Page:Joseph Louis de Lagrange - Œuvres, Tome 12.djvu/84

76

MÉCANIQUE ANALYTIQUE.

et, ces valeurs étant substituées dans l’expression de $da$ de l’article 58, à la place de $\mathrm {X,\,Y,\,Z} ,$ elle deviendra

${\begin{aligned}da=&+\left({\frac {\partial a}{\partial x'}}{\frac {\partial a}{\partial x}}+{\frac {\partial a}{\partial y'}}{\frac {\partial a}{\partial y}}+{\frac {\partial a}{\partial z'}}{\frac {\partial a}{\partial z}}\right){\frac {\partial \Omega }{\partial a}}dt\\&+\left({\frac {\partial a}{\partial x'}}{\frac {\partial b}{\partial x}}+{\frac {\partial a}{\partial y'}}{\frac {\partial b}{\partial y}}+{\frac {\partial a}{\partial z'}}{\frac {\partial b}{\partial z}}\right){\frac {\partial \Omega }{\partial b}}dt\\&+\left({\frac {\partial a}{\partial x'}}{\frac {\partial c}{\partial x}}+{\frac {\partial a}{\partial y'}}{\frac {\partial c}{\partial y}}+{\frac {\partial a}{\partial z'}}{\frac {\partial c}{\partial z}}\right){\frac {\partial \Omega }{\partial c}}dt.\end{aligned}}$

On peut faire disparaître de cette expression les termes multipliés par ${\frac {\partial \Omega }{\partial a}},$ par la considération que, $\Omega$ ne contenantpointles variables $x',\,y',\,z',$ on a

${\begin{aligned}{\frac {\partial \Omega }{\partial x'}}=&{\frac {\partial \Omega }{\partial a}}{\frac {\partial a}{\partial x'}}+{\frac {\partial \Omega }{\partial b}}{\frac {\partial b}{\partial x'}}+{\frac {\partial \Omega }{\partial c}}{\frac {\partial c}{\partial x'}}+\ldots =0,\\{\frac {\partial \Omega }{\partial y'}}=&{\frac {\partial \Omega }{\partial a}}{\frac {\partial a}{\partial y'}}+{\frac {\partial \Omega }{\partial b}}{\frac {\partial b}{\partial y'}}\,+{\frac {\partial \Omega }{\partial c}}{\frac {\partial c}{\partial y'}}+\ldots =0,\\{\frac {\partial \Omega }{\partial z'}}=&{\frac {\partial \Omega }{\partial a}}{\frac {\partial a}{\partial z'}}\,+{\frac {\partial \Omega }{\partial b}}{\frac {\partial b}{\partial z'}}\,+{\frac {\partial \Omega }{\partial c}}{\frac {\partial c}{\partial z'}}+\ldots =0.\end{aligned}}$

Donc, si l’on soustrait de la valeur de $da$ la quantité

$\left({\frac {\partial \Omega }{\partial x'}}{\frac {\partial a}{\partial x}}+{\frac {\partial \Omega }{\partial y'}}{\frac {\partial a}{\partial y}}+{\frac {\partial \Omega }{\partial z'}}{\frac {\partial a}{\partial z}}\right)dt,$

${\begin{aligned}da=&+\left({\frac {\partial a}{\partial x'}}{\frac {\partial b}{\partial x}}+{\frac {\partial a}{\partial y'}}{\frac {\partial b}{\partial y}}+{\frac {\partial a}{\partial z'}}{\frac {\partial b}{\partial z}}-{\frac {\partial a}{\partial x}}{\frac {\partial b}{\partial x'}}-{\frac {\partial a}{\partial y}}{\frac {\partial b}{\partial y'}}-{\frac {\partial a}{\partial z}}{\frac {\partial b}{\partial z'}}\right){\frac {\partial \Omega }{\partial b}}dt\\&+\left({\frac {\partial a}{\partial x'}}{\frac {\partial c}{\partial x}}+{\frac {\partial a}{\partial y'}}{\frac {\partial c}{\partial y}}+{\frac {\partial a}{\partial z'}}{\frac {\partial c}{\partial z}}-{\frac {\partial a}{\partial x}}{\frac {\partial c}{\partial x'}}-{\frac {\partial a}{\partial y}}{\frac {\partial c}{\partial y'}}-{\frac {\partial a}{\partial z}}{\frac {\partial c}{\partial z'}}\right){\frac {\partial \Omega }{\partial c}}dt.\end{aligned}}$

Cette expression de $da$ est, en apparence, plus compliquée que la formule primitive d’où nous sommes partis ; mais elle a, d’un autre côté, le grand avantage que les coefficients des différences partielles ${\frac {\partial \Omega }{\partial b}},\,{\frac {\partial \Omega }{\partial c}},\ldots$ deviennent indépendants du temps $t,$ après la substitution des valeurs de $x,\,y,\,z,\,x',\,y',\,z'$ en $t$ et $a,\,b,\,c,\ldots ,$ données par le mou-