Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/246

Soit prise successivement la somme de leurs produits d’abord par

${\frac {\operatorname {d} x}{\operatorname {d} a}},\quad {\frac {\operatorname {d} y}{\operatorname {d} a}},\quad {\frac {\operatorname {d} z}{\operatorname {d} a}},$

puis par

${\frac {\operatorname {d} x}{\operatorname {d} b}},\quad {\frac {\operatorname {d} y}{\operatorname {d} b}},\quad {\frac {\operatorname {d} z}{\operatorname {d} b}},$

puis enfin par

${\frac {\operatorname {d} x}{\operatorname {d} c}},\quad {\frac {\operatorname {d} y}{\operatorname {d} c}},\quad {\frac {\operatorname {d} z}{\operatorname {d} c}}\,;$

et réduisant à l’aide des relations du n.^o 4, on trouvera

$\left.{\begin{aligned}{\frac {\operatorname {d} \nu }{\operatorname {d} a}}-{\frac {1}{\Delta }}{\frac {\operatorname {d} p}{\operatorname {d} a}}&=\operatorname {d} '^{2}x.{\frac {\operatorname {d} x}{\operatorname {d} a}}+\operatorname {d} '^{2}y.{\frac {\operatorname {d} y}{\operatorname {d} a}}+\operatorname {d} '^{2}z.{\frac {\operatorname {d} z}{\operatorname {d} a}},\\\\{\frac {\operatorname {d} \nu }{\operatorname {d} b}}-{\frac {1}{\Delta }}{\frac {\operatorname {d} p}{\operatorname {d} b}}&=\operatorname {d} '^{2}x.{\frac {\operatorname {d} x}{\operatorname {d} b}}+\operatorname {d} '^{2}y.{\frac {\operatorname {d} y}{\operatorname {d} b}}+\operatorname {d} '^{2}z.{\frac {\operatorname {d} z}{\operatorname {d} b}},\\\\{\frac {\operatorname {d} \nu }{\operatorname {d} c}}-{\frac {1}{\Delta }}{\frac {\operatorname {d} p}{\operatorname {d} c}}&=\operatorname {d} '^{2}x.{\frac {\operatorname {d} x}{\operatorname {d} c}}+\operatorname {d} '^{2}y.{\frac {\operatorname {d} y}{\operatorname {d} c}}+\operatorname {d} '^{2}z.{\frac {\operatorname {d} z}{\operatorname {d} c}}\,;\end{aligned}}\right\}\quad (2)$

forme sous laquelle on devra employer les équations, lorsqu’on voudra déterminer $x,y,z,p,$ en fonction de $a,b,c,t\,;$ mais, si on veut déterminer $p,\operatorname {d} 'x,\operatorname {d} 'y,\operatorname {d} 'z,$ en fonction de $t,x,y,z,$ il faudra mettre les, équations (1) sous la forme

$\left.{\begin{aligned}{\frac {\operatorname {d} \nu }{\operatorname {d} x}}-{\frac {1}{\Delta }}{\frac {\operatorname {d} p}{\operatorname {d} x}}&={\frac {\operatorname {dd} 'x}{\operatorname {d} t}}+{\frac {\operatorname {dd} 'x}{\operatorname {d} x}}\operatorname {d} 'x+{\frac {\operatorname {dd} 'x}{\operatorname {d} y}}\operatorname {d} 'y+{\frac {\operatorname {dd} 'x}{\operatorname {d} z}}\operatorname {d} 'z,\\\\{\frac {\operatorname {d} \nu }{\operatorname {d} y}}-{\frac {1}{\Delta }}{\frac {\operatorname {d} p}{\operatorname {d} y}}&={\frac {\operatorname {dd} 'y}{\operatorname {d} t}}+{\frac {\operatorname {dd} 'y}{\operatorname {d} x}}\operatorname {d} 'x+{\frac {\operatorname {dd} 'y}{\operatorname {d} y}}\operatorname {d} 'y+{\frac {\operatorname {dd} 'y}{\operatorname {d} z}}\operatorname {d} 'z,\\\\{\frac {\operatorname {d} \nu }{\operatorname {d} z}}-{\frac {1}{\Delta }}{\frac {\operatorname {d} p}{\operatorname {d} z}}&={\frac {\operatorname {dd} 'z}{\operatorname {d} t}}+{\frac {\operatorname {dd} 'z}{\operatorname {d} x}}\operatorname {d} 'x+{\frac {\operatorname {dd} 'z}{\operatorname {d} y}}\operatorname {d} 'y+{\frac {\operatorname {dd} 'z}{\operatorname {d} z}}\operatorname {d} 'z,\end{aligned}}\right\}\quad (3)$