Page:Annales de mathématiques pures et appliquées, 1810-1811, Tome 1.djvu/266

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QUESTIONS RÉSOLUES.

Le cas le plus simple que puisse présenter le problème général que nous venons de résoudre, est celui où l’on veut passer de l’hypothèse où $z$ est fonction de $x$ et $y$ à celle où, par exemple, $x$ est fonction de $y$ et $z$ ; on peut poser alors

$x=x,\quad y=u,\quad z=v~;$

d’où

${\begin{array}{lllll}{\tfrac {\mathrm {d} x}{\mathrm {d} u}}={\tfrac {\mathrm {d} x}{\mathrm {d} y}}&,{\tfrac {\mathrm {d} x}{\mathrm {d} v}}={\tfrac {\mathrm {d} x}{\mathrm {d} z}}&,{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} u^{2}}}={\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} y^{2}}}&,{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} u\mathrm {d} v}}={\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} y\mathrm {d} z}}&,{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} v^{2}}}={\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} z^{2}}},\\{\tfrac {\mathrm {d} y}{\mathrm {d} u}}=1&,{\tfrac {\mathrm {d} y}{\mathrm {d} v}}=0&,{\tfrac {\mathrm {d} ^{2}y}{\mathrm {d} u^{2}}}=0&,{\tfrac {\mathrm {d} ^{2}y}{\mathrm {d} u\mathrm {d} v}}=0&,{\tfrac {\mathrm {d} ^{2}y}{\mathrm {d} v^{2}}}=0,\\{\tfrac {\mathrm {d} z}{\mathrm {d} u}}=0&,{\tfrac {\mathrm {d} z}{\mathrm {d} v}}=1&,{\tfrac {\mathrm {d} ^{2}z}{\mathrm {d} u^{2}}}=0&,{\tfrac {\mathrm {d} ^{2}z}{\mathrm {d} u\mathrm {d} v}}=0&,{\tfrac {\mathrm {d} ^{2}z}{\mathrm {d} v^{2}}}=0~;\\\end{array}}$

par suite de quoi les valeurs générales de $p,q,r,s,t,$ deviennent

${\begin{aligned}p&={\tfrac {1}{\tfrac {\mathrm {d} x}{\mathrm {d} z}}},\qquad \qquad q=-{\tfrac {\tfrac {\mathrm {d} x}{\mathrm {d} y}}{\tfrac {\mathrm {d} x}{\mathrm {d} z}}},\\r&=-{\tfrac {\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} z^{2}}}{\left({\tfrac {\mathrm {d} x}{\mathrm {d} z}}\right)^{3}}},\qquad s={\tfrac {{\tfrac {\mathrm {d} x}{\mathrm {d} z}}{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} y\mathrm {d} z}}-{\tfrac {\mathrm {d} x}{\mathrm {d} y}}{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} z^{2}}}}{\left({\tfrac {\mathrm {d} x}{\mathrm {d} z}}\right)^{3}}},\\t&=-{\tfrac {\left({\tfrac {\mathrm {d} x}{\mathrm {d} z}}\right)^{2}{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} y^{2}}}-2{\tfrac {\mathrm {d} x}{\mathrm {d} z}}{\tfrac {\mathrm {d} x}{\mathrm {d} y}}{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} y\mathrm {d} z}}+\left({\tfrac {\mathrm {d} x}{\mathrm {d} y}}\right)^{2}{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} z^{2}}}}{\left({\tfrac {\mathrm {d} x}{\mathrm {d} z}}\right)^{3}}}.&\\\end{aligned}}$