Page:Joseph Louis de Lagrange - Œuvres, Tome 11.djvu/124

106

MÉCANIQUE ANALYTIQUE

Il est clair que

${\frac {\delta \partial z}{\partial x}}={\frac {\partial \delta z}{\partial x}}\qquad {\text{et}}\qquad {\frac {\delta \partial x}{\partial x}}={\frac {\partial \delta x}{\partial x}}\,;$

ainsi l’on aura

$\delta z'={\frac {\partial \delta z}{\partial x}}-z'{\frac {\partial \delta x}{\partial x}}={\frac {\partial (\delta z-z'\delta x)}{\partial x}}+{\frac {\partial z'}{\partial x}}\delta x,$

ou bien

$\delta z'={\frac {\partial (\delta z-z'\delta x-z_{_{'}}\delta y)}{\partial x}}+{\frac {\partial z'}{\partial x}}\delta x+{\frac {\partial z_{_{'}}}{\partial x}}\delta y.$

On aura de même

$\delta z_{_{'}}={\frac {\partial (\delta z-z'\delta x-z_{_{'}}\delta y)}{\partial y}}+{\frac {\partial z'}{\partial y}}\delta x+{\frac {\partial z_{_{'}}}{\partial y}}\delta y,$

à cause de

${\frac {\partial \delta x}{\partial y}}=0\qquad {\text{et}}\qquad {\frac {\partial \delta y}{\partial x}}=0.$

On aura ensuite

$\delta z''=\delta {\frac {\partial z'}{\partial x}}={\frac {\partial \delta z'}{\partial x}}-{\frac {\partial z'}{\partial x}}{\frac {\partial \delta x}{\partial x}}.$

Substituant la valeur de $\delta z',$ on aura

$\delta z''={\frac {\partial ^{2}(\delta z-z'\delta x-z_{_{'}}\delta y)}{\partial x^{2}}}+{\frac {\partial ^{2}z'}{\partial x^{2}}}\delta x+{\frac {\partial ^{2}z_{_{'}}}{\partial x^{2}}}\delta y.$

On aura de même

$\delta z'_{_{'}}=\delta {\frac {\partial z'}{\partial y}}={\frac {\partial \delta z'}{\partial y}}-{\frac {\partial z'}{\partial y}}{\frac {\partial \delta y}{\partial y}}.$

Substituant aussi la valeur de $\delta z',$ on aura, à cause de ${\frac {\partial z_{_{'}}}{\partial x}}={\frac {\partial z'}{\partial y}},$

$\delta z'_{_{'}}={\frac {\partial ^{2}(\delta z-z'\delta x-z_{_{'}}\delta y)}{\partial x\partial y}}+{\frac {\partial ^{2}z'}{\partial x\partial y}}\delta x+{\frac {\partial ^{2}z_{_{'}}}{\partial x\partial y}}\delta y.$

On trouvera pareillement

$\delta z_{_{''}}={\frac {\partial ^{2}(\delta z-z'\delta x-z_{_{'}}\delta y)}{\partial y^{2}}}+{\frac {\partial ^{2}z'}{\partial y^{2}}}\delta x+{\frac {\partial ^{2}z_{_{'}}}{\partial y^{2}}}\delta y,$

et ainsi de suite.

33. Donc, si l’on fait, pour abréger,

$\delta z-{\frac {\partial z}{\partial x}}\delta x-{\frac {\partial z}{\partial y}}\delta y=\delta u,$