Page:Joseph Louis de Lagrange - Œuvres, Tome 12.djvu/288

280

MÉCANIQUE ANALYTIQUE.

ensuite

${\begin{aligned}\beta {\frac {\partial x}{\partial a}}+\beta '{\frac {\partial y}{\partial a}}+\beta ''{\frac {\partial z}{\partial a}}=&0,\\\gamma {\frac {\partial x}{\partial a}}+\gamma '{\frac {\partial y}{\partial a}}+\gamma ''{\frac {\partial z}{\partial a}}=&0,\\\alpha {\frac {\partial x}{\partial a}}+\alpha '{\frac {\partial y}{\partial a}}+\alpha ''{\frac {\partial z}{\partial a}}=&0,\end{aligned}}$

et ainsi de suite ; de sorte que, par ces opérations et ces réductions, on aura les transformées

(D)

$\left\{{\begin{aligned}\Delta \left[\left({\frac {d^{2}x}{dt^{2}}}+\mathrm {X} \right){\frac {\partial x}{\partial a}}+\left({\frac {d^{2}y}{dt^{2}}}+\mathrm {Y} \right){\frac {\partial y}{\partial a}}+\left({\frac {d^{2}z}{dt^{2}}}+\mathrm {Z} \right){\frac {\partial z}{\partial a}}\right]&-{\frac {\partial \lambda }{\partial a}}=0,\\\Delta \left[\left({\frac {d^{2}x}{dt^{2}}}+\mathrm {X} \right){\frac {\partial x}{\partial b}}+\left({\frac {d^{2}y}{dt^{2}}}+\mathrm {Y} \right){\frac {\partial y}{\partial b}}+\left({\frac {d^{2}z}{dt^{2}}}+\mathrm {Z} \right){\frac {\partial z}{\partial b}}\right]&-{\frac {\partial \lambda }{\partial b}}=0,\\\Delta \left[\left({\frac {d^{2}x}{dt^{2}}}+\mathrm {X} \right){\frac {\partial x}{\partial c}}+\left({\frac {d^{2}y}{dt^{2}}}+\mathrm {Y} \right){\frac {\partial y}{\partial c}}+\left({\frac {d^{2}z}{dt^{2}}}+\mathrm {Z} \right){\frac {\partial z}{\partial c}}\right]&-{\frac {\partial \lambda }{\partial c}}=0.\end{aligned}}\right.$

^[1] On aurait pu parvenir directement à ces dernières équations, en introduisant dans les formules de l’article 2, au lieu des variations $\delta x,\,\delta y,\,\delta z,$ celles des coordonnées de l’état initial, $\delta a,\delta b,\delta c\,;$ car, en regardant $x,\,y,\,z$ comme fonctions de $a,\,b,\,c,$ on aura

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

On fera ces substitutions dans la formule (a) de l’article 2, et l’on égalera à zéro les quantités multipliées par $\delta a,\,\delta b,\,\delta c.$

En observant que, $\lambda$ étant fonction de $x,\,y,\,z,$ on a, par rapport à $a,\,b,\,c,$

${\begin{aligned}{\frac {\partial \lambda }{\partial a}}=&{\frac {\operatorname {D} \lambda }{\operatorname {D} x}}{\frac {\partial x}{\partial a}}+{\frac {\operatorname {D} \lambda }{\operatorname {D} y}}{\frac {\partial y}{\partial a}}+{\frac {\operatorname {D} \lambda }{\operatorname {D} z}}{\frac {\partial z}{\partial a}},\\{\frac {\partial \lambda }{\partial b}}=&{\frac {\operatorname {D} \lambda }{\operatorname {D} x}}{\frac {\partial x}{\partial b}}+{\frac {\operatorname {D} \lambda }{\operatorname {D} y}}{\frac {\partial y}{\partial b}}+{\frac {\operatorname {D} \lambda }{\operatorname {D} z}}{\frac {\partial z}{\partial b}},\\{\frac {\partial \lambda }{\partial c}}=&{\frac {\operatorname {D} \lambda }{\operatorname {D} x}}{\frac {\partial x}{\partial c}}+{\frac {\operatorname {D} \lambda }{\operatorname {D} y}}{\frac {\partial y}{\partial c}}+{\frac {\operatorname {D} \lambda }{\operatorname {D} z}}{\frac {\partial z}{\partial c}},\end{aligned}}$

1. Cette fin de l’article 6 ne se trouve pas dans la première édition. (J. Bertrand.)