试证明: 如果质量力有势, 即存在 $\phi$ 使 ${\bf F}=-\n \phi$, 那么理想流体的能量守恒方程的微分形式可写为 $$\bex \cfrac{\rd}{\rd t}\sex{e+\cfrac{u^2}{}+\cfrac{p}{\rho}+\phi} =\cfrac{1}{\rho}\cfrac{\p p}{\p t}+\cfrac{\p \phi}{\p t}. \eex$$
证明: 由 (1. 21), $$\bex \cfrac{\rd }{\rd t}\sex{e+\cfrac{u^2}{2}} +\cfrac{1}{\rho }[p\Div{\bf u}+({\bf u}\cdot\n)p]=-({\bf u}\cdot\n)\phi, \eex$$ 而又 $$\beex \bea \cfrac{1}{\rho}[p\Div{\bf u}+({\bf u}\cdot\n)p] &=\cfrac{p}{\rho^2}\sez{-\cfrac{\rd \rho}{\rd t}} +\cfrac{1}{\rho}\sex{\cfrac{\rd p}{\rd t}-\cfrac{\p t}{\p t}}\\ &=p\cfrac{\rd }{\rd t}\cfrac{1}{\rho} +\cfrac{1}{\rho}\cfrac{\rd p}{\rd t} -\cfrac{1}{\rho}\cfrac{\p p}{\p t}\\ &=\cfrac{\rd }{\rd t}\cfrac{p}{\rho} -\cfrac{1}{\rho}\cfrac{\p p}{\p t},\\ -({\bf u}\cdot\n)\phi&=-\cfrac{\rd\phi}{\rd t}+\cfrac{\p \phi}{\p t}, \eea \eeex$$ 我们有 $$\bex \cfrac{\rd}{\rd t}\sex{e+\cfrac{u^2}{2}+\cfrac{p}{\rho}+\phi} =\cfrac{1}{\rho}\cfrac{\p p}{\p t}+\cfrac{\p \phi}{\p t}. \eex$$
