Page:Joseph Louis de Lagrange - Œuvres, Tome 5.djvu/407

par les formules de ce dernier numéro

${\begin{alignedat}{2}&\ [r,r']&=&\left(r^{2}+r'^{2}\right)(r,r')-rr'(r,r')_{1},\\&\ [r,r']_{1}&=&4rr'(r,r')-\left(r^{2}+r'^{2}\right)(r,r')_{1},\\&{\frac {d[r,r']}{dr}}&=&-r(r,r')+{\frac {1}{2}}r'(r,r')_{1},\\&{\frac {d[r,r']_{1}}{dr}}&=&-2r'(r,r')+{\frac {r'^{2}}{r}}(r,r')_{1},\\&{\frac {d^{2}[r,r']}{dr^{2}}}&=&2(r,r')-{\frac {r'}{2r}}(r,r')_{1},\\&{\frac {d^{2}[r,r']_{1}}{dr^{2}}}&=&{\frac {6r'}{r}}(r,r')-{\frac {2r'^{2}}{r^{2}}}(r,r')_{1},\end{alignedat}}$

les quantités $(r,r'),(r,r')_{1}$ étant les mêmes que nous avons déjà employées ci-dessus.

Différentiant les deux dernières équations par $r,$ on aura

${\begin{aligned}{\frac {d^{3}[r,r']}{dr^{3}}}\ \ =&2{\frac {d(r,r')}{dr}}+{\frac {r'}{2r}}{\frac {d(r,r')_{1}}{dr}},\\{\frac {d^{3}[r,r']_{1}}{dr^{3}}}=&-{\frac {6r'}{r^{2}}}(r,r')+{\frac {6r'}{r}}{\frac {d(r,r')}{dr}}+{\frac {4r'^{2}}{r^{3}}}(r,r')_{1}-{\frac {2r'^{2}}{r^{2}}}{\frac {d(r,r')}{dr}}.\end{aligned}}$

Or, si dans les formules générales du même numéro cité on fait $s={\frac {3}{2}},$ les quantités $\mathrm {A,B}$ deviennent $(r,r'),(r,r')_{1}\,;$ ainsi l’on aura par ces formules

${\begin{aligned}{\frac {d(r,r')}{dr}}\ \ =&-{\frac {6r(r,r')+r'(r,r')_{1}}{2\left(r^{2}-r'^{2}\right)}},\\{\frac {d(r,r')_{1}}{dr}}=&-{\frac {6r'(r,r')+\left(2r-{\cfrac {r'^{2}}{r}}\right)(r,r')_{1}}{r^{2}-r'^{2}}}\,;\end{aligned}}$

de sorte que par ces substitutions les formules précédentes deviendront

${\begin{aligned}{\frac {d^{3}[r,r']}{dr^{3}}}\ \ =&{\frac {\left(-6r+{\cfrac {3r'^{2}}{r}}\right)(r,r')+\left({\cfrac {r'}{2}}-{\cfrac {r'^{3}}{r^{2}}}\right)(r,r')_{1}}{r^{2}-r'^{2}}},\\{\frac {d^{3}[r,r']_{1}}{dr^{3}}}=&{\frac {\left(-24r'+{\cfrac {18r'^{3}}{r^{2}}}\right)(r,r')+\left({\cfrac {5r'^{2}}{r}}-{\cfrac {6r'^{4}}{r^{3}}}\right)(r,r')_{1}}{r^{2}-r'^{2}}}.\end{aligned}}$