试证明: 利用连续性方程及动量方程, 能量守恒方程 (2. 15) 可化为 (2. 21) 的形式.
证明: 注意到 \beex \bea &\quad\cfrac{\p}{\p t}\sex{\cfrac{1}{2}\rho u^2} +\Div\sez{\cfrac{1}{2}\rho u^2{\bf u}-{\bf P}{\bf u}} -\rho {\bf F}\cdot{\bf u}\\ &=\cfrac{u^2}{2}\cfrac{\p \rho}{\p t} +\cfrac{u^2}{2}\Div(\rho {\bf u}) -\Div{\bf P}\cdot{\bf u}-\rho {\bf F}\cdot{\bf u} +\rho \cfrac{\p }{\p t}\cfrac{u^2}{2} +\sex{\rho {\bf u}\cdot\n}\cfrac{u^2}{2} -\tr({\bf P}\cdot\n{\bf u})\\ &={\bf u}\cdot\sez{\rho \cfrac{\p u}{\p t} +(\rho {\bf u}\cdot\n){\bf u}-\Div{\bf P}-\rho {\bf F}}-\tr({\bf P}\cdot\n {\bf u})\\ &=-\tr({\bf P}\cdot\n{\bf u})\\ &=p\Div{\bf u} -\sex{\mu'-\cfrac{2}{3}\mu}|\Div{\bf u}|^2 -\mu \sum_{i,j=1}^3 \sex{\cfrac{\p u_i}{\p x_j}+\cfrac{\p u_j}{\p x_i}}\cfrac{\p u_j}{\p x_i}\\ &\quad \sex{{\bf P}=-p{\bf I}+\sex{\mu'-\cfrac{2}{3}\mu}(\Div{\bf u}){\bf I} +2\mu{\bf S}} \eea \eeex
