Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/252

$\operatorname {d} '^{2}x\delta x+\operatorname {d} '^{2}y\delta y+\operatorname {d} '^{2}z\delta z+\operatorname {d} 'x\operatorname {d} '\delta x+\operatorname {d} 'y\operatorname {d} '\delta y+\operatorname {d} 'z\operatorname {d} '\delta z=\operatorname {d} '\psi ,$

ou encore

$\operatorname {d} '^{2}x\delta x+\operatorname {d} '^{2}y\delta y+\operatorname {d} '^{2}z\delta z=\operatorname {d} '\psi -{\frac {1}{2}}\delta \left(\operatorname {d} 'x^{2}+\operatorname {d} 'y^{2}+\operatorname {d} 'z^{2}\right)\,;$

en observant que

${\begin{aligned}&\operatorname {d} 'x\operatorname {d} '\delta x=\operatorname {d} 'x\delta \operatorname {d} 'x={\frac {1}{2}}\delta .\operatorname {d} 'x^{2},\\\\&\operatorname {d} 'y\operatorname {d} '\delta y=\operatorname {d} 'y\delta \operatorname {d} 'y={\frac {1}{2}}\delta .\operatorname {d} 'y^{2},\\\\&\operatorname {d} 'z\operatorname {d} '\delta z=\operatorname {d} 'z\delta \operatorname {d} 'z={\frac {1}{2}}\delta .\operatorname {d} 'z^{2}.\end{aligned}}$

Au moyen de cette dernière valeur de la quantité

$\operatorname {d} '^{2}x\delta x+\operatorname {d} '^{2}y\delta y+\operatorname {d} '^{2}z\delta z,$

l’équation (9) deviendra

$\delta \nu -{\frac {\delta p}{\Delta }}=\operatorname {d} '\psi -{\frac {1}{2}}\delta \left(\operatorname {d} 'x^{2}+\operatorname {d} 'y^{2}+\operatorname {d} 'z^{2}\right)\,;\qquad$(11)

de sorte que, si l’on détermine $\phi$ au moyen de l’équation

$\nu -\int {\frac {\delta p}{\Delta }}=\operatorname {d} '\phi -{\frac {1}{2}}\left(\operatorname {d} 'x^{2}+\operatorname {d} 'y^{2}+\operatorname {d} 'z^{2}\right),\qquad$(12)

ce qui est toujours possible, au moins par les quadratures, et ce qui donne

$\delta \nu -{\frac {\delta p}{\Delta }}=\delta \operatorname {d} '\phi -{\frac {1}{2}}\delta \left(\operatorname {d} 'x^{2}+\operatorname {d} 'y^{2}+\operatorname {d} 'z^{2}\right),$

on trouvera, en comparant cette dernière avec l’équation (11),