar p_{sj}) =\sum_s \cfrac{\p}{\p f_{ij}}(f_{ks}\bar p_{sl})\\ &\ra \delta_{ik}\bar p_{lj}+\sum_s f_{is}\cfrac{\p \bar p_{sj}}{\p f_{kl}} =\delta_{ik}\bar p_{jl}+\sum_s f_{ks}\cfrac{\p \bar p_{sl}}{\p f_{ij}}\\ &\ra \delta_{ik}\bar p_{lj} +\sum_{m,n,s}f_{is}\cfrac{\p \bar p_{sj}}{\p c_{mn}}\cfrac{\p c_{mn}}{\p f_{kl}}=\cdots\\ &\ra \delta_{ik} \bar p_{lj}+\sum_{m,n,s}f_{is}a_{sjmn}(\delta_{lm}f_{kn}+\delta_{ln}f_{km})=\cdots\\ &\ra \delta_{ik}\bar p_{lj} +\sum_{n,s}f_{is}a_{sjln}f_{kn} +\sum_{m,s}f_{is}a_{sjml}f_{km}=\cdots\\ &\ra \delta_{ik}\bar p_{lj}+2\sum_{n,s} f_{is}a_{sjln}f_{kn}= \delta_{ki}\bar p_{jl}+2\sum_{n,s}f_{ks}a_{sljn}f_{in}\quad(a_{ijkl}=a_{ijlk})\\ &\ra \sum_{n,s} f_{is}a_{sjln}f_{kn}=\sum_{n,s}f_{ks}a_{sljn}f_{in} =\sum_{s,n}f_{in}a_{sljn}f_{ks} =\sum_{n,s}f_{is}a_{nljs}f_{kn}\\ &\ra \sum_{i,k}(f^{-1})_{mi}\sez{\sum_{n,s} f_{is}a_{sjln}f_{kn}} (f^{-T})_{kt}=\sum_{i,k}(f^{-1})_{mi}\sez{\sum_{n,s}f_{is}{a_{nljs}f_{kn}}} (f^{-T})_{kt}\\ &\ra a_{mjlt}=a_{tljm}\\ &\ra a_{mjlt}=a_{tljm}=a_{ltmj}. \eea \eeex$$