Page:Joseph Louis de Lagrange - Œuvres, Tome 12.djvu/82

74

MÉCANIQUE ANALYTIQUE.

dessus, et des valeurs semblables de $db,\,dc,\ldots ,$ car, en faisant ces substitutions dans les termes ${\frac {\partial x}{\partial a}}da+{\frac {\partial x}{\partial b}}db+\ldots$ et ordonnant par rapport aux quantités $\mathrm {X,Y,Z} ,$ on aura

${\begin{aligned}&\left({\frac {\partial x}{\partial a}}{\frac {\partial a}{\partial x'}}+{\frac {\partial x}{\partial b}}{\frac {\partial b}{\partial x'}}+{\frac {\partial x}{\partial c}}{\frac {\partial c}{\partial x'}}+\ldots \right)\mathrm {X} dt\\+&\left({\frac {\partial x}{\partial a}}{\frac {\partial a}{\partial y'}}+{\frac {\partial x}{\partial b}}{\frac {\partial b}{\partial y'}}\,+{\frac {\partial x}{\partial c}}{\frac {\partial c}{\partial y'}}+\ldots \right)\mathrm {Y} dt\\+&\left({\frac {\partial x}{\partial a}}{\frac {\partial a}{\partial z'}}+{\frac {\partial x}{\partial b}}{\frac {\partial b}{\partial z'}}\,+{\frac {\partial x}{\partial c}}{\frac {\partial c}{\partial z'}}+\ldots \right)\mathrm {Z} dt.\end{aligned}}$

Mais, en regardant d’abord $x,\,y,\,z,\,x',\,y',\,z'$ comme fonctions de $a,\,b,\,c,\,h,\,i,\,k,$ et ensuite $a,\,b,\,c,\ldots$ comme fonctions de $x,\,y,\,z,\,x',\ldots ,$ on a

${\begin{aligned}&dx={\frac {\partial x}{\partial a}}da+{\frac {\partial x}{\partial b}}db+{\frac {\partial x}{\partial c}}dc+{\frac {\partial x}{\partial h}}dh+\ldots ,\\&dy={\frac {\partial y}{\partial a}}da\,+{\frac {\partial y}{\partial b}}db\,+{\frac {\partial y}{\partial c}}dc\,+{\frac {\partial y}{\partial h}}dh+\ldots ,\\&.\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;\\&da={\frac {\partial a}{\partial x}}dx+{\frac {\partial a}{\partial y}}dy+{\frac {\partial a}{\partial z}}dz+{\frac {\partial a}{\partial x'}}dx'+\ldots ,\\&db={\frac {\partial b}{\partial x}}dx\,+{\frac {\partial b}{\partial y}}dy\,+{\frac {\partial b}{\partial z}}dz+{\frac {\partial b}{\partial x'}}dx'+\ldots ,\\&..\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;\\\end{aligned}}$

Substituant dans l’expression de $dx,\,dy,\ldots$ ces valeurs de $da,\,db,\ldots ,$ on doit avoir des équations identiques ; par conséquent, il faudra que les termes affectés de $dx',\,dy',\,dz',$ dans les expressions de $dx,\,dy,\,dz,$ deviennent nuls ; ce qui donnera, par rapport à $dx,$ les équations identiques

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

On aura donc simplement

$dx={\frac {\partial x}{\partial t}}dt,$