试证明: 利用连续性方程, 可将动量方程 (2. 14) 及未燃流体质量平衡方程 (2. 16) 分别化为 (2. 19) 与 (2. 20) 的形式.
证明: 注意到 $$\beex \bea \cfrac{\p}{\p t}(\rho{\bf u}) +\Div(\rho{\bf u}\otimes{\bf u})&=\sez{\cfrac{\p\rho}{\p t}+\Div(\rho{\bf u})}{\bf u} +\rho \sez{\cfrac{\p{\bf u}}{\p t}+({\bf u}\cdot\n){\bf u}} =\rho\cfrac{\rd {\bf u}}{\rd t},\\ -\Div{\bf P}&=-\Div\sez{-p{\bf I}+\sex{\mu'-\cfrac{2}{3}\mu}(\Div{\bf u}) {\bf I}+2\mu {\bf S}}\\ &=\n p-\Div[\mu(2{\bf S})]-n\sez{\sex{\mu-\cfrac{2}{3}\mu}\Div{\bf u}};\\ \cfrac{\p}{\p t}(\rho Z)+\Div(\rho Z{\bf u}) &=\sez{\cfrac{\p\rho }{\p t}+\Div(\rho {\bf u})}Z +\rho\sez{\cfrac{\p Z}{\p t}+({\bf u}\cdot\n)Z} =\rho\cfrac{\rd Z}{\rd t} \eea \eeex$$ 即知结论.