Page:Lagrange - Œuvres (1867) vol. 1.djvu/521

(j’écris $\operatorname {d} x{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} x}}$ au lieu de $\operatorname {d} \mathrm {F} ,$ pour dénoter que cette différentielle doit être prise en ne variant que $x$), et à

$\mathrm {F} '\delta x'-{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!x-\mathrm {S} \operatorname {d} x{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} x}}\delta x$

en complétant l’intégrale, suivant la remarque que nous avons faite à la fin de l’Article I du Mémoire précédent ; on changera de même

$\mathrm {S} \operatorname {d} y{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} y}}\delta y\quad {\text{en}}\quad \mathrm {F} '\delta y'-{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!y-\mathrm {S} \operatorname {d} y{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} y}}\delta y,$

et

$\mathrm {S} \operatorname {d} z{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} z}}\delta z\quad {\text{en}}\quad \mathrm {F} '\delta z'-{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!z-\mathrm {S} \operatorname {d} z{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} z}}\delta z\,;$

donc

${\begin{aligned}\mathrm {S^{2}F} \delta f=&\mathrm {S} ^{2}\operatorname {d} y\operatorname {d} z\left(\mathrm {F} '\delta x'-{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!x\right)+\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z\left(\mathrm {F} '\delta y'-{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!y\right)\\&+\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y(\mathrm {F} '\delta z'-{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!z-\mathrm {S} ^{2}\operatorname {d} y\operatorname {d} z\mathrm {S} \operatorname {d} x{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} x}}\delta x\\&-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z\mathrm {S} \operatorname {d} y{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} y}}\delta y-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y\mathrm {S} \operatorname {d} z{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} z}}\delta z\\=&\mathrm {S} ^{2}\operatorname {d} y\operatorname {d} z\mathrm {F} '\delta x'+\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z\mathrm {F} '\delta y'+\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y\mathrm {F} '\delta z'\\&-\mathrm {S} ^{2}\operatorname {d} y\operatorname {d} z{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!x-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!y-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!z\\&-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y\operatorname {d} z\left({\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} x}}\delta x+{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} y}}\delta y+{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} z}}\delta z\right)\,;\end{aligned}}$

donc, substituant dans l’équation $(n),$ au lieu de $\mathrm {S^{2}F} \delta f,$ l’expression qu’on vient de trouver, on aura enfin

${\begin{aligned}\int &\left[\mathrm {S} ^{2}\operatorname {d} y\operatorname {d} z\mathrm {F} '\delta x'+\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z\mathrm {F} '\delta y'+\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y\mathrm {F} '\delta z'\right.\\&\left.-\mathrm {S} ^{2}\operatorname {d} y\operatorname {d} z{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!x-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} z{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!y-\mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y{\grave {\,}}\!\mathrm {F} \delta {\grave {\,}}\!z\right]dt\\\end{aligned}}$

$-\int \mathrm {S} ^{2}\operatorname {d} x\operatorname {d} y\operatorname {d} z\left[\left(\mathrm {D} d{\frac {dx}{dt}}+\mathrm {D} \Pi dt+{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} x}}dt\right)\delta x\right.$

${\begin{aligned}&+\left(\mathrm {D} d{\frac {dy}{dt}}+\mathrm {D} \varpi dt+{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} y}}dt\right)\delta y\\&+\left.\left(\mathrm {D} d{\frac {dz}{dt}}+\mathrm {D} \Psi dt+{\frac {\operatorname {d} \mathrm {F} }{\operatorname {d} z}}dt\right)\delta z\right]=0,\end{aligned}}$