Page:Annales de mathématiques pures et appliquées, 1828-1829, Tome 19.djvu/194

 Les corrections sont expliquées en page de discussion

${\begin{array}{lll}0={\frac {\operatorname {d} \delta x}{\operatorname {d} x}}+{\frac {\operatorname {d} \delta X}{\operatorname {d} X}},&0={\frac {\operatorname {d} \delta x}{\operatorname {d} y}}+{\frac {\operatorname {d} \delta Y}{\operatorname {d} X}},&0={\frac {\operatorname {d} \delta x}{\operatorname {d} z}}+{\frac {\operatorname {d} \delta Z}{\operatorname {d} X}},\\\\0={\frac {\operatorname {d} \delta y}{\operatorname {d} x}}+{\frac {\operatorname {d} \delta X}{\operatorname {d} Y}}\,;&0={\frac {\operatorname {d} \delta y}{\operatorname {d} y}}+{\frac {\operatorname {d} \delta Y}{\operatorname {d} Y}},&0={\frac {\operatorname {d} \delta y}{\operatorname {d} z}}+{\frac {\operatorname {d} \delta Z}{\operatorname {d} Y}},\\\\0={\frac {\operatorname {d} \delta z}{\operatorname {d} x}}+{\frac {\operatorname {d} \delta X}{\operatorname {d} Z}}\,;&0={\frac {\operatorname {d} \delta z}{\operatorname {d} y}}+{\frac {\operatorname {d} \delta Y}{\operatorname {d} Z}}\,;&0={\frac {\operatorname {d} \delta z}{\operatorname {d} z}}+{\frac {\operatorname {d} \delta Z}{\operatorname {d} Z}}.\end{array}}$

Comme les différens termes qui entrent dans ces équations sont indépendans les uns des $x,y,z$ et les autres de $X,Y,Z,$ on en conclura sans peine qu’ils sont tous indépendans tant de $x,y,z$ que de $X,Y,Z,$ et que, par conséquent, on doit avoir identiquement

${\begin{array}{lll}{\frac {\operatorname {d} \delta x}{\operatorname {d} x}}={\frac {\operatorname {d} \delta X}{\operatorname {d} X}},&{\frac {\operatorname {d} \delta x}{\operatorname {d} y}}={\frac {\operatorname {d} \delta X}{\operatorname {d} Y}},&{\frac {\operatorname {d} \delta x}{\operatorname {d} z}}={\frac {\operatorname {d} \delta X}{\operatorname {d} Z}},\\\\{\frac {\operatorname {d} \delta y}{\operatorname {d} x}}={\frac {\operatorname {d} \delta Y}{\operatorname {d} X}}\,;&{\frac {\operatorname {d} \delta y}{\operatorname {d} y}}={\frac {\operatorname {d} \delta Y}{\operatorname {d} Y}},&{\frac {\operatorname {d} \delta y}{\operatorname {d} z}}={\frac {\operatorname {d} \delta Y}{\operatorname {d} Z}},\\\\{\frac {\operatorname {d} \delta z}{\operatorname {d} x}}+{\frac {\operatorname {d} \delta Z}{\operatorname {d} X}}\,;&{\frac {\operatorname {d} \delta z}{\operatorname {d} y}}={\frac {\operatorname {d} \delta Z}{\operatorname {d} Y}}\,;&{\frac {\operatorname {d} \delta z}{\operatorname {d} z}}={\frac {\operatorname {d} \delta Z}{\operatorname {d} Z}}\,;\end{array}}$

dont la comparaison avec celles qui précédent nous conduit à faire

${\begin{array}{ll}{\frac {\operatorname {d} \delta x}{\operatorname {d} x}}=0,&{\frac {\operatorname {d} \delta y}{\operatorname {d} z}}=-{\frac {\operatorname {d} \delta z}{\operatorname {d} y}}=\lambda ,\\\\{\frac {\operatorname {d} \delta y}{\operatorname {d} y}}=0,&{\frac {\operatorname {d} \delta z}{\operatorname {d} x}}=-{\frac {\operatorname {d} \delta x}{\operatorname {d} z}}=\mu ,\\\\{\frac {\operatorname {d} \delta z}{\operatorname {d} z}}=0,&{\frac {\operatorname {d} \delta x}{\operatorname {d} y}}=-{\frac {\operatorname {d} \delta y}{\operatorname {d} x}}=\nu ,\\\\\end{array}}$