Page:Joseph Louis de Lagrange - Œuvres, Tome 2.djvu/29

et j’en prends les différentielles logarithmiques, en regardant toujours $dt$ comme constante ; j’ai

${\begin{aligned}{\frac {d^{2}x}{dx}}=&\mathrm {\frac {X'}{2X}} dx-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dx}}dx-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dy}}dy,\\{\frac {d^{2}y}{dy}}=&\mathrm {\frac {Y'}{2Y}} dy-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dx}}dx-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dy}}dy,\end{aligned}}$

donc

${\begin{aligned}d^{2}x=&\left(\mathrm {\frac {X'}{2X}} -{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dx}}\right)dx^{2}-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dy}}dxdy,\\d^{2}y=&\left(\mathrm {\frac {Y'}{2Y}} -{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dy}}\right)dy^{2}-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dx}}dxdy,\end{aligned}}$

ou bien, en mettant $\mathrm {\frac {X}{T^{2}}} dt^{2}$ au lieu de $dx^{2},$ et $\mathrm {\frac {Y}{T^{2}}} dt^{2}$ au lieu de $dy^{2},$

${\begin{aligned}d^{2}x=&{\frac {1}{2}}{\frac {d\left(\mathrm {\cfrac {X}{T^{2}}} \right)}{dx}}dt^{2}-{\frac {d\log \mathrm {T} }{dy}}dxdy,\\d^{2}y=&{\frac {1}{2}}{\frac {d\left(\mathrm {\cfrac {Y}{T^{2}}} \right)}{dy}}dt^{2}-{\frac {d\log \mathrm {T} }{dx}}dxdy.\end{aligned}}$

Soit maintenant $\mathrm {Z}$ une fonction quelconque de $x$ et de $y,$ et supposons

$d\mathrm {Z} =\mathrm {P} dx+\mathrm {Q} dy,$

on aura, en différentiant de nouveau,

$d^{2}\mathrm {Z} =\mathrm {P} d^{2}x+\mathrm {Q} d^{2}y+{\frac {d\mathrm {P} }{dx}}dx^{2}+2{\frac {d\mathrm {P} }{dy}}dxdy+{\frac {d\mathrm {Q} }{dy}}dy^{2}\,;$

donc, substituant au lieu de $d^{2}x,d^{2}y,dx^{2}$ et $dy^{2}$ leurs valeurs, on aura

${\begin{aligned}d^{2}Z=&\left[{\frac {1}{2}}\mathrm {P} {\frac {d\left(\mathrm {\cfrac {X}{T^{2}}} \right)}{dx}}+\mathrm {\frac {X}{T^{2}}} {\frac {d\mathrm {P} }{dx}}\right]dt^{2}+\left[{\frac {1}{2}}\mathrm {Q} {\frac {d\left(\mathrm {\cfrac {Y}{T^{2}}} \right)}{dy}}+\mathrm {\frac {Y}{T^{2}}} {\frac {d\mathrm {Q} }{dy}}\right]dt^{2}\\&+\left[2{\frac {d\mathrm {P} }{dy}}-\mathrm {P} {\frac {d\log \mathrm {T} }{dy}}-\mathrm {Q} {\frac {d\log \mathrm {T} }{dx}}\right]dxdy.\end{aligned}}$

Donc, si l’on suppose

(H)

$2{\frac {d\mathrm {P} }{dy}}-\mathrm {P} {\frac {d\log \mathrm {T} }{dy}}-\mathrm {Q} {\frac {d\log \mathrm {T} }{dx}}=0$