另一种形式的欧拉方程

现在对$F(x,y,y^{\prime })$做$x$的全微分： $$\frac{\partial F}{\partial x}=\frac{\partial F}{\partial x}\cdot \frac{\partial x}{\partial x}+\frac{\partial F}{\partial y}\cdot \frac{\partial y}{\partial x}+\frac{\partial F}{\partial y^{\prime }}\cdot \frac{\partial y^{\prime }}{\partial x}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\cdot y^{\prime }+\frac{\partial F}{\partial y^{\prime }}\cdot y^{\prime \prime }$$

考虑这样一个函数的微分 $$\frac{\partial }{\partial x}\left(y^{\prime }\cdot \frac{\partial F}{\partial y^{\prime }}\right)=y^{\prime \prime }\frac{\partial F}{\partial y^{\prime }}+y^{\prime }\cdot \frac{\partial }{\partial x}(\frac{\partial F}{\partial y^{\prime }})$$ 由全微分可推知 $$\frac{\partial F}{\partial y^{\prime }}\cdot y^{\prime \prime }=\frac{\partial F}{\partial x}-\frac{\partial F}{\partial x}-\frac{\partial F}{\partial y}\cdot y^{\prime }$$ 将其代入得 $$\frac{\partial }{\partial x}\left(y^{\prime }\cdot \frac{\partial F}{\partial y^{\prime }}\right)=\left(\frac{\partial F}{\partial x}-\frac{\partial F}{\partial x}-\frac{\partial F}{\partial y}\cdot y^{\prime }\right)+y^{\prime }\cdot \frac{\partial }{\partial x}(\frac{\partial F}{\partial y^{\prime }})$$ 拆括号化简可得 $$\frac{\partial }{\partial x}\left(y^{\prime }\cdot \frac{\partial F}{\partial y^{\prime }}\right)=\frac{\partial F}{\partial x}-\frac{\partial F}{\partial x}+y^{\prime }\left[\frac{\partial F}{\partial y}-\frac{\partial }{\partial x}(\frac{\partial F}{\partial y^{\prime }})\right]$$ $\left[\frac{\partial F}{\partial y}-\frac{\partial }{\partial x}(\frac{\partial F}{\partial y^{\prime }})\right]$实际上就是上面所推导出的欧拉方程，在最优路径上其值为 0，因此该式可化为 $$\frac{\partial F}{\partial x}-\frac{\partial }{\partial x}(F-y^{\prime }\cdot \frac{\partial F}{\partial y^{\prime }})=0$$ 这就是另一种欧拉方程的形式。在某些情况下若能给定$\frac{\partial F}{\partial x}=0$，这意味着$x$并不显含于$F$（也就是$F(x,y,y^{\prime })$实际可能是$F(y,y^{\prime })$，$x$ 仍隐含于$F$），所有其他项在关于$x$偏微分时都视为常数，也就是说$F-y^{\prime }\cdot \frac{\partial F}{\partial y^{\prime }}$是一个常数$c$。 $$F-y^{\prime }\cdot \frac{\partial F}{\partial y^{\prime }}=C$$