Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/243

${\frac {\operatorname {dd} 'u}{\operatorname {d} t}}=\operatorname {d} '{\frac {\operatorname {d} u}{\operatorname {d} t}}.$

6. Nous supposerons que la caractéristique $\delta$ des variations est uniquement relative aux coordonnées primitives $a,b,c,$ et non au tems $t.$ Dès lors nous aurons

${\begin{aligned}\delta x&={\frac {\operatorname {d} x}{\operatorname {d} a}}\delta a+{\frac {\operatorname {d} x}{\operatorname {d} b}}\delta b+{\frac {\operatorname {d} x}{\operatorname {d} c}}\delta c,\\\\\delta y&={\frac {\operatorname {d} y}{\operatorname {d} a}}\delta a+{\frac {\operatorname {d} y}{\operatorname {d} b}}\delta b+{\frac {\operatorname {d} y}{\operatorname {d} c}}\delta c,\\\\\delta z&={\frac {\operatorname {d} z}{\operatorname {d} a}}\delta a+{\frac {\operatorname {d} z}{\operatorname {d} b}}\delta b+{\frac {\operatorname {d} z}{\operatorname {d} c}}\delta c,\\\\\end{aligned}}$

et en général

$\delta u={\frac {\operatorname {d} u}{\operatorname {d} x}}\delta x+{\frac {\operatorname {d} u}{\operatorname {d} y}}\delta y+{\frac {\operatorname {d} u}{\operatorname {d} z}}\delta z={\frac {\operatorname {d} u}{\operatorname {d} a}}\delta a+{\frac {\operatorname {d} u}{\operatorname {d} b}}\delta b+{\frac {\operatorname {d} u}{\operatorname {d} c}}\delta c\,;$

d’où il est facile de conclure

$\delta \operatorname {d} 'u=\operatorname {d} '\delta u,\qquad \delta {\frac {\operatorname {d} u}{\operatorname {d} t}}={\frac {\operatorname {d} \delta u}{\operatorname {d} t}},$

et toutes les autres relations qu’on en peut déduire par des différentiations répétées.

Si l’on avait

$\psi =\delta \phi +H,$

on en tirerait

$\operatorname {d} '\psi =\operatorname {d} '\delta \phi +\operatorname {d} 'H,$

et

${\frac {\operatorname {d} \psi }{\operatorname {d} t}}={\frac {\operatorname {d} \delta \phi +}{\operatorname {d} t}}+{\frac {\operatorname {d} H}{\operatorname {d} t}}.$