# 貝爾特拉米等式

${\displaystyle I(u)=\int _{a}^{b}L(x,u,u')\,dx}$

${\displaystyle {\frac {d}{dx}}\left(L-u'{\frac {\partial L}{\partial u'}}\right)-{\frac {\partial L}{\partial x}}=0.}$

L是力學系統中的拉格朗日量，且L並非x顯函數，即拉格朗日量並非時間的顯函數，那麼，貝爾特拉米等式表明其哈密頓量是一守恆能量。

## 証明

${\displaystyle p={\partial L \over \partial u'}}$

${\displaystyle {dp \over dx}-{\partial L \over \partial u}=0}$

${\displaystyle p'={\partial L \over \partial u}}$

${\displaystyle H=pu'-L\,}$

${\displaystyle H'={dH \over dx}=p'u'+pu''-{\partial L \over \partial u'}u''-{\partial L \over \partial u}u'-{\partial L \over \partial x}}$

${\displaystyle H'=-{\partial L \over \partial x}}$

## 應用

L獨立於x，則貝爾特拉米等式說明H為一常數：

${\displaystyle -H'={\frac {d}{dx}}\left(L-u'{\frac {\partial L}{\partial u'}}\right)=0}$

${\displaystyle \int _{0}^{1}{{\sqrt {1+y'^{2}}} \over {\sqrt {y}}}dx}$

${\displaystyle L(y,y')={{\sqrt {1+y'^{2}}} \over {\sqrt {y}}}}$

${\displaystyle H=py'-L={y'^{2} \over {\sqrt {1+y'^{2}}}{\sqrt {y}}}-{{\sqrt {1+y'^{2}}} \over {\sqrt {y}}}={-1 \over {\sqrt {1+y'^{2}}}{\sqrt {y}}}}$
${\displaystyle {\sqrt {1+y'^{2}}}{\sqrt {y}}={\text{constant}}}$