Page:Lagrange - Œuvres (1867) vol. 1.djvu/499

de l’Article I du Mémoire précédent, que la valeur complète du terme $\mathrm {U} dt\delta y$ sera $\mathrm {U} 'dt\delta y'-{\grave {\,}}\!\mathrm {U} dt\delta {\grave {\,}}\!y.$

Mais pour peu qu’on réfléchisse sur la nature de nos formules, il est aisé de voir que quand $\mathrm {U} ={\grave {\,}}\!\mathrm {U}$ l’intégrale $\mathrm {S} \operatorname {d} x\mathrm {D} \left(d{\frac {dx}{dt}}+\Pi \delta t\right)$ est nulle, et que quand $\mathrm {U=U'} ,$ cette intégrale est précisément égale à $\mathrm {T} dt\,;$ c’est pourquoi l’on aura ${\grave {\,}}\!\mathrm {U=T}$ et $\mathrm {U} '=0$ donc enfin

$\mathrm {S} \operatorname {d} y{\frac {\mathrm {U} dt\operatorname {d} \delta y}{\operatorname {d} y}}=-\mathrm {T} \delta {\grave {\,}}\!y-\mathrm {S} \operatorname {d} y{\frac {\operatorname {d} (\mathrm {U} dt)}{\operatorname {d} y}}\delta y.$

On trouvera par des opérations et des raisonnements semblables

$\mathrm {S} \operatorname {d} z{\frac {-\mathrm {U} dt\operatorname {d} \delta z}{\operatorname {d} z}}=-\mathrm {T} \delta {\grave {\,}}\!z-\mathrm {S} \operatorname {d} z{\frac {\operatorname {d} (\mathrm {U} dt)}{\operatorname {d} z}}\delta z\,;$

donc

$\mathrm {S} ^{3}\operatorname {d} x\operatorname {d} y\operatorname {d} z\mathrm {U} dt\left({\frac {\operatorname {d} \delta y}{\operatorname {d} y}}+{\frac {\operatorname {d} \delta z}{\operatorname {d} z}}\right)$

se changera en

${\begin{aligned}&-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z\mathrm {T} dt\delta {\grave {\,}}\!y-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z\mathrm {S} \operatorname {d} y{\frac {\operatorname {d} (\mathrm {U} dt)}{\operatorname {d} y}}\delta y\\&-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y\mathrm {T} dt\delta {\grave {\,}}\!z-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y\mathrm {S} \operatorname {d} z{\frac {\operatorname {d} (\mathrm {U} dt)}{\operatorname {d} z}}\delta z,\end{aligned}}$

ou, en réduisant,

$-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z\mathrm {T} dt\delta {\grave {\,}}\!y-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z\mathrm {T} dts\delta {\grave {\,}}\!z-\mathrm {S} ^{3}\operatorname {d} x\operatorname {d} y\operatorname {d} z\left({\frac {\operatorname {d} (\mathrm {U} dt)}{\operatorname {d} y}}\delta y+{\frac {\operatorname {d} (\mathrm {U} dt)}{\operatorname {d} z}}\delta z\right)\,;$

donc

$\mathrm {S} ^{3}\operatorname {d} x\operatorname {d} y\operatorname {d} z\mathrm {D} \left(d{\frac {dx}{dt}}+\Pi dt\right)\delta x=\mathrm {S} ^{2}\operatorname {d} y\operatorname {d} z\mathrm {T} dt\delta {\grave {\,}}\!x-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z\mathrm {T} dt\delta {\grave {\,}}\!y$

$-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y\mathrm {T} dt\delta {\grave {\,}}\!z+\mathrm {S} ^{3}\operatorname {d} x\operatorname {d} y\operatorname {d} z\left({\frac {\operatorname {d} (\mathrm {U} dt)}{\operatorname {d} y}}\delta y+{\frac {\operatorname {d} (\mathrm {U} dt)}{\operatorname {d} z}}\delta z\right),$

donc l’équation $(a)$ deviendra

$(d)\quad \left\{{\begin{aligned}\int &\left(\mathrm {S} ^{2}\operatorname {d} y\operatorname {d} z\mathrm {T} dt\delta {\grave {\,}}\!x+\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z\mathrm {T} dt\delta {\grave {\,}}\!y+\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y\mathrm {T} dt\delta {\grave {\,}}\!z\right)\\&+\int \mathrm {\mathrm {S} } ^{3}\operatorname {d} x\operatorname {d} y\operatorname {d} z\left\{\left[{\frac {\operatorname {d} (\mathrm {\mathrm {U} } dt)}{\operatorname {d} y}}+\mathrm {D} \left(d{\frac {dy}{dt}}+\varpi dt\right)\right]\delta y\right.\\&\qquad \qquad \qquad \qquad \left.+\left[{\frac {\operatorname {d} (\mathrm {\mathrm {U} } dt)}{\operatorname {d} z}}+\mathrm {D} \left(d{\frac {dz}{dt}}+\Psi dt\right)\right]\delta z\right\}=0\,;\end{aligned}}\right.$