物質表示した連続体の運動方程式

記述方法が二つある

$ \Rho J\ddot{\bm\phi}=\bm P\cdot\overleftarrow{\bm\nabla}+J\bm K

$ \Rho\ddot{\bm\phi}\cdot\bm F=\left.\bm\sigma\right|_{\bm x=\bm\phi(\bm X,t)}\cdot\overleftarrow{\bm\nabla}+\bm K\cdot\bm F

ここで

$ \Rho:=\left.\rho\right|_{\bm x=\bm\phi(\bm X,t)}

$ \bm K:=\left.\bm k\right|_{\bm x=\bm\phi(\bm X,t)}

数式に不安があるので検算

$ \mathrm d\bm p=\bm\sigma\cdot\mathrm d\bm a

$ \mathrm d\bm p(\bm\phi(\bm X,t),t)=\left.\bm\sigma\cdot\mathrm d\bm a\right|_{\bm x=\bm\phi(\bm X,t)}

$ = \left.\bm\sigma\right|_{\bm x=\bm\phi(\bm X,t)}\cdot J{\bm F^{-1}}^\top\cdot\mathrm d\bm A

$ =\bm P\cdot\mathrm d\bm A

導出(上)

$ \rho\frac{\mathrm D\bm v}{\mathrm Dt}=\bm\sigma\cdot\overleftarrow{\bm\nabla}+\bm k

$ \iff\frac{\partial}{\partial t}\int_{B_t}\rho\bm v\mathrm dv=\int_{\partial B_t}\bm\sigma\cdot\mathrm d\bm a+\int_{B_t}\bm k\mathrm dv\quad\text{.for }\forall B_t

$ \iff\frac{\partial}{\partial t}\int_{B_0}\Rho\dot{\bm\phi} J\mathrm dV=\int_{\partial B_0}\left.\bm\sigma\right|_{\bm x=\bm\phi(\bm X,t)}\cdot J{\bm F^{-1}}^\top\cdot\mathrm d\bm A+\int_{B_0}\bm KJ\mathrm dV\quad\text{.for }\forall B_0

$ \iff\int_{B_0}\Rho J\ddot{\bm\phi}\mathrm dV=\int_{B_0}\bm P\cdot\overleftarrow{\bm\nabla}\mathrm dV+\int_{B_0}\bm KJ\mathrm dV\quad\text{.for }\forall B_0

$ \iff\Rho J\ddot{\bm\phi}=\bm P\cdot\overleftarrow{\bm\nabla}+\bm KJ

導出(下)

$ \rho\frac{\mathrm D\bm v}{\mathrm Dt}=\bm\sigma\cdot\overleftarrow{\bm\nabla}+\bm k

$ \iff\Rho\ddot{\bm\phi}=\left.\bm\sigma\cdot\overleftarrow{\bm\nabla}\right|_{\bm x=\bm\phi(\bm X,t)}+\bm K

$ \iff\Rho\ddot{\bm\phi}\cdot\bm F=\left.\bm\sigma\cdot\overleftarrow{\bm\nabla}\right|_{\bm x=\bm\phi(\bm X,t)}\cdot\bm F+\bm K\cdot\bm F

$ \iff\Rho\ddot{\bm\phi}\cdot\bm F=\left.\bm\sigma\right|_{\bm x=\bm\phi(\bm X,t)}\cdot\overleftarrow{\bm\nabla}+\bm K\cdot\bm F

2式の関係を調べる

$ \Rho\ddot{\bm\phi}-\bm K=\frac1J\bm P\cdot\overleftarrow{\bm\nabla}

$ \Rho\ddot{\bm\phi}-\bm K=\left.\bm\sigma\right|_{\bm x=\bm\phi(\bm X,t)}\cdot\overleftarrow{\bm\nabla}\cdot\bm F^{-1}

$ \therefore\bm P\cdot\overleftarrow{\bm\nabla}=\left.\bm\sigma\right|_{\bm x=\bm\phi(\bm X,t)}\cdot\overleftarrow{\bm\nabla}\cdot J\bm F^{-1}

$ \bm P\cdot\overleftarrow{\bm\nabla}=\left.\bm\sigma\right|_{\bm x=\bm\phi(\bm X,t)}\cdot\left(J{\bm F^{-1}}^\top\cdot\overleftarrow{\bm\nabla}\right)+\left(\left.\bm\sigma\right|_{\bm x=\bm\phi(\bm X,t)}\overleftarrow{\bm\nabla}\right):J{\bm F^{-1}}^\top

$ \bm\nabla J=J\overleftarrow{\bm\nabla}=J{\bm F^{-1}}^\top:\bm F\overleftarrow{\bm\nabla}

$ {\bm F^{-1}}^\top\cdot\overleftarrow{\bm\nabla}=-{\bm F^{-1}}^\top\cdot(\bm F^\top\cdot\overleftarrow{\bm\nabla})\cdot{\bm F^{-1}}^\top

真ん中がvectorになるが、左右どちらのtensorと先に縮約しても結果は同じ

$ J{\bm F^{-1}}^\top\cdot\overleftarrow{\bm\nabla}=J{\bm F^{-1}}^\top:\bm F\overleftarrow{\bm\nabla}\cdot\bm F^{-1}-J{\bm F^{-1}}^\top\cdot(\bm F^\top\cdot\overleftarrow{\bm\nabla})\cdot{\bm F^{-1}}^\top

$ =J(F^{-1}_{ji}(\partial_kF_{ij})F^{-1}_{kl}-F^{-1}_{kl}F^{-1}_{kj}\partial_iF_{ij})

$ =J{\bm F^{-1}}^\top\cdot({\bm F^{-1}}^\top:\bm F\overleftarrow{\bm\nabla})-J{\bm F^{-1}}^\top\cdot\bm F^{-1}\cdot(\bm F^\top\cdot\overleftarrow{\bm\nabla})

$ = J{\bm F^{-1}}^\top\cdot({\bm F^{-1}}^\top:\bm F\overleftarrow{\bm\nabla}-\bm F^{-1}\cdot(\bm F^\top\cdot\overleftarrow{\bm\nabla}))

ええ……本当に一致するのかこれ？