Page:Annales de mathématiques pures et appliquées, 1813-1814, Tome 4.djvu/188

178

PROBLÈMES

$2\varpi F^{2}\left({\frac {\operatorname {d} ^{2}t}{\operatorname {d} \lambda \operatorname {d} \mu }}\right)=F\left({\frac {\operatorname {d} M}{\operatorname {d} \mu }}\right)-M\left({\frac {\operatorname {d} F}{\operatorname {d} \mu }}\right)=-F\left({\frac {\operatorname {d} N}{\operatorname {d} \lambda }}\right)+N\left({\frac {\operatorname {d} F}{\operatorname {d} \lambda }}\right),$

$2\varpi F^{2}\left({\frac {\operatorname {d} ^{2}t}{\operatorname {d} \mu ^{2}}}\right)=-\left({\frac {\operatorname {d} N}{\operatorname {d} \mu }}\right)+N\left({\frac {\operatorname {d} F}{\operatorname {d} \mu }}\right).$

35. Ainsi donc, pour trouver les coefficiens $C,C',C'',$ de nos termes du second ordre, il faudra voir ce que deviennent ces rapports différentiels partiels, dans le cas de $\lambda =0,\mu =0.$ On tire de cette supposition $P=1,\ Q-1,$$\phi ={\frac {2\varpi A}{p}},\ \psi ={\frac {2\varpi A}{q}}\,;$ et en continuant, par abréviation, d’employer les lettres $\phi$ et $\psi$ à la place de leurs valeurs, on aura, dans la même supposition de $\lambda =0,\ \mu =0$,

$F=q-p,\qquad M=2pq\operatorname {Sin} .\phi ,\qquad N=2pq\operatorname {Sin} .\psi ,$

$\left({\frac {\operatorname {d} \phi }{\operatorname {d} \lambda }}\right)={\frac {2p\operatorname {Sin} .\phi }{q-p}},\qquad \left({\frac {\operatorname {d} \psi }{\operatorname {d} \mu }}\right)=-{\frac {2q\operatorname {Sin} .\psi }{q-p}}\,;$

et ensuite

$\left({\frac {\operatorname {d} F}{\operatorname {d} \lambda }}\right)=-2q\operatorname {Cos} .\phi ,\qquad \left({\frac {\operatorname {d} F}{\operatorname {d} \mu }}\right)=+2p\operatorname {Cos} .\psi \,;$

$\left({\frac {\operatorname {d} M}{\operatorname {d} \lambda }}\right)=-{\frac {pq(q-5p)\operatorname {Sin} .\phi \operatorname {Cos} .\phi }{q-p}},\quad \left({\frac {\operatorname {d} M}{\operatorname {d} \mu }}\right)=-{\frac {4pq^{2}\operatorname {Cos} .\phi \operatorname {Sin} .\psi }{q-p}}\,;$

$\left({\frac {\operatorname {d} N}{\operatorname {d} \mu }}\right)=-{\frac {pq(5q-p)\operatorname {Sin} .\psi \operatorname {Cos} .\psi }{q-p}},\quad \left({\frac {\operatorname {d} N}{\operatorname {d} \lambda }}\right)=+{\frac {4p^{2}q\operatorname {Cos} .\psi \operatorname {Sin} .\phi }{q-p}}.$

36. De là on pourra passer immédiatement aux rapports différentiels du second ordre ${\frac {\operatorname {d} ^{2}t}{\operatorname {d} \lambda ^{2}}},{\frac {\operatorname {d} ^{2}t}{\operatorname {d} \lambda \operatorname {d} \mu }},{\frac {\operatorname {d} ^{2}t}{\operatorname {d} \mu ^{2}}}$. On aura, toujours dans le cas de $\lambda =0,\ \mu =0,$

${\frac {\operatorname {d} ^{2}t}{\operatorname {d} \lambda ^{2}}}={\frac {pq(5p+3q)}{2\varpi (q-p)^{2}}}\operatorname {Sin} .\phi \operatorname {Cos} .\phi ,$

${\frac {\operatorname {d} ^{2}t}{\operatorname {d} \lambda \operatorname {d} \mu }}=-{\frac {4pq}{2\varpi (q-p)^{2}}}(p\operatorname {Sin} .\phi \operatorname {Cos} .\psi +q\operatorname {Sin} .\psi \operatorname {Cos} .\phi ),$