Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/302

XXXVII.

On trouvera ensuite

${\begin{aligned}\mathrm {H} \ \,=&\mathrm {L} r^{2}-\left({\frac {d\left(r^{2}\right)}{2dt}}\right)^{2}-{\frac {i\mathrm {D} \left(15z^{3}-9r^{2}z\right)r^{2}}{2\mathrm {R} ^{4}}}-\ldots ,\\\mathrm {H} '\,=&{\frac {1}{i^{4}}}\left[\mathrm {MR} ^{2}-\left({\frac {d\left(\mathrm {R} ^{2}\right)}{2dt}}\right)^{2}\right]+\ldots ,\\\mathrm {H} ''=&{\frac {1}{i^{4}}}\left[\mathrm {MR} ^{2}-\left({\frac {d\left(\mathrm {R} ^{2}\right)}{2dt}}\right)^{2}\right]\\&+{\frac {1}{i^{3}}}\left({\frac {\mathrm {BR} ^{3}z}{r^{3}}}-\mathrm {NR^{2}-2MR} z+{\frac {d(\mathrm {R} ^{2})d(\mathrm {R} z)}{dt^{2}}}\right)+\ldots ,\end{aligned}}$

et, à cause de

${\begin{aligned}v\ \,=&{\frac {\mathrm {M} }{i^{2}}}+{\frac {1}{i}}\left({\frac {\mathrm {BR} z}{r^{3}}}-{\frac {\mathrm {N} }{2}}\right)+\ldots ,\\v'\,=&-{\frac {\mathrm {N} }{2i}}+\mathrm {L} -{\frac {\mathrm {B} }{2r}}+{\frac {3\mathrm {D} \left(3z^{2}-r^{2}\right)}{4\mathrm {R} ^{3}}}+\ldots ,\\v''=&{\frac {\mathrm {N} }{2i}}+{\frac {\mathrm {B} }{2r}}-{\frac {3\mathrm {D} \left(3z^{2}-r^{2}\right)}{4\mathrm {R} ^{3}}}+\ldots ,\end{aligned}}$

on aura de même (Article XXII)

${\begin{aligned}\Psi \ \,=&{\frac {1}{i^{4}}}\left[\mathrm {MR} ^{2}-\left({\frac {d\left(\mathrm {R} ^{2}\right)}{2dt}}\right)^{2}\right]\\&+{\frac {1}{i^{3}}}\left[{\frac {\mathrm {BR} ^{3}z}{r^{3}}}-\mathrm {{\frac {NR^{2}}{2}}-MR} z+{\frac {d\left(\mathrm {R} ^{2}\right)d(\mathrm {R} z)}{dt^{2}}}\right]+\ldots ,\\\Psi '\,=&{\frac {1}{i^{2}}}\left[{\frac {\mathrm {NR} z}{2}}-\left({\frac {d(\mathrm {R} z)}{2dt}}\right)^{2}+{\frac {\sigma ^{2}}{4}}\right]\\&+{\frac {1}{i}}\left[-\left(\mathrm {L} -{\frac {\mathrm {B} }{2r}}+{\frac {3\mathrm {D} \left(3z^{2}-r^{2}\right)}{4\mathrm {R} ^{3}}}\right)\mathrm {R} z-{\frac {\mathrm {N} r^{2}}{2}}+{\frac {d(\mathrm {R} z)d\left(r^{2}\right)}{2dt^{2}}}\right]+\ldots ,\\\Psi ''=&{\frac {1}{i^{2}}}\left[{\frac {\mathrm {NR} z}{2}}-\left({\frac {d(\mathrm {R} z)}{2dt}}\right)^{2}+{\frac {\sigma ^{2}}{4}}\right]+{\frac {1}{i}}\left({\frac {\mathrm {B} }{2r}}-{\frac {3\mathrm {D} \left(3z^{2}-r^{2}\right)}{4\mathrm {R} ^{3}}}\right)\mathrm {R} z+\ldots .\end{aligned}}$

Donc, si l’on substitue ces valeurs dans les expressions de ${\frac {d\varphi }{dt}}$ et ${\frac {d\varphi '}{dt}},$