Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/238

IV.

Or on a, en différentiant les valeurs de $u^{2},u'^{2},u''^{2},$

${\begin{aligned}u\ \,du\,\ =&{\frac {dxd^{2}x+dyd^{2}y+dzd^{2}z}{dt^{2}}},\\u'\,du'\,=&{\frac {dx'd^{2}x'+dy'd^{2}y'+dz'd^{2}z'}{dt^{2}}},\\u''du''=&{\frac {dx''d^{2}x''+dy''d^{2}y''+dz''d^{2}z''}{dt^{2}}}\,;\end{aligned}}$

donc, si l’on fait ici les mêmes substitutions que ci-dessus, et qu’on suppose pour un moment

${\begin{aligned}d\mathrm {V} \,\ =&x'\,dx\ +y'\,dy\ +z'\,dz,\\d\mathrm {V} '\,=&x\,dx'\ +y\,dy'\ +z\,dz',\\d\mathrm {V} ''=&x'dx''+y'dy''+z'dz'',\end{aligned}}$

à cause de

$xdx+ydy+zdz=rdr,\quad x'dx'+y'dy'+z'dz'=r'dr',$

$x''dx''+y''dy''+z''dz''=r''dr'',$

on aura

${\begin{aligned}u\ \,du\ \,=&-\left({\frac {\mathrm {A+B} }{r^{3}}}+{\frac {\mathrm {C} }{r''^{3}}}\right)r\ dr\ \ -\mathrm {C} \left({\frac {1}{r'^{3}}}-{\frac {1}{r''^{3}}}\right)d\mathrm {V} ,\\u'\,du'\,=&-\left({\frac {\mathrm {A+C} }{r'^{3}}}+{\frac {\mathrm {B} }{r''^{3}}}\right)r'dr'\ -\mathrm {B} \left({\frac {1}{r^{3}}}-{\frac {1}{r''^{3}}}\right)d\mathrm {V} ',\\u''du''=&-\left({\frac {\mathrm {B+C} }{r''^{3}}}+{\frac {\mathrm {A} }{r^{3}}}\,\ \right)r''dr''-\mathrm {A} \left({\frac {1}{r'^{3}}}-{\frac {1}{r^{3}}}\right)d\mathrm {V} ''.\end{aligned}}$

Soit, pour abréger,

${\begin{aligned}d\mathrm {R} \ \,=&\,\ {\frac {rdr}{r''^{3}}}\ \,+2\left({\frac {1}{r'^{3}}}-{\frac {1}{r''^{3}}}\right)d\mathrm {V} ,\\d\mathrm {R} '\,=&\ {\frac {r'dr'}{r''^{3}}}+2\left({\frac {1}{r^{3}}}\ -{\frac {1}{r''^{3}}}\right)d\mathrm {V} ',\\d\mathrm {R} ''=&{\frac {r''dr''}{r^{3}}}+2\left({\frac {1}{r'^{3}}}-{\frac {1}{r^{3}}}\ \right)d\mathrm {V} '',\end{aligned}}$