Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/793

Comme cette expression de $\mathrm {C}$ résulte de celle de $\mathrm {B} ,$ en y changeant simplement $b$ en $c,$ on aura pareillement celles de $\mathrm {F,G,H} ,$ en changeant successivement $b,$ en $f,g,h.$

19. Ainsi la valeur de ${\frac {d\Omega }{da}}dt$ (10) deviendra, par ces substitutions,

${\begin{aligned}{\frac {d\Omega }{da}}dt&=(a,b)db+(a,c)dc+(a,f)df+(a,g)dg+(a,h)dh\\+&\left({\frac {d^{2}\mathrm {R} }{dr\,ds'}}-{\frac {d^{2}\mathrm {R} }{ds\,dr'}}\right){\frac {dr}{da}}\,\left({\frac {ds}{db}}db\,+{\frac {ds}{dc}}dc+{\frac {ds}{df}}df+{\frac {ds}{dg}}dg+{\frac {ds}{dh}}dh\right)\\-&\left({\frac {d^{2}\mathrm {R} }{dr\,ds'}}-{\frac {d^{2}\mathrm {R} }{ds\,dr'}}\right){\frac {ds}{da}}\,\left({\frac {dr}{db}}db\,+{\frac {dr}{dc}}dc+{\frac {dr}{df}}df+{\frac {dr}{dg}}dg+{\frac {dr}{dh}}dh\right)\\+&\left({\frac {d^{2}\mathrm {R} }{dr\,du'}}-{\frac {d^{2}\mathrm {R} }{du\,dr'}}\right){\frac {dr}{da}}\left({\frac {du}{db}}db+{\frac {du}{dc}}dc+{\frac {du}{df}}df+{\frac {du}{dg}}dg+{\frac {du}{dh}}dh\right)\\-&\left({\frac {d^{2}\mathrm {R} }{dr\,du'}}-{\frac {d^{2}\mathrm {R} }{du\,dr'}}\right){\frac {du}{da}}\left({\frac {dr}{db}}db+{\frac {dr}{dc}}dc+{\frac {dr}{df}}df+{\frac {dr}{dg}}dg+{\frac {dr}{dh}}dh\right)\\+&\left({\frac {d^{2}\mathrm {R} }{ds\,du'}}-{\frac {d^{2}\mathrm {R} }{du\,ds'}}\right){\frac {ds}{da}}\left({\frac {du}{db}}db+{\frac {du}{dc}}dc+{\frac {du}{df}}df+{\frac {du}{dg}}dg+{\frac {du}{dh}}dh\right)\\-&\left({\frac {d^{2}\mathrm {R} }{ds\,du'}}-{\frac {d^{2}\mathrm {R} }{du\,ds'}}\right){\frac {du}{da}}\left({\frac {ds}{db}}db+{\frac {ds}{dc}}dc+{\frac {ds}{df}}df+{\frac {ds}{dg}}dg+{\frac {ds}{dh}}dh\right).\end{aligned}}$

20. Si maintenant on se rappelle que les équations $\delta r=0,\delta s=0,\delta u=0$ du n^o 7 donnent

${\begin{alignedat}{6}&{\frac {dr}{db}}db&&+{\frac {dr}{dc}}dc&&+{\frac {dr}{df}}df&&+{\frac {dr}{dg}}dg&&+{\frac {dr}{dh}}dh&&=-{\frac {dr}{da}}da,\\&{\frac {ds}{db}}db&&+{\frac {ds}{dc}}dc&&+{\frac {ds}{df}}df&&+{\frac {ds}{dg}}dg&&+{\frac {ds}{dh}}dh&&=-{\frac {ds}{da}}da,\\&{\frac {du}{db}}db&&+{\frac {du}{dc}}dc&&+{\frac {du}{df}}df&&+{\frac {du}{dg}}dg&&+{\frac {du}{dh}}dh&&=-{\frac {du}{da}}da,\end{alignedat}}$

on voit tout de suite que cette expression de ${\frac {d\Omega }{da}}dt$ se réduit à la forme très-simple

${\frac {d\Omega }{da}}dt=(a,b)db+(a,c)dc+(a,f)df+(a,g)dg+(a,h)dh\,;$