Page:Joseph Louis de Lagrange - Œuvres, Tome 14.djvu/214

Pour cet effet, considérant une fonction quelconque de $t$ et $x$ qui soit $v,$ puisque

$dv={\frac {\partial v}{\partial t}}dt+{\frac {\partial v}{\partial x}}dx$

et par les nouvelles variables

$dv={\frac {\partial v}{\partial p}}dp+{\frac {\partial v}{\partial q}}dq,$

en substitutuant ici pour $dp$ et $dq$ leur valeur, j'aurai

${\begin{aligned}dv=&\alpha {\frac {\partial v}{\partial p}}dx+\beta {\frac {\partial v}{\partial p}}dt+\gamma {\frac {\partial v}{\partial q}}dx+\delta {\frac {\partial v}{\partial q}}dt,\\=&\left(\alpha {\frac {\partial v}{\partial p}}+\gamma {\frac {\partial v}{\partial q}}\right)dx+\left(\beta {\frac {\partial v}{\partial p}}+\delta {\frac {\partial v}{\partial q}}\right)dt\,;\end{aligned}}$

d’où il s'ensuit, pour les substitution dont j’ai besoin,

${\frac {\partial v}{\partial t}}=\beta {\frac {\partial v}{\partial p}}+\delta {\frac {\partial v}{\partial q}}\qquad {\text{et}}\qquad {\frac {\partial v}{\partial x}}=\alpha {\frac {\partial v}{\partial p}}+\gamma {\frac {\partial v}{\partial q}}\,;$

donc

${\begin{alignedat}{2}{\frac {\partial z}{\partial t}}=&\beta {\frac {\partial z}{\partial p}}+\delta {\frac {\partial z}{\partial q}}\qquad {\text{et}}\qquad &{\frac {\partial ^{2}z}{\partial t^{2}}}=&\beta ^{2}{\frac {\partial ^{2}z}{\partial p^{2}}}+2\beta \delta {\frac {\partial ^{2}z}{\partial p\partial q}}+\delta ^{2}{\frac {\partial ^{2}z}{\partial q^{2}}},\\{\frac {\partial z}{\partial x}}=&\alpha {\frac {\partial z}{\partial p}}+\gamma {\frac {\partial z}{\partial q}}\qquad {\text{et}}\qquad &{\frac {\partial ^{2}z}{\partial x^{2}}}=&\alpha ^{2}{\frac {\partial ^{2}z}{\partial p^{2}}}+2\alpha \gamma {\frac {\partial ^{2}z}{\partial p\partial q}}+\gamma ^{2}{\frac {\partial ^{2}z}{\partial q^{2}}}.\end{alignedat}}$

Maintenant je pose

$\beta ^{2}-\alpha ^{2}a=0\qquad {\text{et}}\qquad \delta ^{2}-\gamma ^{2}a=0$

ou

$\beta =\alpha {\sqrt {a}}\qquad {\text{et}}\qquad \delta =-\gamma {\sqrt {a}},$

pour avoir cette équation

$2(\beta \delta -\alpha \gamma a){\frac {\partial ^{2}z}{\partial p\partial q}}=0\qquad {\text{ou}}\qquad {\frac {\partial ^{2}z}{\partial p\partial q}}=0.$

À présent M. d’Alembert ne saurait disconvenir que l’intégration de cette formule, en ne prenant que $p$ variable, ne donne

${\frac {dz}{dq}}=\varphi '(q)$