Page:Joseph Louis de Lagrange - Œuvres, Tome 11.djvu/366

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348

MÉCANIQUE ANALYTIQUE

sorte qu’on aura les équations

{\begin{aligned}&d\delta {\frac {\partial \mathrm {Z} }{\partial \xi '}}-\delta {\frac {\partial \mathrm {Z} }{\partial \xi }}dt=0,\\&d\delta {\frac {\partial \mathrm {Z} }{\partial \psi '}}-\delta {\frac {\partial \mathrm {Z} }{\partial \psi }}dt=0,\\&d\delta {\frac {\partial \mathrm {Z} }{\partial \varphi '}}-\delta {\frac {\partial \mathrm {Z} }{\partial \varphi }}dt=0,\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{aligned}}

De même, si, pour représenter des variations différentes des mêmes constantes arbitraires, on emploie la caractéristique $\Delta ,$ on aura

{\begin{aligned}&d\Delta {\frac {\partial \mathrm {Z} }{\partial \xi '}}\,-\Delta {\frac {\partial \mathrm {Z} }{\partial \xi }}dt=0,\\&d\Delta {\frac {\partial \mathrm {Z} }{\partial \psi '}}-\Delta {\frac {\partial \mathrm {Z} }{\partial \psi }}dt=0,\\&d\Delta {\frac {\partial \mathrm {Z} }{\partial \varphi '}}-\Delta {\frac {\partial \mathrm {Z} }{\partial \varphi }}dt=0,\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}

5. Multiplions maintenant les premières équations respectivement par $\Delta \xi ,\,\Delta \psi ,\,\Delta \varphi ,\,\ldots ,$ et retranchons de leur somme celle des dernières équations multipliées respectivement par $\delta \xi ,\,\delta \psi ,\,\delta \varphi ,\,\ldots \,;$ on aura

{\begin{aligned}&\Delta \xi \,d\delta {\frac {\partial \mathrm {Z} }{\partial \xi '}}+\Delta \psi \,d\delta {\frac {\partial \mathrm {Z} }{\partial \psi '}}+\Delta \varphi \,d\delta {\frac {\partial \mathrm {Z} }{\partial \varphi '}}+\ldots \\-&\delta \xi \,d\Delta {\frac {\partial \mathrm {Z} }{\partial \xi '}}-\delta \psi \,d\Delta {\frac {\partial \mathrm {Z} }{\partial \psi '}}-\delta \varphi \,d\Delta {\frac {\partial \mathrm {Z} }{\partial \varphi '}}+\ldots \\-&\left(\Delta \xi \,\delta {\frac {\partial \mathrm {Z} }{\partial \xi }}+\Delta \psi \quad \delta {\frac {\partial \mathrm {Z} }{\partial \psi }}+\Delta \varphi \quad \delta {\frac {\partial \mathrm {Z} }{\partial \varphi }}+\ldots \right)dt\\+&\left(\delta \xi \,\Delta {\frac {\partial \mathrm {Z} }{\partial \xi }}+\delta \psi \quad \Delta {\frac {\partial \mathrm {Z} }{\partial \psi }}+\delta \varphi \quad \Delta {\frac {\partial \mathrm {Z} }{\partial \varphi }}+\ldots \right)dt=0.\end{aligned}}

Or $\Delta \xi \,d\delta {\frac {\partial \mathrm {Z} }{\partial \xi '}}=d(\left(\Delta \xi \,\delta {\frac {\partial \mathrm {Z} }{\partial \xi '}}\right)-d\,\Delta \xi \,\delta {\frac {\partial \mathrm {Z} }{\partial \xi '}}\,;$ mais $d\,\Delta \xi =\Delta \,d\xi$ $=\Delta \xi 'dt,$ à cause de $d\xi =\xi 'dt\,;$ donc

\Delta \xi \,d\delta {\frac {\partial \mathrm {Z} }{\partial \xi '}}=d\left(\Delta \xi \,\delta {\frac {\partial \mathrm {Z} }{\partial \xi '}}\right)-\Delta \xi '\delta {\frac {\partial \mathrm {Z} }{\partial \xi '}}dt.