Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 9.djvu/133

Dans la question présente, nous avons

$\varepsilon =-2lc,\qquad r=0,\qquad f=6l\qquad {\text{et}}\qquad p=6l\,;$

partant,

${\begin{aligned}\mathrm {C} \Delta =&\left[b\sin(\varphi -\varpi )-a\cos(\varphi -\varpi )\right]{\frac {24}{5}}\pi \Delta l\cos \theta \\&+\left[b'\sin(2\varphi -2\varpi )-a'\cos(2\varphi -2\varpi )\right]{\frac {48}{5}}\pi \Delta l\sin \theta \end{aligned}}$

ou

$\mathrm {C} \Delta ={\frac {24}{5}}\pi \Delta l\left(\cos \theta {\frac {\partial z}{\partial \varpi }}+\sin \theta {\frac {\partial x}{\partial \varpi }}\right)\,;$

donc

$\alpha \mathrm {C} \Delta ={\frac {24}{5}}\pi \alpha \Delta lv\sin \theta \,;$

on aura pareillement

$\mathrm {B} \Delta ={\frac {24}{5}}\pi \Delta lc\sin 2\theta +\int _{0}^{\pi }{\frac {\partial .\mathrm {C} \sin \theta }{\partial \theta }}d\varpi \,;$

d’où l’on tirera facilement

$\alpha \mathrm {B} \Delta ={\frac {24}{5}}\alpha \pi \Delta lu\,;$

les équations (12) et (13) deviendront ainsi

${\begin{aligned}{\frac {d^{2}u}{dt^{2}}}=&lg{\frac {\partial ^{2}u}{\partial \theta ^{2}}}+lg{\frac {\partial ^{2}v}{\partial \varpi \partial \theta }}+lg{\frac {\partial {\cfrac {u\cos \theta }{\sin \theta }}}{\partial \theta }}+{\frac {24}{5}}\pi \Delta lu\\&+\mathrm {K} \left\{\sin 2\theta \left[{\frac {1}{2}}\sin ^{2}\nu -\cos ^{2}\nu +{\frac {1}{2}}\sin ^{2}\nu \cos(2\varphi -2\varpi )\right]\right.\\&\qquad \qquad \qquad \qquad \qquad \qquad \left.+2\cos 2\theta \sin \nu \cos \nu \cos(\varphi -\varpi ){\begin{aligned}\\\\\end{aligned}}\right\}\\\end{aligned}}$

${\begin{aligned}{\frac {d^{2}v}{dt^{2}}}\sin ^{2}\theta =&lg{\frac {\partial ^{2}v}{\partial \varpi ^{2}}}+lg{\frac {\partial ^{2}u}{\partial \varpi \partial \theta }}+lg{\frac {\cos \theta }{\sin \theta }}{\frac {\partial u}{\partial \varpi }}\\&\qquad +{\frac {24}{5}}\pi \Delta lv\sin ^{2}\theta +\mathrm {K} \sin \nu \cos \nu \sin 2\theta \sin(\varphi -\varpi )\\&\qquad \qquad \qquad +\mathrm {K} \sin ^{2}\nu \sin ^{2}\theta \sin(2\varphi -2\varpi )\,;\end{aligned}}$

en intégrant ces équations par la méthode de l’article VII, on parviendra aux deux suivantes :

(19)

${\begin{aligned}0=&{\frac {d^{2}x}{dt^{2}}}+\left(6lg-{\frac {24}{5}}\pi \Delta l\right)x\\&\qquad -\mathrm {K} \left[{\frac {1}{2}}\sin ^{2}\nu -\cos ^{2}\nu +{\frac {1}{2}}\sin ^{2}\nu \cos(2\varphi -2\varpi )\right],\end{aligned}}$

(20)

$0={\frac {d^{2}z}{dt^{2}}}+\left(6lg-{\frac {24}{5}}\pi \Delta l\right)z-2\mathrm {K} \sin \nu \cos \nu \cos(\varphi -\varpi ),$

équations qui ne diffèrent des équations (16) et (17) qu’en ce que $g$ se