Pour cet effet, considérant une fonction quelconque de
et
qui soit
puisque
![{\displaystyle dv={\frac {\partial v}{\partial t}}dt+{\frac {\partial v}{\partial x}}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb49425d64b869872068e647bb5464c45c8e8a88)
et par les nouvelles variables
![{\displaystyle dv={\frac {\partial v}{\partial p}}dp+{\frac {\partial v}{\partial q}}dq,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb9364eb0c131bb63b139f00a7b5b0ccf507fc3e)
en substitutuant ici pour
et
leur valeur, j'aurai
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/812ab9f0c43ccd69d0c6314482e09c134d3f2159)
d’où il s'ensuit, pour les substitution dont j’ai besoin,
![{\displaystyle {\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}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4c525596b097d9401a33507dd5c52c063cda93a)
donc
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a43246c438fe9561a1188812a3b00334fd12df92)
Maintenant je pose
![{\displaystyle \beta ^{2}-\alpha ^{2}a=0\qquad {\text{et}}\qquad \delta ^{2}-\gamma ^{2}a=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/710c20dae1806232073f8760bd89464d35f52735)
ou
![{\displaystyle \beta =\alpha {\sqrt {a}}\qquad {\text{et}}\qquad \delta =-\gamma {\sqrt {a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09a810310b3797154cfc7749bb33d5aee18346a6)
pour avoir cette équation
![{\displaystyle 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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4efc27c58089977ecb289c6851aa7dce8d2060a3)
À présent M. d’Alembert ne saurait disconvenir que l’intégration de cette formule, en ne prenant que
variable, ne donne
![{\displaystyle {\frac {dz}{dq}}=\varphi '(q)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/798e09274648bdf5c5c6f90ad30b4860ae30bebb)