Les opérateurs différentiels : formules locales
Nous allons démontrer trois formules locales relatives aux opérateurs différentiels dont nous devrons nous servir dans la suite du cours.
Ces formules sont démontrées en coordonnées cartésiennes.
\(\mathrm{div} (\stackrel{\hookrightarrow}{rot}\vec B) = 0\)
En coordonnées cartésiennes on a :
\(\stackrel{\hookrightarrow}{rot}\vec B = (\mathrm{rot}\vec B)_{x} \vec e_x + (\mathrm{rot}\vec B)_y \vec e_y + (\mathrm{rot}\vec B)_z \vec e_z\)
avec \(\stackrel{\hookrightarrow}{rot} \vec B = \left| \begin{array}{c c c} \frac{\partial}{\partial x} \\\\ \frac{\partial}{\partial y} \\\\ \frac{\partial}{\partial z} \end{array}\right. \wedge \left|\begin{array}{c c c} B_x \\ B_y \\ B_z \end{array}\right. = \left|\begin{array}{c c c} \frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z} = (\mathrm{rot}\vec B)_x \\\\ \frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x} = (\mathrm{rot}\vec B)_y \\\\ \frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y} = (\mathrm{rot}\vec B)_z \end{array}\right.\)
d'où, \(\mathrm{div} (\stackrel{\hookrightarrow}{rot} \vec B) = \frac{∂}{∂x}\mathrm(\mathrm{rot}\vec B)_x + \frac{∂}{∂y}(\mathrm{rot}\vec B)_y + \frac{∂}{∂z}(\mathrm{rot}\vec B)_z\)
soit : \(\mathrm{div}(\stackrel{\hookrightarrow}{rot}\vec B) = \Big(\frac{∂^2B_z}{∂_x∂_y} - \frac{∂^2B_y}{∂_x∂_z}\Big) + \Big(\frac{∂^2B_x}{∂_y∂_z} - \frac{∂^2B_z}{∂_y∂_x}\Big) + \Big(\frac{∂^2B_y}{∂_z∂_x} - \frac{∂^2B_x}{∂_z∂_y}\Big) = 0\)
\(\stackrel{\hookrightarrow}{rot}(\overrightarrow{\mathrm{grad}} f) = \vec 0\)
En coordonnées cartésiennes :
\(\overrightarrow{\mathrm{grad}} f = (\mathrm{grad} f)_x \vec e_x + (\mathrm{grad} f)_y \vec e_y + (\mathrm{grad} f)_z \vec e_z = \frac{∂f}{∂x}\vec e_x + \frac{∂f}{∂y}\vec e_y + \frac{∂f}{∂z}\vec e_z\)
et
\(\stackrel{\hookrightarrow}{rot}(\overrightarrow{\mathrm{grad}} f) = \left|\begin{array}{c c c} \frac{\partial}{\partial x} \\\\ \frac{\partial}{\partial y} \\\\ \frac{\partial}{\partial z} \end{array}\right. \wedge \left|\begin{array}{c c c} \frac{\partial f}{\partial x} \\\\ \frac{\partial f}{\partial y} \\\\ \frac{\partial f}{\partial z} \end{array}\right. = \left|\begin{array}{c c c} \frac{\partial}{\partial y}\Big(\frac{\partial f}{\partial z}\Big) - \frac{\partial}{\partial z}\Big(\frac{\partial f}{\partial y}\Big) =0 \\\\ \frac{\partial}{\partial z} \Big(\frac{\partial f}{\partial x}\Big)- \frac{\partial}{\partial x}\Big(\frac{\partial f}{\partial z}\Big) =0 \\\\ \frac{\partial}{\partial x}\Big(\frac{\partial f}{\partial y}\Big) - \frac{\partial}{\partial y}\Big(\frac{\partial f}{\partial x}\Big) = 0\end{array}\right.\)
Cette relation permet de déterminer si un champ vectoriel dérive d'un potentiel scalaire.
\(\stackrel{\hookrightarrow}{rot}(\stackrel{\hookrightarrow}{rot}\vec B) = \overrightarrow{\mathrm{grad}} (\mathrm{div}\vec B) - \bigtriangledown^2 \vec B\)
En coordonnées cartésiennes et compte tenu de ce que l'on a écrit précédemment, on a :
\(\stackrel{\hookrightarrow}{rot}(\stackrel{\hookrightarrow}{rot}\vec B) = \left|\begin{array}{c c c} \frac{\partial }{\partial x} \\\\ \frac{\partial }{\partial y} \\\\ \frac{\partial }{\partial z} \end{array}\right. \wedge \left|\begin{array}{c c c} \frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z} \\\\ \frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x}\\\\ \frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y} \end{array}\right.\) \(= \left|\begin{array}{c c c} \frac{\partial }{\partial y}\Big(\frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y}\Big) - \frac{\partial }{\partial z}\Big(\frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x}\Big) \\\\ \frac{\partial }{\partial z}\Big(\frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z}\Big) - \frac{\partial }{\partial x}\Big(\frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y}\Big) \\\\ \frac{\partial }{\partial x}\Big(\frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x}\Big) - \frac{\partial }{\partial y}\Big(\frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z}\Big)\end{array}\right.\)
Considérons la première composante, elle s'écrit : \(\frac{∂}{∂x}\Big(\frac{∂B_y}{∂y} + \frac{∂B_z}{∂z}\Big) - \Big(\frac{∂^2B_x}{∂y^2} + \frac{∂^2B_x}{∂z^2}\Big)\)
On peut encore écrire en ajoutant \(\Big(\frac{∂^2B_x}{∂x^2} - \frac{∂^2B_x}{∂x^2}\Big)\) à cette expression :
\(\frac{∂}{∂x}\Big(\frac{∂B_x}{∂x} + \frac{∂B_y}{∂y} + \frac{∂B_z}{∂z}\Big) - \Big(\frac{∂^2B_x}{∂x^2} + \frac{∂^2B_x}{∂y^2} + \frac{∂^2B_x}{∂z^2}\Big) = \frac{∂}{∂x}(\mathrm{div}\vec B) - \bigtriangledown^2B_x\)
On obtient enfin : \(\stackrel{\hookrightarrow}{rot}(\stackrel{\hookrightarrow}{rot}\vec B) =\left|\begin{array}{c c c} \frac{\partial }{\partial x} (\mathrm{div} \vec B) \\ \frac{\partial }{\partial y}(\mathrm{div} \vec B) \\ \frac{\partial }{\partial z}(\mathrm{div} \vec B) \end{array}\right. ~-~\left|\begin{array}{c c c} \bigtriangledown^2B_x \\ \bigtriangledown^2B_y \\ \bigtriangledown^2B_z \end{array}\right.\)
soit : \(\stackrel{\hookrightarrow}{rot}(\stackrel{\hookrightarrow}{rot}\vec B) = \overrightarrow{\mathrm{grad}}(\mathrm{div}\vec B) - \bigtriangledown^2\vec B\)
\(\bigtriangledown^2\vec B\) étant le laplacien[1] de \(\vec B\) qui s'écrit en repérage cartésien :
\(\bigtriangledown^2\vec B = \left|\begin{array}{ccc}\bigtriangledown^2B_x = \frac{\partial ^2B_x}{\partial x^2} + \frac{\partial ^2B_x}{\partial y^2} + \frac{\partial ^2B_x}{\partial z^2}\\ \bigtriangledown^2B_y = \frac{\partial ^2B_y}{\partial x^2} + \frac{\partial ^2B_y}{\partial y^2} + \frac{\partial ^2B_y}{\partial z^2}\\ \bigtriangledown^2B_z = \frac{\partial ^2B_z}{\partial x^2} + \frac{\partial ^2B_z}{\partial y^2} + \frac{\partial ^2B_z}{\partial z^2}\end{array}\right.\)