Page:Joseph Louis de Lagrange - Œuvres, Tome 12.djvu/97

89

SECONDE PARTIE. — SECTION VII.

${\begin{aligned}\delta X'=&-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\delta \theta }{1-e}}={\frac {\mathrm {g} }{a^{2}}}{\frac {dc}{(1-e)^{2}}},\\\delta Y'=&{\sqrt {\frac {\mathrm {g} }{a}}}{\sqrt {1-e^{2}}}{\frac {de}{(1-e)^{2}}}+{\frac {d\left({\sqrt {\cfrac {\mathrm {g} }{a}}}{\sqrt {1-e^{2}}}\right)}{1-e}}\\=&d\left({\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\sqrt {1-e^{2}}}{1-e}}\right)={\frac {\sqrt {1-e^{2}}}{1-e}}d{\sqrt {\frac {\mathrm {g} }{a}}}+{\sqrt {\frac {\mathrm {g} }{a}}}{\frac {de}{(1-e){\sqrt {1-e^{2}}}}}.\end{aligned}}$

69. Ici nous avons conservé la quantité $e,$ qui est l’excentricité ; mais si, à sa place, on veut employer le paramètre $b=a\left(1-e^{2}\right),$ on aura, par la différentiation,

$de={\frac {\left(1-e^{2}\right)da-db}{2ae}},$

et les expressions de $\delta X$ et de $\delta Y',$ qui contiennent $de,$ deviendront

${\begin{aligned}\delta X=&{\frac {-\left(1-e^{2}\right)da+db}{2e}},\\\delta Y'=&{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\sqrt {1-e^{2}}}{2ae}}da-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {db}{2ae(1-e){\sqrt {1-e^{2}}}}}.\end{aligned}}$

De là nous aurons les différences partielles

${\begin{alignedat}{2}{\frac {\partial X}{\partial a}}\ =&-{\frac {\left(1-e^{2}\right)}{2e}},&{\frac {\partial X}{\partial b}}\ =&{\frac {1}{2e}},\quad {\frac {\partial X}{\partial c}}\ =0,\\{\frac {\partial Y}{\partial a}}\ =&0,&{\frac {\partial Y}{\partial b}}\ =&0,\qquad {\frac {\partial Y}{\partial c}}\ =-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\sqrt {1-e^{2}}}{1-e}},\\{\frac {\partial X'}{\partial a}}=&0,&{\frac {\partial X'}{\partial b}}=&0,\qquad {\frac {\partial X'}{\partial c}}={\frac {\mathrm {g} }{a^{2}}}\ {\frac {1}{(1-e)^{2}}},\\{\frac {\partial Y'}{\partial a}}=&{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\sqrt {1-e^{2}}}{2ae}},\quad &{\frac {\partial Y'}{\partial b}}=&-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {1}{2ae(1-e){\sqrt {1-e^{2}}}}},&\quad {\frac {\partial Y'}{\partial c}}=0,\end{alignedat}}$

et l’on trouvera, par la substitution de ces valeurs dans les expressions des symboles $[a,b],\,[a,c],\,[b,c]$ de l’article 67,

${\begin{aligned}\,[a,b]=&0,\\\,[a,c]=&-{\frac {g}{2a^{2}e}}+{\frac {g\left(1-e^{2}\right)}{2a^{2}e(1-e)}}={\frac {g}{2a^{2}}},\\\,[b,c]=&{\frac {g}{2a^{2}e}}\left[{\frac {1}{\left(1-e^{2}\right)}}-{\frac {\sqrt {1-e^{2}}}{(1-e)^{2}{\sqrt {1-e^{2}}}}}\right]=0.\end{aligned}}$