レイノルズ輸送定理

Ω _{0 を}領域Ω( t )の基準配置とする。運動と変形勾配は次のように表される。 $${\displaystyle \mathbf {x} ={\boldsymbol {\varphi }}(\mathbf {X} ,t),}$$ $${\displaystyle {\boldsymbol {F}}(\mathbf {X} ,t)={\boldsymbol {\nabla }}{\boldsymbol {\varphi }}.}$$

J ( X , t ) = det F ( X , t )とします。定義する と、現在の構成と参照構成の積分は次のように関係付けられます。 $${\displaystyle {\hat {\mathbf {f} }}(\mathbf {X} ,t)=\mathbf {f} ({\boldsymbol {\varphi }}(\mathbf {X} ,t),t).}$$$${\displaystyle \int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV=\int _{\Omega _{0}}\mathbf {f} ({\boldsymbol {\varphi }}(\mathbf {X} ,t),t)\,J(\mathbf {X} ,t)\,dV_{0}=\int _{\Omega _{0}}{\hat {\mathbf {f} }}(\mathbf {X} ,t)\,J(\mathbf {X} ,t)\,dV_{0}.}$$

この導出が物質要素に対するものであることは、基準構成の時間不変性に暗黙的に含まれており、物質座標では一定である。体積積分の時間微分は次のように定義される。 $${\displaystyle {\frac {d}{dt}}\left(\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right)=\lim _{\Delta t\to 0}{\frac {1}{\Delta t}}\left(\int _{\Omega (t+\Delta t)}\mathbf {f} (\mathbf {x} ,t+\Delta t)\,dV-\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right).}$$

基準構成上の積分に変換すると、次のようになります。 $${\displaystyle {\frac {d}{dt}}\left(\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right)=\lim _{\Delta t\to 0}{\frac {1}{\Delta t}}\left(\int _{\Omega _{0}}{\hat {\mathbf {f} }}(\mathbf {X} ,t+\Delta t)\,J(\mathbf {X} ,t+\Delta t)\,dV_{0}-\int _{\Omega _{0}}{\hat {\mathbf {f} }}(\mathbf {X} ,t)\,J(\mathbf {X} ,t)\,dV_{0}\right).}$$

Ω _0は時間に依存しない ので、$${\displaystyle {\begin{aligned}{\frac {d}{dt}}\left(\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right)&=\int _{\Omega _{0}}\left(\lim _{\Delta t\to 0}{\frac {{\hat {\mathbf {f} }}(\mathbf {X} ,t+\Delta t)\,J(\mathbf {X} ,t+\Delta t)-{\hat {\mathbf {f} }}(\mathbf {X} ,t)\,J(\mathbf {X} ,t)}{\Delta t}}\right)\,dV_{0}\\&=\int _{\Omega _{0}}{\frac {\partial }{\partial t}}\left({\hat {\mathbf {f} }}(\mathbf {X} ,t)\,J(\mathbf {X} ,t)\right)\,dV_{0}\\&=\int _{\Omega _{0}}\left({\frac {\partial }{\partial t}}{\big (}{\hat {\mathbf {f} }}(\mathbf {X} ,t){\big )}\,J(\mathbf {X} ,t)+{\hat {\mathbf {f} }}(\mathbf {X} ,t)\,{\frac {\partial }{\partial t}}{\big (}J(\mathbf {X} ,t){\big )}\right)\,dV_{0}.\end{aligned}}}$$

Jの時間微分は次のように与えられる: ^[6] $${\displaystyle {\begin{aligned}{\frac {\partial J(\mathbf {X} ,t)}{\partial t}}&={\frac {\partial }{\partial t}}(\det {\boldsymbol {F}})\\&=(\det {\boldsymbol {F}})\operatorname {tr} \left({\boldsymbol {F}}^{-1}{\frac {\partial {\boldsymbol {F}}}{\partial t}}\right)\\&=(\det {\boldsymbol {F}})\operatorname {tr} \left({\frac {\partial {\boldsymbol {X}}}{\partial {\boldsymbol {\varphi }}}}{\frac {\partial }{\partial t}}\left({\frac {\partial {\boldsymbol {\varphi }}}{\partial {\boldsymbol {X}}}}\right)\right)\\&=(\det {\boldsymbol {F}})\operatorname {tr} \left({\frac {\partial {\boldsymbol {X}}}{\partial {\boldsymbol {\varphi }}}}{\frac {\partial }{\partial {\boldsymbol {X}}}}\left({\frac {\partial {\boldsymbol {\varphi }}}{\partial t}}\right)\right)\\&=(\det {\boldsymbol {F}})\operatorname {tr} \left({\frac {\partial }{\partial {\boldsymbol {x}}}}\left({\frac {\partial {\boldsymbol {\varphi }}}{\partial t}}\right)\right)\\&=(\det {\boldsymbol {F}})({\boldsymbol {\nabla }}\cdot \mathbf {v} )\\&=J(\mathbf {X} ,t)\,{\boldsymbol {\nabla }}\cdot \mathbf {v} {\big (}{\boldsymbol {\varphi }}(\mathbf {X} ,t),t{\big )}\\&=J(\mathbf {X} ,t)\,{\boldsymbol {\nabla }}\cdot \mathbf {v} (\mathbf {x} ,t).\end{aligned}}}$$

したがって、 ここで はfの物質時間微分です。物質微分は次のように与えられます。 $${\displaystyle {\begin{aligned}{\frac {d}{dt}}\left(\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right)&=\int _{\Omega _{0}}\left({\frac {\partial }{\partial t}}\left({\hat {\mathbf {f} }}(\mathbf {X} ,t)\right)\,J(\mathbf {X} ,t)+{\hat {\mathbf {f} }}(\mathbf {X} ,t)\,J(\mathbf {X} ,t)\,{\boldsymbol {\nabla }}\cdot \mathbf {v} (\mathbf {x} ,t)\right)\,dV_{0}\\&=\int _{\Omega _{0}}\left({\frac {\partial }{\partial t}}\left({\hat {\mathbf {f} }}(\mathbf {X} ,t)\right)+{\hat {\mathbf {f} }}(\mathbf {X} ,t)\,{\boldsymbol {\nabla }}\cdot \mathbf {v} (\mathbf {x} ,t)\right)\,J(\mathbf {X} ,t)\,dV_{0}\\&=\int _{\Omega (t)}\left({\dot {\mathbf {f} }}(\mathbf {x} ,t)+\mathbf {f} (\mathbf {x} ,t)\,{\boldsymbol {\nabla }}\cdot \mathbf {v} (\mathbf {x} ,t)\right)\,dV.\end{aligned}}}$$${\displaystyle {\dot {\mathbf {f} }}}$$${\displaystyle {\dot {\mathbf {f} }}(\mathbf {x} ,t)={\frac {\partial \mathbf {f} (\mathbf {x} ,t)}{\partial t}}+{\big (}{\boldsymbol {\nabla }}\mathbf {f} (\mathbf {x} ,t){\big )}\cdot \mathbf {v} (\mathbf {x} ,t).}$$

したがって、 または、 $${\displaystyle {\frac {d}{dt}}\left(\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right)=\int _{\Omega (t)}\left({\frac {\partial \mathbf {f} (\mathbf {x} ,t)}{\partial t}}+{\big (}{\boldsymbol {\nabla }}\mathbf {f} (\mathbf {x} ,t){\big )}\cdot \mathbf {v} (\mathbf {x} ,t)+\mathbf {f} (\mathbf {x} ,t)\,{\boldsymbol {\nabla }}\cdot \mathbf {v} (\mathbf {x} ,t)\right)\,dV,}$$$${\displaystyle {\frac {d}{dt}}\left(\int _{\Omega (t)}\mathbf {f} \,dV\right)=\int _{\Omega (t)}\left({\frac {\partial \mathbf {f} }{\partial t}}+{\boldsymbol {\nabla }}\mathbf {f} \cdot \mathbf {v} +\mathbf {f} \,{\boldsymbol {\nabla }}\cdot \mathbf {v} \right)\,dV.}$$

この アイデンティティを使うと $${\displaystyle {\boldsymbol {\nabla }}\cdot (\mathbf {v} \otimes \mathbf {w} )=\mathbf {v} ({\boldsymbol {\nabla }}\cdot \mathbf {w} )+{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {w} ,}$$$${\displaystyle {\frac {d}{dt}}\left(\int _{\Omega (t)}\mathbf {f} \,dV\right)=\int _{\Omega (t)}\left({\frac {\partial \mathbf {f} }{\partial t}}+{\boldsymbol {\nabla }}\cdot (\mathbf {f} \otimes \mathbf {v} )\right)\,dV.}$$

発散定理と恒等式( a⊗b ) · n =( b · n ) aを用いると、 QEDが得られる。$${\displaystyle {\begin{aligned}{\frac {d}{dt}}\left(\int _{\Omega (t)}\mathbf {f} \,dV\right)&=\int _{\Omega (t)}{\frac {\partial \mathbf {f} }{\partial t}}\,dV+\int _{\partial \Omega (t)}(\mathbf {f} \otimes \mathbf {v} )\cdot \mathbf {n} \,dA\\&=\int _{\Omega (t)}{\frac {\partial \mathbf {f} }{\partial t}}\,dV+\int _{\partial \Omega (t)}(\mathbf {v} \cdot \mathbf {n} )\mathbf {f} \,dA.\end{aligned}}}$$