Page:Joseph Louis de Lagrange - Œuvres, Tome 12.djvu/86

78

MÉCANIQUE ANALYTIQUE.

On trouve de la même manière

${\begin{aligned}d{\frac {\partial a}{\partial y}}=&\left({\frac {\partial a}{\partial x'}}{\frac {\partial ^{2}\mathrm {V} }{\partial x\partial y}}+{\frac {\partial a}{\partial y'}}{\frac {\partial ^{2}\mathrm {V} }{\partial y^{2}}}+{\frac {\partial a}{\partial z'}}{\frac {\partial ^{2}\mathrm {V} }{\partial y\partial z}}\right)dt,\\d{\frac {\partial a}{\partial z}}=&\left({\frac {\partial a}{\partial x'}}{\frac {\partial ^{2}\mathrm {V} }{\partial x\partial z}}+{\frac {\partial a}{\partial y'}}{\frac {\partial ^{2}\mathrm {V} }{\partial x\partial z}}+{\frac {\partial a}{\partial z'}}{\frac {\partial ^{2}\mathrm {V} }{\partial z^{2}}}\right)dt.\end{aligned}}$

On aura ensuite

$d{\frac {\partial a}{\partial x'}}=\left\{{\begin{aligned}&{\frac {\partial ^{2}a}{\partial t\partial x'}}+{\frac {\partial ^{2}a}{\partial x\partial x'}}x'+{\frac {\partial ^{2}a}{\partial y\partial x'}}y'+{\frac {\partial ^{2}a}{\partial z\partial x'}}z'\\&\quad -{\frac {\partial ^{2}a}{\partial x'^{2}}}{\frac {\partial \mathrm {V} }{\partial x}}-{\frac {\partial ^{2}a}{\partial x'\partial y'}}{\frac {\partial \mathrm {V} }{\partial y}}-{\frac {\partial ^{2}a}{\partial x'\partial z'}}{\frac {\partial \mathrm {V} }{\partial z}}\end{aligned}}\right\}dt.$

Mais, en faisant varier $x'$ dans l’équation identique $da=0$ et observant que la fonction $\mathrm {V}$ est supposée ne pas contenir les variables $x',\,y',\,z',$ on a

${\begin{aligned}&{\frac {\partial ^{2}a}{\partial t\partial x'}}+{\frac {\partial ^{2}a}{\partial x\partial x'}}x'+{\frac {\partial ^{2}a}{\partial y\partial x'}}y'+{\frac {\partial ^{2}a}{\partial z\partial x'}}z'\\&\quad -{\frac {\partial ^{2}a}{\partial x'^{2}}}{\frac {\partial \mathrm {V} }{\partial x}}-{\frac {\partial ^{2}a}{\partial x'\partial y'}}{\frac {\partial \mathrm {V} }{\partial y}}-{\frac {\partial ^{2}a}{\partial x'\partial z'}}{\frac {\partial \mathrm {V} }{\partial z}}=0\,;\end{aligned}}$

donc on aura simplement

$d{\frac {\partial a}{\partial x'}}=-{\frac {\partial a}{\partial x}}dt,$

et l’on trouvera de la même manière

${\begin{aligned}d{\frac {\partial a}{\partial y'}}=&-{\frac {\partial a}{\partial y}}dt\\d{\frac {\partial a}{\partial z'}}=&-{\frac {\partial a}{\partial z}}dt.\end{aligned}}$

On aura des expressions semblables pour les différentielles $d{\frac {\partial b}{\partial x}},\,d{\frac {\partial b}{\partial y}},$ $d{\frac {\partial b}{\partial z}},\,d{\frac {\partial b}{\partial x'}},\,d{\frac {\partial b}{\partial y'}},\,d{\frac {\partial b}{\partial z'}},$ en changeant seulement $a$ en $b,$ et ainsi pour les autres quantités semblables.

Si maintenant on différentie le coefficient de ${\frac {\partial \Omega }{\partial b}}dt$ dans l’expression de $da$ de l’article 60, et qu’on y substitue les valeurs qu’on vient de