Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/241

d’où

${\begin{alignedat}{3}d\mathrm {R} \ \ =&{\frac {2rdr}{r''^{3}}}&+&q'&(dp''+d\rho ),\\d\mathrm {R} '\ =&{\frac {2r'dr'}{r''^{3}}}&+&q'&(dp''-d\rho ),\\d\mathrm {R} ''\,=&{\frac {2r''dr''}{r^{3}}}&+&q''&(dp\ -d\rho ).\end{alignedat}}$

Mais

${\frac {1}{r''^{3}}}={\frac {1}{r^{3}}}-q',\quad 2rdr=dp'+dp''\,;$

donc

${\frac {2rdr}{r''^{3}}}={\frac {2dr}{r^{2}}}-q'(dp'+dp'')\,;$

on trouvera de même

${\begin{aligned}{\frac {2r'\,dr'\,}{r''^{3}}}=&{\frac {2dr'}{r'^{2}}}\,-q\ (dp\,+dp''),\\{\frac {2r''dr''}{r^{3}}}=&{\frac {2dr''}{r''^{2}}}+q'(dp'+dp)\,;\end{aligned}}$

de sorte qu’en substituant ces valeurs, et faisant pour plus de simplicité

(I)

$\left\{{\begin{aligned}d\mathrm {Q} \ \,=&q'dp'-q''dp''-qd\rho ,\\d\mathrm {Q} '\,=&q\ \,dp\,+q''dp''+q'd\rho ,\\d\mathrm {Q} ''=&-qdp+q'dp'+q''d\rho ,\end{aligned}}\right.$

on aura

$\mathrm {R} =-{\frac {2}{r}}-\mathrm {Q} ,\quad \mathrm {R} '=-{\frac {2}{r'}}-\mathrm {Q} ',\quad \mathrm {R} ''=-{\frac {2}{r''}}-\mathrm {Q} '',$

et de là

(J)

$\left\{{\begin{aligned}u^{2}\ \,=&{\frac {2(\mathrm {A+B+C} )}{r}}+\mathrm {CQ} ,\\u'^{2}\,=&{\frac {2(\mathrm {A+B+C} )}{r'}}+\mathrm {BQ'} ,\\u''^{2}=&{\frac {2(\mathrm {A+B+C} )}{r''}}+\mathrm {AQ''} .\end{aligned}}\right.$