Page:Annales de mathématiques pures et appliquées, 1821-1822, Tome 12.djvu/13

9

ET DES SURFACES COURBES.

Dans l’hypothèse actuelle, on a, en vertu de l’équation (3),

${\frac {\operatorname {d} ^{2}u}{\operatorname {d} t^{2}}}={\frac {\operatorname {d} \left({\frac {\operatorname {d} u}{\operatorname {d} t}}\right)}{\operatorname {d} t}}={\frac {\operatorname {d} \left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)}{\operatorname {d} t}}={\frac {\frac {\operatorname {d} ^{2}y}{\operatorname {d} x^{2}}}{\frac {\operatorname {d} t}{\operatorname {d} x}}}\,;$

mais, en différentiant l’équation (1), il vient

${\frac {\operatorname {d} t}{\operatorname {d} x}}-1+\left({\frac {\operatorname {d} u}{\operatorname {d} x}}-{\frac {\operatorname {d} y}{\operatorname {d} x}}\right){\frac {\operatorname {d} y}{\operatorname {d} x}}+(u-y){\frac {\operatorname {d} ^{2}y}{\operatorname {d} x^{2}}}=0\,;$

nous avons d’ailleurs trouvé tout à l’heure

${\frac {\operatorname {d} u}{\operatorname {d} x}}={\frac {\operatorname {d} y}{\operatorname {d} x}}{\frac {\operatorname {d} t}{\operatorname {d} x}},$

ce qui donnera en substituant

$\left({\frac {\operatorname {d} t}{\operatorname {d} x}}-1\right)\left\{1+\left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)^{2}\right\}+(u-y){\frac {\operatorname {d} ^{2}y}{\operatorname {d} x^{2}}}=0\,;$

d’où

${\frac {\operatorname {d} t}{\operatorname {d} x}}=1-{\frac {(u-y){\frac {\operatorname {d} ^{2}y}{\operatorname {d} x^{2}}}}{1+\left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)^{2}}}\,;$

substituant cette valeur dans celle de ${\frac {\operatorname {d} u^{2}}{\operatorname {d} t^{2}}}$ elle deviendra

${\frac {\operatorname {d} ^{2}u}{\operatorname {d} t^{2}}}={\frac {\frac {\operatorname {d} ^{2}y}{\operatorname {d} x^{2}}}{1-{\frac {(u-y){\frac {\operatorname {d} ^{2}y}{\operatorname {d} x^{2}}}}{1+\left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)^{2}}}}}$

et telle est la relation cherchée.