Page:Lagrange - Œuvres (1867) vol. 1.djvu/514

deviennent, en divisant par $\mathrm {D} ,$

${\frac {\operatorname {d} \Pi }{\operatorname {d} y}}={\frac {\operatorname {d} \varpi }{\operatorname {d} x}},\qquad {\frac {\operatorname {d} \Pi }{\operatorname {d} z}}={\frac {\operatorname {d} \Psi }{\operatorname {d} x}},$

lesquelles renferment les conditions de l’équilibre des fluides homogènes.

Supposons que le fluide soit composé de différentes couches, dont, chacune soit d’une densité uniforme, et qu’on en cherche l’équation ; soient $x,y,z$ les coordonnées de chacune de ces couches, on aura par hypothèse

${\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} x}}\operatorname {d} x+{\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} y}}\operatorname {d} y+{\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} z}}\operatorname {d} z=0.$

Or les équations

${\frac {\operatorname {d} (\mathrm {D} \Pi )}{\operatorname {d} y}}={\frac {\operatorname {d} (\mathrm {D} \varpi )}{\operatorname {d} x}},\qquad {\frac {\operatorname {d} (\mathrm {D} \Pi )}{\operatorname {d} z}}={\frac {\operatorname {d} (\mathrm {D} \Psi )}{\operatorname {d} x}}$

donnent

${\begin{aligned}\Pi {\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} y}}+\mathrm {D} {\frac {\operatorname {d} \Pi }{\operatorname {d} y}}=&\varpi {\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} x}}+\mathrm {D} {\frac {\operatorname {d} \varpi }{\operatorname {d} x}}\\\Pi {\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} z}}+\mathrm {D} {\frac {\operatorname {d} \Pi }{\operatorname {d} z}}=&\Psi {\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} x}}+\mathrm {D} {\frac {\operatorname {d} \Psi }{\operatorname {d} x}}\,;\end{aligned}}$

substituant dans l’équation ci-dessus les valeurs de ${\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} y}},{\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} z}}$ tirées de celles-ci, et ordonnant les termes, il viendra

${\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} x}}\left(\operatorname {d} x+{\frac {\varpi }{\Pi }}\operatorname {d} y+{\frac {\Psi }{\Pi }}\operatorname {d} z\right)+{\frac {\mathrm {D} }{\Pi }}\left[\left({\frac {\operatorname {d} \varpi }{\operatorname {d} x}}-{\frac {\operatorname {d} \Pi }{\operatorname {d} y}}\right)\operatorname {d} y+\left({\frac {\operatorname {d} \Psi }{\operatorname {d} x}}-{\frac {\operatorname {d} \Pi }{\operatorname {d} z}}\right)\operatorname {d} z\right]=0,$

savoir, en multipliant par ${\frac {\Pi }{\mathrm {D} }},$

${\frac {1}{\mathrm {D} }}{\frac {\operatorname {d} \mathrm {D} }{\operatorname {d} x}}\left(\Pi \operatorname {d} x+\varpi \operatorname {d} y+\Psi \operatorname {d} z\right)+\left({\frac {\operatorname {d} \varpi }{\operatorname {d} x}}-{\frac {\operatorname {d} \Pi }{\operatorname {d} y}}\right)\operatorname {d} y+\left({\frac {\operatorname {d} \Psi }{\operatorname {d} x}}-{\frac {\operatorname {d} \Pi }{\operatorname {d} z}}\right)\operatorname {d} z=0,$

équation qui exprimera la figure de chaque couche où la densité est uniforme.

Si l’on a

${\frac {\operatorname {d} \Pi }{\operatorname {d} y}}={\frac {\operatorname {d} \varpi }{\operatorname {d} x}},\qquad {\frac {\operatorname {d} \Pi }{\operatorname {d} z}}={\frac {\operatorname {d} \Psi }{\operatorname {d} x}},$