Annales de mathématiques pures et appliquées/Tome 14/Analise transcendante, article 7

Note sur les conditions d’intégrabilité ; par M. B. D. C.

Annales de mathématiques pures et appliquées, Tome 14, 1823-1824 (p. 319-323).

ANALISE TRANSCENDANTE.

Note sur les conditions d’intégrabilité ;

Par M. B. D. C.

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Soient $x$ et $y$ des fonctions quelconques d’une troisième variable $t\,;$ et soient représentés, en général, pour abréger,

${\frac {\operatorname {d} ^{k}x}{\operatorname {d} t^{k}}}\quad$par$\quad x_{k},$

${\frac {\operatorname {d} ^{k}y}{\operatorname {d} t^{k}}}\quad$par$\quad y_{k},$

quel que puisse être d’ailleurs le nombre entier positif $k.$ Soit $V$ une fonction quelconque de

$x,x_{1},x_{2},x_{3},\ldots x_{m},$

$y,y_{1},y_{2},y_{3},\ldots y_{m},$

où on suppose que $m$ soit tout au plus égal à $n,$ et posons

${\begin{aligned}&X=\left({\frac {\operatorname {d} V}{\operatorname {d} x}}\right)-{\frac {\operatorname {d} \left({\frac {\operatorname {d} V}{\operatorname {d} x_{1}}}\right)}{\operatorname {d} t}}+{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} V}{\operatorname {d} x_{2}}}\right)}{\operatorname {d} t^{2}}}-{\frac {\operatorname {d} ^{3}\left({\frac {\operatorname {d} V}{\operatorname {d} x_{3}}}\right)}{\operatorname {d} t^{3}}}+\ldots \pm {\frac {\operatorname {d} ^{m}\left({\frac {\operatorname {d} V}{\operatorname {d} x_{m}}}\right)}{\operatorname {d} t^{m}}}\\\\&Y=\left({\frac {\operatorname {d} V}{\operatorname {d} y}}\right)-{\frac {\operatorname {d} \left({\frac {\operatorname {d} V}{\operatorname {d} y_{1}}}\right)}{\operatorname {d} t}}+{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} V}{\operatorname {d} y_{2}}}\right)}{\operatorname {d} t^{2}}}-{\frac {\operatorname {d} ^{3}\left({\frac {\operatorname {d} V}{\operatorname {d} y_{3}}}\right)}{\operatorname {d} t^{3}}}+\ldots \pm {\frac {\operatorname {d} ^{n}\left({\frac {\operatorname {d} V}{\operatorname {d} y_{n}}}\right)}{\operatorname {d} t^{n}}}\end{aligned}}$

Si $V\operatorname {d} t$ est une différentielle exacte, on aura, comme l’on sait^[1], $X=0,\ Y=0\,;$ et réciproquement, si ces deux équations ont lieu, $V\operatorname {d} t$ sera une différentielle exacte ; et telles sont les conditions nécessaires et suffisantes pour l’intégrabilité de la fonction différentielle $V\operatorname {d} t,$ lorsque $x$ et $y$ sont deux variables indépendantes l’une de l’autre.

Mais si, au contraire, $y$ était une fonction de $x,$ c’est-à-dire, si l’on avait $y=\varphi (x),$ il est clair qu’alors une seule condition serait nécessaire et suffisante pour rendre intégrable la fonction différentielle $V\operatorname {d} t.$ Mais quelle devrait être cette condition unique ? C’est ce que nous nous proposons ici de rechercher.

Désignons en général par $\varphi _{k}(x)$ la dérivée de l’ordre $k$ de la fonction $\varphi (x),$ prise par rapport à $x,$ quel que soit le nombre entier positif $k\,;$ à cause de $y=\varphi (x),$ on aura, en différentiant,

${\begin{aligned}y\ \,&=\varphi (x)\\\\y_{1}&=\varphi _{1}(x).x_{1}\\\\y_{2}&=\varphi _{2}(x).x_{1}^{2}+\varphi _{1}(x).x_{2}\\\\y_{3}&=\varphi _{3}(x).x_{1}^{3}+3\varphi _{2}(x).x_{1}x_{2}+\varphi _{1}(x).x_{3}\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\\end{aligned}}$

Désignons par $V'$ ce que devient $V$ lorsqu’on y substitue ces valeurs en $x,x_{1},x_{2},\ldots x_{n}$ pour $y,y_{1},y_{2},\ldots y_{n}\,;$ en posant

$X'=\left({\frac {\operatorname {d} V'}{\operatorname {d} x}}\right)-{\frac {\operatorname {d} \left({\frac {\operatorname {d} V'}{\operatorname {d} x_{1}}}\right)}{\operatorname {d} t}}+{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} V'}{\operatorname {d} x_{2}}}\right)}{\operatorname {d} t^{2}}}-{\frac {\operatorname {d} ^{3}\left({\frac {\operatorname {d} V'}{\operatorname {d} x_{3}}}\right)}{\operatorname {d} t^{3}}}+\ldots \pm {\frac {\operatorname {d} ^{n}\left({\frac {\operatorname {d} V'}{\operatorname {d} x_{n}}}\right)}{\operatorname {d} t^{n}}},$

$X'=0$ sera évidemment l’équation de condition qu’il s’agit de déterminer.

Or, on a

${\begin{aligned}{\frac {\operatorname {d} V'}{\operatorname {d} x}}&={\frac {\operatorname {d} V}{\operatorname {d} x}}+{\frac {\operatorname {d} y}{\operatorname {d} x}}{\frac {\operatorname {d} V}{\operatorname {d} y}}+{\frac {\operatorname {d} y_{1}}{\operatorname {d} x}}{\frac {\operatorname {d} V}{\operatorname {d} y_{1}}}+{\frac {\operatorname {d} y_{2}}{\operatorname {d} x}}{\frac {\operatorname {d} V}{\operatorname {d} y_{2}}}+\ldots +{\frac {\operatorname {d} y_{n}}{\operatorname {d} x}}{\frac {\operatorname {d} V}{\operatorname {d} y_{n}}},\\\\{\frac {\operatorname {d} V'}{\operatorname {d} x_{1}}}&={\frac {\operatorname {d} V}{\operatorname {d} x_{1}}}+{\frac {\operatorname {d} y_{1}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{1}}}+{\frac {\operatorname {d} y_{2}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{2}}}+\ldots +{\frac {\operatorname {d} y_{n}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{n}}},\\\\{\frac {\operatorname {d} V'}{\operatorname {d} x_{2}}}&={\frac {\operatorname {d} V}{\operatorname {d} x_{2}}}+{\frac {\operatorname {d} y_{2}}{\operatorname {d} x_{2}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{2}}}+{\frac {\operatorname {d} y_{3}}{\operatorname {d} x_{2}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{3}}}+\ldots +{\frac {\operatorname {d} y_{n}}{\operatorname {d} x_{2}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{n}}},\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\{\frac {\operatorname {d} V'}{\operatorname {d} x_{n}}}&={\frac {\operatorname {d} V}{\operatorname {d} x_{n}}}+{\frac {\operatorname {d} y_{n}}{\operatorname {d} x_{n}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{n}}}\,;\end{aligned}}$

ce qui donnera, en substituant,

$X'={\frac {\operatorname {d} V}{\operatorname {d} x}}+{\frac {\operatorname {d} y}{\operatorname {d} x}}{\frac {\operatorname {d} V}{\operatorname {d} y}}+{\frac {\operatorname {d} y_{1}}{\operatorname {d} x}}{\frac {\operatorname {d} V}{\operatorname {d} y_{1}}}+{\frac {\operatorname {d} y_{2}}{\operatorname {d} x}}{\frac {\operatorname {d} V}{\operatorname {d} y_{2}}}+\ldots {\frac {\operatorname {d} y_{i}}{\operatorname {d} x}}{\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}+\ldots$

${\begin{aligned}&-{\frac {\operatorname {d} \left({\frac {\operatorname {d} V}{\operatorname {d} x_{1}}}\right)}{\operatorname {d} t}}-{\frac {\operatorname {d} \left({\frac {\operatorname {d} y_{1}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{1}}}\right)}{\operatorname {d} t}}-{\frac {\operatorname {d} \left({\frac {\operatorname {d} y_{2}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{2}}}\right)}{\operatorname {d} t}}-\ldots -{\frac {\operatorname {d} \left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t}}-\ldots \\\\&+{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} V}{\operatorname {d} x_{2}}}\right)}{\operatorname {d} t^{2}}}\qquad \qquad +{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} y_{2}}{\operatorname {d} x_{2}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{2}}}\right)}{\operatorname {d} t^{2}}}+\ldots +{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{2}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t^{2}}}+\ldots \\\\&-{\frac {\operatorname {d} ^{3}\left({\frac {\operatorname {d} V}{\operatorname {d} x_{3}}}\right)}{\operatorname {d} t^{3}}}\qquad \qquad \qquad -\ldots -{\frac {\operatorname {d} ^{3}\left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{3}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t^{3}}}-\ldots \\&+\ldots \ldots \ldots \ldots \qquad \qquad +\ldots \ldots \ldots \ldots \ldots \ldots \\&+{\frac {\operatorname {d} ^{i+1}\left({\frac {\operatorname {d} V}{\operatorname {d} x_{i+1}}}\right)}{\operatorname {d} t^{i+1}}}\end{aligned}}$

Considérons, dans ce développement, la colonne dont le rang est $i+2,$ laquelle est, comme on le voit,

${\frac {\operatorname {d} y_{i}}{\operatorname {d} x}}{\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}-{\frac {\operatorname {d} \left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t}}+{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{2}}}{\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t^{2}}}-\ldots \pm {\frac {\operatorname {d} ^{i+1}\left({\frac {\operatorname {d} V}{\operatorname {d} x_{i+1}}}\right)}{\operatorname {d} t^{i+1}}}\,;$

en la développant et ordonnant suivant les différentielles successives de ${\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\,;$ on pourra lui donner la forme suivante :

${\begin{array}{r|r|r|r}{\frac {\operatorname {d} y_{i}}{\operatorname {d} x}}&{\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}-{\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{1}}}&{\frac {\operatorname {d} \left({\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t}}+{\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{2}}}&{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t^{2}}}-\ldots \pm {\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{i}}}{\frac {\operatorname {d} ^{i}\left({\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t^{i}}}\\\\-{\frac {\operatorname {d} \left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{1}}}\right)}{\operatorname {d} t}}&+2{\frac {\operatorname {d} \left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{2}}}\right)}{\operatorname {d} t}}&-3{\frac {\operatorname {d} \left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{3}}}\right)}{\operatorname {d} t}}&+\ldots \\\\+{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{2}}}\right)}{\operatorname {d} t^{2}}}&-3{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{3}}}\right)}{\operatorname {d} t^{2}}}&+\ldots \ldots &\\\\-{\frac {\operatorname {d} ^{3}\left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{3}}}\right)}{\operatorname {d} t^{3}}}&+\ldots \ldots &\pm {\frac {i}{1}}.{\frac {i-1}{2}}{\frac {\operatorname {d} ^{i-2}\left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{i}}}\right)}{\operatorname {d} t^{i-2}}}&\\\\+\ldots \ldots &\pm i{\frac {\operatorname {d} ^{i-1}\left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{i}}}\right)}{\operatorname {d} t^{i-1}}}&&\\\\\pm {\frac {\operatorname {d} ^{i}\left({\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{i}}}\right)}{\operatorname {d} t^{i}}}&&&\end{array}}$

Or, les coefficiens des termes,

${\frac {\operatorname {d} V}{\operatorname {d} x_{i}}},\quad {\frac {\operatorname {d} \left({\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t}},\quad {\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t^{2}}},\ldots {\frac {\operatorname {d} ^{i-1}\left({\frac {\operatorname {d} V}{\operatorname {d} y_{i}}}\right)}{\operatorname {d} t^{i-1}}},$

de ce développement, égalés séparément à zéro, ne sont autre chose que les équations de condition qui exprimeraient que $y_{i}\operatorname {d} t^{i}$ est une différentielle exacte de l’ordre $i\,;$ ces coefficiens doivent donc être nuls d’eux-mêmes, puisque $y_{i}\operatorname {d} t^{i}$ est en effet une telle différentielle ; ce développement se réduit donc simplement à son dernier terme ${\frac {\operatorname {d} y_{i}}{\operatorname {d} x_{i}}}{\frac {\operatorname {d} ^{i}\left({\frac {\operatorname {d} V}{\operatorname {d} y_{1}}}\right)}{\operatorname {d} t^{i}}}\,;$ c’est donc aussi à ce dernier terme que se réduit la $(i+2)$^me colonne du développement de $X'\,;$ d’où il suit ce développement lui-même se réduit à

$X'={\frac {\operatorname {d} V}{\operatorname {d} x}}-{\frac {\operatorname {d} \left({\frac {\operatorname {d} V}{\operatorname {d} x_{1}}}\right)}{\operatorname {d} t}}+{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} V}{\operatorname {d} x_{2}}}\right)}{\operatorname {d} t^{2}}}-\ldots \pm {\frac {\operatorname {d} ^{n}\left({\frac {\operatorname {d} V}{\operatorname {d} x_{n}}}\right)}{\operatorname {d} t^{n}}}$

$+{\frac {\operatorname {d} y}{\operatorname {d} x}}{\frac {\operatorname {d} V}{\operatorname {d} y}}-{\frac {\operatorname {d} y_{1}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} \left({\frac {\operatorname {d} V}{\operatorname {d} y_{1}}}\right)}{\operatorname {d} t}}+{\frac {\operatorname {d} y_{2}}{\operatorname {d} x_{2}}}{\frac {\operatorname {d} ^{2}\left({\frac {\operatorname {d} V}{\operatorname {d} y_{2}}}\right)}{\operatorname {d} t^{2}}}-\ldots \pm {\frac {\operatorname {d} y_{n}}{\operatorname {d} x_{n}}}{\frac {\operatorname {d} \left({\frac {\operatorname {d} V}{\operatorname {d} y_{n}}}\right)}{\operatorname {d} t^{n}}}\,;$

mais il est visible que

${\frac {\operatorname {d} y}{\operatorname {d} x}}={\frac {\operatorname {d} y_{1}}{\operatorname {d} x_{1}}}={\frac {\operatorname {d} y_{2}}{\operatorname {d} x_{2}}}=\ldots ={\frac {\operatorname {d} y_{n}}{\operatorname {d} x_{n}}}\,;$

donc, on aura

$X'=X+Y{\frac {\operatorname {d} y}{\operatorname {d} x}}\,;$

puis donc que la condition d’intégrabilité est, dans le cas présent $X'=0,$ cette condition sera

$X\operatorname {d} x+X\operatorname {d} y=0\quad$ou$\quad Xx_{1}+Yy_{1}=0.$

Cette conclusion est exactement celle de Lagrange dans sa 21.^e leçon sur le Calcul des fonctions^[2].

1. Voyez la page 197 du présent volume.
2. Voyez la page 412 de l’édition in-8.^o ou la page 12 du supplément in-4^o.