Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/260

 Les corrections sont expliquées en page de discussion

$x=a+x_{1},\qquad y=b+y_{1},\qquad z=c+z_{1},$

en désignant par $x_{1},y_{1},z_{1},$ des fonctions de $x,y,z,$ nulles en même temps que $t$ et convenablement déterminées, dès-lors nous aurons

${\begin{aligned}&\operatorname {d} 'x=\operatorname {d} 'x_{1}={\frac {\operatorname {d} x}{\operatorname {d} t}}+{\frac {\operatorname {d} x_{1}}{\operatorname {d} x}}\operatorname {d} 'x+{\frac {\operatorname {d} x_{1}}{\operatorname {d} y}}\operatorname {d} 'y+{\frac {\operatorname {d} x_{1}}{\operatorname {d} z}}\operatorname {d} 'z,\\\\&\operatorname {d} 'y=\operatorname {d} 'y_{1}={\frac {\operatorname {d} y}{\operatorname {d} t}}+{\frac {\operatorname {d} y_{1}}{\operatorname {d} x}}\operatorname {d} 'x+{\frac {\operatorname {d} y_{1}}{\operatorname {d} y}}\operatorname {d} 'y+{\frac {\operatorname {d} y_{1}}{\operatorname {d} z}}\operatorname {d} 'z,\\\\&\operatorname {d} 'z=\operatorname {d} 'z_{1}={\frac {\operatorname {d} z}{\operatorname {d} t}}+{\frac {\operatorname {d} z_{1}}{\operatorname {d} x}}\operatorname {d} 'x+{\frac {\operatorname {d} z_{1}}{\operatorname {d} y}}\operatorname {d} 'y+{\frac {\operatorname {d} z_{1}}{\operatorname {d} z}}\operatorname {d} 'z,\end{aligned}}$

dans le cas actuel $x_{1},y_{1},z_{1},$ étant très-petits, nous pouvons négliger les termes de plus d’une dimension par rapport à ces quantités ; nous aurons donc

$\operatorname {d} 'x={\frac {\operatorname {d} x_{1}}{\operatorname {d} t}},\quad \operatorname {d} 'y={\frac {\operatorname {d} y_{1}}{\operatorname {d} t}},\quad \operatorname {d} 'z={\frac {\operatorname {d} z_{1}}{\operatorname {d} t}},$

et par suite, au moyen des équations (26)

${\frac {\operatorname {d} x_{1}}{\operatorname {d} t}}={\frac {\operatorname {d} \phi }{\operatorname {d} x}}+P,\quad {\frac {\operatorname {d} y_{1}}{\operatorname {d} t}}={\frac {\operatorname {d} \phi }{\operatorname {d} y}}+Q,\quad {\frac {\operatorname {d} z_{1}}{\operatorname {d} t}}={\frac {\operatorname {d} \phi }{\operatorname {d} z}}+R.$

d’où l’on tire, par l’élimination successive de $R,Q$ et $P,$

${\begin{aligned}&{\frac {\operatorname {d} ^{2}x_{1}}{\operatorname {d} t\operatorname {d} y}}-{\frac {\operatorname {d} ^{2}y_{1}}{\operatorname {d} t\operatorname {d} x}}={\frac {\operatorname {d} P}{\operatorname {d} y}}-{\frac {\operatorname {d} Q}{\operatorname {d} x}},\\\\&{\frac {\operatorname {d} ^{2}y_{1}}{\operatorname {d} t\operatorname {d} z}}-{\frac {\operatorname {d} ^{2}z_{1}}{\operatorname {d} t\operatorname {d} y}}={\frac {\operatorname {d} Q}{\operatorname {d} z}}-{\frac {\operatorname {d} R}{\operatorname {d} y}},\\\\&{\frac {\operatorname {d} ^{2}z_{1}}{\operatorname {d} t\operatorname {d} x}}-{\frac {\operatorname {d} ^{2}x_{1}}{\operatorname {d} t\operatorname {d} z}}={\frac {\operatorname {d} R}{\operatorname {d} x}}-{\frac {\operatorname {d} P}{\operatorname {d} z}}\,;\end{aligned}}$

et de là en intégrant par rapport à $t,$