Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/239

et l’on aura

${\begin{aligned}u^{2}\ \,=&{\frac {2(\mathrm {A+B} )}{r}}-\mathrm {CR} ,\\u'^{2}\,=&{\frac {2(\mathrm {A+C} )}{r'}}-\mathrm {BR'} ,\\u''^{2}=&{\frac {2(\mathrm {B+C} )}{r''}}-\mathrm {AR''} \,;\end{aligned}}$

de sorte que les équations $(\mathrm {F} )$ deviendront

(G)

$\left\{{\begin{aligned}{\frac {d^{2}(r^{2})}{2dt^{2}}}&-{\frac {\mathrm {A+B} }{r}}\\&+\mathrm {C} \left[{\frac {r^{2}}{r''^{3}}}+{\frac {1}{2}}\left({\frac {1}{r'^{3}}}-{\frac {1}{r''^{3}}}\right)\left(r^{2}+r'^{2}-r''^{2}\right)+\mathrm {R} \right]=0,\\{\frac {d^{2}(r'^{2})}{2dt^{2}}}&-{\frac {\mathrm {A+C} }{r'}}\\&+\mathrm {B} \left[{\frac {r'^{2}}{r''^{3}}}+{\frac {1}{2}}\left({\frac {1}{r^{3}}}-{\frac {1}{r''^{3}}}\right)\left(r^{2}+r'^{2}-r''^{2}\right)+\mathrm {R} '\right]=0,\\{\frac {d^{2}(r''^{2})}{2dt^{2}}}&-{\frac {\mathrm {B+C} }{r''}}\\&+\mathrm {A} \left[{\frac {r''^{2}}{r^{3}}}+{\frac {1}{2}}\left({\frac {1}{r'^{3}}}-{\frac {1}{r^{3}}}\right)\left(r'^{2}+r''^{2}-r^{2}\right)+\mathrm {R} ''\right]=0\,;\end{aligned}}\right.$

et il ne restera plus qu’à trouver les valeurs de

$d\mathrm {V} ,\quad d\mathrm {V} ',\quad d\mathrm {V} ''.$

Pour cela, je fais

$d\rho =x'dx+y'dy+z'dz-xdx'-ydy'-zdz',$

et comme l’on a

$xx'+yy'+zz'={\frac {r^{2}+r'^{2}-r''^{2}}{2}},$

on aura, en différentiant,

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

donc

${\begin{aligned}d\mathrm {V} \,=&{\frac {rdr+r'dr'-r''dr''+d\rho }{2}},\\d\mathrm {V} '=&{\frac {rdr+r'dr'-r''dr''-d\rho }{2}},\end{aligned}}$

et ensuite

$d\mathrm {V} ''=r'dr'-d\mathrm {V} ,$