et j’en prends les différentielles logarithmiques, en regardant toujours
comme constante ; j’ai
![{\displaystyle {\begin{aligned}{\frac {d^{2}x}{dx}}=&\mathrm {\frac {X'}{2X}} dx-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dx}}dx-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dy}}dy,\\{\frac {d^{2}y}{dy}}=&\mathrm {\frac {Y'}{2Y}} dy-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dx}}dx-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dy}}dy,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdb04f2fd20a219b3c70acd2d861b418f0851f78)
donc
![{\displaystyle {\begin{aligned}d^{2}x=&\left(\mathrm {\frac {X'}{2X}} -{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dx}}\right)dx^{2}-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dy}}dxdy,\\d^{2}y=&\left(\mathrm {\frac {Y'}{2Y}} -{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dy}}\right)dy^{2}-{\frac {1}{\mathrm {T} }}{\frac {d\mathrm {T} }{dx}}dxdy,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36cd7b95cb6db5944222b48b7d67bfa47b62f930)
ou bien, en mettant
au lieu de
et
au lieu de ![{\displaystyle dy^{2},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8ec9c40ba4cb650e537575a13b0b2e637a6ba08)
![{\displaystyle {\begin{aligned}d^{2}x=&{\frac {1}{2}}{\frac {d\left(\mathrm {\cfrac {X}{T^{2}}} \right)}{dx}}dt^{2}-{\frac {d\log \mathrm {T} }{dy}}dxdy,\\d^{2}y=&{\frac {1}{2}}{\frac {d\left(\mathrm {\cfrac {Y}{T^{2}}} \right)}{dy}}dt^{2}-{\frac {d\log \mathrm {T} }{dx}}dxdy.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9a2484bb8507332a28536658080921eb11c442d)
Soit maintenant
une fonction quelconque de
et de
et supposons
![{\displaystyle d\mathrm {Z} =\mathrm {P} dx+\mathrm {Q} dy,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dab9b4c9f6f522fea18a24bc9411cb62919d86b3)
on aura, en différentiant de nouveau,
![{\displaystyle d^{2}\mathrm {Z} =\mathrm {P} d^{2}x+\mathrm {Q} d^{2}y+{\frac {d\mathrm {P} }{dx}}dx^{2}+2{\frac {d\mathrm {P} }{dy}}dxdy+{\frac {d\mathrm {Q} }{dy}}dy^{2}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfcd5071c3d7bfc565e9baecef64402271ff1b43)
donc, substituant au lieu de
et
leurs valeurs, on aura
![{\displaystyle {\begin{aligned}d^{2}Z=&\left[{\frac {1}{2}}\mathrm {P} {\frac {d\left(\mathrm {\cfrac {X}{T^{2}}} \right)}{dx}}+\mathrm {\frac {X}{T^{2}}} {\frac {d\mathrm {P} }{dx}}\right]dt^{2}+\left[{\frac {1}{2}}\mathrm {Q} {\frac {d\left(\mathrm {\cfrac {Y}{T^{2}}} \right)}{dy}}+\mathrm {\frac {Y}{T^{2}}} {\frac {d\mathrm {Q} }{dy}}\right]dt^{2}\\&+\left[2{\frac {d\mathrm {P} }{dy}}-\mathrm {P} {\frac {d\log \mathrm {T} }{dy}}-\mathrm {Q} {\frac {d\log \mathrm {T} }{dx}}\right]dxdy.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ea4e9f93ae33406aa2b26f718206f4e0ab10c6e)
Donc, si l’on suppose
(H)
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