Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/788

${\begin{alignedat}{5}+&d{\frac {d^{2}\mathrm {R} }{dr'du'}}&&(\Delta r\,d\delta u'&&-\delta r\,d\Delta u'&&+\Delta u\,d\delta r'-\delta u\,d\Delta r')\\+&d{\frac {d^{2}\mathrm {R} }{ds'du'}}&&(\Delta s\,d\delta u'&&-\delta s\,d\Delta u'&&+\Delta u\,d\delta s'-\delta u\,d\Delta s')\\+&d{\frac {d^{2}\mathrm {R} }{dr\,dr'}}&&(\Delta r\,d\delta r&&-\delta r\,d\Delta r&&-\Delta r\delta r'dt+\delta r\Delta r'dt)\\+&d{\frac {d^{2}\mathrm {R} }{ds\,dr'}}&&(\Delta r\,d\delta s&&-\delta r\,d\Delta s&&-\Delta s\delta r'dt+\delta s\Delta r'dt)\\+&d{\frac {d^{2}\mathrm {R} }{du\,dr'}}&&(\Delta r\,d\delta u&&-\delta r\,d\Delta u&&-\Delta u\delta r'dt+\delta u\Delta r'dt)\\+&d{\frac {d^{2}\mathrm {R} }{dr\,ds'}}&&(\Delta s\,d\delta r&&-\delta s\,d\Delta r&&-\Delta r\delta s'dt+\delta r\Delta s'dt)\\+&d{\frac {d^{2}\mathrm {R} }{ds\,ds'}}&&(\Delta s\,d\delta s&&-\delta s\,d\Delta s&&-\Delta s\delta s'dt+\delta s\Delta s'dt)\\+&d{\frac {d^{2}\mathrm {R} }{du\,ds'}}&&(\Delta s\,d\delta u&&-\delta s\,d\Delta u&&-\Delta u\delta s'dt+\delta u\Delta s'dt)\\+&d{\frac {d^{2}\mathrm {R} }{dr\,du'}}&&(\Delta u\,d\delta r&&-\delta u\,d\Delta r&&-\Delta r\delta u'dt+\delta r\Delta u'dt)\\+&d{\frac {d^{2}\mathrm {R} }{ds\,du'}}&&(\Delta u\,d\delta s&&-\delta u\,d\Delta s&&-\Delta s\delta u'dt+\delta s\Delta u'dt)\\+&d{\frac {d^{2}\mathrm {R} }{du\,du'}}&&(\Delta u\,d\delta u&&-\delta u\,d\Delta u&&-\Delta u\delta u'dt+\delta u\Delta u'dt)=0,\end{alignedat}}$

15. Si maintenant on fait attention que $r'dt=dr,$ et par conséquent

$\delta r'dt=\delta dr=d\delta r,$

à cause de l’indépendance des caractéristiques $d$ et $\delta ,$ et par la même raison

$\delta s'dt=d\delta s,\quad \delta u'dt=d\delta u,$

ainsi que

$\Delta r'dt=d\Delta r,\quad \Delta s'dt=d\Delta s,\quad \Delta u'dt=d\Delta u,$

on verra d’abord que les coefficients de ${\frac {d^{2}\mathrm {R} }{drdr'}},{\frac {d^{2}\mathrm {R} }{dsds'}},{\frac {d^{2}\mathrm {R} }{dudu'}}$ se détruisent