Page:Lagrange - Œuvres (1867) vol. 1.djvu/262

et de même

$\int \mathrm {Q} xdx,\quad {\text{d’où}}\quad u={\frac {\int \mathrm {Q} xdx}{x}}.$

Que $\varphi$ et $\psi$ représentent deux fonctions quelconques régulières ou irrégulières, telles que

${\begin{aligned}&\mathrm {MN=\varphi (AM)} =\varphi (x),\\&\mathrm {MQ=\psi (AM)} =\psi (x),\end{aligned}}$

on aura

${\begin{aligned}&\mathrm {M'N'=\varphi (AM')} =\varphi \left(x+t{\sqrt {c}}\right),\\&\mathrm {{\grave {\,}}M\,{\grave {\,}}N=\varphi (A\,{\grave {\,}}M)} =\varphi \left(x-t{\sqrt {c}}\right),\\&\mathrm {M'Q'=\psi (AM')} =\psi \left(x+t{\sqrt {c}}\right),\\&\mathrm {{\grave {\,}}M\,{\grave {\,}}Q=\psi (A\,{\grave {\,}}M)} =\psi \left(x-t{\sqrt {c}}\right),\end{aligned}}$

et

${\begin{aligned}\mathrm {M} n=&{\sqrt {c}}{\frac {d\varphi (\mathrm {AM} )}{d\mathrm {AM} }}={\sqrt {c}}{\frac {d\varphi (x)}{dx}},\\\mathrm {M} q=&{\frac {1}{\sqrt {c}}}\int \psi (\mathrm {AM} )d\mathrm {AM} ={\frac {1}{\sqrt {c}}}\int \psi (x)dx.\end{aligned}}$

et par conséquent

${\begin{aligned}\mathrm {M} 'n'=&{\sqrt {c}}{\frac {d\varphi \left(x+t{\sqrt {c}}\right)}{dx}},\\{\grave {\,}}\mathrm {M} \,{\grave {\,}}n=&{\sqrt {c}}{\frac {d\varphi \left(x-t{\sqrt {c}}\right)}{dx}},\\\mathrm {M} 'q'=&{\frac {1}{\sqrt {c}}}\int \psi \left(x+t{\sqrt {c}}\right)dx,\\{\grave {\,}}\mathrm {M} \,{\grave {\,}}q=&{\frac {1}{\sqrt {c}}}\int \psi \left(x-t{\sqrt {c}}\right)dx\,;\end{aligned}}$

donc

${\begin{aligned}\mathrm {P} =&{\frac {1}{2}}\left[\varphi \left(x+t{\sqrt {c}}\right)+\varphi \left(x-t{\sqrt {c}}\right)\right]\\&+{\frac {1}{2{\sqrt {c}}}}\left[\int \psi \left(x+t{\sqrt {c}}\right)dx-\int \psi \left(x-t{\sqrt {c}}\right)dx\right],\\\mathrm {Q} =&{\frac {1}{2}}\left[\psi \left(x+t{\sqrt {c}}\right)+\psi \left(x-t{\sqrt {c}}\right)\right]\\&+{\frac {\sqrt {c}}{2}}\left[{\frac {d\varphi \left(x+t{\sqrt {c}}\right)}{dx}}-{\frac {d\varphi \left(x-t{\sqrt {c}}\right)}{dx}}\right].\end{aligned}}$